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FOR Engineers 


Tr.i.cJu o^^jr r.i^ 
















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This book describes what has for many years been the 
most important part of the regular course in the Calculus 
for Mechanical and Electrical Engineering students at the 
Finsbury Technical College. It was supplemented by easy 
work involving Fourier, Spherical Harmonic, and Bessel 
Functions which I have been afraid to describe here because 
the book is already much larger than I thought it would 

The students in October knew only the most elementary 
mathematics, many of them did not know the Binomial 
Theorem, or the definition of the sine of an angle. In July 
they had not only done the work of this book, but their 
knowledge was of a practical kind, ready for use in any 
such engineering problems as I give here. 

One such student, Mr Norman Endacott, has corrected 
the manuscript and proofs. He has worked out many of 
the exercises in the third chapter twice over. I thank him 
here for the care he has taken, and I take leave also to 
say that a system which has, year by year, produced many 
men with his kind of knowledge of mathematics has a 
good deal to recommend it. I say this through no vanity 
but because I wish to encourage the earnest student. Besides 
I cannot claim more than a portion of the credit, for I 
do not think that there ever before was such a complete 

J 2/ 12b 


harmony in the working of all the departments of an 
educational institution in lectures and in tutorial, labora- 
tory, drawing office and other practical work as exists in 
the Finsbury Technical College, all tending to the same 
end ; to give an engineer such a perfect acquaintance with 
his mental tools that he actually uses these tools in his 

Professor Willis has been kind enough to read through 
the proofs and I therefore feel doubly sure that no important 
mistake has been made anywhere. 

An experienced friend thinks that I might with advantage 
have given many more illustrations of the use of squared 
paper just at the beginning. This is quite possible, but if 
a student follows my instructions he will furnish all this sort 
of illustration very much better for himself Again I might 
have inserted many easy illustrations of integration by 
numerical work such as the exercises on the Bull Engine 
and on Beams and Arches which are to be found in my book 
on Applied Mechanics. I can only say that I encourage 
students to find illustrations of this kind for themselves; 
and surely there must be some limit to spoon feeding. 


EoYAL College of Science, 
16th March, 1897. 



Introductory Remarks ..,.,,.. 1 

Chap. I. The Study of ^** . 6 

Chap. II. The Compound Interest Law and the Harmonic 

Function 161 

Chap. III. General Differentiation and Integration . 267 



1. The Engineer has usually no time for a general mathe- 
matical training — mores the pity — and those young engineers 
who have had such a training do not always find their mathe- 
matics helpful in their profession. Such men will, I hope, 
find this book useful, if they can only get over the notion 
that because it is elementary, they know already all that it 
can teach. 

But I write more particularly for readers who have had 
very little mathematical training and who are willing to work 
very hard to find out how the calculus is applied in Engineer- 
ing problems. I assume that a good engineer needs to know 
only fundamental principles, but that he needs to know these 
very well indeed. 

2. My reader is supposed to have an elementary know- 
ledge of Mechanics, and if he means to take up the Electrical 
problems he is supposed to have an elementary knowledge of 
Electrical matters. A common-sense knowledge of the few 
fundamental facts is what is required; this knowledge is 
seldom acquired by mere reading or listening to lectures; 
one needs to make simple experiments and to work easy 
numerical exercises. 

In Mechanics, I should like to think that the mechanical 
engineers who read this book know what is given in the 
elementary parts of my books on Applied Mechanics and the 
Steam and Gas Engine. That is, I assume that they know 

P 1 


the elementary facts about Bending Moment in beams, Work 
done by forces and the Efficiency of heat engines. Possibly 
the book may cause them to seek for such knowledge. I take 
almost all my examples from Engineering, and a man who 
works these easy examples will find that he knows most of 
what is called the theory of engineering. 

3. I know men who have passed advanced examinations 
in Mathematics who are very shy, in practical work, of the 
common formulae used in Engineers' pocket-books. How- 
ever good a mathematician a student thinks himself to be, 
he ought to practise working out numerical values, to find 
for example the value of a^ by means of a table of logarithms, 
when a and h are any numbers whatsoever. Thus to find 
V'014, to find 2*36o~<''2«, &c., to take any formula from a 
pocket-book and use it. He must not only think he knows ; 
he must really do the numerical work. He must know that 
if a distance 2 4 5 4 has been measured and if one is not sure 
about the last figure, it is rather stupid in multiplying or 
dividing by this number to get out an answer with many 
significant figures, or to say that the indicated power of an 
engine is 324"65 Horse power, when the indicator may be in 
error 5 per cent, or more. He must know the quick way of 
finding 3'216 x 4571 to four significant figures without using 
logarithms. He ought to test the approximate rule 

(1 +a)"=l + ?ia, 

or (1 + a)" (1 + ^)'« = 1 + 7ia + m^, 

if a and /S are small, and see for himself when a = '01 or 

— -01, or yS = ± -025 and ti = 2 or ^ or — 1^, and m = 4 or 2 

or — 2 or J or any other numbers, what errors are involved in 

the assumption. 

f As to Trigonometry, the definitions must be known. For 

example, Draw BA C an angle of, say, 35°. Take any point B 

and drop the perpendicular. Measure AB and ^0 and AC 

as accurately as possible. Is AC^ -{- BC^ = AB^ 1 Work 


this out numerically. Now -r^ = sin 35°, -rv^ = cos 35°, 

BC ^^ ^^ 

-j-^ = tan 35°. Try if the answers are those given in the 

tables. Learn how we calculate the other sides of the 


triangle ABC when we know one side and one of the acute 
angles. Learn also that the sine of 130° is positive, and the 
cosine of 130° is negative. Also try with the book of tables if 

sin {A-\- B) — sin A , cos B + cos A . sin B, 

where A and B are any two angles you choose to take. 
There are three other rules like this. In like manner the 
four which we obtain by adding these formulae and subtract- 
ing them, of which this is one, 

2 sin a . cos /S = sin (a + y8) + sin (a — /8) ; 

also cos 2^ = 1 - 2 sin^ A =2 cos^ ^ - 1. 

Before readers have gone far in this book I hope they will 
be induced to take up the useful (that is, the elementary 
and interesting) part of trigonometry, and prove all rules for 
themselves, if they haven't done so already. 

Calculate an angle of 1*6 degrees in radians (1 radian is 
equal to 57'296 degrees) ; see how much the sine and tangent 
of this angle differ from the angle itself Remember that 
when in mathematics we say sin x, x is supposed to be in 

I do not expect a man to know much about advanced 
algebra, but he is supposed to be able to give the factors of 
a?^ + 7^ 4- 12 or of x"^ — o? for example ; to be able to simplify 
expressions. It is not a knowledge of permutations or com- 
binations or of the theory of equations, of Geometrical Conies 
or tangent planes to quadrics, that the Engineer wants. 
Happy is the Engineer who is also a mathematician, but 
it is given to only a few men to have the two so very different 

A prolonged experience of workshops, engineers and 
students has convinced me that although a Civil Engineer 
for the purposes of surveying may need to understand the 
solution of triangles, this and many other parts of the 
Engineer's usual mathematical training are really useless to 
the mechanical or electrical engineer. This sounds un- 
orthodox, but I venture to emphasise it. The young engi- 
neer cannot be drilled too much in the mere simplification 
of algebraic and trigonometrical expressions, including ex- 
pressions involving J —1, and the best service done by 



elementary calculus work is in inducing students to again 
undergo this drilling. 

But the engineer needs no artificial mental gymnastics 
such as is furnished by Geometrical Conies, or the usual 
examination-paper puzzles, or by evasions of the Calculus 
through infinite worry with elementary Mathematics. The 
result of a false system of training is seen in this, that not 
one good engineer in a hundred believes in what is usually 
called theory. 

4. I assume that every one of my readers is thoroughly 
well acquainted already with the fundamental notion of the 
Calculus, only he doesn't know it in the algebraic form. He 
has a perfect knowledge of a rate, but he has never been 

accustomed to write -^ ; he has a perfect knowledge of an 

area, but he has not yet learnt the symbol used by us, 

lf(x).dos. He has the idea, but he does not express his 

idea in this form. 

1 assume that some of my readers have passed difficult 
examinations in the Calculus, that they can differentiate any 
function of x and integrate many ; that they know how 'to 
work all sorts of difficult exercises about Pedal Curves and 
Roulettes and Elliptic Integrals, and to them also I hope to 
be of use. Their difficulty is this, their mathematical know- 
ledge seems to be of no use to them in practical engineering 
problems. Give to their afs and y's a physical meaning, 
or use p's and vs instead, and what was the easiest book 
exercise becomes a difficult problem. I know such men 
who hurriedly skip in reading a book when they see a 

~ y or a sign of integration. 

5. When I started to write this book I thought to put 
the subject before my readers as I have been able, I think — 
I have been told — very successfully, to bring it before some 
classes of evening students; but much may be done in 
lectures which one is unable to do in a cold-blooded fashion 
sitting at a table. One misses the intelligent eyes of an 
audience, warning one that a little more explanation is needed 


or that an important idea has already been grasped. An 
idea could be given in the mere drawing of a curve and 
illustrations chosen from objects around the lecture-room. 

Let the reader skip judiciously; let him work up no 
problem here in which he has no professional interest. The 
problems are many, and the best training comes from the 
careful study of only a few of them. 

The reader is expected to turn back often to read again 
the early parts. 

The book would be unwieldy if I included any but the 
more interesting and illustrative of engineering problems. I 
put off for a future occasion what would perhaps to many 
students be a more interesting part of my subject, namely, 
illustrations from Engineering (sometimes called Applied 
Physics) of the solution of Partial Differential Equations. 
Many people think the subject one which cannot be taught 
in an elementary fashion, but Lord Kelvin showed me long 
ago that there is no useful mathematical weapon which an 
engineer may not learn to use. A man learns to use the 
Calculus as he learns to use the chisel or the file on actual 
concrete bits of work, and it is on this idea that I act in 
teaching the use of the Calculus to Engineers. 

This book is not meant to supersede the more orthodox 
treatises, it is rather an introduction to them. In the 
first chapter of 160 pages, I do not attempt to differentiate 
or integrate any function of x, except x^. In the second 
chapter I deal with e^^, and sin (aw + c). The third chapter 
is more difficult. 

For the sake of the training in elementary Algebraic 
work, as much as for use in Engineering problems, I have 
included a set of exercises on general dinerentiation and 

Parts in smaller type, and the notes, may be found too 
difficult by some students in a first reading of the book. An 
occasional exercise may need a little more knowledge than 
the student already possesses. His remedy is to skip. 



6. Everybody has already the notions of Co-ordinate 
Geometry and uses squared paper. Squared paper may 
be bought at sevenpence a quire : people who arc ignorant 
of this fact and who pay sevenpence or fourteen pence a 
sheet for it must have too great an idea of its value to use 
it properly. 

When a merchant has in his office a sheet of squared 
paper with points lying in a curve which he adds to day by 
day, each point showing the price of iron, or copper, or cotton 
yarn or silk, at any date, he is using Co-ordinate Geometry. 
Now to what uses does he put such a curve ? 1. At any 
date he sees what the price was. 2. He sees by the slope of 
his curve the rate of increase or fall of the price. 3. If he 
plots other things on the same sheet of paper at the same 
dates he will note what effect their rise and fall have upon 
the price of his material, and this may enable him to pro- 
phesy and so make money. 4. Examination of his curve for 
the past will enable him to prophesy with more certainty 
than a man can do who has no records. 

Observe that any point represents two things; its 
horizontal distance from some standard line or axis is called 
one co-ordinate, we generally call it the x co-ordinate and it 
is measured horizontally to the right of the axis of y ; some 
people call it the abscissa ; this represents time in his case. 
The other co-ordinate (we usually call it the 3/ co-ordinate or 
tlie ordinate, simply), the vertical distance of the point above 
some standard line or axis; this represents his price. In the 
newspaper you will find curves showing how the thermometer 
and barometer are rising and falling. I once read a clever 
article upon the way in which the English population and 
wealth and taxes were increasing; the reasoning was very 


difficult to follow. On taking the author s figures however 
and plotting them on squared paper, every result which he 
had laboured so much to bring out was plain upon the 
curves, so that a boy could understand them. Possibly this 
is the reason why sDme writers do not publish curves: if they 
did, there would be little need for writing. 

7. A man making experiments is usually finding out 
how one thing which I shall call y depends upon some other 
thing which I shall call x. Thus the pressure p of saturated 
steam (water and steam present in a vessel but no air or 
other fluid) is always the same for the same temperature. 
A curve drawn on squared paper enables us for any given 
temperature to find the pressure or vice versa, but it shows 
the rate at which one increases relatively to the increase of 
the other and much else. I do not say that the curve is always 
better than the table of values for giving information ; some 
information is better given by the curve, some by the table. 
Observe that when we represent any quantity by the length 
of a line we represent it to some scale or other ; 1 inch 
represents 10 lbs. per square inch or 20 degrees centigrade 
or something else ; it is always to scale and according to a 
convention of some kind, for of course a distance 1 inch is a 
very different thing from 20 degrees centigrade. 

When one has two columns of observed numbers to plot 
on squared paper one does it, 1. To see if the points lie in 
any regular curve. If so, the simpler the curve the simpler 
is the law that we are likely to find. 2. To correct errors of 
observation. For if the points lie nearly in a simple regular 
curve, if we draw the curve that lies most evenly among the 
points, using thin battens of w^ood, say, then it may be taken 
as probable that if there were no errors of observation the 
points would lie exactly in such a curve. Note that when 
a point is — 5 feet to the right of a line, we mean that it is 
5 feet to the left of the line. I have learnt by long ex- 
perience that it is worth while to spend a good deal of time 
subtracting from and multiplying one's quantities to fit the 
numbers of squares (so that the whole of a sheet of paper is 
needed for the points) before beginning to plot. 

Now let the reader buy some squared paper and without 
asking help from anyone let him plot the results of some 


observations. Let him take for example a Whitaker's 
Almanack and plot from it some sets of numbers ; the average 
temperature of every month last year; the National Debt 
since 1688; the present value of a lease at 4 per cent, for 
any number of years ; the capital invested in Railways since 
1849 ; anything will do, but he had better take things in 
which he is interested. If he has made laboratory observa- 
tions he will have an absorbing interest in seeing what sort 
of law the squared paper gives him. 

8. As the observations may be on pressure jj and tem- 
perature t, or p and volume v, or v and t, or Indicated Horse 
Power and Useful Horse Power of a steam or gas engine, or 
amperes and volts in electiicity, and we want to talk generally 
of any such pair of quantities, I shall use x and y instead 
of the p'q and v^ and ^ s and all sorts of letters. The short 
way of saying that there is some law connecting two variable 
quantities x and y is F (x, y) =0...(1), or in words "there 
is some equation connecting x and y." Any expression 
which contains x and y (it may contain many other letters 
and numbers also) is said to be a function of x and y and we 
use such symbols as F(x, y),f{x, y), Q(x, y) etc. to represent 
functions in general when we don't know what the expres- 
sions really are, and often when we do know, but want to 
write things shortly. Again we use F{x) or f{x) or any 
other convenient symbol to mean "any mathematical ex- 
pression containing x" and we say " let f(x) be any function 
of x!* Thus y =f(x) ... (2) stands for any equation which 
would enable us when given x to calculate y. 

X' ifi 

The law ^ + ^t^ = 1 comes under the form (1) given 

above, whereas if we calculate y in terms of x and get 
y = 4- 1 \/25 — x'^ we have the form (2). But in either case 
we have the same law connecting y and x. In pure mathe- 
matics X and y are actual distances ; in applied mathematics 
X and y stand for the quantities which we are comparing and 
which are represented to scale. 

9. ' Graph ' Exercises. 

I. Draw the curve y = 2 + -^j-^x-. 

Take ^ = and we find 2/ =2; take x = l, then 7/ = 20333 ; 

'graph' exercises. 9 

take X = 2, then y = 2 + IS^o = 2-133; and so on. Now plot 
these values of x and y on your sheet of squared paper. The 
curve is a parabola. 

II. Draw the curve y = 1 — ^cc-\- -J^x^ which is also a 
parabola, in the same way, on the same sheet of paper. 

III. Draw the curve xy = 120. Now if x = l, y=120; 
if X = 2, y = 60 ; if x = S, y = 4<0; if a; = 4, y = 30 and so on : 
this curve is a rectangular hyperbola. 

IV. Draw yx'-'^' = 100 or y = 100^-i-«^ If the student 
cannot calculate y for any value of x, he does not know how 
to use logarithms and the sooner he does know how to use 
logarithms the better. 

V. Draw y = ax''^ where a is any convenient number. I 
advise the student to spend a lot of time in drawing members 
of this great family of useful curves. Let him try ?i = — 1 
(he drew this in III. above), 7i=— 2, ?i=— 1-J, n=—^, ?i=— 0*1, 
n = 0, n = "I, ri = f , n = l, n=l\,n = '2 (this is No. I. above), 
?i = 3, w = 4 &c. 

YI. Draw y = a sin (bx + c) taking any convenient 
numbers for a, h and c. 

Advice. As hx-{-c is in radians (one radian is 57'2958 
degrees) and the books of tables usually give augles in 
degrees, choose numbers for h and c which will make the 
arithmetical work easy. Thus take 6 = 1-7- 114"6, take c the 
number of radians which correspond to say 30° 


this is ^ or '5236 

Let a = 5 say. Now let x — 0, 10, 20, &;c., and calculate y. 

Thus when «?= 6, 3/ = 5 sin [yrxTp + "5236) ; but if the 
angle is converted into degrees we have 

2/ = 5 sin (16 + 30 degrees) = b sin 33° = 2723. 

Having drawn the above curve, notice what change would 

occur if c were changed to or ^ or ^ or ^ . Again, if a 

were changed. More than a week may be spent on this curve, 
very profitably. 


VII. Draw y = ae***. Try 6 = 1 and a = 1 ; try other 
values of a and h ; take at least two cases of negative values 
for h. 

In the above work, get as little help from teachers as 
possible, but help from fellow students will be very useful 
especially if it leads to wrangling about the subject. 

The reason why I have dwelt upon the above seven cases 
is this: — Students learn usually to differentiate and integrate 
the most complicated expressions: but when the very simplest 
of these expressions comes before them in a real engineering 
problem they fight shy of it. Now it is very seldom that an 
engineer ever has to face a problem, even in the most intri- 
cate part of his theoretical work, which involves a knowledge 
of more functions than these three 

y = ax^y y = a sin {hx + c), y = ad^, 

but these three must be thoroughly well understood and the 
engineering student must look upon the study of them as his 
most important theoretical work. 

Attending to the above three kinds of expression is a 
student's real business. I see no reason, however, for his not 
having a little amusement also, so he may draw the curves 

X' 4- y^ = 25 (Circle), ^ + f^ = 1 (Ellipse), 

|g-^ = l (Hyperbola), 

and some others mentioned in Chapter, III., but from the 
engineer's point of view these curves are comparatively un- 

10. Having studied y = e~^^ and y = 6 sin (ex + ^) a 
student will find that he can now easily understand one of 
the most important curves in engineering, viz : 

y = be~** sin (ex + g). 

He ought first to take such a curve as has already been 
studied by him, y = h sin (ex + g) ; plot on the same sheet of 
paper y = e~^^ ; and multiply together the ordinates of the 
two curves at many values of x to find the ordinate of the 



new curve. The curve is evidently wavy, y reaching maximum 
and minimum values; y represents the displacement of 
a' pendulum bob or pointer of some measuring instrument 
whose motion is damped by fluid or other such friction, x being 
the time, and a student will understand the curve much better 
if he makes observations of such a motion, for example with 
a disc of lead immersed in oil vibrating so slowly under 
the action of torsional forces in a wire that many ob- 
servations of its angular position (using pointer and scale 
of degrees) which is called y, x being the time, may be 
made in one swing. The distance or angle from an extreme 
position on one side of the zero to the next extreme position 
on the other side is called the length of one swing. The 
Napierian logarithm of the ratio of the length of one swing to 
the next or one tenth of the logarithm of the ratio of the 
first swing to the eleventh is evidently a multiplied by half 
the periodic time, or it is a multiplied by the time occupied 
in one swing. This logarithmic decrement as it is called, 
is rather important in some kinds of measurement. 

11. When by means of a drawing or a model we are able to find the 
path of any point and where it is in its path when we know the 
position of some other point, we are always able to get the same 
information algebraically. 

Example (1). A point jPand a straight line DD being given ; what 
is the path of a point P when it moves so that its distance from the 
point F is always in the same ratio to its distance from the straight line? 

Thus in the figure let FF=exPI> ... (I), where e is a constant. 
Draw FFX at right angles to DD. y 
If the distance FD is called a; and 
the perpendicular FG is y ; our 
problem is this ; — What is the equa- 
tion connecting x and y? Now all 
we have to do is to express (1) in 
terms of x and ^. Let FF be called a. 


FF= \/FG^+FG^= Vy^ + (^'-a)2 

so that, squaring (1) we have 

This is the answer. If e is 1 the 
curve is called a parabola. If e is 
greater than 1, the curve is called an 

Fig. 1. 

hyperbola. If e is less than I, the curve is called an ellipse. 



Example (2). The circle APQ rolls on the straight line OX, 
What is the Path of any point P on. the circumference? If whenP 

touched the line it was at 0, let OX and OF bo the axes, and let SP 
be X and PT be y. Let the radius of the circle be a. Let the angle 
PCQ be called <^. Draw CB, peq^endicular to PT. Observe that 

PB=a . sin PCB=a sin (</> - 90) = - a cos (/>, 

BC= a cos PCB = a sin (/>. 

Now the arc QP= a .cf) = OQ. Hence SiS x=OQ — BC, and 

/c have 

y = BT^-PB, 
x=axf)- a sin 0| 
y = a —acoscfij 


If from (3) we eliminate we get one equation connecting x and y. 
I^ut it is better to retain </> and to have two equations because of the 
greater simplicity of calculation. In fact the two equations (3) may 
be called the equation to the curve. The curve is called the oycloid 
as all my readers know already. 

Example (3). A crank and connecting rod work a slider 

in a straight path. Where is the slider for any position of the crank ? 

Let the path be in the direction of the centre of the crank shaft. 


Fig. 3. 

If A is the end of the path, evidently ^0 is equal to l-\-r, r being 
length of crank. 



It is well to remember in all such problems that if we project all 
the sides of a closed figure upon any two straight lines, we get two in- 
dependent equations. Projecting on the horizontal we see that 


s+l cos (f)+r cos 6=1 + r\ 
Projecting on the vertical ^ sin ^ = r sin ^ J 

If we eliminate from these equations we can calculate s for any 
value of 6. The student ought to do this for himself, but I am weak 
enough to do it here. We see that from the second equation of (1) 


so that the first becomes 



s=l\l-^l-^^^sm^0J+r{l-cosd)..* (2). 

Students ought to work a few exercises, such as; — 1. The ends 
A and B of a rod are guided by two straight slots OA and OB which 
are at right angles to one another ; find the equation to the path of 
any point P in the rod. 2. In Watt's parallel motion there is 
a point which moves nearly in a straight path. Find the equation to 
its complete path. 

In fig. 4 the Mean Position is shown as OABC. The best place 
for P is such that BP/PA = OAICB. Draw the links in any other 

Fig. 4. 

* Note that if as is usual, -j^isa. small fraction, then slnca ,Jl-a=l-\a 

■when a is small, we can get an approximation to the value of s, which can 
be expressed in terms of d and 2d. This is of far more importance than it 
here seems to be. When the straight path of Q makes an angle a with the 
Hne joining its middle point and O, if a is not large, it is evident that s is 
much the same as before, only divided by cos a. When a is large, the 
algebraic expression for s is rather complicated, but good approximations 
may always be found which will save trouble in calculation. 


position. The complete path of P would be a figure of 8. 3. Find 
the equation to the path of a point in the middle of an ordinaiy con- 
necting rod. 4. A^ the end of a link, moves in a straight path COC\ 
being the middle of the path, with a simple haraaonic motion 
OA=asmpti where t is time ; the other end B moves in a straight 
path OBD which is in a direction at right angles to COC ; what is ^'s 
motion ? Show that it is approximately a simple harmonic motion of 
twice the frequency of ^. 5. In any slide valve gear, in which there 
are several links, &c. driven from a uniformly rotating crank ; note 
this fact, that the motion of any point of any link in any particular 
direction consists of a fundamental simple harmonic motion of the 
same frequency as the crank, together with an octave. The proper 
study of Link Motions and Radial valve gears from this 
point of view is' worth months of one's life, for this contains the secret 
of why one valve motion gives a better diagram than another. 
Consider for example the Hackworth gear with a curved and with a 
straight slot. What is the difference ? See Art. 122. 

12. Plotted points l3ring in a straight line. Proofs 
will come later; at first the student ought to get well ac- 
quainted with the thing to be proved. I have known boys 
able to 'prove mathematical propositions who did not really 
know what they had proved till years afterwards. 

Take any expression like y = a + hx, where a and h are 
numbers. Thus let y = 2 + l^x. Now take ^7 = 0, a?=l, 
a; = 2, a? = 3, <Src. and in each case calculate the corresponding 
value of y. Plot the corresponding values of a; and y as the 
co-ordinates of points on squared paper. You will find that 
they lie exactly in a straight line. Now take say y = 2-\-Sx 
or 2 + ^a; or 2 — ^a; or 2 — Sx and you will find in every case 
a straight line. Men who think they know a little about 
this subject already will not care to take the trouble and if 
you do not find yourselves interested, I advise you not to 
take the trouble either; yet I know that it is worth your 
while to take the trouble. Just notice that in every case I 
have given you the same value of a and consequently all 
your lines have some one thing in common. What is it? 
Take this hint, a is the value of y when a? = 0. 

Again, try y=:2 + l^x, y = 1 +l^x, y = 0-i-lix, 

y = ^l + lix, 2/ = -2 + l^a?, 

and so see what it means when h is the same in every case. 
You will find that all the lines with the same b have the 


same slope and indeed I am in the habit of calling h the 
slope of the line. 

If y = a 4- 6^, when oc = x-^, find y and call it y^y 

when a; = a?! + 1, find y and call it y^. 
It is easy to show that y.^—y^ = h. So that what I mean by 
the slope of a straight line is its rise for a horizontal distance 
1. (Note that when we say that a road rises ^ or 1 in 20, 
we mean 1 foot rise for 20 feet along the sloping road. Thus 
■jjj is the sine of the angle of inclination of the road to the 
horizontal; whereas our slope is measured in a different way). 
Our slope is evidently the tangent of the inclination of the 
line to the horizontal. Looking upon y as a quantity whose 
value depends upon that of x, observe that the rate of in- 
crease of y relatively to the increase of x is constant, being 
indeed h, the slope of the line. The symbol used for this 

rate is -p. Observe that it is one symbol; it does not mean 

/Y V 7/ 

-^ — ~ . Try to recollect the statement that if y= a + hx, 
a ^ X 

-~—h, and that if -,— = h, then it follows that y:=A-\-hx, 
dx dx 

where A is some constant or other. 

Any equation of the first degree connecting x and y 

such as Ax-\-By = G where A, B and C are constants, can 

G A 
be put into the shape y = ^ — ^x, so that it is the equation 


to a straight line whose slope is — ^ and which passes 

through the point whose x = 0, whose y = -n, called point 

( 0, p ) . Thus 4^ + 2^/ = 5 passes through the point x = 0, 

y = 2^ and its slope is — 2. That is, y diminishes as x in- 
creases. You are expected to draw this line y = 2|^ — 2^7 and 
distinguish the difference between it and the line 3/ = 2 J + 2x. 
Note what is meant by positive and what by negative slope. 
Draw a few curves and judge approximately by eye of the 
slope at a number of places. 

13. Problems on the straight line. 

1. Given the slope of a straight line; if you are also 



told that it passes through the point whose x = 3, and whose 
y = 2, what is the equation to the line ? Let the slope be 

The equation is y = a + 0'35.'^^^. where a is not known. 

But (3, 2) is a point on the line, so that 2 = a + 035 x 3, 
or a = 0"95 and hence the line is y = 0'95 + 35a?. 

2. What is the slope of any line at right angles to 
y=za-\-bxl Let -45 be the given line, cutting OX in C. Then 

h = tan BOX. If DE is any line at right angles to the first, 
its slope is tan DEX or — tan DEC or - cot BGE or — ^ . 

So that y = A—j-a)is typical of all lines at right angles 
to y = a-{-h.v] A being any constant. 

3. Where do the two straight lines Ax -\-By-\-C—0 
and Mx + Ny -{■8 = meet ? Answer, In the point whose x 
and y satisfy both the equations. We have therefore to do 
what is done in Elementary Algebra, solve simultaneous 

4. When tan a and tan ^ are known, it is easy to find 
tan (a — /S), and hence when the straight lines y=a + hx 
and y = m-\-nx are given, it is easy to find the angle between 

5. The line y = a-\-bx passes through the points a?=l, 
y = 2, and a? = 3, y = 1, find a and h. 


6. A line y = a-{-bx is at right angles to y = 2 + Sa) and 
passes through the point x = l, y = l. Find a and h. 

14. Obtaining Empirical Formulae. 

When in the laboratory we have made measurements of two 
quantities which depend upon one another, we have a table showing 
corresponding values of the two, and we wish to see if there is a simple 
relation between them, we plot the values to convenient scales as the 
co-ordinates of points on squared paper. If some regular curve (a 
cm*ve without singular points as I shall afterwards call it) seems as if 
it might pass through all the points, save for possible errors of 
measurement, we try to obtain a formula y = f(x), which we may call 
the law or rule connecting the quantities called y and x. 

If the points appear as if they might lie on a straight line, a 
stretched thread may be used to help in finding its most probable 
position. There is a tedious algebraic method of finding the straight 
line which represents the positions of the points with least error, but 
for most engineering purposes the stretched string method is suffi- 
ciently accurate. 

If the ciu-ve seems to follow such a law as y~a-\-hx^, plot y and 
the square of the observed measurement, which we call x^ as the co-or- 
dinates of points, and see if they lie on a straight line. If the curve 


seems to follow such a law as y= j ■ (1), which is the same as 

^.\.hy = a, divide each of the quantities which you call y by the corre- 
sponding quantity x ; call the ratio X. Now plot the values of X and 
of y on squared paper ; if a straight line passes through the plotted 

points, then we have such a law as X=A-{-By, or - — A-\-By^ or 

, so that (1) is true. 

^ l-Bx' 

Usually we can apply the stretched thread method to find the 
probability of truth of any law containing only two constants. 

Thus, suppose measurements to be taken from the expansion part 
of a gas engine indicator diagram. It is im]3ortant for many purposes 
to obtain an empirical formula connecting p and v, the pressure and 
volume. I always find that the following rule holds with a fair amount 
of accuracy pv" = C where s and C are two constants. We do not 
much care to know C, but if there is such a rule, the value of s is very 
important*. To test if this rule holds, plot log jo and log v as the co- 

* There is no known physical reason for expecting such a rule to hold. 
At first I thought that perhaps most curves drawn at random approximately 
like hyperbolas would approximately submit to such a law as yx^= C, but I 
found that this was by no means the case. The following fact is. worth 
mentioning. When my students find, in carrying out the above rule that 
log p and log v do not lie in a straight line, I find that they have 

P. 2 


ordinates of points on squared pa^^er (common logarithms will do). 
If they lie approximately in a straight line, we see that 

a constant, and therefore the rule holds. 

When we wish to test with a formula containing three independent 
constants we can often i*educe it to such a shape as 

Av+Bw+Cz=l (2), 

where y, w, z contain x and y in some shajje. Thus to test if y=—- — , 

we have y + cxy —a-\-hx^ ov- + -xy — x=\. Here y itself is the old v, 

xy is the old w>, and x itself is the old z. 

If (2) holds, and if t?, \o and z were plotted as the three co-ordinatea 
of a ix)int in space, all the points ought to lie in a plane. By means 
of three sides of a wooden box and a number of beads on the ends of 
pointed wires this may be tried directly ; immersion in a tank of water 
to try whether one can get the beads to lie in the plane of the siuface 
of the water, being used to find the plane. I have also used a descrip- 
tive geometry method to find the plane, but there is no method yet 
used bv me which compares for simplicity with the stretched thread 
method in the other case. 

But no hard and fast rules can be given for trying all sorts of em- 
pirical formulae upon one's observed numbers. The student is warned 
that his formula is an empirical one, and that he must not deal with it 
as if he had discovered a natural law of infinite exactness. 

AVhen other formulae fail, we try 


because we know that with sufficient terms this will satisfy any curve. 
When there are more than two constants, we often find them by a 
patient application of what is called the method of least squares. To 
test if the pressure and temperature of saturated steam follow the 
rule jo = a(^+/3)"... (3), where B is temperature, Centigrade, say, three 
constants have to be found. The only successful plan tried by me is 
one in which I guess at /3. I know that /3 is nearly 40. I ask one 
student to try ^ = 40, another to try i3 = 41, another ^=39 and so on; 

made a mistake in the amount of clearance. Too much clearance and too 
little clearance give results which depart in opposite ways from the straight 
hne. It is convenient in many calculations, if there is such an empirical 
formula, to use it. If not, one has to work with rules which instruct us to 
draw tangents to the curve. Now it is an excellent exercise to let a number 
of students trace the same curve with two points marked upon it and to let 
them all independently draw tangents at those points to their curve, and 
measure the angle between them. It is extraordinary what very different 
lines they will draw and what different angles they may obtain. Let them 
all measure by trial the radius of curvature of the curve at a point ; in this 
the discrepancies are greater than before. 



they are asked to liiid the rule (3) which most accurately represents 
p and 6 between, say jo = 71b. per sq. inch, and p = 150. He who gets 
a straight Hne lying most evenly (judging by the eye) among the points, 
when \ogp and log(^-f /3) are used as co-ordinates, has used the best 
value of 4. The method may be refined upon by ingenious students. 

(See end of Chap. I.) 

15. We have now to remember that if y = a-^hx, then 

-^ — h, and if -^ = h, then y = A -{■ hx, Avhere A is some 
ax ax ^ 


Let us prove this algebraically. 

\iy = a-\-hx. Take a particular value of x and calculate 
y. Now take a new value of x, call it ^ + hx, and calculate 
the new y, call it y -t- Sy, 

y ^^y — a-\-h{x-\- hx). 

Subtract y—a-\-hx and we get 

hy = hhx, or -^ = 6, 

and, however small hx or Zy may become their ratio is 6, we 
therefore say ~ = h. 

Fig. 6. 

16. In the curve of fig. 6 there is positive slope 

{^ increases as x increases) in the parts AB, DF and HI and 




Fig. 7. 

negative slope {y diminishes as x increases) in the parts 
BD and FH. The slope is at 5 and F which are called 
points of maximum or points where i/ is a maximum; and it 
is also at D and H which are points of minimum. The 
point E is one in which the slope ceases to increase and 
begins to diminish : it is a point of inflexion. 

Notice that if we want to know the slope at the point P 
we first choose a point F which 
is near to P. (Imagine that 
in fig. 6 the little portion of 
the curve at P is magnified a 
thousand times.) Call PS=x, 
sotha,tPM=Bx,F3I==Si/. Now 
FM/PM or By/Bx is the average 
slope between P and F. It is 
tan FPM. Imagine the same 
sort of figure drawn but for a 
point F' nearer to P. Again, 
another, still nearer P. Ob- 
serve that the straight line FP or F'P or F"P gets gi-adually 
more and more nearly what we mean by the tangent to the 
curve at P. In every case hyjhx is the tangent of the angle 
which the line FP or F'P or F'P makes with the horizontal, 
and so we see that in the limit the slope of the line or dy/dx 
at P is the tangent of the angle which the tangent at P 
makes with the axis of X. If then, instead of judging 
roughly by the eye as we did just now in discussing fig. 6, we 
wish to measure very accurately the slope at the point P ; — 
Note that the slope is independent of where the axis of X is, 
so long as it is a horizontal line, and I take care in using my 
rule here given, to draw OX below the part of the curve 
where I am studying the slope. Draw a tangent PR to 
the curve, cutting OX in P. Then the slope is tan PRX. 
If drawn and lettered according to my instructions, observe 
that PRX is always an acute angle when the slope is 
positive and is always an obtuse angle when the slope is 

Do not forget that the slope of the curve at any point 
means the rate of increase of y there with regard to a?, and 


that we may call it the slope of the curve or tan PRQ or by 
the symbol -^- or "the differential coefficient of y with 
regard to x," and all these mean the same thing. 

Every one knows what is meant when on going up a 
hill one says that the slope is changing, the slope is diminish- 
ing, the slope is increasing ; and in this knowledge he already 
possesses the fundamental idea of the calculus. 

17. We all know what is meant when in a railway train we 
say ^'we are going at 30 miles per hour.^' Do we mean 
that we have gone 30 miles in the last hour or that we are 
really going 30 miles in the next hour ? Certainly not. We 
may have only left the terminus 10 minutes ago ; there may be 
an accident in the next second. What we mean is merely this, 
that the last distance of 3 miles was traversed in the tenth 
of an hour, or rather, the last distance of 0*0003 miles was 
traversed in 0*00001 hour. This is not exactly right ; it is 
not till we take still shorter and shorter distances and divide 
by the times occupied that we approach the true value of 
the speed. Thus it is known that a body falls freely 
vertically through the following distances in the following 
intervals of time after two seconds from rest, at London. 
That is between 2 seconds from rest and 2*1 or 2*01 or 2001, 
the distances fallen through are given. Each of these 
divided by the interval of time gives the average velocity 
during the interval. 

Intervals of time in seconds 
Distances in feet fallen through 
Average velocities 

We see that as the interval of time after 2 seconds is 
taken less and less, the average velocity during the interval 
approaches more and more the true value of the velocity at 
2 seconds from rest which is exactly 64*4 feet per second. 

We may find the true velocity at any time when we know 
the law connecting s and t as follows. 

Let s = 16'lt2, the well known law for bodies falling 
freely at London. If t is given of any value we can calculate 











s. If t has a slightly greater value called t-\- St (here St 
is a symbol for a small portion of time, it is not S "Xt, but a 
very different thing), and if we call the calculated space 
s + Ss, then s -^ hs = U'l (t + Btf or l&l {f + 2t. St + (Sty]. 
Hence, subtracting, Ss = 16*1 {2t . St 4- (Sty], and this formula 
will enable us to calculate accurately the space Ss passed 
through between the time t and the time t+St. The average 
velocity during this interval of time is Ss -^ St or 

^ = 32-2^+ 10- 1 8/. 

Please notice that this is absolutely correct ; there is no 
vagueness about it. 

Now I come to the important idea; as St gets smaller 

and smaller, ^ approaches more and more nearly 32*2^, the 

other term 16'1S^ becoming smaller and smaller, and hence 
we say that in the limit, Ss/St is truly 32-2^. The limiting 

value of ^ as St gets smaller and smaller is called , or the 

rate of change of s as t increases, or the differential coefficient 
of s with regard to t, or it is called the velocity at the time t 

Now surely there is no such great difficulty in catching 
the idea of a limiting value. Some people have the notion 
that we are stating something that is only approximately 
true ; it is often because their teacher will say such things as 
"reject 16'1S^ because it is small," or "let dt be an infinitely 
small amount of time" and they proceed to divide something 
by it, showing that although they may reach the age of 
Methuselah they will never have the common sense of an 

Another trouble is introduced by people saying " let 

Ss ds 
St — and kt or -y- is so and so." The true statement is, " as 

of at . . Ss 

St gets smaller and smaller without limit, ^ approaches more 

and more nearly the finite value 322^," and as I have already 
said, everybody uses the important idea of a limit every day 
of his life. 



From the law connecting s and t, if we find -r- or the 

velocity, we are said to differentiate s with regard to the 
time t. When we are given ds/dt and we reverse the above 
process we are said to integrate. 

If I were lecturing I might dwell longer upon the correct- 
ness of the notion of a rate that one already has, and by 
making many sketch'es illustrate my meaning. But one may 
listen intently to a lecture which seems dull enough in a 
book. I will, therefore, make a virtue of necessity and say 
that my readers can illustrate my meaning perfectly well to 
themselves if they do a little thinking about it. After all 
my great aim is to make them less afraid than they used to be 
of such symbols as dy/dx and fy . dx. 

18. Given s and t in any kind of motion, as a set of num- 
bers. How do we study the motion ? For example, imagine 
a Bradshaw's Railway Guide which not merely gave a few 
stations, but some hundred places between Euston and Rugby. 
The entries might be like this: s would be in miles, t in 
hours and minutes. .9 = would mean Euston. 

10 o'clock 


10. .10 


10. .15 


10.. 20 


10. .23 


10.. 28 


10. .33 


One method is this : plot t (take times after 10 o'clock) 
horizontally and s vertically on a sheet of squared paper and 
draw a curve through the points. 

The slope of this curve at any place represents the velocity 
of the train to some scale which depends upon the scales for 
s and t. 


Observe places where the velocity is great or small. 
Between ^=10.20 and i = 10.23 observe that the velocity 
is 0. Indeed the train has probably stopped altogether. 
To be absolutely certain, it would be necessary to give s 
for every value of t, and not merely for a few values. A 
curve alone can show every value. I do not say that the 
table may not be more valuable than the curve for a great 
many purposes. 

If the train stopped at any place and travelled towards 
Euston again, we should have negative slope to our curve 
and negative velocity. 

Note that acceleration being rate of change of velocity 
with time, is indicated by the rate of change of the slope of 
the curve. Why not on the same sheet of paper draw a curve 
which shows at every instant the velocity of the train ? 
The slope of this new curve would evidently be the accelera- 
tion. I am glad to think that nobody has yet given a name 
to the rate of change of the acceleration. 

The symbols in use are 

s and t for space and time ; 

velocity i; or -^ , or Newton's symbol s ; 

acceleration -j- or -yz , or Newton's s. 
at dv 

Rate of change of acceleration would be -r- . 

Note that ^ is one symbol, it has nothing whatever to do 

d^ X s 
with such an algebraic expression as , — -— . The symbol is 

Gt X I 

supposed merely to indicate that we have differentiated s 
twice mth regard to the time. 

I have stated that the slope of a curve may be found by 
drawing a tangent to the curve, and hence it is easy to find 
the acceleration from the velocity curve. 



19. Another way, better than by drawing tangents, is 
illustrated in this Table : 







feet per 

in feet per 
second per 
second or 

or 8sJ8t 































In a new mechanism it was necessary for a certain 
purpose to know in every position of a point A what its 
acceleration was, and to do this I usually find its velocity 
first. A skeleton drawing was made and the positions of 
A marked at the intervals of time t from a time taken 
as 0. In the table I give at each instant the distance 
of A from a fixed point of measurement, and I call it s. 
If I gave the table for all the positions of A till it gets 
back again to its first position, it would be more instructive, 
but any student can make out such a table for himself 
for some particular mechanism. Thus for example, let s 
be the distance of a piston from the end of its stroke. 
Of course the all-accomplished mathematical engineer will 
scorn to take the trouble. He knows a graphical rule 
for doing this in the case of the piston of a steam 
engine. Yes, but does he know such a rule for every 


possible mechanism ? Would i^t be worth while to seek 
for such a graphical nile for every possible mechanism ? 
Here is the straightforward Engineers' common-sense way 
of finding the acceleration at any point of any mechanism, 
and although it has not yet been tried except by myself 
and my pupils, I venture to think that it will commend 
itself to practical men. For beginners it is invaluable. 

Now the mass of the body whose centre moves like the 
point A, being m (the weight of the body in pounds at 
London^ divided by 32*2)*, multiply the acceleration in 
feet per second per second which you find, by m, and you 
have the force which is acting on the body increasing the 
velocity. The force will be in pounds. 

* I have given elsewhere my reasons for using in books intended for 
engineers, the units of force employed by all practical engineers. I have 
used this system (which is, after all, a so-called absolute system, just as 
much as the c. a. s. system or the Foundal system of many text books) for 
twenty years, with students, and this is why their knowledge of mechanics 
is not a mere book knowledge, something apart from their practical work, 
but fitting their practical work as a hand does a glove. One might as well 
talk Choctaw in the shops as speak about what some people call the 
English system, as if a system can be English which speaks of so many 
poundals of force and so many foot-poundals of work. And yet these same 
philosophers are astonished that practical engineers should have a contempt 
for book theory. I venture to say that there is not one practical engineer 
in this country, who thinks in Poundals, although all the books have used 
these units for 30 years. 

In Practical Dynamics one second is the unit of time, one foot is the 
unit of space, one ijoiind (what is called the weight of 1 lb. in London) is the 
unit of force. To satisfy the College men who teach Engineers, I would say 
that "The unit of Mass is that mass on which the force of 1 lb. produces 
an acceleration of 1 ft. per sec. per see." 

We have no name for unit of mass, the Engineer never has to speak 
of the inertia of a body by itself. His instructions are "In all Dynamical 
calculations, divide the weight of a body in lbs. by 32'2 and you have its 
mass in Engineer's units — in those units which will give all your answers in 
the units in which an Engineer talks." If you do not use this system evei-y 
answer you get out will need to be divided or multiplied by something before 
it is the language of the practical man. 

Force in pounds is the space-roXe at which work in foot-pounds is done, 
it is also the f/we-rate at which momentum is produced or destroyed. 

Example 1. A Kaxnxner head of 2^ lbs. moving with a velocity of 40 ft. 
per sec. is stopped in '001 sec. What is the average force of the blow? 
Here the mass being 2^-^32•2, or -0776, the momentum (momentum is 
mass X velocity) destroyed is 3-104. Now force is momentum per sec. and 
hence the average force is 3•104-^•001 or 3104 lbs. 

Example 2. Water in a jet flows with the linear velocity of 20 ft. per sec. 


20. We considered tlic case of falling bodies in which 
space and time are connected by the law 6' = iu^^? where cf 
the acceleration due to gravity is 32 "2 feet per second per 
second at London. But many other pairs of things are con- 
nected by similar laws and I will indicate them generally by 

y = ax2. 

Let a particular value of a be taken, say a = -^. Now take 
a) = 0, x = l, x==2, 00 = 3, &c. and in every case calculate y. 

Plot the corresponding points on squared paper. They 
lie on a parabolic curve. At any point on the curve, say 

where x = S, find the slope of the curve ( I call it ~j^] , do 

the same at a; = 4, ^ = 2, &c. Draw a new curve, now, with the 

same values of x but with ~- as the ordinate. This curve 


shows at a glance (by the height of its ordinate) what is 

the slope of the first curve. If you ink these curves, let 

the y curve be black and the -~ curve be red. Notice that 

the slope or -~ at any point, is 2a multiplied by the x of the 


We can investigate this algebraically. As before, for 
any value of x calculate y. Now take a greater value of x 
which I shall call x + Bx and calculate the new y, calling it 
y + 8y. We have then 

y -\- By = a (x -{- Bxf 

= a[x^^+2x.Bx + (8xy}. 

Subtracting ; By = a [2x . Bx + (Bx)-]. 

Divide by Bx, -— = 2ax + a . Bx. 

(relatively to the vessel from which it flows), the jet being 0*1 sq. ft. in cross 
section ; what force acts upon the vessel ? 

Here we have 20 x -1 cu. ft. or 20 x -1 x 62-3 lbs. of water per sec. or 
a mass per second in Engineers' units of 20 x -1 x 62*3-1- 32-2. This mass is 
3-^7, its momentum is 77 '4, and as this momentum is lost by the vessel 
every second, it is the force acting on the vessel. 

A student who thinks for himself will see tliat this force is the same 
whether a vessel is or is not in motion itself. 


Imagine Bw to get smaller and smaller without limit and 
use the symbol —^ for the limiting value of -A > ^^^ we have 
-p = 2aic, a fact which is known to us already from our 


squared paper*. 

21. Note that when we repeat the process of differentiation 
we state the result as -^ and tho answer is 2a. You must 
become familiar with these symbols. If y is a function of 
X, ,- is the rate of change of y with regard to a?; t4 is the 
rate of change of -^ with regard to x. 

Or, shortly; ^^ is the differential coefficient of ^ 

with regard to a? ; ;p is the differential coefficient of y with 
regard to x. 

Or, again ; integrate ~; and our answer is ~- ; inte- 
grate ~ and our answer is y. 

You will, I hope, get quite familiar with these symbols 
and ideas. I am only afraid that when we use other letters 
than x& and y^ you may lose your familiarity. 

* Symbolically. Let 2/=/(a;)..,(l), where /(a;) stands for any expression 
containing x. Take any value of x and calculate y. Now take a slightly 
greater value of x say x + 8x and calculate the new y ; call it y + 8y 

then y + dy^f{x + 8x) (2). 

Subtract (1) from (2) and divide by dx. 

Sy_ f{x + dx)-f{ x) 

5^- 5i ^^^• 

,^ , dw . f(x + 8x)-f(x) 

What we mean by ^~ is the limiting value of —— — ^-^-^ as dx is 
ax ox 

made smaller and smaller without limit. This is the exact definition of -^ . 

It is quite easy to remember and to write, and the most ignorant person may 
get full marks for an answer at an examination. It is easy to see that the 
differential coefficient of of {x) is a times the differential coefficient of / (x) 
and also that the differential coefficient of/ {x) + F.{x) is the sum of the two 
differential coefficients. 


The differential coefficient of 
where a, b and c are constants, is 


The integral of + b-\-kx with regard to w is A-\-bx + ^kaf, 
where A is any constant whatsoever. 

Similarly, the integral of 6 + A;^ with regard to z is 

A+bz-i- ^kz\ 
The integral oib-^kv with regard to v/i& A + bv-\- ^kv"^. 

It is quite easy to work out as an exercise that if y=aa^^ then 
~-=Mx^, and again that if y = ax^^ then -p=4a^. All these are 

examples of the fact that if y = ax\ then -^—naod^'K 

In working out any of these examples we take it that —■ becomes -— 


or that ^y=8xx-^ more and more nearly as dx gets smaller and 

smaller without limit. 

This is sometimes written y-\-by=y-{-bx . -f ,ov 

f{x+bx)^f{x) + bx'^.. (I). 

22. Uniformly accelerated motion. 


If acceleration, ^2~^ (!)• 

Integrate and we have ~z=b-\-at— velocity v. Observe 

that we have added a constant 6, because if we differentiate 
a constant the answer is 0. There must be some information 
given us which will enable us to find what the value of the 
constant b is. Let the information be v = Vo, when i = 0. 
Then b is evidently v^. 

So that velocity v— ^- =Vo + at (2). 


Again integrate and 5 = c + ?;„< + ^at-. Again you will notice 
that we add an unknown constant, when wc integrate. Some 
information must be given us to find the value of the constant 
c. Thus if s = So when ^ = 0, this 5o is the value of c, and so 
we have the most complete statement of the motion 

s = So + Vot + ^at^ (3). 

If (3) is differentiated, we obtain (2) and if (2) is differen- 
tiated we obtain (1). 

23. We see here, then, that as soon as the student is 
able to differentiate and integrate he can work the fol- 
lowing kinds of problem. 

I. If s is given as any function of the time, differentiate 
and the velocity at any instant is found ; differentiate again 
and the acceleration is found. 

II. If the acceleration is given as some function of the 
time, integrate and we find the velocity; integrate again 
and we find the space passed through. 

Observe that s instead of being mere distance may be 
the angle described, the motion being angular or rotational. 

Better then call it 0. Then 6 or -^- is the angular velocity 

and S or J— is the angular acceleration. 

24. Exercises on Motion with constant Accele- 

1. The acceleration due to gravity is doiunwards and is 
usually called g, g being 32*2 feet per second per second at 
London. If a body at time is thrown vertically upwards 
with a velocity of F© feet per second; where is it at the end 
of t seconds ? If s is measured upwards, the acceleration is 
— g and s^V^Jt — \gt^. (We assume that there is no resist- 
ance of the atmosphere and that the true acceleration is g 
downwards and constant.) 

Observe that v=Vo—gt and that v=0 when Vo—gt = 

or t = ~- . When this is the case find s. This eives the 

highest point and the time taken to reach it. 

When is s = again ? What is the velocity then ? 


2. The body of Exercise 1 has been given, in addition 
to its vertical velocity, a horizontal velocity Uo which keeps 
constant. If oo is the horizontal distance of it away from 

(I?X (17' 

the origin at the time ^, — = and -7- = Uq, x — ujt. If we 
call s by the new name ?/, we have at any time t, 
y = V,t - \gt\ 

iXj ^^ UQVy 

V of 

and if we eliminate t, we find y=—^x — ^g-- which is a 


3. If the body had been given a velocity V in the 
dii'ection a above the horizontal, we may use V sin a for Vq 
and V cos a for Uf^ in the above expressions, and from them 
we can make all sorts of useful calculations concerning pro- 

Plot the curve when V= 1000 feet per second and a = 45°. 
Again plot with same V when a = 60°, and again when 
a = 30°. 

25. Kinetic Energy. A small body of mass vi is at 
5 = when ^ = and its velocity is Vq, and a force F acts 
upon it causing an acceleration F/ni. As in the last case 
at any future time 

v^Vf, + - t...(l), and s = -^Vot + I ~ t" (2), 

(2) maybe written s^^tilv^A — t\ and it is easy to see 

from (1) that this is s = \t{vQ-\-v), and that the average 

velocity in any interval is half the sum of the velocities at 

the beginning and end of the interval. Now the work done 

by the force F in the distance s is Fs. Calculating F from 

(1), F={v — Vq)— and multiplying upon s we find that the 

work is ^m (v^ — Vq^) which expresses the work stored up in a 

moving body in terms of its velocity. In fact the work 

done causes ^mvo^ to increase to imv^ and this is the reason 

why ^mv'^ is called the kinetic energy of a body. 


Otherwise. Let a small body of mass iii and velocity v 

pass through the very small space hs in the time ht gaining 

velocity hv and let a force F be acting upon it. Now 

F= m X acceleration or ^= 77i -^ and Bs=^v , Bt so that 

F,hs=^ nivBt -^ =1^1' vBv = BE, 

if BE stands for the increase in the kinetic energy of the body 


-K- = m . -y. But our equations are only entirely true when 

Bs, Bt, &c., are made smaller and smaller without limit : 

Hence as -Y- = mv, or in words, "the differential coefficient of 

E with regard to v is mv" if we integrate with regard to v, 

E = \mv^ + c where c is some constant. Let E — when 

v==0 m that c = and we have E — |mt;l 

Practise differentiation and integration using other letters 

than X and y. In this case -^ stands for our old -^^ If we 
•^ dv dx 

had had —■ = mx it might have been seen more easily thaf 

y = Jmar* + c, but you must escape from .the swaddling bands 
of X and y. 

26. Exercise. If x is the elon/^ation of a spring 

when a force F is applied and if a; = - , a representing the 

stiffness of the spring ; F .Bx'is the work done in elongating 
the spring through the small distance Bx. If F is gradually 
increased from to i^ and the elongation from to x, what 
strain energy is stored in the spring ? 

The gain of energy from x to x-\-Bx is BE = F . Bx, or 
rather ~=— = F = ax, hence E = ^ax^ + c. Now if ^ = when 

x=^0, we see that c = 0, so that the energy E stored is 
E=iax' = iFx..,{l). 

It is worth noting that when a mass M is vibrating at the 
end of a spiral spring ; when it is at the distance x from its 


"... ^< ■ '"'' ■ ^". 

position of equilibrium, the potential energy is ^ax^ and the 
kinetic energy is Jilfy- or the ^/tot^l, energy is ^Mv^ + ^ax"^.... 

Note that when a forc0 F is required to pi-oduce an 
elongation or compression ^ in a rod, or a deflexion ^r in a beam, 
and ii F=^ ax where a is some constant, the energy stored up 
as strain energy or potential energy is ^ax^ or ^Fx. 

Also if a Torque T is required to produce a turning 
through the angle ^ in a shaft or spring or other structure, 
and if r = aO, the energy stored up as strain energy or 
potential energy is |a^ or ^T6. If T is in pound-feet and 
is in radians, the answer is in foot-pounds. 

Work done = Force x distance, or Torque x angle. 

27. If the student knows anything about electricity let 
him translate into ordinary language the improved Ohm's law 

V = RC + L.dC/dt (1). 

Observe that if R (Ohms) and L (Henries) remain con- 

stant, if C and -^ are known to us, we know F, and if the 

law of F, a changing voltage, is known you may see that there 
must surely be some means of finding C the changing current. 
Think of L as the back electromotive force in volts when the 
current increases at the rate of 1 ampere per second. 

If the current in the primary of a transformer, and there- 
fore the induction in the iron, did not alter, there would be 
no electro-motive force in the secondary. In fact the E.M.F. 
in the secondary is, at any instant, the number of turns of 
the secondary multiplied by the rate at which the induction 
changes per second. Rate of increase of / per second is what 
we now call the differential coefficient of / with regard to 
time. Although L is constant only when there is no iron or 
else because the induction is small, the correct formula being 

V= BC-h /V ,- (2), it is found that, practically, (1) with 

L constant is of nearly universal application. See Art. 183. 

28. If y = ax" and you wish to find -,- , I am afraid that 


I must assume that you know the Binomial Theorem 
which is : — 

(x + by = a?« + nbo)''-^ + ^ ^^^ " ^ 62a;«-2 

+ -^ ^ ^ h'x''-^ + &c. 

It is easy to show by multiplication that the Binomial 
Theorem is true when n = 2 or 3 or 4 or 5, but when n = ^ or 
J or any other fraction, and again, when n is negative, you 
had better perhaps have faith in my assertion that the 
Binomial Theorem can be proved. 

It is however well that you should see what it means by 
working out a few examples. Illustrate it with n = 2, then 
?i = 3, 71 = 4, &c., and verify by multiplication. Again try 
n = — l, and if you want to see whether your series is correct, 

just recollect that (x + h)~^ is j and divide 1 by a? + 6 in 

the regular way by long division. 

Let us do with our new function of co as we did with aao^. 
Here y = aoo'\ y-\-hy = a{x + hxY =a[x^-\- n.hx . x^-^ 

Tl ill/ ~~ W 

H — ^-^ — - {hx)' x^~^ + terms involving higher powers of hx]. 
Now subtract and divide by hx and you will find 


n . ^"-1 + ^^\ ^^ {hx) X''-' + &c. 

We see now that as 8x is made smaller and smaller, in 
the limit we have only the first term left, all the others 
having in them Bx or (8xy or higher powers of Bx, and they 
must all disappear in the limit, and hence, 

^ = nax"-i. (See Notes p. 159.) 

Thus the differential coefficient of a^ is 6x^, of x^^ it is 
2^x^^, and of x~^ it is — |^~i 

When we find the value of the differential coefficient of 
any given function we are said to differentiate it. When 

given -^ to find y we are said to integrate. The origin of 


. dec. 35 

the words differential and integral need not be considered. 
They are now technical terms. 

Differentiate ax^ and we find nax'^~^. 

Integrate naaf^^ and we find ax^ -\- c. We always add a 
constant when we integrate. 

Sometimes we write these, ^- (ax'^) = nax'^~^ and 

dx^ ^ 

\nax'^~^ . dx = aaf\ 

Obsei've that we write / before and dx after a function 

when we wish to say that it is to be integrated with regard 
to X. Both the symbols are needed. At present you ought 
not to trouble your head as to why these particular sorts of 
symbol are used*. 

You will find presently that it is not difficult to learn 
how to differentiate any known mathematical function. You 
will learn the process easily ; but integration is a process of 
guessing, and however much practice we may have, ex- 
perience only guides us in a process of guessing. To some 
extent one may say that differentiation is like multiplica- 
tion or raising a number to the 5t]i power. Integration is 
like division, or extracting the 5th root. Happily for the 
engineer he only needs a very few integi-als and these are 

* When a great number of things have to be added together in an 
engineer's office — as when a clerk calculates the weight of each little bit of a 
casting and adds them all up, if the letter w indicates generally any of the 
little weights, we often use the symbol I,io to mean the sum of them all. 
When we indicate the sum of an infinite number of little quantities we 

replace the Greek letter s or S by the long English s or / . It will be seen 

presently that Integration may be regarded as finding a sum of this kind. 

Thus if y is the ordinate of a curve ; a strip of area is y . dx and jy . dx 

means the sum of all such strips, or the whole area. Again, if 8m stands for 
a small portion of the mass of a body and r is its distance from an axis, then 
r^ . 57/1 is called the moment of inertia of dm about the axis, and Sr^ . dm or 


r^ . dm indicates the moment of inertia of the whole body about the axis. 
Or if 5 F is a small element of the volume of a body and m is its mass per 

unit volume, then ir^m . dV is the body's moment of inertia. 



well known. As for the rest, lie can keep a good long list of 
them ready to refer to, but he had better practise working 
them out for himself. 

Now one is not often asked to integrate n(Mf~^. It is too 
nicely arranged for one beforehand. One is usually asked to 

inteoi-ate 6a?"*...(l). I know that the answer is -,...(2). 

How do I prove this ? By differentiating (2) I obtain (1), there- 
fore I know that (2) is the integral of (1). Only I ought to 
add a constant in (2), any constant whatever, an arbitrary 
constant as it is called, because the differential coefficient of 
a constant is 0. Students ought to work out several 
examples, integrating, say, a*^, hsc^, hx^, ax~^, ci^, auo^. When 
one has a list of differential coefficients it is not wise to use 
them in the reversed way as if it were a list of integrals, for 
things are seldom given so nicely arranged. 

For instance J4>a^ .da; = a^. But one seldom is asked to 

integrate ^oc^, more likely it will be Sx^ or 5./"^, that is given. 

We now have a number of interesting results, but this 
last one includes the others. Thus if 

y = (f? or y — x^ or y = a?^ or 3/= a;"*, 

we only have examples of y — x*^, and it is good for the 
student to work them out as examples. Thus 


If 7i = 1 this becomes \x^ or 1. If ?i = it becomes ()x~'^ 
or 0. But we hardly need a new way of seeing that if y is a 
constant, its differential coefficient is 0. We know that if 

y = a-^hx + cx^ + ex^-\- &c. + (/x^\ 
f = + h + 2cx-^ Sex" + &c. + ng 

with this knowledge we have the means of working quite 

* I suppose a student to know that anj^thing to the power is unit}-. It 
is instructive to actually calculate by logarithms a high i-oot of any number 
to see how close to 1 the answer comes. A high root means a small power, 
the higher the root the more nearly does the power approach 0. 

Then ^ = + 6 + 2c^ + Sex" + &c. + ngx''-^ 


half the problems supposed to be difficult, that come before 
the engineer. , 

The two important things to remember now, are : If 

y = ax'K then -~ = nax'^~^ ; and if -:,*- = bx^, then 
-^ ' dx ' dx ' 

y = i^''"+i + c, 

^ m + 1 

where c is some constant, or 


bx'^dx = *•'«+! + c. 

m + 1 

I must ask students to try to discover for themselves 

illustrations of the fact that if y = x'\ then -i^ = ??a'"-^ I do 

-^ dx 

not give here such illustrations as happened to suit myself; 

they suited me because they were my own discovery. I 

would suggest this, however : 

Take y = af\ Let x = l'()2, calculate y by logarithms. 
Now let X = 1*03 and calculate y. Now divide the increment 
of y by *01, which is the increment of x. 

Let the second x be 1"021, and repeat the process. 

Let the second x be 1*0201, and repeat the process. 

It will be found that ^^ is approaching the true value of 

^^ which is 5 (l-02y. 

Do this again when y = x^"^ for example. A student need 
not think that he is likely to waste time if he works for 
weeks in manufacturing numerical and graphical illustrations 
for himself. Get really familiar with the simple idea that 

if V = j;" then -,- = nx'^-^ : 
^ dx ' 

that \ax^ . dx — — — - ip*+i + constant : 

J 5 + 1 

that \av^ . dv = \f^^ + constant. 

j sH-1 

Practise this with s= '7 or "8 or I'l or —5 or —"8, and use 
other letters than x or v. 



29. Exercises. Find the following Integrals. The 
constants are not added. 

Ix". dx. Answer, \x^. \v^ . dv. Answer, ^if. 

/v~*. dv. The answer i.> - — v^~^- 
J 1-5 

j\^V' .dv or jv^ , dv. Answer, ^v^, 
r^.dt. Answer, 2e^ 

j-dx or lx~^.dx. Here the rule fails to help us for 

of . . 
we get -zr which is x , and as we can always subtract an in- 
finite constant our answer is really indeterminate. In our 
work for some time to come we need this integral in only 
one case. Later, we shall prove that 

I— dx = log X, and 1 ^— dx = log (x + a), 

and if y = log a?, -— = - and l-dv = log v. 

If j9 = av^, then ~ = Sav\ 

If t; = mrK then ^ = _ ^^^rl 

30. If pv = Rt, where i^ is a constant. Work the 
following exercises. Find -— , if v is constant. Answer, — . 

Find ,- , if « is constant. Answer, — . 
at ^ p 

The student knows already that the three variables p, v 

and t are the pressure volume and absolute temperature of a 

gas. It is too long to Avrite -^ when v is constant." We 




shall use for this the synibol [-4i) , the brackets indicating that 
the variable not there mentioned, is constant. 

Find (;/ ) • Answer, As p=Rt . v~^ we have [-T-) = — Rtv~\ 

and this simplifies to —pv~^. 
Find f ^' ] . Answer, As v = Rt . p~^ we have (;t- ) = — Rtp~^, 

and this simplifies to — vp''\ 

Find (I) . Answer, As i = | . /, we have Q = ^ . 

Find the continued product of the second, fifth, and third of 
the above answers and meditate upon the fact that 

fdv\ fdt\ (dp\___ 
[dtj [dp) [dv)" 

Generally we may say that if u is a function of two 
variables x and y, or as we say 


efficient of u with regard to x when y is considered to 
be constant. 

These are said to be partial differential coefiKcients. 

31. Here is an excellent exercise for students : — 
Write out any function of x and y ; call it u. 

Find ( ^7- ) • Now differentiate this with regard to y, 

assuming that x is constant. The symbol for the result is 

dy .dx' 

It will always be found that one gets the same answer if 
one differentiates in the other order, that is 

d^u d^u 


then we shall use the symbol ( t- ) to mean the differential co- 

dy . dx dx . dy 


Thus iry 11 = 0,^ + if+ a(i?ij + hx]f. 

Again, \-^\ = + 3?/= 4- ax- + ^hxy, 

and r/""V = ^ + ^ + 2aa; + 2bij, 

which is the same as before. 

A student ought not to get tired of doing this. Use other 
letters than w and y, and work many examples. The fact 
stated in (3) is of enormous importance in Thermodynamics 
and other applications of Mathematics to engineering. A 
proof of it will be given later. The student ought here to 
get familiar with the importance of what will then be 

32. One other thing may be mentioned. Suppose we 
have given us that ii is a function of x and y, and that 

^ =aa? + hf-\-co(?y-\-gxy\ 

Then the integral of this is 

11 = lax* + by^x + Jc^y + ^gxy -\-f( y), 

where f{y) is some arbitrary function of y. This is added 
because we always add a constant in integration, and as ?/ is 

regarded as a constant in finding (-7- J we add f(y), which 

may contain the constant y in all sorts of forms multiplied 
by constants. 

33. To illustrate the fact, still unproved, that if y = log x, 

then ^- = - . A student ouo^ht to take such values of x as 3, 
dv X ^ 

3001, 3*002, 3"003 &c., find y in every case, divide increments 

of y by the corresponding increments of x, and see if our rule 

holds good. 


Note that when a mathematician writes logo) he always 
means the Napierian logarithm of x. 

34. Example of / — =log « + constant. 

It is proved in Thermodynamics that if in a heat engine the work- 
ing stuff' receives heat // at temperature t, and if t^ is the temperature 
of the refrigerator, then the work done by a perfect heat engine would 

be ir.'-Jo,orH(l-f). 

If one pound of water at Iq is heated to ^j, and wc assume that 
the heat received per degree is constant, being 1400 foot-lbs. ; what is 
the work which a j^erfect heat engine would give out in equivalence 
for the totiil heat ? Heat energy is to be expressed in foot-pounds. 

To raise the temperature from t to t-\-dt the heat is 14005^ in foot-lb. 
This stands for JI in the above expression. Hence, for this heat we 

have the equivalent work 8 W= l-iOOdt ( 1 - -^ j , or, rather, 

"^^ =1400-1400^. 
at t 

Hence W= \AQOt - UOOIq log t + constant. 

Now W=0 when t=tQ, 

= 1400^0 - 1400^0 log ^0 + constant, 
therefore the constant is known. Using this value we find equivalent 
work for the heat given from t^ to ^i= 1400 (t^ - to) - 1400^0 ^^S j" • 

If now the pound of water at t^ receives the heat X^ foot-lb. (usually 
called Latent Heat) and is all converted into steam at the constant 
temperature t^ , the work which is thermodynamically equivalent to this 

is L^ll--]. We see then that the work which a perfect steam 

engine would give out as equivalent to the heat received, in raising 
the pound of water from t^ to t^ and then evaporating it, is 

1400 {t, - 1,) - urn, loge ^^ + X, (^1 - ^f^ . 

Exercise. What work would a perfect steam engine perform per 
pound of steam at ^^ = 439 (or 102 lb. per sq. inch), or 165^0, if ^q = 374 
or 100' C. Here L^ = 681 ,456 foot-pounds. 

The work is found to be 107,990 ft.-lb. per lb. of steam. Engineers 
usually wish to know how many pounds of steam are used per hour 

per Indicated Horse Power, w lb. per hour, means pr! ^^ ft.-lb. per 

minute. Putting this equal to 3.3(KM) we find 2V to be 18-35 lb. of steam 
per hour per Indicated Horse Power, as the requirement of a perfect 
steam engine working between the temperatm'es of 165° C and 100° C. 



35. Exercises. It is proved iii Thermodynamics when 
ice and water or water and steam are together at the same 
temperature, if Si is the volume of unit mass of stuff in the 
higher state and Sq is the volume of unit mass of stuff in the 
lower state. Then 

where t is the absolute temperature, being 274 + 6" C, L 
being the latent heat in unit mass in foot-pounds. If we 
take L as the latent heat of 1 lb. of stuff, and Si and Sq are 
the volumes in cubic feet of 1 lb. of stuff, the formula is 
still correct, p being in lb. per sq. foot. 

I. In Ice-water, So = '01747, 8^ = '01602 at ^=274 (cor- 
responding to 0° C), p being 2116 lb. per sq. foot, and 

X = 79 X 1400. Hence ^ = - 278100. 


And hence the temperature of melting ice is less as the 
pressure increases ; or pressure lowers the melting point of 
ice ; that is, induces towards melting the ice. Observe the 

quantitative meaning of -^ ; the melting point lowers at the 

rate of '001 of a degree for an increased pressure of 278 lb. 
per sq. foot or nearly 2 lb. per sq. inch. 

II. Water Steam. It seems almost impossible to 
measure accurately by experiment, Si the volume in cubic 
feet of one pound of steam at any temperature. So for water 
is known. Calculate s^ — Sq from the above formula, at a few 
temperatures having from Regnault's experiments the follow- 
ing table. I think that the figures explain themselves. 



pressure in 

lb. per sq. 



lb. per 
sq. foot 



in foot- 




















It is here assumed that the vakie of dpjdt for IDS'" C. is 
half the sum of 81 "5 and 94. The more correct way of 
proceeding would be to plot a great number of values of 
hplht on squared paper and get dpjdt for 105° C. more 
accurately by means of a curve, f 

s, - So for 105° C. = 740710 -- (379 x 87*8) = 22-26. Now 
So = "OIG for cold water and it is not worth while making any 
correction for its warmth. Hence we may take s^ = 22*28 
which is sufficiently nearly the correct answer for the present 

Example. Find Si for 275° F. from the following, L being 

t°F. 248° 


257° 266° I 275' 

284° ; 293° 


4854 I 5652 6551 

7563 8698 


Example. If the formula for steam pressure, p=a6^ where a and b 
are known numbers, and 6 is the temperature measured from a certain 
zero which is known, is found to be a useful but incorrect formula 
ibr re])reseiiting Reguault's experimental results ; deduce a fornuila for 
the volume s^ of one p<nnid of steam. We have also the well known 
formula for latent heat L—c-et^ where t is the absolute temperature 

and c and e are known numbers. Hence, as i? which is the same as 
dv • ' de 

^ is haB^-\ s^-SQ=(c-et)^tbae^-\ 

After subjecting an empirical formula to mathematical operations 
it is wise to test the accuracy of the result on actual experimental 
numbers, as the formula represents facts only approximately, and the 
small and apparently insignificant terms in which it difl:ers from fact, 
may become greatly magnified in the mathematical operations. 

36. Study of Curves. When the equation to a new 
curve is given, the practical man ought to rely first upon his 
power of plotting it upon squared paper. 

Very often, if we find ~ or the slope, everywhere, it 

gives us a good deal of information. 

If we are told that a.\, y^ is a point on a curve, and we are 
asked to find the equation to the tangent there, we have 
simply to find the straight line which lias the same slope 
as the curve there and which passes through x^, y^. The 
normal is the straight line which passes through x^, y^ and 
whose slope is minus the reciprocal of the slope of the curve 
there. See Art. 13. 


P (fig. 8) is a point in a curve APB at which the 
tangent FS and the normal FQ are drawn. OX and Y are 

the axes. OR — x, RF — y, tan FSR = ;,- •, the distance SR 
is called the subtangent ; prove that it is equal to ?/ -i- -^ . 

The distance RQ is called the subnormal; it is evidently 

equal to y -^ . The length of the tangent FS will be found 

to be y\J\-\-{,\ , the length of the normal FQ is 
y W^^ ^£i • ^^^ Intercept OS is ^- - y '^ . 

Example 1. Find the length of the sub tangent and sub- 
normal of the Parabola y = mx'^, 

- dy 




Subtangent = mx- -=- 2?/iif or ^x. 
Subnormal = y x 2nix or 2m^x^. 
Example 2. Find the length of the subtangent of y = Wio;", 



Subtangent = nix'^ -i- vinx'^~^ — xju. 


Example 3. Find of Avhat curve the subnormal is constant 

in length, 

dy d^ 1 

y-f- = a or -r = ~ n. 
^ dx dy a-^ 

The integral of - ?/ Avith regard to y is a* = ~ 3/^ + a con- 

Cl Act 

stant h, and this is the equation to the curve, where h may 
have any value. It is evidently one of a family of parabolas. 
(See Art. 9 where xs and y's are merely interchanged.) 

Example 4. The point ^ = 4, ?/ = 3 is a point in the para- 
bola y = '^a^. Find the equation to the tangent there. The 

slope is -^ = 1- X f A'"^ or, as ;r = 4 there, the slope is f x ^ or -| . 

The tangent is then, y = m+ g^. To find m we have ^ = 3 
when a? = 4 as this point is in the tangent, or 3 = 9/1 + | x 4, 
so that ??i is 1 J and the tangent is 3/ = li|- + %x. 

Example 5. The point ^ = 32, 1/ = 3 is evidently a point in 
the curve y = 2-{-^x-'. Find the equation to the normal there. 

The slope of the curve there is ^ = -^x' * = yj^ and the 

slope of the normal is minus the reciprocal of this or — 160. 
Hence the normal is y = m — \ Q>Ox. But it passes through 
the point x=S2, y = S and hence S = m— KiO x 32. 
Hence m = 5123 and the normal is y = 5123 — 160;r. 

Example 6. At what point in the curve y = aa?""" is there 
the slope 6? dy 

-f = - nax-'^-K 

The point is such that its x satisfies — nax~^^~'^ = h or, 

— j-j . Knowing its x we know its y from the 

equation to the curve. It is easy to see and well to remem- 
ber that if ^1, 3/1 is a point in a straight line, and if the slope 
of the line is 6, then the equation to the line most quickly 
written is 

x — x^ 



Hence the equation to the tangent to a curve at the 
point a?i, 2/1 on the curve is 

•^^ y^ = the -^ at the point. 

X — Xx dx 

And the equation to the normal is 

x — Xi aij 

- = — the ^ at the point. 

Exercise 1. Find the tangent to the curve x^y^^ = a at 
the point x^, y^ on the curve. Answer, — X'\- - y=m-\- n. 

Exercise 2. Find the normal to the same curve. 

Answer, — (a? — a?,) 



(!/ - y.) = 0. 

Exercise 3. Find the tangent and normal to the parabola 
y" = ^ax at the point where x = a. 

Answer, y = x -\- a, y = Sa — x. 
Exercise 4. Find the tangent to the curve 
y =^ a -{-hx ■\- ca^ + ex^ 
at a point on the curve x^, y^. 

Answer, ^ ~^-^ = 6 + 2cXi -f ^ex^^ 

X — Xi 

37. When y increases to a certain value and then 
diminishes, this is said to be a maximum value of y; 
when y diminishes to a certain value and then increases, 
this is said to be a minimum value of y. It is evident 

that for either case -^ = 0. See Art. 16 and fief. 6. 
dx ^ 

Example 1. Divide 12 into two parts such that the 
product is a maximum. The practical man tries and easily 
finds the answer. He tries in this sort of way. Let x be one 
part and 12— x the other. He tries x = 0, x = l, x = 2, &c., 
in every case finding the product. Thus 

a^|0|l|2i3 4 5 6 7 8 9 
Product I i 11 i 20 27 32 35 36 35 32 27 


It seems as if ^' = 6, giving the product 36, were the correct 
answer. But if we want to be more exact, it is good to get 
a sheet of squared paper; call the product y and plot the 
corresponding values of x and y. The student ought to do 
this himself 

Now it is readily seen that where y has a maximum or a 
minimum value, in all cases the slope of a curve is 0. Find 
then the point or points where dy\dx is 0. 

Thus if a number a is divided into two parts, one of them 
X and the other a — x, the product is y = ^ (a — a;) or ax — x^, 

and £ =a — 2x. Find where this is 0. Evidently where 

2x — a or x= \a. 

The practical man has no great difficulty in any of his 
problems in finding whether it is a maximum or a minimum 
which he has found. In this case, let a = 12. Then x = Q 
gives a product 36. Now if a? = 5 999, the other part is 6*001 
and the product is 35-999999, so that a? = 6 gives a greater 
product than x = 5*999 or x= 6*001, and hence it is a maxi- 
mum and not a minimum value which we have found. This 
is the only method that the student will be given of dis- 
tinguishing a maximum from a minimum at so early a period 
of his work. 

Example 2. Divide a number a into two parts such 
that the sum of their squares is a minimum. If x is one 
part, a—x is the other. The question is then, if 

y = ar' + (a — xf, when is y a minimum ? 

2/ = 2^-2 + a^ - 2ax, 

-^= 4^ — 2a, and this is when x — ^a. 
dx 2 

Examjile 3. When is the sum of a number and its 

reciprocal a minimum ? Let x be the number and y =x + -. 

When is 2/ a minimum ? 

The differential coefficient of - or x~'^ being — x~^, we have 

dy -, 1 11..^. ^ . 
-IT" = 1 -, and this is when a? = 1. 



The student ought to take numbers and a sheet of squared 
paper and try. Trying a;= 100, 10, 4 &c. we have 


















Now let him plot x and i/ and he will see that y is a 
minimum when a; = 1. 

Example 4. The strength of a rectangular beam of 

given length, loaded and supported in any 2:)articular way, is 
proportional to the breadth of the section multiplied by the 
square of the depth. If the diameter a is given of a cylindric 
tree, what is the strongest beam which may be cut from it ? 
Let X be its breadth. Then if you draw the rectangle inside 
the circle, you will see that the depth is Ja'^ — x-. Hence 
the strength is a maximum when y is a maximum if 

y = x((r--x-), 

or 3/ = a\x — x-\ 

j^ = a- — 3a", and this is when x—~=^, 
dx V3 

In the same way find the siiffest beam which may be cut 
from the tree by making the breadth x the cube of the depth 
a maximum. 

This, however, mav wait till the student has read 
Chap. III. 

Example 5. Experiments on the explosion of mixtures 

(at atmospheric pressure) lead to a roughly correct rule 

p = 83-3-2a?,t 

where p is the highest pressure produced in the explosion, 
and X is the volume of air together with products of previous 
combustions, added to one cubic foot of coal gas before ex- 
plosion. Taking px as roughly proportional to the work done 
in a gas engine during explosion and expansion ; what value 
of X will make this a maximum ? 


That is, when is 83^3 — S'2x^ a maximum ? Answer, When 
83 — 6'4fX= 0, or x is about 13 cubic feet. 

I am afraid to make Mr Grover responsible for the above 
result which I have drawn from his experiments. His most 
interesting result was, that of the above 13 cubic feet it is 
very much better that only 9 or 10 should be air than that 
it should all be air. 

Example 6. Prove that ax — x^ is a maximum when 

x — \a. 

Example 7. Prove that x — x^ is a maximum when 


, Example 8. The volume of a circular cylindric cistern 
being given (no cover) when is its surface a minimum ? 
Let X be the radius and y the length ; the volume is 

irafy = a, say (1). 

The surface is ttx^ + 27rxi/ (2). 

When is this a minimum ? 

From (1), y is — ; using this in (2) we see that we 


must make 

, 2a . . 
TTX^ H a mmimum, 


27rx r=0 or x^ = - . 

a? TT 

„ iTX^y 
x^= — — or x=^y. 


The radius of the base is equal to the height of the cistern. 

Example 9. Let the cistern of Ex. 8 be closed top and 
bottom, find it of minimum surface and given volume. 

The surface is 2irx- + ^irxy, and proceeding as before we 
find that the diameter of the cistern is equal to its height. 

Example 10. If v is the velocity of water in a river and 
X is the velocity against stream of a steamer relatively to the 
luater, and if the fliel burnt per hour is a + bx^; find the 

p. 4 


velocity a? so as to make the consumption of fuel a minimum 
for a given distance m. The velocity of the ship relatively 
to the bank of the river is ac—v, the time of the passage is 

, and therefore the fuel burnt during the passage is 

OS —~ V 

m(a-\- ha?) 

x — v 

Observe that a -\- hx^ with proper values given to a and h 
may represent the total cost per hour of the steamer, in- 
cluding interest and depreciation on the cost of the vessel, 
besides wages and provisions. 

You cannot yet differentiate a quotient, so I will assume 


a — 0, and the question reduces to this: when is a 

x — v 

minimum ? Now this is the same question as : — when is 

X "^ V 

— r— a maximum ? or when is ar^ — vx~^ a maximum ? The 

differential coefficient is — 2ar^ + ^vx~\ Putting this equal 

to we find x = '^v, or that the speed of the ship relatively to 

the water is half as great again as that of the current. 

Notice here as in all other cases of maximum and mini- 
mum that the engineer ought not to be satisfied merely with 
such an answer. x='^v is undoubtedly the best velocity, it 
makes a?j{x — v) a minimum. But suppose one runs at less 
or more speed than this, does it make much difference ? Let 
V = 6, the best a? is 9, 

— x-=243 if «; = 9. 
.-r — 6 

= 250 if a; = 10. 

= 256 if a;=8; 

and these figures tell us the nature of the extra expense in 
case the theoretically correct velocity is not adhered to*. 

* Assuming that you know the rule for the differentiation of a quotient 
— usually learnt at the very beginning of one's work in the Calculus, and 
without assuming a to be as above, we have 

(a;-i?)36a;*=a + 6x', 
26x3_3tv«2_rt (1). 


Example 11. The sum of the squares of two factors of a 
is a minimum, find them. If x is one of them, - is the 

Qj^ (it/ ZiO' 

other, and y'=x^-{-— is to be a minimum, -^ = 2x , and 

x^ ax x^ 

this is when x^ = ct- ^r x — \'a. 

Example 12. To arrange n voltaic cells so as to obtain 
the maximum current through a resistance R. Let the E.M.F. 
of each cell be e and its internal resistance r. If the cells 
are arranged as x in series, n-jx in parallel, the E.M.F. of the 

battery is xe, and its internal resistance is - - . Hence the 

current G == xe-^ I \- R] 


As the student cannot yet differentiate a quotient, we 

shall say that (7 is a maximum when its reciprocal is a 

... fx'r . „\ xr R 

minimum, so we ask when is h R] -^^e or 1 — 

\ n J n X 

Given the values of a, 6 and v the proper value of x can be found by trial. 
Thus let the cost per day in pounds be 30 + ^V^^ so that a = 30, 6 = ^ and 
let r = 6. Find x from (1) which becomes 

.r3_ 9^2- 300 = (2). 

I find that a; = 11-3 is about the best answer. 

Tliis is a cubic equation and so has three roots. But the engineer needs 
only one root, he knows about how much it ought to be and he only wants 
it approximately. He solves any equation whatsoever in the following 
sort of way. 

Let x^-9x--S00 be called f{x). The question is, what value of x 
makes this 0? Try a;= 10, f{x) turns out to be - 200, 







11-3 1 


-200 I 


+ 176 



whereas we want it to be 0. Now I try a; = 8, this gives -360 which is 
further wrong. Now I try 12 and I get 176 so that x evidently lies between 
10 and 12. Now I try 11 and find - 57. It is now worth while to use 
squared paper and plot the curve y=f{x) between a; = 10 and a;=12. One 
can find the true answer to any number of places of decimals by repeating 
this process. In the present case no great accm*acy is wanted and I take 
a: = 11*3 as the best answer. Note that the old answer obtained by assuming 
a = is only 9. A practical man will find much food for thought in thinking 
of these two answers. Note that the captain of a river steamer must always 
be making this sort of calculation although he may not put it down on 



. . ?' jR 
minimum ? Its differential coefficient is and this is 

when R = — , which is the internal resistance of the battery. 

Hence we have the rule : Arrange the battery so that its 
internal resistance shall be as nearly as possible equal to the 
external resistance. 

Example 13. Voltaic cell of E.M.F. = e and internal re- 

sistance r; external resistance i?. The current is (7= ^z. 

The power given out is P = RC\ What value of R will make 

P a maximum ? P = R 

{r + Rf 

To make this suit such work as we have already done we may 

(?• + Ry 
say, what value of R will make ^^ — j~^ a minimum, or 

7^ + 2Rr + jR^ 

D ~ <^r ')'^R~^ + 2r + P a minimum ? 

Putting its diiferential coefficient with regard to R equal 
to we have - i-^R-^ -t- 1 = so that R = r, or the external 
resistance ought to be equal to the internal resistance. 

Example 14. What is the volume of the greatest box 
which may be sent by Parcels post ? Let x be the length, 
y and z the breadth and thickness. The P. 0. regulation is 
that the length plus girth must not be greater than 6 feet. 
That is, we want v = xyz to be a maximum, subject to the 
condition that x-\-2{y + z) = Q. It is evident that y and z 
enter into our expressions in the same way, and hence y = z. 
So that a; + 4i/ = 6 and v = xy^ is to be a maximum. Here as 
a; = 6 — 4^ we have -y = (6 ■- 4^) y^ or 6^* — 4^' to be a maxi- 

mum. Putting -T- = we have 12^ — 123/^ = 0. Rejecting 

v = for an obvious reason, y = 1, and hence our box is 2 feet 

long, 1 foot broad, 1 foot thick, containing 2 cubic feet. 

Find the volume of the greatest cylindric parcel which may 

be sent by Post. Length being I and diameter d,l + ird — Q 

IT . . 4 

and -r Idr is to be a maximum. Answer, I = 2 feet, d = — feet, 

4 TT 

Volume = 8 -^ TT or 2'55 cubic feet. 


Example 15. Ayrton-Perry Spring. Prof. Ayrton and 
the present writer noticed that in a spiral spring fastened at 
one end, subjected to axial force F, the free end tended to 
rotate. Now it was easy to get the general formula for the 
elongation and rotation of a spring of given dimensions, and 
by nothing more than the above principle we found what these 
dimensions ought to be for the rotation to be great. 

Thus for example, the angle of the spiral being a the 
rotation was proportional to sin a. cos a. It at once followed 
that a ought to be 45°. 

Again, the wire being of elliptic section, x and y being 
the principal radii of the ellipse, we found that the rotation 
was proportional to 

To make this a maximum, the section (which is propor- 
tional to xy) being given. Let xy = 5, a constant, then the 
above expression becomes 

"-— ^ — r, and this is to be a maximum. 
s^ o sy^ 

Here we see that there is no true maximum. The larger 
we make y or the smaller we make y (for small values of y 
the rotation is negative but we did not care about the direc- 
tion of our rotation, that is, whether it was with or against 
the usual direction of winding up of the coils) the greater is 
the rotation. This is how we were led to make springs of 
thin strips of metal wound in spirals of 45°. The amount 
of rotation obtained for quite small forces and small axial 
elongations is quite extraordinary. The discovery of these very 
useful springs was complete as soon as we observed that any 
spring rotated when an axial force was applied. Students 
who are interested in the practical application of mathematics 
ought to refer to the complete calculations in our paper 
published in the Proceedings of the Royal Society of 1884. 

Example 16. From a Hypothetical Indicator Diagram 

the indicated work done per cubic foot of steam is 

w — 144^1 (1 -f- log r) — 144?"j[)3 — x. 

54 CALCULUS FOR engineehs. 

where jh and jXi arc the initial and back pressures of the 

steam ; r is the ratio of cut off (that is, cut off is at -th of 

the stroke) and a? is a loss due to condensation in the 
cylinder, w depends upon r. 

1st. If X were 0, what value of r would give most 
indicated work per cubic foot of steam ? 

We must make .— = 0, and we find — =^ — 144«.i = 

or ?' = — . If it is brake energy which is to be a maximum 

per cubic foot of steam, we must add to p^ a term represent- 
ing engine friction. 

2nd. Mr Willans found by experiment in non-condensing 
engines that ?• = — ^— gave maximum indicated w. Now 

if we put in the above -,— = we have ^ — 144^^ — i =0. 

So that ~~ = ^^^ (p, + 10) - 144^3 or ^^^ = 1440. So 

that .r = 1440r + constant. Hence Mr Willans' practical rule 
leads us to the notion that the work lacking per cubic foot of 
steam is a linear function of 7\ 

This is given here merely as a pretty exercise in maxima 
and*minima. As to the practical engineering value of the 
result, much might be said for and against. It really is as if 
there was an extra back pressure of 10 lb. per sq. inch which 
represented the effect of condensation. 

Mr Willans found experimentally in a non-condensing 
engine that the missing water per Indicated Horee Power 
hour is a linear function of ?• using the same steam in the 
boiler, but this is not the same as our .^. We sometimes 
assume the ratio of condensed steam to indicated steam to 
be proportional to log r, but a linear function of r will agree 
just as well with such experimental results as exist. 

Example 17. The weight of gas which will flow 
per second through an orifice from a vessel where it is at 


pressure />o into another vessel where it is at the pressure jfj 

is proportional to «> V 1 — « ^ ; where a is pjpf^ and 7 is a 
known constant, when is this a maximum ? That is, when is 

ay — a y a. maximum ? See Art. 74, where this example 
is repeated. 

Differentiating with regard to a and equating to we 
find j^ 

In the case of air 7 = 1*41 and we find p = "527^0 > that is, 
there is a maximum quantity leaving a vessel per second 
when the outside pressure is a little greater than half the 
inside pressure. 

Example 18. Taking the waste going on in an electric 
conductor as consisting of (1) the ohmic loss ; the value of 
G^r watts, where r is the resistance in ohms of a mile of going 
and coming conductor and C is the current in amperes; 
(2) the loss due to interest and depreciation on the cost of 
the conductor. I have taken the price lists of manufacturers 
of cables, and contractors' prices for laying cables, and I find 
that in every case of similar cables, similarly laid, or suspended 
if overhead, the cost of a mile of conductor is practically 
proportional to the weight of copper in it, that is, inversely 
proportional to the resistance, plus a constant. The cost of 
it per year will depend upon the cost of copper per ton, 
multiplied by the number taken as representing rate per 
cent, per annum of interest and depreciation. We can state 
this loss per year or per second, in money per year or per 
second and the ohmic loss is in watts. We cannot add them 
together until we know the money value per year or per 
second of 1 watt. There are three things then that decide 
the value of the quantity which we call t^. I prefer to 
express the total waste going on in watts rather than in 

pounds sterling per annum and I find it to hey— C^r -\ f- 6, 

where h is some constant. The value of t may be taken as 
anything from 17 to 40 for the working of exercises, but 


students had better take figures of their own for the cost ot 
power, copper and interest*. 

For a given current (7, when is the y, the total waste, a 
minimum ? that is, what is the most economical conductor for 

a given current ? ^~^^~';^ ^^^ *^^^ ^^ ^ when r — -^. 

Thus if ^ = 40, r = ^. 

Now if a is the cross section of the conductor in square 

inches, r = — nearly, so that (7= 1000a, or it is most econom- 
ical to provide one square inch of copper for every 1000 
amperes of current. 

When w is a function of more than one independent 

* The weight of a mile of copper, a square inches in cross section, is to 
be figured out. Call it ma tons. If p is the price in pounds sterling of a 
ton of copper, the price of the cable may be taken as, nearly, pma + some 
constant. If R is the rate per cent, per annum of interest and depreciation, 
then the loss per annum due to cost of cable may be expressed in pounds as 


-r^l'wa + some constant. If £1 per annum is the value of %v watts, (ob- 

serve that this figure w must be evaluated with care. If the cable is to 
have a constaiat current for 24 hours a day, every day, w is easily evaluated), 

then the cost of the cable leads to a perpetual loss of zrj-r. wpma + some 


constant. Now taking a= — , we see that our t^ is -^7=7^77 . 
** r 2000 

Men take the answer to this problem as if it gave them the most econom- 
ical current for any conductor under all circumstances. But although the 
above items of cost are most important, perhaps, in long cables, there are 
other Items of cost which are not here included. The cost of nerves and 
eyesight and comfort if a light blinks ; the cost in the armature of a dynamo 
of the valuable space in which the current has to be carried. 

If a man will only write down as a mathematical expression the total 
cost of any engineering contrivance as a function of the size of one or more 
variable parts, it is quite easy to find the best size or sizes; but it is not 
always easy to write down such a function. And yet this is the sort of 
problem that every clever engineer'is always working in his head; increasing 
something has bad and good effects; what one ought to do is a question in 
' maxima and minima. 

Notice also this. Suppose we find a value of x which makes y a maxi- 
mum ; it may be, that quite different values of x from this, give values of 
y which are not very different from the maximum value. The good practical 
engineer will attend to matters of this kind and in such cases he will not 
insist too strongly upon the use of a particular value of x. 


variable^ say x and y. Then | — - J = 0, y being, considered 

constant during the differentiation, and f — j = 0, x being 

considered constant during the differentiation, give two equa- 
tions which enable the values of x and y to be found which 
will make u a maximum or a minimum. Here, however, 
there is more to be said about whether it is a minimum or 
a maximum, or a maximum as to x and a minimum as to y, 
which one has found, and we cannot here enter into it. 

Sometimes in the above case although it is a function of 
X and y, there may be a law connecting x and y, and a little 
exercise of common sense will enable an engineer to deal 
with the case. All through our work, that is what is wanted, 
no mere following of custom; a man's own thought about his 
own problems will enable him to solve very difficult ones 
with very little mathematics. 

Thus for example, if we do not want to find the best conductor 
for a given current of Electricity: if it is the Power to be 
delivered at the distant place that is fixed. If the distance is n miles, 
and the conductors have a resistance of r ohms per mile (go and return), 
if \\ is the potential, given, at the Generating end, and C is the 
current. Then the potential at the receiving end being T' ; F^ - T^= Cnr. 

CV=PiH fixed, and the cost per mile is y = C^r + -...{l), where t^ is 

known. When is y a minimum? 

Here both C and r may vary, but not independently. V= V^ - Cnr 

and P=(7F, — (7%r ... (2). One simple plan is to state ?/ in terms of r 

CV —P 
alone or of C alone. Thus r from (2) is r= — ^ — (3). 

Substituting for this jn (1), we get 

CV,-P . f-a^n 

y=-ir-^cv;^p ^^^- 

Here everything is constant except (7, so we can find the value of C 
to make y a minimum, and when we know C we also know r from (3). 

At present the student is supposed to be able to difibrentiate only 
^", so he need not proceed with the problem until he has worked a few 
exercises in Chap. III.* 

* To differentiate (4) is a very easy exercise in Chap. III. and leads to 

j>, = — i + - — - — -/xr^ — 5svo > and on putting this equal to we obtain 

d(J n {CVi~Py JT o 

the required value of C. It would not be of much use to proceed further 


In ray Cantor lectures on Hydraulic Machinery, I wrote out an 
expression for the total loss in pounds per annum in Hydraulic 
transmission of povirer by a piiwj. I gave it in terms of the 
maximum pressure, the power sent in, and the diameter d of the pipe. 
It was easy to choose d to make the total cost a minimum. If how- 
ever I had chosen jo, the pressure at the receiving end as fixed, and the 
power delivei-ed as fixed, and therefore Q, the cubic feet of water per 
second, and if I had added the cost of laying as proportional to the 
square of the diameter, I should have had an expression for the total 
cost like i(^ pqi 

when the values of a, h and c depend upon the cost of power, the 
interest on the cost of iron, &c. This is a minimum when its differen- 
tial coefficient with regard too? is zero or ^cld—hal(^d'^-\-Zh^^M~^^ 
and d can be obtained by trial. The letters 6 and c also involv^ the 
strength of the material, so that it wjxs possible to say whether wrought 
iron or ca»t iron was on the whole the cheaper. But even here a term 
is neglected, the cost of the Engine and Pumps. 

The following example comes in conveniently here, although it is 
not an example of Maximum or Minimum. 

An Electric Conductor gives out continuously a amperes 
of current in every mile of its length. Let x be the distance of any 
jKiint in miles from the end of the line remote from the generator, let 
C be the current there and V the voltage. Let r ohms per mile be the 
resistance of the conductor (that is, of one mile of going and one mile 
of coming conductor). The current given out in a distance hx is dC, 

or rather hx -j- , and the power is bx . V '-f- y «o that if P is the 

power per mile (observe the meaning of per), 

''='-f :•••(')• 

Also if V is voltage at x and V+bVat x+8x] — 

As the i-esistance is i' . 8x the current is dV-rr . dx^ or rather, since 
these expressions are not correct until 8x is supposed smaller and 
smaller witliout limit, 

^-l^ (^)- 

unless we had numerical values given us, 

91 = 10 miles, P= 20000 watts, f^=H)00, find G and then r. 

Consult a Paper in the Joimial of the Institution of tJie Society of Tele- 
graph Engineers, p. 120, Vol. xv. 1886, if there is any further difficulty. 

It has not yet been sufficiently noted that if V^ and P and r are given, 
there is a limiting length of line 


and when this is the case P is exactly equal to the ohmic loss in the con- 



As -T- = ^h C=t<.<,- if C is when .r = 0. 
Hence if r is constant (2) becomes 


rax = , so that T^ = Fq + i ^^-^-^ 


^o being the voltage at the extremity of the line. . 

(1) becomes P^aV^+^ra'^x'^ (4). 

Taking ro = 200 volts, a = 25 amperes per mile, r=l ohm per mile, 
it is easy to see by a numerical example, how the power dispensed per 
mile, and the voltage, diminish as we go away from the generator. 


















If Fi is the voltage at the Dynamo and the line is n miles long 

The power per mile at the extremity being l\ = aVQ, if we are given 
V^ and 1\ to find T'q, we shall find that n cannot be greater than 


and this gives the limiting length of the line. 

If we wish, as in Electric traction to get a nearer approach 
to uniform P, let us try 

C=ax-hx^ (5), 

where a. h, c are constants. 

1 ^y 7^ 

- - - = ax - oyf\ 
r ax 



As P=V ,- , or V {a - cbx^-'^), we can easily detennine the three 

constants a, b, c so that P shall be the same at any three points of 
thehne. Thus let r=l ohm, Fo=100 volts, and let P= 10000 watts, 
where .^;=0, x=l mile, .r=l| miles. 

We find by trial that 


and from this it is eas}'^ to calculate Cat any point of the line 



Example 19. A machine costs ax-\-hy/\t'& value to mc is 
proportional to xy, find the best values of x and y if the cost 
is fixed. Here xy is to be a maximum. Let c = cw; + 63/, so 

that V = J -~ j.^> ^^d ^y is r ^ — T ^^. This is a maximum 

Hence ax = hy = 0/2 makes xy a 

when j = 2 r X or ax — -^ . 
b 2 


Examj)le 20. The electric time constant of a cylindric 
coil of wire is approximately 

II = mxyzl{ax +hy + cz), 

where x is the mean radius, y is the difference between the 
internal and external radii, z is the axial length and m, a, b, c 
are known constants. 

The volume of the coil is 27rxyz. 

Find the values of x, y, z to make u a maximum if the 
volume of the coil is fixed. Let then 27r . xyz = g ; when is 

— I a minimum ? That is, substituting for z, when 

yz xz 



is ax + bi/ 4- ^ — = v, say, a minimum ? As x and y are p(ir- 
fectly independent we put f ^ J = and ( ;t~ ) = 0, 


ft + 



Fig. 9. 

and + 6 - s^^^, = 0, 

so that A'^y = ^^^ , 





- or y — T, .1 - -7- 


x^ = -^ or 

^ V a=27r 

W ¥2'ir' 

and 2^ 


or -? 


V C»27i^ 



38. The chain of a suspension bridge supports a load 
by means of detached rods ; the loads are about equal and 
equally spaced. Suppose a chain to be really continuously 
loaded, the load being w per unit length horizontally. Any 
veiy flat uniform chain or telegraph wire is nearly in this 
condition. What is its shape ? Let be the lowest point. 
OX is tangential to the chain 
and horizontal at 0. OF is 
vertical. Let P be any point 
in the chain, its co-ordinates 
being x and y. Consider the 
equilibrium of the portion 
OP. OP is in equilibrium, 
under the action of Tq the 
horizontal tensile force at 0, 
Tthe inclined tangential force 
at P and wx the resultant 
load upon OP acting ver- -p. ^^ 

tically. We employ the laws 

of forces acting upon rigid bodies. A rigid body is a body 
which is acted on by forces and is no longer altering its 

If we draw a triangle whose sides are parallel to these 
forces they are proportional to the forces, 
and if 6 is the inclination of T to the c. 5^ ^a 


\ I ^M \ , 





cos 0. 





but tan ^ is j^ , so that ^ = 7^ ;«;... (3); 
dx dx Tq 

Pig. 11. 

1 w 

hence, integrating, y = ^7rr^ + constant. 

- -to 

Now we see that y is when x is 0, so that the constant 
is 0. Hence the equation to the curve is 

1 ^ ^ 



and it is a parabola. Now tan ^ is ^ o^, so that sec- 6 is 

1-f^,^^ Andasr = rosec^, r=ToA/l+^U-^...(5). 

From this, all sorts of calculations may be made. Thus 
if I is the span and D the dip of a telegraph wire, if the whole 
curve be drawn it will be seen that we have only to put in 
(4) the information that when x—\l,y = D, 

and the greater tension elsewhere is easy to find. 

In the problem of the shape of any uniform chain, loaded 
only with its own weight, the integi'ation is not so easy. I 
give it in a note*. When it is so flat that we may take the 

* The integration in this note requires a knowledge of Chapter in. 

If the weight of the portion of chain OP, instead of being wx is %cs, 
where s is the length of the curve from to P, the curve y is called the 
Catenary, Equation (3) above becomes 

| = ^^.or,e«ingr,=.c.^V.i (X). 

If bs is the length of an elementary bit of chain, we. see that in the limit 

60 that "i- = A / 3- + If 

and hence -^ = . — . This being integrated gives y + c=v^c2 + «-...( 2), 

"« 'slc' + s^ 
the constant added in integration being such that « = when y = 0. From (2) 
we find 6-2=^2^ 2j/c... (3), and using this in (1), we have 

dx _ c 
the integral of which is 

as when y=0, a?=0, if is the origin, no constant is to be added. Putting 
this in the exponential form 

c^lc _ y ^ ^. ^. JyT. ^ 2yc, 

transposing and squaring we find 



load on any piece of it as proportional to the horizontal pro- 
jection of it, we have the parabolic shape. | 

39. Efficiency of Heating Surface of Boiler. 

If 1 lb. of gases in a boiler flue would give out the heat 
in cooling to the temperature of the water (0 maybe taken 
as proportional to the difference of temperature between 
gases and water, but this is not quite correct), we find from 
Peclet's experiments that the heat per hour that flows through 
a square foot of flue surface is, roughly, md\ Let ^ = ^i at 
the furnace end of a flue and 6 = 0^ at the chimney end. 
Let us study what occurs at a place in the flue. 

The gases having passed the area S in coming from the 
furnace to a certain place where the temperature is 6, pro- 

Or changing the origin to a point at the distance c below O, as at O in fig. 12 
>Yhere SP is y' and RP is x, we have 

This is sometimes called 

Using (1) we find 
sometimes called 

y' = c cosh xjc. 

, I xlc -x!c\ 

s = c sinh xjc. 


Nv R 





Fig. 12. 

Note that tables of the values of sinh u and cosh u have been publighed. 
Returning to the original figure, the tension at P being T, 




-p , and from (3), 



so that 

W8 S 

Hence T=zio{y + c) or T=wy' 


ceed further on to a place where S has become aS' 4- ^S and 
has become 6 + B0 (really SO is negative as will be seen). 
A steady state is maintained and during one hour the gase i 
lose the heat 7n0^ . BS through the area hS. If during the 
hour W lb. of gases lost at the place the amount of heat 
-W.Bd, then 

orrather ^ = -^-g. «• 

That is, integrating with regard to 0, 

where c is some constant. 

Putting in 6 = ^i the temperature at the furnace end 

when S=0, we hav©' 

' Wl Wl 

0= —-pf-^c orc = ^, 

711 Ui m Oi • 

80 that (2) becomes 

^•=^^U^J (3). 

This shows how 6 diminishes as S increases from the 
furnace end, and it is worth a student's while to plot the 
curve connecting S and 6. If now S is the whole area of 
heating surface and ^ = ^2 at the smoke-box end, 

»-^a-i) <•> 

The heat which one pound of gases has at the furnace 
end is 6^, it gives up to the water the amount 61—62. 
Therefore the efficiency of the heating surface may be 
taken as 

. j^t_ ^i- 62 . . 

^'^~dr ^^^' 

and it follows from (4) that 


^ 6,mS 


Now if W is the weight of coals burnt j^er hour ; 
W=13W' if air is admitted just sufficient for complete 
combustion ; W = about 20 W in the case of ordinary 
forced draught ; W = about 26 W in the case of chimney 
draught. In these cases 0^ does not seem to alter inversely 
as W, as might at first siglit appear : but we do not know 
exactly how 0^ depends upon the amount of excess of air ad- 
mitted. We can only say that if W'-^S is the weight of coal per 
hour per square foot of heating surface and we call it tu, there 

seems to be some such law as B = z, , where a clepends 

1 -h aw ^ 

upon the amount of air admitted. In practice it is found 

that a = 0*5 for chimney draught and 0'3 for forced draught, 

give fairly correct results. Also the numerator may be taken 

as greater than 1 when there are special means of heating 

the feed water. 

Instead of the law given above (the loss of heat by 
gases in a flue oc 6^^), if we take what is probably more likely, 
that the loss is proportional to 0, 

Then (1) above becomes 

de mO ^ ^' 

or S= loe: ^ + constant (2). 

Let 0= 6j at furnace end or when S=0 so that our con- 

. W 

stant IS — log Oi and (2) becomes 



'i'-^ai) »■ 

If S is the area of the whole flue and 6^ is the temperature 
at the smoke-box end, then 

^ = Zlog^i (4), 

Sm ^ 


The efficiency E J-^^ (5) 

becomes E=^\-~^ = l-e ^^ (6). 

Or if w is the weight of fuel per square foot of heating 
surface as above (6) becomes 

j5'=l-e"a«' (7). 

40. Work done by Expanding Fluid*. If p is the 

Eressure and v the volume at any instant, of a fluid which 
as already done work W in expanding, one good definition of 

pressure is p—'-r- •..(!)) or in words, pressure is the rate at which 

work is done per unit change of volume. Another way of putting this 
is: if the fluid expands through the volume hv there is an increment 

8 IF of work done so that 'p .hv=hW^ or jo=-— , but this is only 

strictly tiiie when hv is made smaller and smaller without li«iit, and so 

(1) is absolutely true. Now if the fluid expands according to the law 

p%}'=c. a constant ... (2) : p = cv~'j and this is the differential coefficient 

of W with regai*d to v or, as we had better write it down, -j— = cv'. 

We therefore integrate it according to our rule and we have 

T^=Z7^i^-'-*-^+<^' ^^^' 

where C is some constant. To find C\ let us say that we shall only 
begin to count W from v=Vi. That is, W=0 when v^Vi, Then 

0= -^ v.^-' + C, so that C= - ^ v.^-'. 

Insert this value of G in (3) and we have 

F=^^(^i--V-) (4), ■ 

which is the work done in expanding from v^ to r. 
Now if we want to know W when ^ = ^2) we have 

^^'i2=rr-/^2'"'-^^'"'^ ^^^- 

* Observe that if for p and v we write y and x this work becomes very 



This answer may be put in other shapes. Thus from (2) we know that 
c^p^v^' or pav/, 

so that 

iir —P_^.u, 1-8. 

a formula much used in gas engine and steam engine calculations. 

There is one case in which this answer turns out to be useless ; try 
it when s — \. That is, find what work is done from Vy to Vg by ^ fluid 
expanding according to the law (it would be the isothermal law if the 
fluid were a gas) 


If you have noticed how it fails, go back to the statement 



You will find that when you integrate x"^ with regard to m, the 
general answer has no meaning, cannot be evaluated, if m= - 1. But 
I have already said, and I mean to prove presently that the integral of 
x"^ is log^. So the integral of (7) is 


Proceeding as before we find that, in this particular case. 



41. Hypothetical Steam Engine Diagram. 

Let steam be admitted to a cylinder at the constant pressure jt?,. 
the volume increasing from to 
Vi in the cylinder. The work done 
is VjjOj. Let the steam expand 
to the volume Vg according to the 
law 'pijP^c. The work done is 
given by (6) or (8). Let the back 
pressure bejOg, then the work done 
in driving out the steam in the 
back stroke is p^^. We neglect 
cushioning in this hypothetical 
diagram. Let v^-~v^ be called r 
the ratio of cut-off. Then the 
nett work done altogether is 

Fig. 13. 




If pe is the effective i)re.s.sure so that ^e^^ ^^ equal to the above iiett 
work Ipe is measured from actual indicator diagrarus, as the average 
pressure) ; putting it equal and dividing by v^ we have on simplifying 



In the special case of s = l we find pe=Pi p^ in the same 


42. Definite Integral. 

Definition. The symbol I /(a?) . dx 

tells us: — " Find the general integral of/ (a?); insert in it the 
value a for ic, insert in it the value 6 for a? ; subtract the 
latter from the former value*." This is said to be the 

(1), tells us to integrate u (which is a 
function of x and ?/), with regard 
to y, as if x were constant ; then 
insert F{x) for y and also f(x) for 
y and subtract. This result is to 
be integrated with regard to x, 
and in the answer a and b are in- 
serted for X and the results sub- 

I. If w = 1, dx . dy evidently 
means an element of area, a little 
rectangle. The result of the first 
process leaves 

Fig. 14. 


{F{x)-f{x)}dx (2) 

still to be done. Evidently we 
have found the area included between the curves y = F{x) and y~f{x) and 
two ordinates at a: = a and x = b. Beginners had better always use form (2) 
in finding areas, see fig. 14. 

II. If u is, saA', the weight of gold per unit area upon the above men- 
tioned area, then u . dx . dy is the weight upon the little elementary area 
dx .dy, and our integral means the weight of all the gold upon the area I 
have mentioned. 

When writers of books wish to indicate generally that they desire to 
integrate some jwoperty u (which at any place is a function of x, y, z), 
throughout some volume, they will write it with a triple integral, 


u .dx.dy. dz, 

and summation over a surface by 

I j V . dx . dy. 



integral of f{x) between the limits a and h. Observe now 
that any constant which may be found in the general 
integral simply disappears in the subtraction. 

Iq integrating between limits we shall find it convenient 
to work in the following fashion. 

Example, to find I x^ . dx. The general integral is ^xf^ 

J h 

and we write I x"^ .dx—\ ^a^ — ^d^ — ^bK 

Symbolically. If F{x) is the general integral of /{x) 
then rf(x) . dx = P F(x)] =F(a) - F(b), 

J b U J 

Note as evidently true from our definition, that 
I f(x),dx = - I f(x).dx, 

J b J a 

and also that 

T/w • ^^' = f Vw • ^^' + f "/(^) • ^^• 

43. Area of a curve. 

some function of x and 
let PS be the curve. It 
is required to find the 
area MPQT. 

be called A and OT=x, 
QT=y, OW = X -ir Sx, 
WR = y-\-8y, and the area 
MPRWheA + 8A then 
BA = siTesiTQIiW. 

Let y of the curve be known as 




, 15. 

Indeed some writers use 

V over an area, and 

jto . 


dS to mean generally the summation of 

els to mean the summation of 2a along a line or 

what is often called "the line integral of ic. The line integral of the pull 
exerted on a tram car means the work done. The surface integral of the 
normal velocity of a fluid over an area is the total volume flowing per 
second. Engineers are continually finding line, surface and volume integrals 
in their practical work and there is nothing in these symbols which is not 
already perfectly well known to them. 




If the short distance QR were straight, 

Therefore ^ = y + i^y» as ha) gets smaller and smaller 

and in the limit -y— =?/ 

Hence A is such a function of x that y is its differential 
coefficient, or A is the integral of y. 

In fig. 16 CQD is the curve y=a-\- hx^ and EMGF is the 

curve showing 

so that A is the integral of 
y. In what sense does A 
represent the area of the 
curve CD ? The ordinate 
of the A curve, GT^ repre- 
sents to some scale or other, 
the area of the y curve 
MPQT from some standard 
Fig. 16. ordinate MP. 

The ordinate TQ represents to scale^ the slope of 
EF at G. Observe, however, that if we diminish or increase 
all the ordinates of the A curve by the same amount, we do not 
change its slope anywhere, and y, which is given us, only tells 
us the slope of A. Given the y curve we can therefore find 
any number of A curves ; we settle the one wanted when we 
state that we shall reckon area from a particular ordinate 
such as MP. Thus, in fig. 16 if the general integral of y is 
F(x)-{- c. If we use the value x = OM we have, area up to 
MP from some unknown standard ordinate =F(OM)-i-c. 

Taking x = ON, we have area up to NR from some 
unknown standard ordinate =F{ON)-\-c. And the area 
between MP and NR is simply the difference of these 
F(ON) — F{OM), the constant disappearing. 

Now the symbol I y .dx tells us to follow these instruc- 


tions : — integrate y ; insert ON for x in the integral ; insert 
OM for ^ in it ; then subtract the latter. We see therefore 



that the result of such an operation is the area of the curve 
between the ordinate at OM and the ordinate at ON. 

If y and a; represent any quantities whatsoever, and a 
curve be drawn with y as ordinate and x as abscissa, then the 

integral j ; 

y .dx is represented by the area of the curve, and 

we now know how to proceed when we desire to find the sum 
of all such terms rs y . 800 between the limits x=b and a)=a 
when Bw is supposed to get smaller and smaller without 

Example. Find the area enclosed between the parabolic 
curve OA, the ordinate AB 
and the axis OB. Let the 
equation to the curve be 

y = ax^ (1), 

where PQ = y and OQ = x. 
Let QR = Bx. 

The area of the strip PQRS 
is more and more nearly 

ax^ . Sx, 

as Bx is made smaller and 

smaller; or rather the whole 

area is I aa^ . dx, whic'h is 









' G 


k B 

Fig. 17. 

OB -\ 


Now what is a in terms o{ AB and 
x==OB. Hence by (1) 

AB = a.OB^, so that a = 




When y^AB, 

Therefore the area = 5 =^, OJ^ ^AB . OB ; 

that is, frds of the area of the rectangle OMAB. 

Observe that the area of a very flat segment of a circle is 
like that of a parabola when OB is very small compared 
with BA, 



Exercise 1. Find the area between the curve y = mx " 
and the two ordinates at a? = a and x — h. 

The answer is 


irix"''^ . dx = 





Observe (as in Art. 40) that this fails when n — 1; that 
is, in the rectangular hyperbola. 

In this case the answer is 




= m log - . 

The equation to any curve being 

y = a -\- hx -{- ca^ -{■ ex^ 4-/^, 

the area is A = ax f ^bx^ + ^cx^ 4- \ex* + i/ar*. 

Here the area up to an ordinate at x is really measured from 
the ordinate where x = 0, because A =0 Avhen x=0. We 
can at once find the area between any two given ordinates. 

Exercise 2. Find the area of the curve y — a^x between 
the ordinates at *• = a and x — ^. 

aTx^ .dx = a\ ix^'] = ^ (y8^ - a^). 

Exercise 3. Find the area of the curve yx" = a between 
the ordinates at a; = a and x — fi. 

Answer : a \ x~'- .dx = a 

— x~ 

= a(a-'-^-'). 

44. Work done by Expanding Fluid. When we use 

definite integrals the work is somewhat shorter than it was in Art. 40. 
For i{p = cv~', the work done from volume v^ to volume V2 is 

/V2 Fv-i 1 "1 c 

cy~*. dv or c\ y^"* or _- — -(^'2^"'- V') 

The method fails when s = l and then the integral is 

c\ "logeV =cl0ge— . 

L»i J ^1 



45. Centre of Gravity, Only a few bodies have 
centres of gravity. We usually 
mean the centre of mass of a 
body or the centre of an area. 

Fig. 18. 

If each little portion of a 
mass be multiplied by its dis- 
tance from any plane, and the 
results added together, they are 
equal to the whole mass multi- 
plied by the distance of its centre, 
X, from the same plane. Expressed algebraically this is 

If each little portion of a plane area, as in fig. 18, be 
multiplied by its distance from any line in its plane and the 
results added together, they are equal to the whole area 
multiplied by the distance of its centre x from the same line. 
Expressed algebraically this is Saaf = xl^a. 

Example. Find the centre of mass of a right cone. It 
is evidently in the axis OB of the 
cone. Let the line OA rotate 
about OX, it will generate a 
cone. Consider the circular slice 
PQR of thickness hx. Let 
OQ = x, then PQ or 


The mass of PR multiplied 
by the distance from to its 
centre is equal to the sum of the 
masses of all its parts each mul- 
tiplied by its distance from the 
plane YOY^. The volume of the slice PR being its area iry'^ 
multiplied by its thickness hx ; multiply this by rn the mass 
per unit volume and we have its mass miry- . hx. As the 
slice gets thinner and thinner, the distance of its centre from 
gets more and more nearly x. Hence we have to find the 
sum of all such terms as mirxy- . hx, and put it equal to the 
whole mass {^irm . ABr . OB) multiplied by x, the distance of 

Fig. 19. 



its centre of gravity from 0. Putting in the value of y^ in 
terms of x we have 

roB fAB\^ 

I rmr ( -^f^ ] ^'^ • dx equated to ^TrmAB^ . OB . x. 

rOB FOB -] 

Now af,dx== i^ = iOB\ 

and hence '"^'^ (Tfn] iOB^^^ir^ii . AB^.OB.x. 

Hence x = J OB. That is, the centre of mass is f of the 
way along the axis from the vertex towards the base. 

46. It was assumed that students knew how to find the 
volume of a cone. We shall now prove the rule. 

The volume of the slice PR is it .y"^ ^ hx and the whole 
volume is 

£ V • d^ = \y (oi)' *■= • '^- = - i^) [p] 

or J of the volume of a cylinder on the same base AG and of 
the same height OB, If we had taken y — ax all the work 
would have looked simpler. 

Example. Find the volume and centre of mass of 

uniform material (of mass m per 
unit volume) bounded by a para- 
boloid of revolution. 

Let PQ = y, OQ=^x, QS=Bx. 

Let the equation to the curve 
OFA hey=aa^ (1). 

The volume of the slice PSR 
is Try^ . Bx; so that the whole 

volume is 1 tt . a^x . dx or 









B ^ 

Fig. 20. 

l-iraKOB^ (2). 


Now what is a ? When 

7/ = AB, a)=: OB, 

so that from (1), AB = a . OB^ and a is ^-^ . Hence the 

volume is ^ir -7y^ OB'^ or 

^TT.A&.OB (3). 

That is, half the area of the circle AC multiplied by the 
height OB. Hence the volume of the paraboloid is half the 
volume of a cylinder on the same base and of the same height. 
(The volumes of Cylinder, Paraboloid of revolution, and Cone 
of same bases and heights are as 1 : i : J.) 

Now as to the centre of mass of the Paraboloid. It is 

evidently on the axis. We must find I jriir . y^x . da), or 

jmira) . a^js . da;, or 7/t7ra- Ix^ . dx, 


or mTTcv^ • i^M ^^^^ *his is ^mira^ . 0B\ Inserting as before 

the value of «- or -jTW ^^ have the integral equal to 

IniTT.OB^.AB^. This is equal to the whole mass multiplied 
by the x of the centre of mass, x, or m^Tr . AB^ . OB . x, so that 
x = ^OB. The centre of mass of a paraboloid of revolution 
is |rds of the way along the axis towards the base from the 

Example. The curve y = ax^ revolves about the axis 
of X, find the volume enclosed by the surface of revolution 

between x = and x = h. 

The volume of any surface of revolution is obtained by 
integrating iry"^ . dx. Hence our answer is 


{'a'x-^ . dx = J^ Vx'^-^A = ^-^ 6^+\ 
J 2n 4- 1 Lo J 2/1 + 1 

Find its centre of mass if in is its mass per unit volume. 
For any solid of revolution we integrate m . xiry^ . dx and 


divide by the whole mass which is the integral of mir'fdx. 
If m is constant we have 

rb rb 

niTT I a;a"iv-"' . clx = niirtv^ I a;2"+i dx 
Jo Jo 

~2n + 2Lo J~2n + 2^ ' 

and the whole mass is _ 6-"+\ so that Jj = _ ^ b. 

2/1 +1 2?i -I- 2 

Suppose m is not constant but follows the law 

'ill = //i„ 4- caf. 

To find the mass and centre of mass of the above solid. Our 
first integral is 

TT |(?/toA' + caf+') ttU-'^ . dw, or (/ V [(vi^''+' + cx''''+'+^) clx, 

|_o 2n + 2 2/1 + 5 + 2 J ^ 

ass is a^TT I (?yio + co."*) x^ . c?a; 

aV \^, af''^' +-« ^ ^ ar^+'-^»] (2). 

L2m+1 2n + 5+l J ^'^ 

Substituting h for ^' iu both of these and dividing (1) by (2), 
we find x. 

An ingenious student can manufacture for himself many 
exercises of this kind which only involve the integration 

An arc of the ellipse ~ + ^^ = l revolves about the axis of 

X, find the volume of the portion of the ellipsoid of revolu- 
tion between the two planes where x = and where x = c. 

Here y^ = — (a^ — x-). The integral of tti/^ is 






The volume of the whole ellipsoid is -t- iira; aud of ^ 

it IS -^ a^ ^^ 



47. Lengths of Curves. In fig. 21 the co-ordinates of 
P are x and y and of Q they 
are a; -\- Sx and 3/ + S^. If we 
call the length of the curve 
from some fixed place to P by 
the name s and the length PQ, 
8s, then (8sy=(Bxy-\-(8i/y more 
and more nearly as 8x gets 
smaller, so that 



or rather, in the limit 

Fig. 21. 




To find 5 then, we have only to integi'ate 


It is unfortunate that we are only supposed to know as 
yet j x^ . dx, because this does not lend itself much to exer- 
cises on the lengths of curves. 

Example. Find the length of the curve y — a-\-hx (a 
straight line) between the limits x = Q and x = c. 

t^ ds [^ — ~ 

5=1 j^.dx=\ ^/l+6^(^a; = 





Exercise. There is a curve whose slope is »Ja^x^—l, 
find an expression for its length. Answer: s= -«;'**+->/-. 

Other exercises on lengths of curves will be given later. 

48. Areas of Surfaces of Revolution. When 

the curve APB revolving 
about the axis OX de- 
scribes a surface of revo- 
lution, we have seen that 
the volume between the 
ends AC A' and BDB' is 
the integral of iry^ with 
regard to x between the 
limits OG and OD. 

Again the elementary 
area of the surface is what 
is traced out by the ele- 
mentary length PQ or hs 
and is in the limit 2iry . ds. 
Hence we have to integrate 

Fig. 22. 

roD ^ 

I liTV .-T-.dx, and as the law of the curve is known, y . 
Joe dx 



can be expressed in terms of x. 

Example. The line y = a-\-bx revolves about the axis of 
X ; find the surface of the cone between the limits x=0 and 

-^ = 6, so that the area is 27r 
dx JO 


= 2'7r*Jl+b^r{a-\-bx)dx = 2'ir's/l -^¥ I ax + ^boA 

= 27rVITF(ac + Jtc^. 

The problem of finding the area of a spherical surface is 
here given in small printing because the beginner is supposed 
to know only how to differentiate x'^ and this problem 
requires him to know that the differential coefficient of y^ 



with regard to x is the differential coefficient with regard to 

y multiplied hy -~ , or 2y .-^ . As a matter of fact this is 

not a real difficulty to a thinking student. The student can 

however find the area in the following way. Let V be the 

volume of the sphere of radius r, F= -^ r^, Art. 46. Let 


V-{- SVhe the volume of a sphere of radius r*+ 8r, then 

which is only true when Sr is supposed to be smaller and 
smaller without limit. Now if S is the surface of the 
spherical shell of thickness Br, its volume is Bi^ . S. Hence 
Br .S = Br . ^irr^ and hence the area of a sphere is ^iri^. 

Example. Find the area of the surface of a sphere. That 
is, imagine the quadrant of a circle AB oi radius a, fig. 23, to 

Fig. 23. 

revolve about OX and take double the area generated. We 
have as the area, 47r j y a / \ -\- {-^\ dx. 

In the circle x'^ +y^ = <^% or y = \/«^ - x-, 



..^..,.|=0.or|,= -^- 



Hence as 

l + m^^l+^ 


r y 

Guldinus's Theorems. 


49. If each elementary portion hs of the length of a 
curve be multiplied by x its distance from a plane (if the 
curve is all in one plane, x may be the distance to a line in 
the plane) and the sum be divided by the whole length of 
the curve, we get the x of the centre of the curve, or as it is 
sometimes called, the centre of gravity of the curve. Observe 
that the centre of gravity of an area is not necessarily the 
same as the centre of gravity of the curved boundary. 

Volume of a Ring^. 

BC, fig. 24, is any plane 
area ; if it revolves about 
an axis 00 lying in its 
own plane it will generate 
a ring. The volume of 
this ring is equal to the 
area of BG multiplied by 
the circumference of the 
circle passed through by 
the centre of area of BC. 
Imagine an exceed- 
Fig' 24. ingly small portion of the 

area a at a place P at the distance r from the axis, the 
volume of the elementary ring generated by this is a . 27rr 
and the volume of the whole ring is the sum of all such 
terms or F= 27r2ar. But ^ar = rA, if A is the whole area 
of BG. The student must put this in words for himself; r 
means the r of the centre of the area. Hence V= 2irf x A 
and this proves the proposition. 

II. Area of a Ring. The area of the ring surface is 
the length of the Perimeter or boundary of BG multiplied 
by the circumference of the circle passed through by the 
centre of gravity of the boundary. 

Imagine a very short length of the boundary, say hs^ at 


the distance r from the axis ; this generates a strip of area 
of the amount hs x 27rr. Hence the whole area is ^irZZs . r. 
But 2^5 . r = r X 6- if r is the distance of the centre of gi-avity 
of the boundary from the axis and s is the whole length 
of the boundary. Hence the whole area of the ring is 
27rr X 8. 

Example. Find the area of an anchor ring whose sec- 
tion is a circle of radius a, the centre of this circle being 
at the distance JR, from the axis. Answer : — the perimeter 
of the section is 27ra and the circumference of the circle- 
described by its centre is 27rR, hence the area is 47r-ajR. 

Exercise. Find the volume and area of the rim of a fly- 
wheel, its mean radius being 10 feet, its section being a 
square whose side is 1'3 feet. Answer: 

Volume = (1-3)2 x 27r x 10 ; Area = 4 x 1-3 x 27r x 10. 

50. If every little portion of a mass be. multiplied by the 
square of its distance from an axis, the sum is called the 
moment of inertia of the whole mass about the axis. 

It is easy to prove that the moment of inertia about any 
axis is equal to the moment of inertia about a parallel axis 
through the centre of gravity together with the whole mass 
multiplied by the square of the distance between the two 
axes. Thus, let the plane of the paper be at right angles 
to the axes. Let there be a little mass 
m at '^P in the plane of the paper. 
Let be the axis through the centre of 
gravity and 0' be the other axi-s. We want 
the sum of all such terms as m . (O'Py. 

Now (O'py = (ovy + OP' +2. 00'. oq, 

where Q is the foot of a perpendicular from 

P upon 00\ the plane containing the two 

axes. Then calling Xm. (O'Py by the name 

/, calling Xm . OP^ by the name /q, the 

moment of inertia about the axis through 

the centre of gravity of the whole mass, 

then, / = (0'0)^Sm + /o + 2.00'.2:m.OQ. But tm.OQ 

means that each portion of mass m is multiplied by its 

distance from a plane at right angles to the paper through 

P. 6 



the centre of gravity, and this must be by Art. 45. So 
that the proposition is proved. Or letting 2m be called M 
the whole mass 

I = Io + M.(0'0)2. 
Find the moment of inertia of a circular cylinder of 

length I about its axis. 
Let fig. 26 be a section, 
I the axis being 00. Con- 

— o sider an elementary ring 
shown in section at TQPR 
of inside radius r, its out- 
side radius being r + hr. 
Its sectional area is I .hr 



O— - 

Fig. 26. 

SO that the volume of the ring is 2irr . 1 . 8r and its mass is 

m^irrl . h\ Its moment of inertia about 00 is ^irml .t-^.dr 

and this must be integrated between the limits r — R 

the outside radius and ?• = to give the moment of inertia 

of the whole cylinder. The answer is /q = lirmlRK The 

whole mass M=ml7rR\ So that 7o = -«— • If we define the 

radius of gyration as Jc, which is such that Mk^ = Io, we 

have here k^ = ii2^ or A; = -p^ R. 

The moment of inertia about 
the axis NS is 

I=Io + M.R' = ^3IR% 

so that the radius of gyration about 

jsrs is R Vf . 

Moment of inertia of a circle 
about its centre. Fig. 27. Con- 
sider the ring of area between the 
Fig. 27. circles of radii r and r + Br, its area 

is 27rr . Br, more and more nearly 
as Br is smaller and smaller. Its moment of inertia is 27rr^ . dr 
and the integral of this between and R is ^ttR* where R 
is the radius of the circle. The square of the radius of 

gyration is JttjR'^-t- the area =-^ 



At any point in an area, fig. 28, draw two lines OX 
and OF at right angles to one 
another. Let an elementary area 
a be at a distance x from one of 
the lines and at a distance i/ from 
the other and at a distance r from 
0. Observe that ax^ 4- ay- = «/'^ 
so that if the moments of inertia 
of the whole area about the two 
lines be added together the sum 
is the moment of inertia about the 
point 0. Hence the moment of 
inertia of a circle about a diameter is half the above, or 
JttjR*. The square of its radius of gyration is ^R-. 

The moment of inertia of an ellipse about a principal 
diameter AOA. Let OA = a, OB = b. 



y f 

\ > 

/ / 

\ \ 



Fig. 29. 

The moment of inertia of each strip of length BT is t 

times the monaent of inertia of each strip PQ of the circle, 

MT a 
because it is at the same distance from A OA and ytfi = y • 

MQ h 

This is a property of ellipse and circle well known to all 

engineers. But the moment of inertia of the circle of radius 

h about AOA is \it¥, so that the moment of inertia of the 

ellipse about AOA is \'Tr¥a. Similarly its moment of inertia 

about BOB is \'iTa%. 

The above is a mathematical device requiring thought, 
not practical enough perhaps for the engineer's every-day 
work; it is given because we have not yet reached the inte- 



gral which is needed in the straightforward working. The 
integral is evidently this. The area of the strip of length 
ST and breadth % is 2x . Sy, the equation to the ellipse 

beins: -„ + '7. = 1, so that ic=^T "Jh^ — y^ 

Then 2J if.2x.dy or 4^[ ^fs/b^ -if .dy=^ I. 

The student ought to return to this as an example in 
Chap. III. 

51. Moment of Inertia of Rim of Fly-wheel. If 

the rim of a fly-wheel is like a hollow cylinder of breadth 
I, the inside and outside radii being R^ and R.,, the moment 

of inertia is ^irml I 1-^ .dr or ^irml — =i7rml(R/—Ri*). 

J jR, \_Ri ^ J 

The mass is 7r(jR2'-i^f)^//i'=^irsay,_so that I=\{R.^-\-R^^M. 
The radius of gyration is V^ (^R^ + Ri^). It is usual to 
calculate the moment of inertia of the rim of a fly-wheel as 
if all its mass resided at the mean radius of the rim or 


--^ — ^ . The moment of inertia calculated in this way is 

to the true moment of inertia as ^y^- — ^- . Thus if 

R.2 = R + a, Ri = R — a, the pretended I divided by the true 

/ is 1 -^ (1 -f 0:2) ^^^ i^ ^ is small, this is 1 — -^^ nearly. If 

the whole mass of a fly-wheel, including arms and central 
boss, be M, there is usually no very great error in assuming 
that its moment of inertia is / = R'^M. 

52. A rod so thin that its thickness may be neglected 
is of length I, its mass being m per unit length, what is 
its moment of inertia about an axis at right angles to it, 
through one end ? Let w be the distance of a point 
from one end. An elementary portion of length Sw of mass 
in . Boo has a moment of inertia x^ ,m .dx and the integral of 
this from ^ = to x = l \^ i^nP, which is the answer. As 7nl 

^ .X £83s>Q is the whole mass, the 

^U ' --1 ^^^^ square of the radius of 

Fig. 30. gyration is ^l^ /© the 



moment of inertia about a parallel axis through the middle 
of the rod, at right angles to its length, is 

^ml^ — tnl.liz) or ml^ (^ — {) or -^mlK 

So that the square of this radius of gyration is -^P. 

We shall now see what error is involved in neglecting the 
thickness of a cylindric rod. 

If 00 is an axis in the plane of the paper at right angles to the 
axis of a circular cylinder, 
through one end, and OP is <-» 
a; and R is the radius of the 
cylinder, its length being I ; 
if p is the mass of the cylin- 
der per unit volume ; the 
moment of inertia about 00 
of the disc of radius It and 
thickness 8.v is irR^pbx . x*- + 
the moment of inertia of the disc about its own diameter. Now we 
saw that the radius of gyration of a circle about its diameter was 

— , and the radius of gyration of the disc is evidently the same. Hence 

its moment of inertia about its diameter is \ WirR^ . hx . p, or | irpTt^ . 8x. 
Hence the moment of inertia of the disc about is 


If OAj the length of the rod, is I, we must integrate between and I, 
and so we find 



Fig. 31. 



The mass m per unit length is irR'^pj so that 


3 IP 


This is the moment of inertia about an axis through the centre of 
gravity parallel to 00. 

53. Example. Where is the Centre of Area of the 

parabolic segment shown in fig. 20 ? The whole area is 

The centre of area is evidently in the axis. 



The area of a strip PSR is 2i/ . Bx and we musb 
integrate 2xy . Bx. Now y — ax^ where a — AB ^ OBK 

rOB ^^ 
Hence 2 1 x vym^* . dx = ^AO . OB . x. The integral is 

2 g^^ Ux^l or f -^j 0J5t, so that iAB . OB' = ^AB.OB.x 

or x = ^OB, 

Find the centre of area of the segment of the sym- 
metrical area bounded by + ?/ = ax'^ between x = b and x = c. 


We must divide the integral 2 1 x . ax"^ . dx by the 


2 I ax'' . dx. 

2a r"-^"" 

X = 

C«+i _ hn+i ' n + 2' 

Many interesting cases may be taken. Observe that if 
the dimensions of the figure be given, as in fig. 20 : thus if 
AB and PQ and BQ are given, we may find the position of the 
centre of the area in terms of these magnitudes. 

54. Moment of Inertia of a Rectangle. 

The moment of inertia of a rectangle about the line 00 


through its centre, parallel to one side. 

Consider the strip of area between 
OP = y and OQ = y + hj. Its area is b . By 
and its moment of inertia about 00 is 
b .y^ . By, so that the moment of inertia of 
the whole rectangle is 


rhd rid-] 

b f.dyorb\if\ 

J -id L-idA 

"' li 

This is the moment of inertia which is 
so important in calculations on beams. Fig. 33. 

55. Force of Gravity. A uniform spherical shell of 
attracting matter exercises no force upon a body inside it. 
On unit mass outside, it acts as if all its mass were gathered 
at its centre. 

The earth then exercises a force upon unit mass at any 
point P outside it which is inversely proportional to the 
square of r the distance of P from the centre. But if P is 
inside the earth, the attraction there upon unit mass is the 
mass of the sphere inside P divided by the square of r. 

1. If the earth were homogeneous. If m is the mass per 
unit volume and R is the radius of the earth, the attraction 

on any outside point is — mR^ -r- r\ 

The attraction on any inside point is -^ mi^ -h t^ or 

— mr. The attraction then at the surface being called 1, at 

any outside point it is R^ -r- r- and at any inside point it is 

r -f- R. Students ought to illustrate this by a diagram, 

2. If m is greater towards the centre, say m — a — br, 
then as the area of a shell of radius r is 4<TTf\ its mass is 

4mr^ . Ill . 5r, so that the whole mass of a sphere of radius r 
rr ^^ 

47r r^ (a — br) dr, or -^ ar^ — irbr^. Hence on any 
Jo ^ 

inside point the attraction is -;r- a?' — irbr^ and on any 


outside point it is f ^ aR^ — irbR^j r^. 




Dividing the whole mass of the earth by its volume -^ jK^ 


we find its mean density to be a — ^hR, and the ratio of its 

mean density to the density at the surface is 


56. Strength of thick Cylinders. 

The first part of the following is one way of putting the 
well known theory of what goes on in a thin cylindric shell 
of a boiler. It prevents trouble with + and — signs after- 
wards, to imagine the fluid pressure to be greater outside 
than inside and the material to be in compression. 

Consider the elementary thin cylinder of radius r and of 

thickness 8?\ Let the pressure 
inside be p and outside p + Bp 
and let the crushing stress at 
right angles to the radii in the 
material be q. Consider the 
portion of a ring PQSR which 
is of imit length at right angles 
to the paper. 

Radially we have p + 8p 
from outside acting on the area 
RS or (r + Br) BO if QOP = BO, 
because the arc RS is equal to 
radius multiplied by angle; and 
2:> .r. BO from inside or 

{p + Bp){r-\-B7^)Bd-pr.Be 
the whole the radial force from the outside more 
and more nearly as B6 is smaller and smaller. This is 
balanced by two forces each q . Br inclined at the angle B6, 

and just as in page 165 if we 
draw a triangle, each of whose 
sides GA and ^J5 is pamllel 
to q . Br, the angle BAG being 
Bd, and BG representing the 
radial force, we see that this radial force is q . Br . Bd, and 


Fig. 34. 

is on 

Fig. 35. 


expression is more 

smaller and smaller. Hence 

more nearly true as BO is 

(p + Bp) (r + Br) B0-pr.Be = q.Br, B0, 


or p .^7^ -\- r.Bp+ 8p.Sr = q.Sr, 
or rather p-\- r -¥- = <! (1), 

since the term Bp is in the limit. 

When material is subjected to crushing stresses p and q 
in two directions at right angles to one another in the plane 
of the paper, the dimensions at right angles to the paper 
elongate by an amount which is proportional to p + q. 

We must imagine the elongation to be independent of r 
if a plane cross section is to remain a plane cross section, and 
this reasonable assumption we make. Hence (1) has to be 

combined with p-\-q = 2A (2) 

where 2A is a constant. 

Substituting the value of q from (2) in (1) we have 

dp 2A 2p 

or -f- = ^ . 

a?' r r 

Now it will be found on trial that this is satisfied by 

P = ^+J (3). 

and hence from (2), q = A — -; (4). 

To find these constants A and B. In the case of a gun 
or hydraulic press, subjected to pressure po inside where 
r = ro and pressure outside where r = 7\. Inserting these 
values of p in (3) we have 

sothat ^=i^o.(i-l) = Po,;g^, 


The compressive stress — q may be called a tensile stress f, 

/-^V,'-V r^ ^^^. 

/ is greatest at r = 7\ and is then 

fo = Po-^—', (6). 

This is the law of strength for a cylinder which is initially 
unstrained. Note that po can never be equal to the tensile 
strength of the material. We see from (5) that as r increases, 
/ diminishes in proportion to the invei-se square of the 
radius, so that it is easy to show its value in a curve. Thus 
a student ought to take 7\ = 1% r^ = 0'8, p^ — 1500 lb. per sq. 
inch, and graph / from inside to outside. / will be in the same 
units as p. (5) may be taken as giving the tensile stress 
in a thick cylinder to resist bui^ting pressure if it is initially 
unstrained. If when p^ — there are already strains in the 
material, the strains produced by (5) are algebraically added 
to those already existing at any place. Hence in casting 
a hydraulic press we chill it internally, and in making a gun, 
we build it of tubes, each of which squeezes those inside it, 
and we try to produce such initial compressive strain at 
r = ro and such initial tensile strain at r = ri, that when the 
tensile strains due to Pq come on the material and the cylinder 
is about to burst there shall be much the same strain in the 
material from r^ to i\* 

* In the case of a eylindric body rotating with angular velocity a, if p is 
the mass per unit volume ; taking into account the centrifugal force on the 
element whose equilibrium is considered, above the equation (1) becomes 

p + r ^ - r^pa? = q and the solution of this is found iohep = A+ Br~^ + ^pa^r^ 

and by inserting the values of p for two values of 7- we find the constants A 
and B ; q is therefore known. If we take p~0 when r=rQ and also when 

This is greatest when r = r^. 

If the cylinder extends to its centre we must write out the condition that 
the displacement is where r=0, and it is necessary to write out the valuer 


TJim cylinder. Take ro = R and 7\=Il-\- 1, where t is very 
small compared with R, 

. 2Rr' + 2Rt + t' PoR /^ t t-\l(^ t\ 

Fig. 86. 

Now -jz and ^r-^^ and ^r^ become all smaller and smaller 
li ZH- ZK 

as t is thought to be smaller 

and smaller. We may take f=V3 . .. (7) 

(7) as a formula to be used ' t ' " 

when the shell is exceedingly ^j^ ^ 

thin and (8) as a closer ap- ^~^"^2 ^^^' 

proximation, which is the 

same as if we used the average radius in (7). In actual 
boiler and pipe work, there is so much uncertainty as to the 
proper value of/ for ultimate strength, that we may neglect 
the con-ection of the usual formula (7). 

57. Gas Engine Indicator Diagram. It can be 

proved that when a perfect gas (whose law is pv = Rt for a 
pound of gas, R being a constant and equal to K—k the 
difference of the important specific heats; 7 is used to 
denote Kjk) changes in its volume and pressure in any 

of the strains, Eadial strain =pa - g/3 if a is the reciprocal of Young's 
Modulus and ^/a is Poisson's ratio, generally of the value 0-25. 

In this way we find the strains and stresses in a rotating solid cylinder, 
but on applying our results to the case of a thin disc we see that equation 
(2) above is not correct. That is, the solution is less and less correct as the 
disc is thinner. Dr Chree's more correct solution is not difficult. 


way, the rate of reception of heat by it per unit change 
of volume, which we call h (in work units) or -^— , is 

{jpv) + 'p (1), 


'-~i[t^A ^'^- 

Students ouorht to note that this -~ is a very different 

thing from [-i] , because we may give to it any value we 


We always assume Heat to be expressed in work units so as 
to avoid the unnecessary introduction of J^ for Joule's equiva- 

Exercise 1. When gas expands according to the law 
ptf = c...{^) a constant, find li. 

Answer : // = ^^— :. p (4). 

y- 1 

Evidently when 5 = 7, k = 0, and hence we have pyy = con- 
stant as the adiabatic law of expansion of a perfect gas. 
7 is 1*41 for air and 1'37 for the stuff inside a gas or oil 
engine cylinder. When s= 1, so that the law of expansion is 
pv constant, we have the isothermal expansion of a gas, and 
we notice that here h=p, or the rate of reception of heat 
energy is equal to the rate of the doing of mechanical energy. 
Notice that in any case where the law of change is given by 
(3), h is exactly proportional to p. If .9 is greater than 7 the 
stuff is having heat withdrawn from it. 

If the equation (1) be integrated with regard to v we 

have J^oi = — 3T (piVi - p^Vo) + T^oi- • -(5). Here H^^ is the heat 

given to a pound of perfect gas between the states po, Vq, t^ 
and jOi, -yi, ti, and TFoi is the work done by it in expanding 
from the first to the second state. 

This expression may be put in other forms because we 
have the connection pv = Rt...(6). It is very useful in cal- 
culations upon gas engines. Thus, if the volume keeps 


constant Wqi is and the change of pressure due to ignition 
and the gift of a known amount of heat may be found. If 
the pressure keeps constant, TFoi is p(vi — Vo) and the change 
of volume due to the reception of heat is easily found. 

/J JT rit 

Another useful expression is -p =1^~r +'P"'0) where k 

is a constant, being the specific heat at constant volume. 
Integrating this with regard to v we find 

Ho, = k(t,-to)+W,, (8). 

This gives us exactly the same answer as the last method, 
and may at once be derived from (5) by (6). In this form 
one sees that if no work is done, the heat given is k (ti — to) 
and also that if there is no change of temperature the heat 
given is equal to the work done. 

58. Elasticity is defined as increase of stress -r increase 
of strain. Thus, Young's modulus of elasticity is tensile or 
compressive stress (or load per unit of cross section of a tie 
bar or strut) divided by the strain or fractional change of 
length. Modulus of rigidity or shearing elasticity is shear 
stress divided by shear strain. Volumetric elasticity e is 
fluid stress or increase of pressure divided by the fractional 
diminution of volume produced. Thus if fluid at p and v, 
changes to p-\-Bp, v-\-Sv: then the volumetric stress is 
Bp and the volumetric compressive strain is — Sv/v, so 

that by definition e = — Bp-. — , or e = — v J- . The 
•^ ^ V ov 

definition really assumes that the stress and strain are 

smaller and smaller without limit and hence e — — v -/-. . .(1). 


Now observe that this may have any value whatsoever. 
Thus the elasticity at constant pressure is 0. The elasticity 
at constant volume is — oo . To find the elasticity at con- 
stant temperature, we must find ( i^ ) • see Art. 30. As 
pv = Rt, p^ Rtv-\ Here Rt is to be constant, so that 

^^- Rtv-^ and e=Rtv^^^p. 



It is convenient to write this et and we see that Kt^ the 
elasticity at constant temperature, is p. This was the value 
of the elasticity taken by Newton ; by using it in his calcu- 
lation of the velocity of sound he obtained an answer which 
was very different from the experimentally determined velo- 
city of sound, because the temperature does not remain 
constant during quick changes of pressure. 

Exercise. Find the elasticity of a perfect gas when the 
gas follows the law pvy = c, some constant. This is the 
adiabatic law which we found Art. 57, the law connecting j9 and 
V when there is no time for the stuff to lose or gain heat by 

conduction, p = cv~y, so that -t- = — 7C^;~>'~^ and 

6? = + vycv~y~^ or ycv~y, or yp. 

It is convenient to write this e^, and we see that in a 
perfect gas en = yet. When this value of the elasticity of 
air is taken in Newton's calculation, the answer agrees with 
the experimentally found velocity of sound. 

59. Friction at a Plat Pivot. If we have a pivot of 
radius R carrying a load W and the load is uniformly 
distributed over the surface, the load per unit area is 
w=^ W-T-irR^ Let the angular velocity be a radians per 
second. On a ring of area between the radii r and r-\-8r 
the load is w27rr . 8r, and the friction is fjLiv27rr . Sr, where 
fjb is the coefficient of friction. The velocity is v = otr, so that 
the work wasted per second in overcoming friction at this 
elementary area is ^^irwoLr^ . Br, so that the total energy 
wasted per second is 

^irwoLfi \ ?'^ . dr = ^irwafxR^ = |« ^ WR, 

On a collar of internal radius Ri and external R^ we have 
27rwafi r'.dr^ ^irwdfi (R^^ - R^% W^irw {R^^ - R^) and 

J Ri Jl3 _ J^ 3 

hence, the energy wasted per second is '{afiW j^\ _ pV 

60. Exercises in the Bending of Beams. When 
the Bending moment ilf at a section of a beam is known, 
we can calculate the curvature there, if the beam was 


straight when unloaded, or the change of curvature if the 
unloaded beam was originally curved. This is usually written 

1 1 1 if 
- or = -^^ , 

where / is the moment of inertia of the cross section 
about a line through its centre of gravity, perpendicular 
to the plane of bending, and E is Young's modulus for the 
material. Thus, if the beam has a rectangular section of 
breadth h and depth d, then I=^hd^ (see Art. 54); if the 

beam is circular in section, I=-:R\ if J2 is the radius of 

tbe section (see Art. 50). If the beam is elliptic in section, 

I — T ^^^> if ^ aiid h are the radii of the section in and at 


right angles to the plane of bending (see Art. 50). t 

Curvature. The curvature of a circle is the reciprocal 

of its radius, and of any curve it is the curvature of the circle 

which best agrees with the curve. The curvature of a curve 

is also " the angular change (in radians) of the direction of 

the curve per unit length." Now draw a very flat curve, with 

very little slope. Observe that the change in -^ in going 

from a point P to a point Q is almost exactly a change of 

angle change in -^ is really a change in the tangent of 

an angle, but when an angle is very small, the angle, its sine 


and its tangent are all equal . Hence, the increase in -^ 

from P to Q divided by the length of the curve PQ is the 
average curvature from P to Q, and as PQ is less and less 
we get more and more nearly the curvature at P. But the 
curve being very flat, the length of the arc PQ is really hx^ 

and the change in -/ divided by hx, as hx gets less and less, 

is the rate of change of -f- with regard to x, and the symbol 

^2y ° ax ^2y 

for this is -—^ . Hence we may take -y^ as the curvature of 
dx^ '' ^ dx^ 

a curve at any place, when its slope is everywhere small. 


If the beam was not straight originally and if y' was its small 
deflection from straightness at any point, then ,^ was its original 
curvature. We may generalize the following work by using ^^ (y -y') 
instead of --j— everywhere. 

It is easy to show, that a beam of uniform strength, that is a beam 
in which the maximum stress / (if compressive ; positive, if tensile, 
negative), in every section is the same, has the same curvature every- 
where if its depth is constant. 

If d is the depth, the condition for constant strength is that 

-J '\d=±f a constant. But -y=^x curvature, hence curvature 


Exercise. In a beam of constant strength if c^= , . 

d^V '2f 
Then ^-^ = -^ {a-\-hx). Integrating we find 


2?' ^=<' + ^+i^'^^ ■^^.y = e^rCX-\-\ax''--V\.hx^, 
where e and c must be determined by some given condition. Thus 
if the beam is fixed at the end, where ^=0, and -^- = there, and 


also y = there, then c = and e — 0. 

In a beam originally straight we know now that, if 
X is distance measured from any place along the beam to 
a section, and if y is the deflection of the beam at the 
section, and / is the moment of inertia of the section, then 

^ = ^ a) 

dx2 EI ^ ^' 

where M is the bending moment at the section, and E is 
Young's modulus for the material. 

We give to -^ the sign which will make it positive if 

M is positive. If M would make a beam convex upwards 
and y is measured downwards then (1) is correct. Again, 
(1) would be right if M would make a beam concave up- 
wards and y is measured upwards. 



Example /. Uniform beam of length 1 fixed at one 
end, loaded with weight W at the other. Let x be the 
distance of a section from the fixed end of the beam. Then 
M= W{l — x), so that (1) becomes 





■l-x— ^ 

Fig. 37. 

Integrating, we have, as E and / are constants, 
EI dy J . , 

From this we can calculate the slope everywhere. 

To find c, we must know the slope at some one place. 
Now we know that there is no slope at the fixed end, and 

0, hence c = 0. Integrating again, 

To find G, we know that ^ = when x = 0, and hence 
(7=0, so that we have for the shape of the beam, that is, the 
equation giving us y for any point of the beam, 

hence -y^ = where x 




We usually want to know y when x = l, and this value of y 
is called D, the deflection of the beam, so that 




Example II. A beam of length I loaded with W at 
the middle and supported at the ends. Observe that if 
half of this beam in its loaded condition has a casting of 

P. 7 



cement made round it so that it is rigidly held ; the other 
half is simply a beam of length ^l, fixed at one end and 



Fig. 38. 

loaded at the other with ^W, and, according to the last 
example, its deflection is 

ZEI ^ 4:8. EI 


The student ought to make a sketch to illustrate this method 
of solving the problem. 

Example III. Beam fixed at one end with load w per 
unit length spread over it uniformly. 

The load on the part FQ is wx PQ or w{l — x). 

P Q 

— X >* ^'1-X ■*■ 

Fig. 39. 

The resultant of the load acts at midway between P and Q, 
so, multiplying hy ^{1 — x), we find M at P, or 

M=\w{l-xy (6). 

Using this in (1), we have 

w dx^ 


Integrating, we have 
This gives us the slope everywhere 

^0 ax 

BEAMS. 99 

Now -r =0 where ^ = 0, because the beam is fixed 

there. Hence c = 0. 
Again integrating, 


y = ^px" - ^laf + yijja;* + G, 

and as ?/ = where a; = 0, = 0, and hence the shape of the 
beam is 

y=^((ilV-ila^ + a,-) (7). 

1/ is greatest at the end where a; — I, so that the deflection is 

^=M&7^' "'^ = 8 EI <«>• 

if Tr='Z(;^, the whole load on the beam. 

Example IV. Beam of length I loaded uniformly with 
lu per unit length, supported at the ends. 

Each of the supporting forces is half the total load. The 
moment about P of ^zvl, p q 

at the distance PQ, is -------- 

-<,— X — *^-il-x-^ 


against the hands of a > 

watch, and I call this i^i 

direction positive; the 

moment of the load Fig. 40. 

'^^ (2 ^ ~ ^) ^t the average distance ^PQ is therefore negative, 

and hence the bending moment at P is 

^lul (hi — x) — ^w ( JZ — xy, or ^^Z- — ^wx'^ . . .(9), 

so that, from (1), EI ,^ = ^wP - \wx'^, 

y being the vertical height of the point P above the middle 
of the beam, see Art. 60. Integrating we have 


EI -^ = \wl~x — \wa? 4- c, 

a formula which enables us to find the slope everywhere. 



c is determined by our knowledge that ^ = where a; = 0, 
and hence c = 0. Integrating again, 

and C=0, because y = 0, where x = 0. Hence the shape of 
the beam is y = aaW'T^^^^^ " 2^)-"(l^X V is greatest where 
X = i?, and is what is usually called the deflection D of the 
beam, or ^= oQ±pf if" ^= ^^^ the total load. 

61. Beams Fixed at the Ends. Torques applied 
at the ends of a beam to fix them (that is, to keep the end 
sections in vertical planes) are equal and opposite if the 
loading is symmetrical on the two sides of the centre of the 
beam. The torques being equal, the supporting forces are 
the same as before. Now if 7ti is the bending moment 
(positive if the beam tends to get concave upwards) which 
the loads and supporting forces would produce if the ends 
were not fixed, the bending moment is now ni — c because 
the end torques c are equal and opposite, and the supporting 
forces are unaltered by fixing. 

^^"^ d^^=^r w- 

If the beam is uniform and we integrate, we find 

EI .-— = ini.dx — ex -\- const (2). 

Take x as measured from one end. We have the Uvo 

conditions : / = where x = 0, and ,- = where x = lAf I 
ax dx 

is the length of the beam. Hence if we subtract the value 

of (2) when ic = from what it is when x=l, we have 

n I ri 

0=1 7n . dx — cl, or c = J j m . dx, 

that is, c is the average value of m all over the beam. 

The rule is then (for symmetric loads): — Draw the diagram 
of bending moment m as if the beam were merely suppoi^ted 


at the ends. Find the average height of the diagram and 
loiuer the curved outline of the diagram by that amount. 
The resulting diagram, which will be negative at the ends, is 
the true diagram of bending moment. The beam is concave 
upwards where the bending moment is positive, and it is 
convex upwards where the bending moment is negative, and 
there are points of inflexion, or places of no curvature, where 
there is no bending moment. 

Example. Thus it is well known that if a beam of length 
I is supported at the ends and loaded in the middle 

with a load W, the bending moment is \Wl at the middle 
and is at the ends, the diagram being formed of two 
straight lines. The student is supposed to draw this diagram 
(see also Example II.). The average height of it is half the 
middle height or ^Wl, and this is c the torque which must 
be applied at each end to fix it if the ends are fixed. 
The whole diagram being lowered by this amount it is 
evident that the true bending moment of such a beam if its 
ends are fixed, is ^Wl at the middle, half-way to each end 
from the middle so that there are points of inflexion there, 
and — iTT^ at each end. A rectangular beam or a beam of 
rolled girder section, or any other section symmetrical above 
and below the neutral line, is equally ready to break at the 
ends or at the middle. 

Example. A uniform beam loaded uniformly with 
load w per unit length, supported at the ends ; the diagram 
for 711 is a parabola (see Example IV., where ilf = ^tul^ — ^wx^); 
the greatest value of m is at the middle and it is ^tul^ ; m is 
at the ends. Now the average value of m is | of its middle 
value (see Art. 43, area of a parabola). Hence c = -^wP. 
This average value of m is to be subtracted from every value 
and we have the value of the real bending moment every- 
where for a beam fixed at the ends. 

Hence in such a beam fixed at the ends the bending 
moment in the middle is i^^ivl-y at the ends — -^i^wP, and the 
diagram is parabolic, being in fact the diagram for a beam 
supported at the ends, lowered by the amount -^wP every- 
where. The points of inflexion are nearer the ends than in 
the last case. The beam is most likely to break at the ends. 

Students ought to make diagrams for various examples of 


symmetrical loading. Find m by the ordinary graphical 
method and lower the diagram by its average height. 

When the beam symmetrically loaded and fixed 
at the ends is not uniform in section^ the integral 
of (1) is 

^t-i>-^lT (^). 

and as before this is between the limits and I, and hence 

to find c it is necessary to draw a diagram showing the value 

of -J everywhere and to find its area. Divide this by the 

area of a diagram which shows the value of y ever3rwhere, 

or the average height of the Mjl diagram is to be 
divided by the average height of the 1// diagram and 
we have c. Subtract this value of c from every value of 
m, and we have the true diagram of bending moment of the 
beam. Graphical exercises are much more varied and interest- 
ing than algebraic ones, as it is so easy, graphically, to draw 
diagrams of m when the loading is known. 

The solution just given is applicable to a beam of which 

the / of every cross section is settled beforehand in any 

arbitrary manner, so long as / and the loading are symmetrical 

on the two sides of the middle. Let us give to / such a 

value that the beam shall be of uniform strength every- 

where ; that is, that -jz=fc or ft . . .(4), where z is the greatest 

distance of any point in the section from the neutral line on 

the compression or tension side and/, and/^ are the constant 

maximum stresses in compression or tension to which the 

material is subjected in every section. Taking /<. as 

numerically equal to ft and z = ^d, where d is the depth 

of the beam, (4) becomes Trc?= ± 2/"... (5), the + sign being 

taken over parts of the beam where if is positive, the — sign 

where 31 is negative. As I -y c^^ = 0, or, using (5-), 

/iff^^^O (6), 



the negative sign being taken from the ends of the beam to 
the points of inflexion, and the positive sign being taken 
between the two points of inflexion. We see then that to 
satisfy (6) we have only to solve the following problem. 
In the figure, EATUGQE is a diagram whose ordinates 

represent the values of -^ or the reciprocal of the depth of 

the beam which may be arbitrarily fixed, care being taken, 
however, that d is the same at points which are at the same 
distance from the centre. EFGE is a diagram of the values 
of m easily drawn when the loading is known. We are re- 
quired to find a point P, such that the area of EPTA = area 

Fig. 41. 

of POO'Tj where is in the middle of the beam. When 
found, this point P is a point of inflexion and PR is 
what we have called c. That is, m — PR is the real 
bending moment M at every place, or the diagram EFG 
must be lowered vertically till R is at P to obtain the 
diagram of M. Knowing M and d it is easy to find / 
through (5). 

It is evident that if such a beam of uniform strength 
is also of uniform depth, the points of inflexion are half- 
way between the middle and the fixed ends. Beams of 
uniform strength and depth are of the same curvature 
ever3rvvhere except that it suddenly changes sign at the 
points of inflexion. 

61. In the most general way of loading, the bending 
moments required at the ends to fix them are different 

flrom one another, and if m^^ is the torque against the hands of a 
watch apphed at the end A, and ?«._, is the torque with the hands of a 


watch at the end /?, and if the bending moment in case the beam were 
merely supported is m : — 

Consider a weightless unloaded beam of the same length with the 
torques oui and wio applied to its ends; to keep it in equilibrium it is 
necessary to introciuce equal and opposite supporting forces P at the 






4--— • 


— ^ 

B ' 



Fig. 42. 
ends as shown in the figure. Then Fl + m2 = m^f the forces &c. being 
as drawn in fig. 42, so that P=— i-^ — ^. 

If then these torques m.^ and wij are exerted they must be balanced 
by the forces P shown ; that is, at -B a downward force must be exerted ; 
this means that the beam at B tends to rise, and hence the ordinary 
supporting force at B must be diminished by amount P. At any place 
C the bending moment will be m (what it would be if the beam were 
merely supported at the ends) — m^-P. BC. . .(1). If one does not care 
to think much, it is sufficient to say : — The beam was in equilibrium 
being loaded and merely supported at the ends ; the bending moment 
at any place was m ; we have introduced now a new set of forces which 
balance, the bending moment at C due to these new forces is 

-(ma + P.iiC). 
So that the tnie bending moment at C is m - m^ - P. BC. 


Suppose r/i2 = 0, then P= j-, and the bending moment at C is 



62. Beam fixed at the end A^ merely supported at 

B which is exactly on the same level as A. As Wg^^ ^"d letting 
BC^x, we have the very case just mentioned, and 

El'^^^^m-Px (2). 

We will first consider a uniform beam uniformly loaded 

as in Example IV., Art. 60. It will be found that when x is measured 
from the end of the beam, the bending moment in = \wlx -\wx'^-^ if the 
beam is merely supported at its ends and \o is the load per unit length. 
Hence (2) is 

EI^ = ^wlx-\wx''--Px (3), 

El-^=\wlx'^-\wx^-\Px^\c (4). 


We have also the condition that -i\=0 where ^=^...(5), for it is to be 

observed that we measure ,v from the unfixed end. 

Again integrating, 

£;ii/=^wla;^-^wa)*-^Pa^ + CA' (6). 

We need noi add a constant because y is when ^ is 0. 

We also have y=0 when x=l. Using this condition and also (5) 
we find 

0=j\v>l^-^PP + c (7), 

0=^^wl^-^PP+d (8), 

and these enable us to determine P and c. 

Divide (8) by I and subtract from (7) and we have 

(d=^wP - ^PP or P= Iwl, 

hence from (7 ), 0'=^wP — ^wP+CfC—- -^g wP. 

We have the true bending moment, 

^wlx - J wx^ - \ ivlxj 

and (6) gives us the shape of the beam. 

63. If the loading is of any kind whatsoever and 
if the section varies in any way a graphic metliod of inte- 
gration must be used in working the above example. Now if the value 
of an ordinate z which is a function of ^ be shown on a curve, we have 
no instrument which can be relied upon for showing in a new curve 


z . dx, that is, the ordinate of the new curve representing the area of 

the z curve up to that value of .r from any fixed ordinate. I have 
sometimes used squared paper and counted the number of the squares. 
I have sometimes used a planimeter to find the areas up to certain 
values of .v, raised ordinates at those places representing the areas to 
scale, and drawn a curve by hand through the ten or twelve or more 
points so found. There are integrators to be bought ; I have not cared 
to use any of them, and perhaps it is hardly fair to say that I do not 
believe in the accuracy of such of them as I have seen. 

A cheap and accurate form of integrator would not only be very 
useful in the solution of graphical problems ; it would, if it were used, 
give great aid in enabling men to understand the calculus. 

Let us suppose that the student has some method of showing the 

value of I z. dx in a new curve ; the loading being of any kind what- 

soever and / varying, since 



we have on integrating, 

"£-} i'^-'-^l~r-' w. 

We see that it ia necessary to make a diagram whose ordinate every- 

where is -j and we mast integrate it. Let I -^ o?^ be culled fi ; when 

x=lf IX becomes the whole area of the -j diagram and we will call 

this ^j. 

It is also necessary to make a diagram whose ordinate everywhere 

is -J and integrate it. Let J -.dxhQ called X, When x=l^ X becomes 


the whole area of the j diagram and we will call this X^. 
Then as in (9), d * when x=lj 

o=f,,-p. x,+c ao). 

Integrating (9) again, we have 

Ey=\ii. dx - P \X . dx + ex + C. 

In this if we use y — O when x=0, we shall find C'=0, and again if 
y=0 when x=l, and if we use Mj and Xj as the total areas of the /x 
and X curves we have 

= Mi-P.Xi + e;* (11), 

from (10) and (11) i^ and c may be found, and of course P enables us 
to state the bending moment everywhere, -c is the slope when .'t- 
is 0. 

64. Example. Beam of any changing section fixed 

* Without using the letters fi, X, yUj , X^ &c. the above investigation is : — 

e:i]--/:t--/:^^ <-'• 

Integrating again between the limits and I and recollecting that y is 
the same at both limits 

f4]=0=f7'"i..-p/"'f'^%.Z (11). 

a;=0 J J J- J J J- 

The integrations in (10) and (11) being performed, the unknowns P and c 
can be calculated ; the true bending moment everywhere is what we started 

m - Px, 


at the ends^ any kind of loading. Measuring x from one 

end where there is the fixing couple ?«2) 

M=m-m,-rj; (1), 

^'g = ?-'?-''7 (^). 

^j-,= I 7 • dx-m^ I ~r -P \ -y-+constant (3), 

Let fi = I — 'y-^ and fi^ the whole area of the -^ curve; 

(dx 1 

Let F= / -y and Fj the whole area of the y curve; Let 

foe dx 1' 

X=l ' and Xi the whole area of the ~ curve; then 

= /Xi-m2Fi-PXi (4). 

Again integrating 

y=j/jL.dx — 111.2 1 Y ,dx — P \X .dx-\- const. 

Calling the integi^als from to I of the //,, F and X curves 
Mj, Yj, and X^, we have 

= Mi-7?i.,Yi-PXi* (5), 

and as rrio and P are easily found from (4) and (5), (1) is 

* We have used the Bymbols /a, A', Y, /j.^, X^, Y^, M, Y, Mi, Xi, Yi fearing 
that students are still a little unfamiliar with the symbols of the calculus ; 
perhaps it would have been better to put the investigation in its proper 
form and to ask the student to make himself familiar with the usual symbol 
instead of dragging in eleven fresh symbols. 

After (3) above, write as follows ; — 

U d}='=} 07"^- '"■'). T-n.-r w- 

Again integrating between Umits 

l-U=N"^a.-.J'l'^^-r['f'i^ (5, 

x=o J J J- J J ■'■ yoyo-' 

The integrations indicated in (4) and (5) being performed, the unknowns m^ 
and P can be calculated and used in (1). The student must settle for 
himself which is the better course to take; to use the formidable looking 
but really easily understood symbols of this note or to introduce the eleven 
letters whose meaning one is always forgetting. See also the previous note. 



65. In Graphical work. Let A GB (fig. 43) represent m, 
the bending moment, if the beam were merely supported at 
the ends; let AD represent nv^ and let BE represent Wg. 
Join DE. Then the difference between the ordinates of 

Fig. 43. 

ACB and of ADEB represents the actual bending moment; 
that is the vertical ordinates of the space between the straight 
line DE and the curve AFCGB. It is negative from A to 
H and from / to B, and positive from H to I. F and G are 
points of inflexion. 

Useftil Analogies in Beam Problems. If io 

is the load per unit length 
on a beam and M is the bend- 
ing moment at a section 
(positive when it tends to 
make the beam convex up- 
wards*), a; being horizontal 
distance, to prove that 

dF = ^ ^^>- 

If at the section at P 
fig. 44, whose distance to the 
right of some origin is oc there 
is a bending moment M in- 
dicated by the two equal and 
p. ^, opposite arrow heads and a 

shearing force S as shown, 
being positive if the material to the right of the section is 

* This convention is necessary only in the following generalization. 


acted on by downward force, and if PQ is hx so that the load 
on this piece of beam between the sections at P and Q is 
w .hx\ if the bending moment on the Q section is M + hM 
and the shearing force S + hS, then the forces acting on this 
piece of beam are shown in the figure and from their equili- 
brium we know that 

hS — io.hx or — - = w (2), 

and taking moments about Q, 

3I+S.Sx + iw (8xy=^M+ MI, 

or -tT- = o + 1^ . ox, 

ox ^ 

and in the limit as hx is made smaller and smaller 

^ = S (3), 

and hence (1) is true. 

Now it is well known that in beams if y is the deflection 
d2y M 



If we have a diagram which shows at every place the 
value of Wy called usually a diagram of loading, it is an 
exercise known to all students that we can draw at once by 
graphical statics a diagram showing the value of M at every 
place to scale ; that is we can solve (1) very easily graphi- 
cally*. We can see from (4) that if we get a diagram 

showing yn- at every place, we can use exactly the same 

method (and we have exactly the same rule as to scale) to 
find the value of y ; that is, to draw the shape of the 
beam. Many of these exercises ought to be worked by all 

* We find M usually for a beam merely supported at the ends. Let it 
be AGB, fig. 43. If instead, there are bending moments at the ends we 
let AD and BE represent these and join DE. Then the algebraic sum of 
the ordinates of the two diagrams is the real diagram of bending moment. 


Example. In any beam whether supported at the ends 
or not : if -m; is constant, integrating (1) we find 

— - — h-\-iux and M = a-\-hx-\- ^tuaf (5). 

In any problem we have data to determine a and b. 

Take the case of a uniform beam uniformly loaded and 
merely supported at the ends. 

Measure y upwards from the middle and x from the 
middle. Then i/= where x = ^l and — ^l, 

and = tt — ^hl + ^wl-. I 

Hence 6 = 0, ti = — Itvl^ and (5) becomes 

M = - iwl" ■}- ^wx"" (6), 

which is exactly what we used in Example IV. (Art. 60) 
where we afterwards divided M by EI and integrated twice 
to find y. 

Let i be -^- or the slope of the beam. 

e,. dy . di M dM ^ dS 

dx ' dx EI dx dx 

we have a succession of curves which may be obtained 

from knowing the shape of the beam y by differentiation, or 

which may be obtained from knowing w, the loading of the 

beam, by integration. Knowing w there is an easy graphical 

rule for finding MjEI, knowing MjEI we have the same 

graphical rule for finding y. Some rules that are obviously 

true in the w to MjEI construction and need no mathematical 

proof, may at once be used without mathematical proof in 

applying the analogous rule from MjEI to y. Thus the area 

of the MjEI curve between the ordinates x^ and x^ is the 

increase of i from x^ to x.2y and tangents to the curve showing 

the shape of the beam at x^ and x^ meet at a point which 

is vertically in a line with the centre of gravity of the 

portion of area of the MjEI curve in question. Thus the 

whole area of the M/EI curve in a span IIJ is equal to the 

increase in -~ from one end of the span to the other, and 


the tangents to the beam at its ends H, J meet in a point 
P which is in the same vertical as the centre of gravity of 
the whole M/EI curve. These two rules may be taken as 
the starting point for a complete treatment of the subject 
of beams by graphical methods. 

If the vertical from this centre of gravity is at the 
horizontal distance HG from H and GJ from «/, then P is 
higher than H by the amount HG x ij^, the symbol % being 
used to mean the slope at ^ ; J" is higher than P by the 
amount GJ x { at J. Hence J is higher than H by the 


a relation which may be useful when conditions as to the 
relative heights of the supports are given, as in continuous 
beam problems. 

67. Theorem of Three Moments. For some time, 
Railway Engineers, instead of using separate girders for 
the spans of a bridge, fastened together contiguous ends 
to prevent their tilting up and so made use of what 
are called continuous girders. It is easy to show that 
if we can be absolutely certain of the positions of the 
points of support, continuous girders are much cheaper than 
separate girders. Unfortunately a comparatively small 
settlement of one of the supports alters completely the 
condition of things. In many other parts of Applied 
Mechanics we have the same difficulty in deciding between 
cheapness with some uncertainty and a gi-eater expense with 
certainty. Thus there is much greater uncertainty as to the 
nature of the forces acting at riveted joints than at hinged 
joints and therefore a structure with hinged joints is pre- 
ferred to the other, although, if we could be absolutely 
certain of our conditions an equally strong riveted structure 
might be made which would be much cheaper. 

Students interested in the theory of continuous girders 
will do well to read a paper published in the Proceedings of 
the Royal Society/, 199, 1879, where they will find a graphical 
method of solving the most general problems.t I will take 
here as a good example of the use of the calculus, a uniform 
girder resting on supports at the same level, with a uniform 



load distribution on each span. Let ABC be the centre line 
of two spans, the girder originally straight, supported at 
A, B and G. The distance from J. to .S is ^i and from B to G 



Fig. 45. 

is I2 and there are any kinds of loading in the two spans. 
Let A, B and G be the bending moments at A, B and G 
respectively, counted positive if the beam is concave upwards. 
At the section at P at the distance x from A let m be 
what the bending moment would have been if the girder 
on each span were quite separate from the rest. We have 
already seen that by introducing couples 77I2 and mi at A and 
B (tending to make the beam convex upwards at A and B) 
we made the bending moment at P really become what 
is given in Art. 61. Our m^ — — A^m^^ — B, and hence the 
bending moment at P is 

B-A_ d'j 

m-i- A -^ X 



where m would be the bending moment if the beam were 
merely supported at the ends, and the supporting force at A 
is lessened by the amount 

^ A-B 



Assume EI constant and integrate with regard to x and 
we have 


B — A 

m .dx -\- Ax + ^x^ — J 1- Ci 

^'■i (^^)- 

Using the sign jltn. dx . dx to mean the integration of 

the curve representing J7a.dx we have 

r r T) A 

\\m.dx.dx'^\Ax^+^x' — ^ + Ci^p -f e = ^Y . 3/ . . . ( 4). 


As 7/ is when ic=0 and it is evident that llm.dx.da; = 

when cc = 0, e is 0. Again y = when a;=lj^. Using the 

symbol //^ to indicate the sum jjm.dx.dic over the whole 


fi, + iAk'-{-ili'(B-A) + cA = (5). 

From (3) let us calculate the value of EI -—- at the point 
B, and let us use the letter a^ to mean the area of the m 
curve over the span, or I m . dx, so that EI ~ at 5 is 

a^-\-Ak + \k{B-A) + c, (6). 

But at any point Q of the second span, if we had let BQ = x 
we should have had the same equations as (1), (3) and (4) 
using the letters B for A and G for B and the constant Ca. 

Hence making this change in (3) and finding EI -^- at the 
point B where a: = 0, we have (6) equal to Co or '^ 

C2-Ci = ai + ^^i4-i^i(5-^) (7), 

and instead of (5) we have 

/i, + \BU 4- ^U {C-B)-¥ cA = (8). 

Subtracting (5) from (8) after dividing by Zj and l.j we 

c,-c, = ^-f^ + iAh-iBk + lh{B-A)-ih{C-B)...(9). 
The equality of (7) and (9) is 

an equation connecting A, B and C, the bending moments 
at three consecutive supports. If we have any number of 
supports and at the end ones we have the bending moments 
because the girder is merely supported there, or if we have 
two conditions given which will enable us to find them in 
case the girder is fixed or partly fixed, note that by writing 

P. 8 


down (10) for every three consecutive supports we have a 
sufficient number of equations to determine all the bending 
moments at the supports. 

Example. Let the loads be w^ and w^ per unit length 
over two consecutive spans of lengths l^ and l^. Then 

m = \wlx — \wa^, jm.dx= \wla^ — \wa?, 

Hence rti = T^^'j ^^^ \\m.dx ,dx = -^wlaf^ — -^wa^. 

Hence /Aj = -^w^k\ fi^ = i^wj^i*. 

Hence ^ + tti — y^ becomes -^-fW.}.^ + r-^ Zi» — ^w^li^, 

or ^\ (ukI/ 4- w^li"), 

and hence the theorem becomes in this case 

Al, + 2B (I, + k) + Gl, + i (wj2^ + w^k') = (10). 

If the spans are similar and similarly loaded then 

A-h4^B-\-C+^wP=^0 (11). 

Case 1. A uniform and uniformly loaded beam rests on 
three equidistant supports. Here A = C = and B = — ^wl\ 
m = \w {Ix -- aP), and hence the bending moment at a point 
P distant x from A is 

\w {Ix - a^) + --^wl\ 

The supporting force at A is lessened from what it would be 
if the part of the beam AB were distinct by the amount 

shewn in (2), — ^ — or ^wl. It would have been ^l, so now 

it is really ^wl at each of the end supports, and as the total 
load is 2wl, there remains ^^-wl for the middle support. 

Case 2. A uniform and uniformly loaded beam rests on 
four equidistant supports, and the bending moments at 
these supports are A, B, G, D. Now A = D=0 and from 
symmetry B = G. Thus (11) gives us 

+ 5J5+irf = or B=^G=^--^^wl\ 



If the span AB had been distinct, the first support would 
have had the load ^wl, it now has ^wl — -^^wl or -^qwI. The 
supporting force at D is also -^-^wl. The other two supports 
divide between them the remainder of the total load which 
is altogether ^wl and so each receives \^wl. The supporting 
forces are then -^^wl, \^wl, \^wl and -^-^wl. 

68. Shear Stress in Beams. Let the distance 
measured from any section of a beam, say at 0, fig. 46, 
to the section at A be x, and let OB — x-\-hx. Let the 
bending moment at G'AG be M and at B'BD be M + hM, 

C D 

E— IF 

C D 



Fig. 46. 

Fig. 47. 

Fig. 48. 

OAB (fig. 46) and A A (fig. 47) represent the neutral 
surface. We want to know the tangential or shear stress / 
at E on tlie plane GAC. Now it is known that this is 
the same as the tangential stress in the direction EF on 
the plane EF which is at right angles to the paper and 
parallel to the neutral surface at AB. Consider the equi- 
librium of the piece of beam ECDF, shown in fig. 47 as ECE, 
and shown magnified in fig. 48. We have indicated only 
the forces which are parallel to the neutral surface or at 
right angles to the sections. The total pushing forces on DF 
are greater than the total pushing forces on CE, the tangential 
forces mi EF making up for the difference. We have only to 
state this mathematically and we have solved our problem. 

At a place like ^ in the plane CAC at a distance y from 
the neutral surface the compressive stress is known to be 

M . , 

p = Y y^ ^^^ ^^ ^ ^® *^^ breadth of the section there, shown 

as IIH (fig. 47), the total pushing force on the area ECE is 

rAC \f Jlf rAC 

P= b'^y.dy or P^'y- hy.dy (1). 

J AE J- -^ J AE 



Observe that if b varies, we must know it as a function of 
y before we can integrate in (1). Suppose we call this total 
pushing force on EC by the name P, then the total push- 
ing force on DF will be P + 8^ . -r* . The tangential force on 
EF is fx area of EFovf.hx. EE, and hence 

f.Bx.EE = Bx.^~ovf= J:^ . i? (2). 

*^ dx -' EE dx ^ ' 

Example. Beam of uniform rectangular section^ of 

constant breadth 6 and constant depth d. Then 

p 12m p^ , i2iifr^ 1 


and hence /= ^ |(t^'3_^/;.)^^^ (8); 

so that /is known as soon as M is known. 

As to M, let us choose a case, say the case of a beam 
supported at the ends and loaded uniformly with iv lb. 
per unit length of the beam. We saw that in this case, x 
being distance from the middle 

M — \wl^ — \wx^. 

Hence - 7- = — wx, so that (3) is 

f=~{^l''-AE^)wx (4). 

If we like we may now use the letter y for the distance -4 -£^, 
and we see that at any point of this beam, x inches measured 
horizontally from the middle, and y inches above the neutral 
line the shear stress is 

/=-5«*-2'')'^ ^■'^- 

The — sign means that the material below EF acts on 
the material above EF in the opposite sense to that of the 
arrow heads shown at EF, fig. 48. 


Observe that where 2/ = the shear stress is greater 
than at any other point of the section, that is, at points 
in the neutral line. The shear stress is at C. Again, the 
end sections of the beam have greatest shear. A student 
has much food for thought in this result (5). It is interest- 
ing to find the directions and amounts of the principal 
stresses at every point of the beam, that is, the interfaces at 
right angles to one another at any point, across which there 
is only compression or only tension without tangential stress. 

We have been considering a rectangular section. The 
student ought to work exercises on other sections as soon as 

he is able to integrate hy with regard to 3/ in (1) where h is 

any function of y. He will notice that I hy.dy is equal to 


the area of EHGHE, fig. 47, multiplied by the distance of its 
centre of gravity from A A. 

Taking a flanged section the student will find that / is 
small in the flanges and gets greater in the web. Even in a 
rectangular section /' became rapidly smaller further out 
from the neutral line, but now to obtain it we must divide by 
the breadth of the section and this breadth is comparatively 
so great in the flanges that there is practically no shearing 
there, the shear being confined to the web ; whereas in the 
web itself / does not vary very much. The student already 
knows that it is our usual custom to calculate the areas of 
the flanges or top and bottom booms of a girder as if they 
merely resisted compressive and tensile forces, and the web 
or the diagonal bracing as if it merely resisted shearing. 
He will note that the shear in a section is great only where 

, or rather -^\-f] is great. But inasmuch as in Art. QQ 

we saw that -j- — S, the total shearing force at the section, 

there is nothing very extraordinary in finding that the actual 

shear stress anywhere in the section depends upon -r- . In 

a uniformly loaded beam -7- is greatest at the ends and gets 

less and less towards the middle and then changes sign, 
hence the bracing of a girder loaded mainly with its own 
weight is much slighter in the middle than at the ends. 


Deflection of Beams. If a bending moment M acts at a 

section of a beam, the part of length bx gets the strain-energy \ — '' , 

because M . dx/FI is the angular change (see Art. 26), and therefore 
the whole strain-energy in a beam due to bending moment is 

m-'^ <«\ 

If / is a shear stress, the shear strain-energy \}er unit volume is 
lf2/2iV...(7), and by adding we can therefore find its total amount for 
the whole beam. 

By equating tlic strain-energy to the loads multiplied by half the 
displacements produced by them we obtain interesting relations. Thus 
in the case of a beam of length /, of rectangular section, fixed at one 
end and loaded at the other with a load W; at the distance x from the 
end, M= Wx and the energy due to bending is 


dx=WH^lQEI (8). 

1 p TF2; 
^Ej^ I 

The above expression (5) gives for the shearing stress 

f=\^iid?-f)W (9). 

The shear strain-energy in the elementary volume h .bx . by is 
b.bx . by.f'^/2K Integrating this with regard to y from - ^c^ to -{-^d 
we find the energy in the slice between two sections to be 


so that the shear strain-energy in the beam is 3 WH/dNbd... (10). 

If now the load W produces the deflection z at the end of the beam 
the work done is ^ Wz,..{\l). 

Equating (11) to the sum of (8) and (10) we find 

.= i^ + iL ^^ (12) 

Note that the first part of this due to bending is the deflection as 
calculated in Art. 60, Example I. We believe that the other part due 
to shearing has never before been calculated. 

If the deflection due to bending is z^ and to shearing is z^^ 


Taking A^= § ^ as being fairly correct, then zjs.^ = Al^j^d^. If a beam 
is 10 inches deep, when its length is 8-6 inches the deflections due to 
bending and shear are equal ; when its length is 86 inches, the deflection 
due to bending is 100 times that due to shear ; when its length is 0-79 
inch, the deflection due to bending is only 1 /100th of that due to 
shear. Probably however our assumed laws of bending do not apply 
to so short a beam. 



69. Springs which Bend. Let fig. 49 show the centre 
line of a sirring fixed at A, 
loaded at B with a small 
load W in the direction 
shown. To find the 
amount of yielding at B. 
The load and the deflec- 
tion are supposed to be 
very small. Consider 
the piece of spring 
bounded by cross sec- 
tions at P and Q. Let FQ = 8s, the length of the spring 
between B and P being called s. 

Fig. 49. 

The bending moment at P is W . PR or W.w if x is, the 

length of the perpendicular from P upon the direction of W. 

Let BR be called y. Consider first that part of the motion 

of B which is due to the change of shape of QP alone ; that 

is, imagine AQ to be perfectly rigid and PB a rigid pointer. 

The section at Q being fixed, the section at P gets an angular 

change equal to Ss x the change of curvature there, or Bs -^rt- 

8s Wx . . 

or — '^p^ ...(1), where E is Young's modulus and / is the 

moment of inertia of the cross section. The motion of B due 
to this is just the same as if PB were a straight pointer ; 
in fact the pointer PB gets this angular motion and the 
motion of B is this angle, multiplied by the straight 
distance PB or 

Ss. Wx 



Now how much of B's motion is in the direction of IF ? 


It is its whole motion x -p^ or x p-^ and hence B'a 

motion in the direction of W is 

hs . Wx" 



Similarly B'a motion at right angles to the direction of 

W is ^-i^ (4). 

In the most general cases, it is easy to work out the 
integrals of (3) and (4) graphically. 

We usually divide the whole length of the spring from B 

to A into a large number of equal parts so as to have all the 

values of Bs the same, and then we may say (s being the 

s W 
whole length of the spring) that we have to multiply —w— 

upon the average values of -^ and -j- for each part. In a well 

made spring if b is the breadth of a strip at right angles to 
the paper and t its thickness so that / = -f^bt^ we usually 
have the spring equally ready to break everywhere or 

~r42~~f> ''^ constant. When this is the case (3) and (4) 

2/.^^ ^ and ^-^'^^ y 

And if the strip is constant in thickness, varying in 
breadth in proportion to x, then 

If X and y are the x and y of the centre of gravity of 
the curve (see Art. 48) 

^- is the total yielding parallel to W, 

•^-^ is the total yielding at right angles to 'W. 
70. Exercises. The curvature of a curve is 

When the equation to a curve is given it is easy to find 


-^ and v^and calculate - where r is the radius of curvature. 
ax da? r 

This is mere exercise work and it is not necessary to prove 

beforehand that the formula for the curvature is correct. 

1. Find the curvature of the parabola y — ax^ at the 
point d? = 0, 3/ = 0. 

2. The equation to the shape of a beam, loaded uni- 

formly and supported at the ends is 3/ = aqtpt {^^^^^ ~ 2^'*), 

see Art. 60, where the origin is at the middle of the beam ; 
I is the whole length of the beam, lu is the load per unit 
length, E is Young's modulus for the material and / is the 
moment of inertia of the cross section. Take I = 200, w = 5, 
E = 29x 10«, / = 80, find the curvature where x = 0. Show 

that in this case f-^) may be neglected, in comparison with 

1, and that really the curvature is represented by -vt,. 

Show that the bending moment of the above beam is 

^^RFT ^^^ " ^'^•^' ^^^^ *^^^ ^^^^ ^^ greatest at the middle 
of the beam. 

3. Find the curvature of the curve i/= a log x +bx-\-c 
at the point where x = Xi. 

71. Force due to Pressure of Fluids. Exercise 1. 
Prove that if p, the pressure of a fluid, is constant, the 
resultant of all the pressure forces on the plane area A is Ap 
and acts through the 
centre of the area. -"t^j ; 

2. The pressure in ''\ 

a liquid at the depth h \ 

being wh, where w is the ^^\^ 

weight of unit volume, 
what is the total force due 
to pressure on any im- 
mei*sed plane area ?- Let 
DE be the surface from 
which the depth h is 
measured and where the ^^S- ^0. 


pressure is 0. Let BG be an edge view of the area ; imagine 
its plane produced to cut the level surface of the liquid DE 
in D. Let the angle EDO be called a. Let the distance 
DP be called x and let DQ be called x + hx, and let the 
breadth of the area at right angles to the paper at P be 
called z. On the strip of area z . hx there is the pressure 
wh if h is PH the depth of P, and h^x sin a, so that the 
pressure force on the strip is 

wx . sin OL . z . hx, 

and the whole force is F=wsuyol\ x .z .dx (1 ). 

J 1)B 

Also if this resultant acts at a point in the area at a distance 
X from D, taking moments about P, 

FX=tu sina x- .z.dx (2). 

Observe in (1) that I x ,z ,dx — Ax, 

if A is the whole area and x is the distance of its centre of 
gravity from P. Hence, the average pressure over the 
area is the pressure at the centre of gravity of the 


Observe in (2) that I x^z .dx — I the moment of inertia 


of the area about P. Letting I — k^A , where k is called the 
radius of gyration of the area about P, we see that 

F=iu sin a . Ax, FX = w sin a . Ah^. 

Hence X = — . . .(3), the distance from D at which the 


resultant force acts. 

Example, If DB = and the area is rectangular, of 
constant breadth h ; then 

CDC h 

I=h x^.dx = ^DC\ 

Jo «5 

and ^ = 6 . PC so that k^ = ^DC\ Also x = J PC. Hence 
X = fPC, that is, the resultant force acts at | of the way 



down the rectangle from i) to (7 and the average pressure is 
the pressure at a point half way down. 

It is an easily remembered relation that we find in (8). 
For if we have a compoimd pendulum, whose radius of 
gyration is k and if x is the distance from the point of 
support to its centre of gravity and if X is the distance to 
its point of percussion, we have the very same equation (3). 
Again, if X is the length of the simple pendulum which 
oscillates in exactly the same time as the compound one, 
we have again this same relation (3). These are merely 
mathematical helps to the memory, for the three physical 
phenomena have no other relation to one another than a 
mathematical one. 

Whirling Fluid. 

72. Suppose a mass of fluid to rotate like a rigid body about 
an axis with the angular velo- 
city of a radians per second. 
Let 00 be the axis. Let P 
be a particle weighing w lbs. 
Let OP 



The centrifugal force in 
pounds of any mass is the 
mass multiplied by the square 
of its angular velocity, multi- 
plied by X. Here the mass 

is - and the centrifugal force 

• "U) „ 

IS - o?x. Fig. 51. 

Make PR represent this to scale and let PS represent w 
the weight, to the same scale, then the resultant force, repre- 
sented by PT, is easily found and the angle RPT which PT 

makes with the horizontal. Thus tan RPT — w^— ot^x or 

g — a^x, being independent oi w: we can therefore apply our 
results to heterogeneous fluid. Now if y is the distance of 
the point P above some datum level, and we imagine a curve 
drawn through P to which PT is (at P) tangential, and if at 



every point of the curve its direction (or the direction of its 
tangent) represents the direction of the resultant force; if 

such a curve were drawn its slope ~ is evidently — y- and 

its equation is y = — -2 log x + constant (1 ). 

The constant depends upon the datum level from which y is 
measured. This curve is called a line of force. Its direction 
at any place shows the direction of the total force there. We 
see that it is a logarithmic curve. 

Level Surfaces. If there is a curve to which PT is a 

Fig. 52. 



normal at the point P, it is evident that its slope is positive 
and in fact 


— —X, 

so that the curve is 

y — ^x^-\- constant (2), 

the constant depending upon the datum level from which y 
is measured. This is a parabola, and if it revolves about the 
axis we have a paraboloid of revolution. Any surface which 
is everywhere at right angles to the force at every point is 
called a level surface and we see that the level surfaces in 
this case are paraboloids of revolution. These level surfaces 
are sometimes called equi-potential surfaces. It is easy to 
prove that the pressure is constant everywhere in such a 
surface and that it is a surface of equal density, so that if 
mercury, oil, water and air are in a whirling vessel, their 
surfaces of separation are paraboloids of revolution. 

The student ought to draw one of the lines of force and 
cut out a template of it in thin zinc, 00 being another edge. 
By sliding along 00 he can draw many lines of Force. Now 
cut out a template for one of the parabolas and with it draw 
many level surfaces. The two sets of curves cut each other 
everywhere orthogonally. Fig. 52 shows the sort of result 
obtainable where aa^ hb, cc are the logarithmic lines of force 
and A A, BB, GO are the level paraboloidal surfaces. 

73. Motion of Fluid. If AB is si stream tube, in the 


vertical plane of the paper, consider the mass of fluid between 
sections at P and Q of length Bs feet along the stream, and 
cross-section a square feet, where a and Bs are in the limit 
supposed to be infinitely small. Let the pressure at F be 
p lbs. per square foot, the velocity v feet per second, and let 
F be at the vertical height h feet above some datum level. 

At Q let these quantities be ^ + Sp, v + Bv and h + Bh. 
Let the fluid weigh w lbs. per cubic foot. 

Find the forces urging PQ along the stream, that is, 
forces parallel to the stream direction at FQ. 

pa acts on one end F in the direction of motion, and 
(p-\-Bp)a acts at Q retarding the motion. The weight of 
the portion between F and Q is a . Bs . w and, as if on an 
inclined plane, its retarding component is 

. , ^ height of plane ^ Bh 

weight X -, — ^Tu— r n — or a. Bs .w ^, 
° length ot plane Bs 

Hence we have altogether, accelerating the motion from F 
towards Q, 

pa — (p + Bp) a — a . Bs . w . ^— . 

But the mass is — — , and -7- is its acceleration, and we 

have merely to put the force equal to — ^ — '- — . -=- . We have 

then, dividing by a, 

5. ^ Bh Bs .w dv 

-bp-8s.w-^= -J- . 

^ Bs g dt 

Now if Bt be the time taken by a particle in going from 

F to Q, v — j- with greater and greater accuracy as Bs is 

. dv . 
shorter and shorter. Also, the acceleration , is more aud 

Bv . ^^ 

more nearly kt . (It is more important to think this matter 

out carefully than the student may at first suppose.) 

Hence if Bs is very small, Bs .-j- — -^.Bv — v ,Bv, so that 

at ot 

we have Bp + w. Bh-^-v .Bv = (1), 


or as we wish to accentuate the fact that this is more and 
more nearly true as Ss is smaller and smaller, we may 
write it as 

^ + dh-\--.dv==0 (2)*, 

w g 

or integrating, ^ + 9- + I = constant (2). 


We leave the sign of integration on the -^ because w may 
vary. In a liquid where w is constant, 

/* + ^ + --= constant (3). 

2g w ^ -^ 

74. In a gas, we have lu ocpif the temperature could be 

kept constant, or we have the rule for adiabatic flow w x py, 
where 7 is the well-known ratio of the specific heats. In either 

of these cases it is easy to find I — and write out the law. This 

law is of universal use in all cases where viscosity may be 
neglected and is a great guide to the Hydi^aulic Engineer- 

Thus in the case of adiabatic flow w=cpy , the inteOTal of -^ is 
f dp 1 /" _i 1 V 1 1 

we have h-{--;r--\ — ©"^constant (4). 

In a great many problems, changes of level are insignificant and we 

* After a little experience with quantities like 5p &c., knowing as we do 
that the equations are not true unless 5/), &c. are supposed to be smaller 

and smaller without limit and then we write their ratios as ^f , &c., we 


get into the way of writing dp, &c. instead of 5p, &c. 

Again, if /(x) . dx-\-F{y) dy + <p [z) dz = (1), 

then 1/H 'dx+ I F {y) . dy + I <p{z) . d2 = a constant (2). 

There is no harm in getting accustomed to the integration of such an 
equation as (1), all across. 


often use i?^^.,^^*^ constant (4) for gases. Thas, if «« is the 


pressure and zvq the weight of a cubic foot of gas inside a vessel at 

places where there is no velocity and if, outside an orifice, the pressure 

isp; the constant in (4) is evidently + — jOq«, and hence, outside the 


surface, v^=-^{pQ'-p') (5), and as c is Wq-^PqV it is easy to make 

all sorts of calculations on the quantity of gas flowing per second. 

Observe that if p is very little less than joq, if we use the approxi- 
mation (1 +«)" = 1 -\-naf when a is small, we find 

v'=^^{Po-p) (6), 

a simple rule which it is well to remember in fan and windmill problems. 
In a Thomson Water Turbine the velocity of the rim of the wheel is 
the velocity due to half the total available pressure ; so in an air turbine 
when there is no great difference of pressure, the velocity of the rim of 
the wheel is the velocity due to half the pressure difference. 

Thus if Pq of the supply is 7000 lbs. per square foot and if p of the 
exhaust is 6800 lbs. per square foot and if we take ?^'q=0*28 lb. per 
cubic foot, the velocity of the rim V is, since the difference of pressure 
is 200 lbs. per square foot, 


% (100) = 151 feet per second. 

Returning to (5) ; neglecting friction, if there is an orifice of area 
A to which the flow is guided so that the streams of air are parallel, Q 
the volume flowing per second is Q=vA and if the pressure is p, the 
weight of stuff flowing per second is 

1 1 

or since w = cp^ ^ and iOQ=cp^ , 

If the student will now substitute the value of v from (5) and put 
a for p/pQ he will obtain 


=^"'-N/r^^:(--^^^) ^^)- 

Problem. Find p the outside pressure so that for a given inside 
pressure there may be a maximum flow. 


It is obvious that as p is diminished more and more, v the velocity 
mcreases more and more and so does Q. But a large Q does not 
necessarily mean a large quantity of gas. We want W^ to be large. 
When is W a maximum 1 That is, what value of a in (7) will make 

a^l-a-^ )ova^-a ^ 
a maximum 1 Differentiating with regard to a and equating to 
2 , / -.v 1 

?„;— _(i+i)„9=o 

dividing by ay we find a = ( ^ ) 


In the case of air 7/= 1-41 and we find p= •527po- 

That is, there is a maximum quantity leaving the vessel per 
second when the outside pressure is a little greater than half the inside 
pressure. » 

Problem. When p is indefinitely diminished what is v ? 
Answer -. v= j,./ -^ — . 

This is greater than the velocity of sound in the ratio a/ — — t- , 
being 2 -21 for air. That is, the limiting velocity in the case of air is 
2413 feet per second x */ — - , where t is the absolute temperature 

inside the vessel and there is a vacuum outside. 

Students ought to work out as an example, the velocity of flow 
into the atmosphere. 

Returning to equations (2) and (4), we assumed h to be of little 
importance in many gaseous problems of the mechanical engineer. But 
there are many physical problems in which it is necessary to take 
account of changes in level. For example if (2) is integrated on the 
assumption of constant temperature and we assume v to keep constant, 
we find that p diminishes as h increases according to the compound 
interest law considered in Chap. 11. Again under the same condition 
as to V, but with the adiabatic law for w we find that p diminishes with 
h according to a law which may be stated as " the rate of diminution 
of temperature with A, is constant." These two propositions seem to 
belong more naturally to the subject matter of Chapter II. 


75. A great number of interesting examples of the use of 
(2) might be given. It enables us to understand the flow of 
fluid from orifices, the action of jet pumps, the attraction of 
light bodies caused by vibrating tuning-forks, why some 
valves are actually sucked up more against their seats 
instead of being forced away by the issuing stream of fluid, 
and many other phenomena which are thought to be very 

Example 1. Particles of water in a basin, flowing very 
slowly towards a hole in the centre, move in nearly circular 
paths so that the velocity v is inversely proportional to the 

distance from the centre^ Take v = - where a is some con- 


stant and x is the radius or distance from the axis. Then (13) 

(Art. 73) becomes 

k^f-, + P = f;. 

2gx- w 

Now at the surface of the water, p is constant, being the 
pressure of the atmosphere, so that, there 

and this gives us the shape of the curved surface. Assume 
c and a, any values, and it is easy to calculate h for any value 
of X and so plot the curve. This curve rotated about the axis 
gives the shape of* the surface which is a surface of revo- 

Example 2. Water flowing spirally in a horizontal plane 

follows the law v = - if ^ is distance from a central point. 

w h^ 
Note that p=:G^—h- - . 
^ g X'' 

The ingenious student ought to study how p and v vary 
at right angles to stream lines. He has ohly to consider 
the equilibrium of an elementary portion of fluid PQ, fig. 53, 
subjected to pressures, centrifugal force and its own weight 
in a direction normal to the stream. 

He will find that if ~- means the rate at which p 


varies in a direction of the radius of curvature away from 
the centre of curvature and if a is the angle QPR, fig. 53, the 
stream being in the plane of the paper, which is vertical, 

dp wv^ . ,^, 

-T = lusma (1). 

ar g r 

If the stream lines are all in horizontal planes 

f = ^^' (2). 

ar g r ^ 

Example 3. Stream lines all circular and in horizontal 
planes in a liquid, so that h is constant. 

If -y = -, where 6 is a constant, 

dp _ lu h- 
dr g ' T^' 

p:=— I , + constant (3). 

^ ^ g r- 

We see therefore that the fall of pressure as we go out- 
ward is exactly the same as in the last example. Show that 
this law, V = hjr, must be true if there is no ' rotation ' (See 
Example 5). 

Example 4. Liquid rotates about an axis as if it were a 
rigid body, so that v = 6/-, then 

dr g ' 

This shows the law of increase of pressure in the wheel of a 
centrifugal pump when full, but when delivering no water. 

Exercise. The pressure at the inside of the wheel of 
a centrifugal pump is 2116 lbs. per sq. foot, the inside radius 
is 0*5 foot, the outside radius 1 foot. The angular velocity 
of the wheel is 6 = 30 radians per second ; draw a curve show- 
ing the law of p and r from inside to outside when very 
little water is being delivered. If the water leaves the wheel 
by a spiral path, the velocity everywhere outside being 



invei-sely proportional to r, draw also the curve showing 
the law of p in the whirlpool chamber outside. 

Example 5. The expression 
ti^ 1 

which remains constant all along a stream line, may be 
called the total store of energy of 1 lb. of water in the 
stream if the motion is steady. 

^^ dE 1 dv 1 dp dh , « ^. ,.,. 

Now -r- =~ V j--\ r-4-T- becomes irom equation (1), 

d?' g or w dr ar . 

r ~ g 2\r dr) 

This expression ^ \- + -r) is called the "average angular 

velocity" or "the rotation" or the 'spin' of the liquid. 


dE 2v 

^j- = — X rotation. 

dr g 

When liquid flows by gravity from a small orifice in a 
large vessel where, at a distance inside the orifice, the liquid 
may be supposed at rest, it is obvious the E is the same in 

d IR 

all stream lines, so that -T7 is 0, and there is no ' rotation ' 
anywhere. ^ 

If when water is flowing from an orifice in a vessel we 
can say that across some section of the stream the velocity is 
everywhere normal to the section and that the pressure is 
everywhere atmospheric, we can calculate the rate of flow. 
It is as well to say at once that we know of no natural 
foundation for these assumptions. However wrong the 
assumptions may be, there is no harm in using them in mere 
exercises on Integration. There being atmospheric pressure 
at the still water level, if v is the velocity at a point at the 
depth A, if a is an element of area of the section, Q= 2a 'J2gh 
the summation being effected over the whole section, Q being 
the volume flowing. Thus if the section is a vertical plane 
and if at the depth h it is of horizontal breadth z, through 


the area z . Sh water is flowing with the velocity '^2gh, so 
that ^2gh . z .Bhis the elementary volume flowing per second, 
and if hi and h^ are the depths of the highest and lowest 

points of the orifice, the total flow is Q = V2^ ( zh^ . dh. 

J hi 

Example 6. Rectangular section, horizontal breadth 6, 
Q = ^¥gh{\^.dh = |6\/2^ M 1 = lb \f2g (h.^ - h^^). 

Example 7. Triangular section, angle at depth h^, base 
horizontal of length b at depth h... Then within the limits 

of integration it will be found that z= j v (— /<i 4- h). 

fl2 — ill 

Hence Q = |^^ f(- hih^^ + A?) dh = ^-^-^f W-^hJi^ + W-)\ 
«2— "W h'i — hi L/t^ J 

If the ratio hJjhi be called r, it will be found that 

(^ = i^^f J6H-10ri+25l. 

When the student has practised integration in Chap. III., he 
may in the same way find the hypothetical flow through 
circular, elliptic and other sections. 

Keturning to the rectangular section, there is no case 
practically possible in which h^ is 0, but as this is a mere 
mathematical exercise let us assume hi = 0, and we have 
Q= |6 s^'lgh^. Now further assume that if there is a rect- 
angular sharp-edged notch through which water flows, its 
edge or sill being of breadth b and at the depth h^, the 
flow through it is in some occult way represented by the 
above answer, multiplied by a fraction called a coefficient of 
contraction, then Q = cb \l2gh^. Such is the so-called theory 
of the flow through a rectangular gauge notch. A true 
theory was based by Prof. James Thomson on his law of flow 
from similar orifices, one of the very few laws which the 
hydraulic engineer has to depend upon. We are sorry to 
think that nearly all the mathematics to be found in standard 
treatises on Hydraulics is of the above character, that is, it 
has only an occult connection with natural phenomena. 


76. Magnetic Field about a straight round wire. 

There are two great laws in Electrical Science. They concern 
the two circuits, the magnetic circuit and the electric circuit, 
which are always linked through one another. 

I. The line integral (called the Gaussage whatever the 
unit may be) of Magnetic Force round any closed 
curve, is equal to the current [multiplied by 47r if the 
current is in what is called absolute C.G.S. units (curious 
kind of absolute unit that needs a multiplier in the most 
important of all laws); multiplied by 47r/10 if the current 
is in commercial units called Amperes]. 

II. The line integral (called the Voltage whatever 
the unit may be) of Electromotive Force round any- 
closed curve is equal to the magnetic current (really, 
rate of change of induction) which is enclosed. [If the in- 
duction is in absolute C.G.S. units, we have absolute Voltage 
in C.G.S.; if the induction is in Webers the Voltage is 
in Volts. 

We are to remember that in a non-conducting medium 
the voltage in any circuit produces electric displacement, and 
the rate of change of this is cun-ent, and we deal with this 
exactly as we deal with currents in conducting material. 
When we deal with the phenomena in very small portions 
of space we speak of electric and magnetic currents per unit 
area, in which case the line integrals are called ' curls/ 
Leaving out the annoying 47r or 47r/10, we say, with Mr 
Heaviside, " The electric current is the curl of the magnetic 
force and the magnetic current is the negative curl of the 
electric force." When we write out these two statements in 
mathematical language, we have the two great Differential 
Equations of Electrical Analysis. 

The Electrical Engineer is continually using these two 
laws. Many examples will be given, later, of the use of the 
second law. We find it convenient to give here the following 
easy example of the first law. 

Field about a round wire. A straight round wire of 

or A amperes, so 
that G 






If H is the magnetic force at a distance r from the centre 
of the wire, the Gaussage round the circle of radius r is 
Hx27rr, because H is evidently, from symmetry, the same 
all round. Hence, as Gaussage = 47rC, 

2 Al 

Inside the wire, a circle of radius r encloses the total 

current — G, and hence H inside the wire at a distance r 

from the axis is 

2rG r 2r A 

^ ['' To ^ 

If BG is a cross section of the round wire of radius a, 
and if OD is any plane n 

through the axis of the f\ p Q p /^"^ 
wire, and \oJ y \0^ 

OP==r,OQ = ri-Br: ^ f". 54. 

then through the strip of area FQ, which is / centimetres 
long at right angles to the paper, and Sr wide, area I.Br, 
there is the induction If per sq. cm. [We take the perme- 
ability as 1. If /A is the magnetic permeability of the me- 
dium, the induction is yS = fiH per sq. cm.], or H .I.Br 
through the strip of area in question. If there are two 
parallel wires with opposite currents, and if OB is the plane 
through the axes of the two wires, the fields due to the two 
currents add themselves together. If 0' is the centre of the 

other wire, the total i^ at P is 2C ( ^p + yyp 

77. Self-induction of two parallel wires. Let the 

radius of each wire be a, and the distance between their centres 
h, the length of each being I between two planes at right angles 
to both. The wires are supposed to be parts of two infinite 
wires, to get rid of difficulties in imagining the circuit com- 
pleted at the ends. 

The total induction from axis to axis is the sum of the 

[^ G dr 
two amounts, 4>l - - — from the outside of each wire to the 


J a 


fa rC 

axis of the other and 41 \ —- dr from the axis of each wire 

Jo Cb' 

to its own surface. This is 

21G j 2 log - + 1 1 , or - . j 2 log - + 1 [ in absolute units. 

Dividing by 10^ we have it in commercial units. 

This total field when the current is 1, is the self-induction 
L of the circuit (we imagine current to be uniformly distri- 
buted over the section of the wire), and 

y = ^ U^g~i + 1^ in c.G.s. units, 

in Henries per centimetre length of the two circuits. 

78. Function of Two Independent Variables. 

Hitherto wo have been studying a function of one variable, 
which we have generally called x. In trying to under- 
stand Natural Phenomena we endeavour to make one 
thing only vary. Thus in observing the laws of gases, we 
measure the change of pressure, letting the volume only 
change, that is, keeping the temperature constant, and we 

find p cc - , Then we keep v constant and let the tempera- 
ture alter, and we find p cc t (where ^ = ^"^ C. + 274). After 

* Notice that one Henry is lO' absolute units of self-induction ; our 

commercial unit of Induction called the Weber is 10* absolute units of 


The Henry suits the law : \o\iB= RA+L — , 

The Weber suits Volts = J?^ + iV . ^ , 

where R is in ohms, A amperes, L Henries, N the number of turns in a 
circuit, I Weber's of Induction. 

In Elementary Work such as is dealt with in this book, I submit to the 
use of 47r and the difficulties introduced by the unscientific system now in 
use. In all my higher work with students, such as may be dealt with in a 
succeeding volume, I always use now the rational units of Heaviside and I 
feel sure that they must come into general use. 


much trial we find, for one pound of a particular gas, the law 
pv = Rt to be very nearly true, R being a known constant. 

Now observe that any one of the three, p, v or t, is a 
function of the other two; and in fact any values what- 
soever may be given to two, and the other can then be found. 

Thus p = R^ (1), 

we can say that p is a, function of the two independent vari- 
ables t and V. 

If any particular values whatsoever of t and v be taken 
in (1) we may calculate p. Now take new values, say t + Bt 
and v + Sv, where St and 8v are perfectly independent of one 
another, then 

. s^ 7? ^ + ^^ J s> n t + Bt j^t 
p-\-6p = R ^ and 6/; = it s R - . 

V -\-ov ^ V ■\-ov V 

We see therefore that the change Sp can be calculated if the 
independent changes Et and Bv are known. 

When all the changes are considered to be smaller and 
smaller without limit, we have an easy way of expressing Bp 
in terms of Bt and Bv. It is 

'^-m^^-o (^>- 

This will be proved presently, but the student ought fii'st 
to get acquainted with it. Let him put it in words and 
compare his own words with these : " The whole change in ^j 
is made up of two parts, 1st the change which would occur 
inp if V did not alter, and 2nd the change in p if ^ did not 
alter." The first of these is St x the rate of increase of p 

with t when v is constant, or as we write it ( -^ ) Bt, and the 

second of these is Bv x the rate of increase of p with v if t 
is constant. 

This idea is constantly in use by every practical 
man. It is only the algebraic way of stating it that is 
unfamiliar, and a student who is anxious to understand the 
subject w^ill manufacture many familiar examples of it for 


Thus when one pound of stuff which is defined by its p^ v and t^ 
changes in state, the change is completely defined by any two of the 
changes bp and hi\ or bv and bt, or bp and bt^ because we are supposed 
to know the characteristic of the stuff, that is, the law connecting p, v 
and t. 

Now the heat bH given to the stuff in any small change of state 
can be calculated from any two of bv, bt and bp, and all the answers 
ought to agree. As we wish to accentuate the fact that the changes 
are supposed to be exceedingly small we say 
dH=k . dt+ l.dv\ 

= K. dt-\-L.dp> (3), 

= r .dp+V.dv) 
where the coefficients k, I, A', A, P and V are 'all functions of the state 
of the stuff, that is of any two of v, t and p. Notice that k . dt is the 
heat required for a small change of state, defined by its change of 
temperature, if the volume is kept constant : hence k is called the 
specific heat at constant volume. In the same way K is called the 
specific heat at constant pressure. As for I and L perhaps they may 
be regarded as some kinds of latent hent, as the temperature is supposed 
to be constant. 

These coefficients are not usually constant, they depend upon the 
state of the body. The mathematical proof that if bH can be calcu- 
lated from bt and bv^ then dH=k . dt + l. dv, where k and I are some 
numl;ers which depend upon the state of the stuff, is this : — If bH can 
be calculated, then bH=L bt-^\-a(bty + b{bv)^ + c{{btf + 
terms of the third and higher degrees in bt and bv, where k, I, a, b, c, e 
&c. are coefficients depending upon the state of the body. Dividing by 
either bt or bv all across, and assuming bt and bv to diminish without 
limit, the proposition is proved. 

Illustration. Take it that for one pound of Air, (1) is 
true and R is, say, 96, p being in lb. per sq. foot and v in 
cubic feet. 

As « = 96 - , 

Hence, from (2), hp = ^^^ .M — - .hv (4) 

Example. Let t = 300, p = 2000, t; = 1 44. 

If t becomes 301 and v becomes 14*5 it is easy to show 
that p will became 199283. But we want to find the change 
in pressure, using (2) or rather (4), 

^ 96 , 2000 , ^^^„ 

^i^= 14^ >< 1 - 14:4 >< •!=- 7-22 lb. per sq.ft., 

whereas the answer ought to be — 7*17. 

'dp\ 96 fdp\ 
.dt)~ V ' \dvj~ 

V ' 

Bp = ^l,Bt-^.Sv.. 

^ V V 


Now try Bt = 1 and Bv = '01 and test the rule. Again, try 
^^ = •01 and Bv = "001, or take any other very small changes. 
In this way the student will get to know for himself what 
the rule (1) really means. It is only true when the changes 
are supposed to be smaller and smaller without limit. 

Here is an exceedingly interesting exercise : — Suppose 
we put hp = in (2). We see then a connection between Bt 
and 8v when these changes occur at constant pressure. Divide 

one of them by the other ; we have ^ when p is constant, 
or rather 

dv\ \dtj 

(dv\ \dtj ,^s 

\Tt)-'W{ '^''^• 


At first sight this minus sign will astonish the student 
and give him food for thought, and he will do well to manu- 
facture for himself illustrations of (5). Thus to illustrate 
it with pv = Rt Here 

'dv\ _ R (dp\ _ R (dp\ _ Rt p 

/dv\ _ K /dp\ _ R 
\dtl ~ ~p ' \dt) ~ V ' 

Ivj t^ V 

and (5) states the truth that 

^^_^^( _P 

The student cannot have better exercises than those 
which he will obtain by expressing hv in terms of ht and Sp, 
or ht in terms of Bp and hv for any substance, and illustrating 
his deductions by the stuff for which pv = Rt t 

79. Further Illustrations. In (.3) we have the same answer 
whether we calculate from dt and dv, or from dt and dp, or from dp 
and c?y. Thus for example, 

Jt . dt+l .dv = K . de + L . dp (6). 

We saw that ^P={-j^) (^i+{-r-] dv, and hence substituting this 
for dp in (6) we have 

k.dt + Ldv = K.dC + Li'^^]dt-{-L{^^)dv: 



This is true for any independent changes dt jind do\ let do=0^ and 
again let dt=0^ and we have 

^-A-+^g) (7), 

^At) («>• 

Again, in (6) substitute ^^=(j:)^^+(j~) ^^P^ ^^^^ we have 

k.dt + l(^dt-\-l(^\dp = K.dt + L.dp. 
Equating coefficients of dt and of dp as before we have 

^•+'60=''^' '-'^ 

<©=^ (^°)- 

Again, putting h .dt-\-l .dv = P .dp+ V . doy and substituting 

we have /: . <lt + Ld,' = p(^£\ dt+p{^ dv+ V. dv, 

-d *=^(|) ("). 

al»o ;=/>(*)+ F (12). 

Again, putting K . dt + L . dp^ P . dp+ V . dv, and substituting 

we have K (j-j dp+K l-r-j dv + L . dp = P . dp+ V . dt\ 

""■> ^(1)+^=^' c-^)' 

^(1)='^ (»>• 

The relations (7), (8), (9), (lO), (ii), (12), (13) and (14) which 

are not really all iudepeiulcnt of one another (and indeed we may get 

others in the same way) are obtained merely mathematically 

and without assuming any laws of Thermodynamics. We have called 


ZT, heat; t temperature &c., but we need not, unless we please, attach 
any physical meaning to th^ettei-s. 

The relations are true for any substance. Find what 
they become in the case of the stuff for which pv = Rt (the mathe- 
matical abstraction called a perfect gas). We know that 

-^ 1 = — , so that (7) becomes k=K-{-L— (7)*, 


/dp\ ^_P ^^ ^j^3^^ (8) becomes l=-L^ 

(dv\_ R 
\dt)~ p 

(dv\ _ V 

\dt)~ V 

\dv) V 



SO that (9) becomes k + l — = K (9)* 

so that (10) becomes -Z- = X (10)*, 

so that (11) becomes ]c=P- (11)* 

so that (12) becomes l=-r(-+y (12)^ 

It is evident that these are not all independent ; thus using (10)* 
in (9)* we obtain (7)*. 

80. Another Illustration. The Elasticity of our stuff is 
defined, see Art. 58, as 


Now if t is constant, we shall write this e^ = — v ( -, - j , or the 

elasticity when the temperature remains constant. 

If it is the adiahatic elasticity e„ which we require, we want to 

know the value of -^ when the stufl neither loses nor gains heat. In 

the last expression of (3) put dH=Oj and the ratio of our dp and our 

dv will then be just what is wanted or l-f) = — p> *^® ^ being 

affixed to indicate that ff is constant or that the stuff neither loses 

nor gains heat. Hence eij=v ^. 

Taking Y from (14) Art. 79 and P from (11), 
€n_ \dvj \dtj 

ei J 7T% 


but we have already seen as in (5) that* f ^^ ) -i- ( -^ ) = — ( -r-; ] and 

hence for any substance -5 = ^- (15). 

This ratio of the two specific Heats is usually denoted by the 
letter y. Note that neither of the two laws of Thermodynamics nor a 
Scale of temperature is referred to in this proof. 

81. General Proof. If u is a function of a- and y, we may 

write the statement in the form u=f(x, y). Take particular values 
of X and y and calculate u. Now take the values x-\-bx and y + by, 
where bx and by are perfectly independent of one another, and calculate 
the new u^ call it u + bu. Now subtract and we can only indicate our 
result by 

bu=f{x+bx, y+by)-f{x, y). 

Adding And subtracting the same thing f{x^ y + by) we have 

bic=f{x+bx, y+by)-fix, y+by)+f{x, y+by)-f{x, y). 

This is the same as 

^^_ /(^+&r,y+dy)-/(a;,y+&y ) g^, ^f{^,y+b y)-f{x,y) ^^^^^^^^ 

Now if bx and by be supposed to get smaller and smaller without 
limit, the coefficient of by 

or/(^+MzZ(^y) becomes ^^^^ or (f), 
by dy \dy) 

the X being constant. In fact this is our definition of a differential 
coefficient (see Art. 20, Note). Again, the coefficient of bx becomes 

the limiting value of — > ^/ J \ ^ if) ^ i^ecause by is evanescent. 

Writing then u instead of /(^, y) we have 

«»"=(i)<^-+(g)'»y (")• 

Thus if w = ax^ + 6y^ + cxy, du = i2ax + cy) dx-\-{2by + ex) dy. 

82. Notice that although we may have 

dz=M. dx-\-JSr. dy (18), 

where M and iV are functions of x and y ; it does not follow that ^ is a 
function of x and y. For example, we had in (3) 


where k and I are functions of t and v. Now H the total heat 
which has been given to a pound of stuff is not a function of v and t ; 
it is not a flinction of the state of the stuff. Stufi' ma^ 


receive enormous quantities of heat energy, being brought back to its 
original state again, and yet not giving out the same amounts of he^at as 
it received. The first law of Thermodynamics states however chat if 
dE = dH — p . dv, where p . dv is the mechanical work done, we can 
give to E the name Intrinsic Energy because it is something 
which is a function of the state of the stuff. It always comes back to 
the same value when the stuff returns to the same state. 

Our E is then some function of t and v, or of t and jo, or of p and v, 
but H is not ! 

The second law of Thermodynamics is this : — If dH be divided by 
t where t is ^°C. + 274, 6°Q. being measured on the perfect gas thermo- 


meter, and if — be called c?0, then ^ is called the Entropy of the 
stuff, and </> is a function of the state of the stuff. 

83. It is very important, if 

dz=M.dx-\-N .d)/ (18), 

where M and N are functions of x and y, to know when z is a function 
of X and y. If this is the case, then (18) is really 

that is, M is ( ;t- ) and iV is (-r-)^ 

^"^ i^^"^« (^) = ( as^) (1^)' 

because it is known that -. — — ,- = -^ ^ . 

dy .dx dx . a// 

d2u d2u 

* Proof that - — —. = -— T= • 

dy . dx dx . dy 

We gave some illustrations of this in Art. 31, and if the student is not yet 
familiar with what is to be proved, he had better work more examples, or 
work the old ones over again. 

Let w=/(a?, y); 

(^) is the limiting value of f^'^ + ^'^^y)-f('^'y^ as 8x gets smaller and 

smaller. Now this is a function of y, so — ( ^ j or ^^ is, by our defini- 
tion of a differential coefiicient, the limiting value of 

1 l f{x + 8x,y + dy)-f{x,y + Sy) f(x + 8x,y)-f{x,y) ] 
by \ 8x $x \ 

as 8y and 8x get smaller and smaller. 


Here we have an exceedingly important rule: — If 

dz=M.dx + N.dy (18), 

and if 2 is a function of x and v (another way of saying that z is a 
function of x and y is to say that dz=M ,da;-\-N .dy is a complete 
differentiaVjy then 

"£f (-)• 

(dM\ _ (o 

Working the reverse way, we find that ^ — -p is the limiting value of 

1^ { f{x+Sx, y + S y )-f{x-\-dx, y ) f{x, y + Sy)-f{x ,y)\ 
8x I Sy dy ) 

as Sy and 5x get smaller and smaller. Now it is obvious that these two are 
the same for all values of 5x and %, and we assume that they remain the 
same in the limit. 

* M .dx + N.dy (1), 

where M and N are functions of x and y, can always be multiplied by 
some fonction of x and y which will make it a complete differential. 

This multiplier is usually called an integrating factor. For, whatever 
functions of x and y, M and N may be, we can write 

dx-~N ^^'r 

and this means that there is some law connecting x and y. Call it 

^,.„.c.t.en(f).(^-)| = ,3, 

and as -^ from (3) is the same as in (2) it follows that (— )"^(*7— ) = xr» 

and hence (—\= /xM, f — j = fiN, where /* is a function of x and y or 
else a constant. 

Multiplying (1) by /* we evidently get 

(^h-i^^y w. 

and this is a complete differential. It is easy to show that not only is 
there an integrating factor /x but that there are an infinite number of them. 
As containing one illustration of the importance of this proposition I will 
state the ^teps in the proof which we have of the 2nd law of Thermo- 

1. We have shown that for any substance, of which the state is defined 
by its t and v, 

dH=k.dt+l.dv (5), 

where k and I are functions of t and v. 

Observe that t may be measured on any curiously varying scale of tempera- 
ture whatsoever. We have just proved that there is some function /u. of t and 



84. The First Law of Thermodynamics is this : 
If dE= dH-p . dv, or dE=- k.dt + {l -p) dv, then dE 

is a complete differential that is, ^returns to its old value when 

V by which if we multiply (5) all across we obtain a complete differential ; 
indeed there are an infinite number of such functions. Then calling the 

result d0, d(f> = fi.dH=fik.dt + fxl.dv (6). 

Let us see if it is possible to find such a value of n that it is a function 
of t only. If so, as the differential coefficient of fik with regard to v {t being 
supposed constant) is equal to the differential coefficient of fil with regard to 
t {v being supposed constant), 

(dk\ _ dti (dX\ 
\di)t~ Tt'^^Utjv 
fdk\ _fdl\ 

or I — J =( ^1 +-. -^ 

» /* at 

But the first law of Thermodynamics (see Art. 84) gives us 


and hence 

/dk\ _ /dl\ _(dp\ 
\dv)t~\dt)v \dt) 
l/dp\ _ 1 djx 
T\dtJ "■ " |I ■ dt 


This then is the condition that /j.. dH is & complete differential, fi being a 
function of temperature only. Obviously for any given substance (9) will 
give us a value of /x which will answer ; but what we really want to know is 
whether there is a value of /* which will be the same for all substances. 

2. Here is the proof that there is such a value. I need not here give to 
students the usual and well- 
known proof that all rever- 
sible heat engines working 
between the temperatures t 
and t-5t have the same 
efficiency. Now let ABCD 
be a figure showing with in- 
finite magnification an ele- 
mentary Caru ot cycle. Stuff 
at A at the temperature 
t-8t\ A I shows the volume 
and AK the pressure. Let 
^D be the isothermal for 
t - dt and BC the isothermal 
for t, AB and CD being 

Notice carefully that the 
distance AG or WB {JV is in 
DA produced to meet the 
ordinate at B) is {dpjdt) dt. 

Now the area of the 
parallelogram ABCD which 
represents the work done, is 
BW X XZ (if parallelograms on the same base and between the same parallels 
be drawn, this will become clear). Call XZ by the symbol 5v (the increase 
of volume in going along the isothermal from B to C), and we see that 











w V 






M e 



A^ . 





: \ 

( 2 

Fig. 55 




the nett work done in the Carnot cycle is (dpldt)5t. 5v. Now the Heat 
t and V return to their old values, (or another way of putting it is that 

dE for a complete cycle is 0). 


We have seen that the differential coefficient of k with regard to v, 
t being constant, is equal to the differential coefficient of l—p with 
regard to ty v being constant, or 

(a-(a-(s « 

This statement, which is true for any kind of stuff, is itself sometimes 
called the first law of Thermodynamics. 

The Second Law of Thermodynamics is this ; -— or 


taken in at the higher temperature is, from (3), Art. 78, equal to l.dv and 
hence — =j— - — = efficiency = -- ( - . j 8t... (10), and this is the same for all 

As it is the same for all substances, let us try to find its value for any 
one substance. A famous experiment of Joule (two vessels, one with gas at 
high pressure, the other at low pressure witli stopcock between, immersed 
in a bath all at same temperature ; after equalization of pressure in the 
vessels, the temperature of the bath keeps its old value) showed that in 
gases, the intrinsic energy is very nearly constant at constant temperature, 
or what is the same thing, that / in gases is very nearly equal to p, and it is 
also well known that in gases at constant volume, p is & linear function of 
the temperature. Whether there really is an actual substance possible for 
which this is absolutely true, is a question which must now be left to the 
higher mathematicians, but we assume that there is such a substance and 
in it 

I \dty " p \dtj " + 274 ^ '* 

if is the Centigrade reading on the Air Thermometer. If then we take 
t = d + 2l4: as our scale of temperature and (11) as the universal value of 

- ( '--; I , then, from (9), -=--.-r-,or — = — ^or log t + log^= a constant, 

I \dt J ^ t fi at t /Jt. o or- 

or Ai=r-, where c is any constant. This being an integrating factor for (5), 

we usually take unity as the value of c or ^=- as Camot's function. 

It is not probable that, even if there is one which is independent of j? or 

V, there really is so simple a multiplier as ^— o;^ (where is the Centigrade 

temperature on the air thermometer) or that there is such a substance as we 
have postulated above. Calling our divisor t the absolute temperature, we 
believe that for ordinary values of 0, t is ^ + 274, and the greater is, the 
more correctly is t represented by ^ + 274; but when is very small, in all 
probability the absolute temperature is a much more complicated function 
of 6. The great discoverers of the laws of Thermodynamics never spoke of 
- 274° C. as the absolute zero of temperature. 


d<f> = -. dt-\-- .dv ...{21), is a complete diftereutial, and hence the 
differential coefficient of - with regard to v, t being considered 


constant, is equal to the differential coefficient of - with regard to t, 
V being constant t. 

1 /^'\ ^\dt)~^ 



(ii-Q.-i » 

This statement, which is true for any kind of stuff, is itself some- 
times called the second law of Thermodynamics. 

Combining (20) and (22), we have for any stufl' 

(t)A (->- 

a most important law, T 

Applying these to the case of a perfect gas we find 
that (23) becomes - = -, or 1= — , or l=^p (24). 

Hence (20) is (-^ ) =0. It is not of much importance perhaps, 

practically, but a student ought to study this last statement as an 
exercise, k is, for any substance, a function of v and t, and here we are 
told that for a perfect gas, however k may behave as to temperature, it 
does not change with change of volume. Combining (24) with (9)* &c. 
(p. 141), already found, we have K-k—R, and as Regnault found that 
K is constant for air and other gases, k is also constant, so tliat 

y-i- y. 

We can now make exact calculations on the Thermodynamics of a 
perfect gas if we know K and R. 

l=p^ L= —V, P= , ]'= ^^-Aj-» where y=x- 

85. The statements of (3) Art. 78 become for a pound of perfect 

dH = k.dt + p.dv >| 

= K . dt - V . dp 

V— 1 -Y— 1 



+ The rule for finding the differential coefficient of a quotient is given in 
Art. 197. 



I often write this la«t in the shape d{pv)-{-p . dv (2), 

also d£J=k .dty or ^=X'^ + constant (3). 

It is easy to obtain from this other forms of E in terms of p and v. 
To the end of this article, I consider the stuff to be a perfect gas. 

Example 1. d(h = k . - +^ . dv. or as - = — 

. t t ' t V 

d<b = i- — + — . do. 

^ t V 

Hence, integrating, 

<f)=k log t-\-R log V + constant, or (f> = log ^v" + constant (4). 

Again d<^ = --.dt-. dpy but - = — . 

t t P 

Hence d<f)= -dt dp. 

<f) =>: K log t-Ii log p + constant, or </> = log t^p " " + constant ... (5). 

Substituting for t its value -., we have (5) becoming 

(/) = log jo^y*'+ constant (6). 

The adiabatic law^ or constant, may be written down at 
once. Keducing from the above forms we find 

or t^-yp = constant, 

■ or pv^ = constant. 

Students ni^,y manufacture other interesting exercises of this kind 
for themselves. 

Example 2. A pound of gas in the state p^^ v^^^ t^ receives the 
amount of heat ^^j^, what change of state occurs? We get our informa- 
tion from (1). 

I. Let the volume % keep constant. Then dH=k.dt 
from (I). 

The integral of this between ;„ and t^ is II^^^ = Jc{t^ — t^^ and we may 
calculate the rise of temperature to t^ . 

Or again, dH= - - dp. 



Hence, the integral, or H^^ = -^ {p^ - p^), and we may calculate 
the rise of pressure. 

II. Let Pq the pressure, keep constant. 

dH= K . dt, hence ^^i = ^^ (^i - ^o)- 

Again dH= -^ dv, hence H^^ = -^"^^ (v^ - v^). 

III. At constant temperature. 

dH=p.dv or Hq]^= \ p.dv=W, the work done by the gas in ex- 

J Vo 


IV. Under any conditions of changing pressure and 

Jf^^ = h (t^ — tf^ -\- work done. 

Also from (2), Hqi = r(i^i^i— iOo^*o) + work done. 

If ir=0, the work done=X*(^Q — ^i) 

We often write the last equation of (1) in the convenient shape 


If in this we have no reception of heat, 

dff ^ ,. dp 

or _ = 0, then v^ + y.^ = 0, 

or -" + y — = or, integrating, log p-\-y log v = constant, 

or j02;'>'=: constant. 

This is the adiabatic law again. 

Example 3. In a well known gas or oil engine cycle of 



Fig. 56. 


operations, a pouiul of gas at p.j,, v-,, t.,, iudicatcd by the point A is 
compressed adiabatically to B, wnere we have p^, v,, ^j. The work 
done upon the gas is evidently (from iv.) Jc{t^ — t^, being indeed the 
gain of intrinsic energy. 

Heat given at constant volume from B to C where we have 

Work done in adiabatic expansion CJ)=k(t^-t^). 
Nett work done = work in Ci>-work in AB= 

-^ = efficiency eJ'-^l' -^^^=1- ^^ (8). 

But we saw that along an adiabatic W~'^ is constant, and hence 

From this it follows that -t = t — \-] > ^"d ^^^^"^ of these 
t —t h h \^h) 

= * — ^ . Using this value in (8) we have 

efficiency = 1 - f^j (9), 

a formula which is useful in showing the gain of efficiency produced by 
diminishing the clearance v^. 

Students will find other good exercises in other cycles of gas engines. 

Change of State. 

86. Instead of using equations (3) Art. V8, let us get out equations 
specially suited to change of state. Let us consider one pound of 
substance, m being vapour, 1 - m being liquid (or, if the change is from 
solid to liquid, m liquid, l-m solid), and let 

^2 = cubic feet of one pound of vapour, 

*i = » )) of one pound of liquid, 

jt>= pressure, t temperature, p is a function of t only. 

Tf V is the volume of stuff in the mixed condition, 
v=ms2 + {l —m)sj 

= (*2 — «i) wi + Si, or v = mu-{-Xi (1), 

if we write u for ,% — Si. 

AVhen heat dJJ is given to the mixture, consider that t and m 
alter. In fact, take t and m as independent variables, noting that t 
and VI define the state. If 0-2 and o-^ be specific heats of vapour and 
li(iuitl, when in the saturated condition (for example, 0-2 is the heat given 
to one pound of vapour to raise it one degree, its pressure rising at 


the same time according to the proper law), then the 7>i lb. of vapour 
needs the heat ma^.dt^ and the 1— m of liquid needs the heat 
{\ — m)<j^.dt and also if dm of liquid becomes vapour, the heat L . dm 
is needed, if L is latent heat. Hence 

dII={(a-2-o-j)m + ai] (2). 

If E is the Intrinsic Energy, the first law of Thermodynamics gives 

dE=^dlI-p,dv (3). 

Now if m and t define the state, v must be a function of m and t^ or 

Using this in (3) and (2) we find 

dE = |(cr2 - o-i) m + o-i -p {^j^ dt-\-\L-p {~y^ dm ... (4). 
Stating that this is a complete differential, or 

^{(..,-,.,)m + .,-^(|)}=||i;-^(^)}. 

we have, noting from (1), that ( -i— ) = ?/, 

dL , dp fdv\ dp ,_. 

-^ + 0-1 — (To = -& . -1— I , or w -4- (5). 

dt^ ^ '" dt \dm)' dt ^ ' 

Now divide (2) by t and state that d(^ = — is a complete dif- 



dm \ ~^t ]~dt \i) 


dL L 

or -n-^f^\-^^2 = -7 ('^)- 

Hence, with (5) we have T = U -s^ (8), 

87. To arrive at the fundamental Equation (8) more 
rapidly. In fig. 57 we have an elementary Carnot cycle for one pound 

* w7 ( T ) ~ g as will be seen later on when we have the rule for 

differentiating a quotient. But indeerl we may as well confess that to 
understand this article on change of state, students must be able to perform 
' differentiation on a product or a quotient, 



of stuff. The co-ordinates of the point B are FB=s^ the volume, and 
BG the pressure jt? of 1 pound of liquid. At constant tem^^ratrire t, and 

Fig. 57. 

also constant pressure, the stuff expands until it is all vapour at FC^s^ ; 
CD is adiabatic expansion to the temperature t-bt at J). J) A is iso- 
thermal compression at t-bt and AB is the final adiabatic operation. 

The vertical height of the parallelogram is bt ~ , and its area, repre- 

senting the nett work, is bt . -£ (s.^- s^). The heat taken in, in the 

: dp 

operation BO is X, and the efficiency is bt -^ («2 ""*i) "^'^- ^^^^ ^^ i^ i'^ 

a Carnot cycle this is equal to — and so we obtain (8). t 

88. The Entropy. From (6) we find o-^,-<ri = ^ 
we can write (2) as 





Hence, the entropy d(l> = - — = ^^ dt + di-^\^ 
t t \ / 



<(- = - + 

c?^-f constant 


In the case of water, o-j is nearly constant, being Joule's equivalent. 
(We have already stated that all our heat is in work units), and 

= — +0-1 log f-j-f constant (11). 

Hence the adiabatic law for water-steam is 
TnT t 

-T- + o"! log £- = constant (12). 

It is an excellent exercise for students to take a numerical example. . 


Let steam at 165°C. (or ^ = 439) expand adiabatically to 85°C. (ori = 359). 
Take a-i = 14(X) and L in work units, or take (t^ = \ and take L in heat 
imits. In any case, use a table of values of t and L. 

1. At the higher ^2=4^9 1^^ '^2='"- (This is chosen at random.) 
Calculate ii\ at, say ^^ = 394, and also m^ at ^^=359. 

Perhaps we had better take L in heat units as the formula 

Z = 796 --695^ 
is easily remembered. 

Then (12) becomes 

r,r, (^ - -695) +log^^= m^ (^ - '695) +log^^ , 


^ /2/ 


If we want m^ we vise t^ instead of t^. 

Having done this, find the coiTesponding values of v. Now try if 
there is any law like 

pi;* = constant, 

which may be approximately true as the adiabatic of this stuff. 
Rej^eat this, starting with m^=-% say, instead of '7. 

The t, <t> diagram method is better for bringing these matters 
most clearly before students, but one or two examples like the above 
ought to be worked. 

89. When a complete differential dn, is zero, to solve 
the equation du = 0. We see that in the case, 

{x" - ^xy - t\f) dx + {if - ^xy - 'Ice") dy = 0, 
we have a complete differential, because 

-J- (xf^ — 4fxy — 2y^) = — 4;x — ^y, 

^ if - 4^y - 2^0 = - 4y - 4a-, 
so that they are equal. Hence it is of the form 

Integrating a?^ — ^xy — 2?/^, since it is ( y j , Avith i-egard to 


X assuming y constant, and adding, instead of a constant, an 
arbitrary function of y, we get ^ as 

u = :^a^-2afy-2ifx + <i>{y). 
To find </) {y\ we know that \~j~]—y^~ ^^V — ^^' 

Hence -2a^- ^yx + ^ <f> {y) = y"" - ^.x-y - 2a^, 


Hence ^ </, (y) = y^ or </> {y) = \f. 

Hence xi — \a? — 2x^y — ^xy"^ -\- ^y^ =zc. 

We have therefore solved the given differential equation when 
we put this expression equal to an arbitrary constant. 
Solve in the same manner, 

1 + ^ j (/^ - 2 ^ c?y = 0. Answer x^-y^= ex. 

., , 2x.dx n Saf\ , 

holve f- ( -; —jdy=0. Answer x^ — y^ — cy^. 

J ^J J ' 

-Solve (3^- + 3y - ^ dx + {^x - 1 + 3/) dy = 0. 

Answer x^y"^ + ot^y^ + 4^^ ^ ^y"^ ^ ^^yz _ ^y%^ 

90. In the general proof of (17) given in Art. 81, we assumed that 
X and y were perfectly independent. We may now if we 
please make them depend either upon one another or 

any third variable z. Thus if when any inde^Hindent quantity z 
becomes z + bz, x becomes x + bx and y becomes y+by, of course u 
becomes u + bu. Let (IG) Art. 81 be divided all across by bzy and let 
bz be diminished without limit, then (17) becomes 

du _ /o?w\ dx /du\ dy 
di ~ \dx) dz "^ XTy) ^ 


Thus let u = ax^-\-hy^+cxy^ 

and let x = ez^, y = yz'^. 

and consequently 

-T- = {2ax + cy) nez:^ ~^-\- {2by + ex) mgz^ ~ ^ 


In this we may, if we please, substitute for x and y in terms of z, 
and so get our answer all in terms of z. 

This sort of example is rather interesting because it can be worked 
out in oiu* earlier way. In the expression for w, substitute for x and y 
in terms of z, and we find ii^aeh'^^+hgh^'^-'rcegz:^'^^^ and 

-^ = 2?iae%2n - 1 + 2 w6^%2m - 1 -I- (,i + m) cegz"" + "*-!. 

It will be found that this is exactly the same as what was obtained 
by the newer method. The student can easily manufacture examples 
of this kind for himself. 

For instance, let y=uv where u and v are functions of ^, then (1) 
tells us that 

dy dv . du 

a5 = "S+^d5E' 

a formula wliich is usually worked out in a very different fashion. 
See Art. 196. 

In (1) if y is really a constant, the formula becomes 

du _ du dx 
dz " dx ' Hz ' 

which again is a formula which is usually worked out in a very different 
fashion. See Art. 198. 

In (1) assume that z=x and that y is a function of x, then 

dx \dx) "^ \dy) dx ^'^^^ 

The student need not now be told that ;,- is a very different thing 


\dx) • 

Example. Let u = ax^ + hy'^ + cxy^ 

and let y=gx'^. 

Hence (2) is, j- = {2ax + cy) + (26y + ex) mga/^-K 

More directly, substituting for y in u, we have 
u = ax^ + 6^2^2m ^ ^^^» + 1^ 

J- = 2ax + 2?»6^2^'2'« - 1 + (771 + 1 ) cgx^, 
and this will be found to be the same as the other answer. 


If M is a function of three independent variables it is easy to prove, 
as in Art. 81, that 

,MJ'-\a.+m,,U^-)^, (3). 

91. Example. When a mass m is vibrating with one degree 
of freedom under the control of a spring of stiffness a, so that if x is 
the displacement of the mass from its position of equiHbrium, then ax 
is the force with which the spring acts upon the mass ; we know that 
the potential energy is ^ ax^ (see Art. 26), and if v is the velocity of the 
mass at the same time t, the kinetic energy is ^tnv^, and we neglect the 
mass of the spring, then the total store of energy is 

When X is 0, r is at its greatest; when v is 0, x is at its greatest. 

1. Suppose this store E to be constant and differentiate with 
regard to ^, then 

= mv^^ + ax--^^ (1)' 

. dx ... d^x « dv , 
or as V IS ~j- , writnig -y-^ for -j- we have 

^1-^^-0 •••■ ^^)- 

which is (see Art. 119) tlie well known law of simple harmonic motion. 

2. If the total store of energy is not constant but diminishes at 
a rate which is proportional to the square of the velocity, as in the case 

rf If 

of Fluid or Electromagnetic friction, that is, if -r = - Ev^ then (1) 
becomes - Fv^^mv y-4- ax—j , or (2) becomes 
d2x F dx a ^^ 

Compare (1) of Art. 142. 

92. Similarly in a circuit with self induction L and resistance R, 

joining the coatings of a condenser of capacity K, if the current is C\ and 

if the quantity of electricity in the condenser at time t is K V so that 

C= -K --J-J ^LC'^ is called the kinetic energy of the system, and \KV^ 

is the potential energy, and the loss of energy by the system per second 
is RC'^. So that if E is the store of energy at any instant • 

E=^LC'' + \KV\ 


or LC"^- V. C+EO^=0, 


or L^-V+RC=0, 


d2V . R dV^ 1 Tr /% 

aF + L^+EK'^=^ ^^^- 

Differentiating this all across and replacing K -r- with C we have a 
similar equation in C. Compare (4) of § 145. 

93. A mass m moving with velocity v has kinetic energy \mv^. 
If this is its total store E^ 

If E diminishes at a rate proportional to the square of its velocity 
as in fluid friction at slow speeds, 

dE jy ., dv 

dv F ,^, 

or -j-= V (5). 

dt m ^ ' 

We have a similar equation for the dying out of current in an 
electric conductor, \LC'*- being its kinetic energy, and RC^- being the 
rate of loss of energy per second. 

94. In (2) of Art. 90, assume that u is a constant and we find for 
example that if u=f{x, y) = G 

(dn^\_y)\ . ( df{x, y) \ di_ 
\ dx )^\ dy ) dx~ ' 

so that if /(^-j y)=o' or =0, we easily obtain g^. 

1. Thus if x^^f = c, 2..+2y . ^^=0, or ^= -^'. 

2. Alsoif^^+f-!-l=0,?|' + |^/ = 0,orf'=-^;i;, 

a^ W- ' a^ h^ dx ' dx a^ ij 

3. Again if u = A a;"' + %", 

Hence i{ u = 0, or a constant, we have 
dy _ mAx'^~^ 
dx~ nBy'^~'^ 


4. If u==Zi + 1:.. 

a,'" y- 

2a; , 2y , 

5. If ^-3 +3,3 _ 3«.^^ = i^ find ^ . Answer : ^, = ''''' — ^ . 

6. If .rlogy— ylogA'=0, 

t/y^y /^log. 

t/y_y ^^logy-y\ 
.v — xj 

95. Example. Find the equations to the tangent and normal 
to the ellipse ' ., + t^ = l, at the point A'l, yi on the curve. 

•^ 4. 2^ ^^-0 or at the tx)int ^^- -?^' ^i 

Hence the equation to the tangent is 

a;—a;i ^ Vi 

''' -^+-52-;^ + ^' 

and as x^ and y^ are the co-ordinates of a point in the curve, this is 1. 

Hence the tangent is — / + '^ = 1. 
a^ b^ 

The slope of the normal is v^ — , and hence the equation to the 

normal is ti^i = " ^J ^ 
x-Xi b^ x^ 


Page 19. In an engineering investigation if one arrives at mathe- 
matical expressions which cannot really be thought about because 
they are too complicated, one can often get a simple empirical formula 
to replace them with small error within the limits between which they 
have to be used. Sometimes even such a simple expression as a-|-6.r, 
or x^ will replace a complicated portion of an expression with small 
error. Expertness in such substitution is easily attained, especially in 
calculations where some of the terms can be expressed numerically or 
when one makes numerical experiments. 



Exercise 1. The following observed numbers are known to follow 
a law like y = a-\-bx, but there are errors of observation. Find by the 
use of squared paper the most probable values of a and h. 


2 3 




9 12 



5-6 6-85 




16-32 20-25 


Ans. y = 2*5 + 1 'hx. 

Exercise 2. The following numbers are thought to follow a law like 
y=zax\{\-^sx). Find by plotting the values of y\x and y on squared 
paper that these follow a law ylx + sy=a and so find the most probable 
values of a and s. 



1 2 






•97 r22 




Ans. y = 3^/(1 + 2.r). 

Exercise 3. If p is the pressiu-e in poimds per square inch and if v 
the volimie in cubic feet of 1 lb. of saturated steam, 

p 6-86 





163-3 250-3 

V 53-92 





2-748 1-853 

Plotting the common logarithms of p and v on squared paper test 
the truth oi pv^'^^^^Ti^. 

Exercise 4. The following are results of experiments each lasting for 
four hours ; / the indicated horse-power of an engine, transmitting 
B horse-power to Dynamo Machines which gave out E horse-power 
(Electrically), the weight of steam used per hour being TFlb., the weight 
of coal us^ per hour being G lb. (the regulation of the engine was by 
changing the pressure of the steam). Show that, approximately, 
fr=800+21/, i5=-95/-18, ^=-935-10, C=4-2/-62. 





























Page 34. It has been suggested to me by many persons that I 
ought to have given a proof without assuming the Binomial Theorem, 
and then the Binomial becomes only an example of Taylor's. In spite 


of the eminence and experience of my critics, I believe that my method 
is the better — to tell a student that although I know he has not proved 
the Binomial, yet it is well to assume that he knows the theorem to be 
correct. The following seems to me the simplest proof which does not 
assume the Binomial. 

Let y—^^i x+bx=x\^ y + ^^=yi« 

(1) Supjjose n a positive integer ; then 

In the limit, when bx is made smaller and smaller, until ultimately 
Xi=x\ the left-hand side is -^ and the right-hand side is x^~'^+x'^~^ 

-f... to n terms : so that j-=7iar»~i. 

(2) Suppose n a positive fraction, and put n= - where I and m are 
positive integers. We have — — ^- = -^ = ~i,i — ;^ where a;'" =2, 

Xi — X X% — X Zt — * 

Xi=Zj^^j and so on. 

^-linut of (h-^)ih'-'+^'h'~'+ +^'-') 

dx~ {zi-z){z{^-i + z.Zi''^-'^ + -|-2'»-i) 

(3) Suppose n any negative number = —m say, where vi is positive, 
then noticing that Xi~^-x~'^= *' 

we have 

■x-'" 1 x\ 


/It wi ^^"^ 

Now the limit of ^ =mx'^-'^ by cases (1) and (2) whether 111 

Xi — x •' 

be integral or fractional. 

.'. -r-=— -^ ' mx'^ - 1 = — 771^- '»- 1 = MA*"" 1. 

Thus we have shown that -j- (^*) = 7w?»»-i, where ?i is any constant, 
positive or negative, integral or fractional. 



e* and sinx. 

97. The Compound Interest Law. The solutions 
of an enormous number of engineering problems depend 
only upon our being able to differentiate x'\ I have given 
a few examples. Surely it is better to remember that 
the differential coefficient of x^^ is nx^~^, than to write 
hundreds of pages evading the necessity for this little bib 
of knowledge. 

We come now to a very different kind of function, e*, 
where it is a constant quantity e (e is the base of the 
Napierian system of logarithms and is 2*7 183) which is 
raised to a variable power. We calculate logarithms and 
exponential functions from series, and it is proved in Algebra 

The continuous product 1 . 2 . 3 . 4 or 24 is denoted by .4 
or sometimes by 4 ! 

Now if we differentiate e* term by term, we evidently 

SO that the differential coefficient of ^ is itself e*. Simi- 
larly we can prove that the differential coefficient of e^^ is 
a^". This is the only function known to us whose rate of 
increase is proportional to itself; but there are a great many 
phenomena in nature which have this property. Lord 
Kelvin's way of putting it is that "they follow the compound 
interest law." 

P. 11 



Notice that if -i = ay (1), 

that is, the rate of increase of y is proportional to y itself, 

y = be" (2), 

where h is any constant whatsoever ; h evidently represents 
the value of y when a; = 0. 

Here again, it will be well for a student to illustrate his 
proved rule by means of graphical and numerical illustrations. 
Draw the curve y = d^, and show that its slope is equal to its 
ordinate. Or take values of a?, say 2, 2-001, 2002, 2003, &c., 
and calculate the corresponding values of y using a table of 
Logarithms. (This is not a bad exercise in itself, for practical 
men are not always quick enough in their use of logarithms.) 
Now divide the increments of y by the corresponding incre- 
ments of X. An ingenious student will find other and 
probably more complex ways of getting familiar with the 
idea. However complex his method may be it will be 
valuable to him, so long as it is his own discovery, but let him 
beware of irritating other men by trying to teach them 
through his complex discoveries. 

98. It will perhaps lighten our study if we work out a 
few examples of tbe Compound Interest Law. 

Our readers are either Electrical or Mechanical En- 
gineers. If Electrical they must also be Mechanical. The 
Mechanical Engineers who know nothing about electricity 
may skip the electrical problems, but they are advised to 
study them ; at the same time it is well to remember that 
one problem thoroughly studied is more instructive than 
thirty carelessly studied. 

Example 1. An electric condenser of constant capacity 
K, fig. 58, discharging through great resistance R. If v is the 
potential difference (at a particular instant) between the 
condenser coatings, mark one coating as v and the other as 
on your sketch, fig. 58. Draw an arrow-head representing 
the current G in the conductor ; then C =v-t-R. 

But q the quantity of electricity in the condenser is Kv 


and the rate oi diminution oiq per second or — ^ or —K-r-. is 
the very same current. Hence ,^ 







= — 



— _ _ ^ 

^'' ~di~~KR'"' Fig. 58. 

That is, the rate of diminution of v per second, is propor- 
tional to V, and whether it is a diminution or an increase we 
call this the compound interest law. We guess therefore that 
we are dealing with the exponential function, and after a little 
experience we see that any such example as this is a case of 
(1), and hence by (2) 

v=^he~^' (8). 

It is because of this that we have the rule for finding the 
leakage resistance of a cable or condenser. 

For (log 6 -log 2;) = ^^. 

So that if Vi is the potential at time ti and if v^ is the 
potential at time fg 

KR{logb-\ogv,) = t^, 

KR (log h — log V2) = ^2. 

Subtracting, KR (log v^ — log Vo) — U — ti 

So that R = {U - t,)/K log ^ . 

It is hardly necessary to say that the Napierian logarithm 
of a number n, \ogn, is equal to the common logarithm 
logio?! multiplied by 2-3026. 

Such an example as this, studied carefully step by step by 
an engineer, is worth as much as the careless study of twenty 
such problems. 

Example 2. Newton's law of cooling. Imagine a 
body all at the temperature v (above the temperature of sur- 



rounding bodies) to lose heat at a rate which is proportional 
to v. 

Thus let -T. = — «^j 


where t is time. Then by (2) 

V = 6e-«« (4), 

or log h — log V = at. 

Thus let the temperature be Vi at the time t^ and v^ at 
the time t^, then log Vi — log I'g = a (^2 — ^i), so that a can be 
measured experimentally as being equal to 

Example 3. A rod (like a tapering winding rope or like 
a pump rod of iron, but it may be like a tie rod made of 
stone to carry the weight of a lamp in a church) tapers 
gradually because of its own weight, so that it may have 
everywhere in it exactly the same tensile stress / lbs. per 
square inch. If 3/ is the cross section at the distance 
X from its lower end, and if y + 5y is its cross section 
at the distance oc-{- Sx from its lower end, then /. By is 
evidently equal to the weight of the little portion between 
OS and w + 8x. This portion is of volume BiV x y, and if w is 
the weight per unit volume 

f.Bi/ = tv.y. B.€ or rather ;r; = rpl/- 

Hence as before, y=ibe^ (5). 

If when a7 = 0, y = yo, the cross section just sufficient to 
support a weight W hung on at the bottom (evidently 
fy^zrzW), then yQ=b because e°=l. 

It is however unnecessary to say more than that (5) is 
the law according to which the rod tapers. 

Example 4. Compound Interest. £100 lent at 
3 per cent, per annum becomes £103 at the end of a year. 
The interest during the second year being charged on the 
increased capital, the increase is greater the second year, and 
is greater and greater every year. Here the addition of 
interest due is made every twelve months ; it might be 



made every six or three months, or weekly or daily or every 
second. Nature's processes are, however, usually more 
continuous even than this. 

Let us imagine compound interest to be added on to the 
piincipal continually, and not by jerks every year, at the rate 
of r per cent, per annum. Let P be the principal at the 

end of t years. Then hP for the time ht is -— ^ P .ht or 

dP r 1^^ 

-zr- — -—■ P, and hence by (2) we have 
dt 100 ^ ^ ^ 

where 6 = Po the principal at the time t = 0. 

Example 5. Slipping of a Belt on a Pulley. When 
students make experiments on this slipping phenomenon, 
they ought to cause the pulley to be fixed so that they may 
see the slipping when it occurs. 

The pull on a belt at W is Ti, and this overcomes not only 
the pull To but also the 
friction between the belt 
and the pulley. Consider 
the tension I' in the belt 
at P, fig. 59, the angle 
QOP being 6] also the 
tension T+BT at >Sf, the 
angle QO>Sf being (9 + 8(9. 

Fig. 60 shows part of 
OPS greatly magnified, 
SO being very small. In 
calculating the force pres- 
sing the small portion of 
belt P>Si against the pulley 
rim, as we think of PS as a shorter and shorter length, we see 
that the resultant pressing force is T. hd*, so that /ju.T.BOis 

* When two equal forces T make a small angle dd with one another, 
find their equilibrant or resultant. The three forces are parallel to the sides 
of an isosceles triangle like fig. 01, where AB=CA represents 2', where 

Fig. 59. 

Fig. 61. 


the friction, if /x, is the coefficient 
of friction. It is this that BT 
is required to overcome. When 
fi.T .BO is exactly equalled by 
Fig. 60. BT sliding is about to begin. 

Then fi . T , B6 = BT or, ^ = fiT, the compound interest law. 

Hence T=^he^^. Insert now T^T, when ^ = 0, and T = T^ 
when e^QOW or 6, , and we have T, = h, T, = ToCf'K 

In calculating the horse-power H given by a belt to a 
pulley, we must remember that H={1\ - T,) F-=- 33000, if 
Ti and T^ are in pounds and V is the velocity of the belt in 
feet per minute. Again, whether a belt will or will not tear 
depends upon T^ ; from these considerations wo have the 
well-known rule for belting. 

Example 6. Atmospheric Pressure. At a place 
which is h feet above datum level, let the atmospheric 
pressure be p lbs. per sq. foot ; at h + Bh let the pressure 
be p-\-Bp {Bp is negative, as will be seen). The pressure 
at h is really greater than the pressure at h+Bh by the 
weight of air filling the volume Bh cubic feet. If w is the 
weight in lbs. of a cubic foot of air, —Bp = w. Bh. But lu = cp, 
where c is some constant if the temperature is constant. 

Hence — Bp= cp. Bh...(l), or, rather Jr = — cp. Hence, as 

before, we have the compound interest law; the rate of fall of 
pressure as we go up or the rate of increase of pressure as we 
come down being proportional to the pressure itself. Hence 
p = ae~'^^y where a is some constant. If ^; = po> when /i = 0, 
then a = j^oj so that the law is 

p=Poe-'^ (2). 


As for c we easily find it to be -- , w^ being the weight 


of a cubic foot of air at the pressure po. If t is the constant 
(absolute) temperature, and Wq is now the weight of a cubic 

w 274 

foot of air at 0° C. or 274° absolute, then c is — — . 

Po t 

BAG =56 and BC represents the equilibrant. Now it is evident that as 
5^ is less and less, BG-^AB is more and more nearly 5^, so that the 
equilibrant is more and more nearly T . 50. 


If w follows the adiabatic law, so that pw-y is 
constant or w=cp^^y where 7 = 1*414 for air. Then (1) be- 
comes — Bp = cp^ly Bh or f~ = cBh or rather — I ^ = ch or 


1 , 1 

^1 P ^ =oh+G. If ^ =po where A = 0, we can find G, and 

we have p y=po y— ch (3), 


as the more usually correct law for pressure diminishing 

upwards in the atmosphere. 

Observe that when we have the adiabatic law pyy = 6, a 

constant, smd pv=Rt; it follows that the absolute temperature 

1— - 
is proportional to p y , 

So that (3) becomes 

1 7-1 

li 7 

So that the rate of diminution of temperature is 
constant per foot upwards in such a mass of gas. 

Compare Art. 74, (4), if v is 0. 

Example 7. Fly-wheel stopped by a Fluid Frictional 

Let a be its velocity in radians per second, / its moment of 
inertia. Let the resistance to motion be a torque proportional 
to the velocity, say Fa, then 

Fa = — Ix angular acceleration (1), 

/ J + i^a=0 (2), 

da F 
* = -/"• 

Here rate of diminution of angular velocity a, is proportional 
to a, so that we have the compound interest law or 

oL — a^e ^ (3), 

where a^ is the angular velocity at time 0. 


Compare this with the case of a fly-wheel stopped by 
solid fHction. Let a be the constant solid-frictional torque. 

(1) becomes a = — Ida/dt, 

or da/dt + a/I = 0, 

or a = — at/T -f a constant, 

or a = ao — a^// (4), 

where Oo is the angular velocity when t = 0. 

Returning to the case of fluid frictional resistance, if 
M is a varying driving torque applied to a fly-wheel, 
we have 

^=^»+4; <«>• 

Notice the analogy here with the following electric circuit 

Example 8. Electric Conductor left to itself. 

Ohm's law is for constant currents and is V= RC, where 
R is the resistance of a circuit, C is the current flowing in it ; 

V the voltage. We usually have R in ohms, C in amperes, 

V in Volts. When the current is not constant, the law 

V = RC + I.g (1). 

where -7- is the rate of increase of amperes per second, and 

L is called the self-induction of the circuit in Henries. It is 
evident that L is the voltage retarding the current when the 
current increases at the rate of one ampere per second. 

1. IfF=Oin(l) 

dt~ L ' 

which is the compound interest law. 

Consequently C = CoC ^ (2). 


so that from (1) V==Ra-\-(Rh- Lgh) e-^K 

Now let R = Lg or ff—Y and we have V= Ra, so that the 

voltage may keep constant although the current alters. 
Putting in the values we have found, and using Vq for the 
constant voltage so that a= Fq -4- R, we find 

G^'^+be'L' (3). 

If we let C=0 when ^ = 0, then 6 = — ^, and hence we may 

write G= -^{l-e~L^) (4). 

The curve showing how G increases when a constant voltage 
is applied to a cii'cuit, ought to be plotted from ^ = for some 
particular case. Thus plot when Fo = 100, -R = l, L—01. 
What is the current finally reached ? 

99. Easy Exercises in the Differentiation and Integration 
of e**^. 

1. Using the formula of Art. 70, find the radius of 
curvature of the curve y = eF, where a; = 0. Answer : r = \/8. 


2. A point ;ri, y^ is in the curve y = hef^, find the 
equation to the tangent through this point. 

Answer: y^Zyi — Vl^ 
X — Xi a 

Find the equation to the normal through this point. 

Answer : - — ^ = . 

X - Xy^ ?/i 

Find the length of the Subnormal. Answer : y -- or y^-ja. 
Find the length of the Subtangent. Answer \y-\--^- or a. 


3. Find the radius of curvature of the catenary 

y = - (e^'c + g-*/''), at any place. Answer : r — y^jc. 

At the vertex when a? = 0, r = c. 

4. If y = Ae''^^ where i stands for V — 1. Show that if 
i behaves as an algebraic quantity so that i^ = — 1, i' = — i, 

{* = !, i^ = I, &c. then ^ = - a^y. 

5. Find a so that y = Ael^ may be true when 

Show that there are two values of a and that 
y = Ae-"^ + 5e-3«. 

6. Find the subtangent and subnormal to the 

Catenary y = ^ {^"^ + e"*'"), or, as it is sometimes written, 

y = c cosh xjc. 
Answer : 


the subtangent is c coth - or c (e^^" + e~^l*')l{d^'*' — e~^l*^), 


C 2ul7 c 

the subnormal is ^ sinh — or 2 (e^^^ — g-^^/c). 

7. The distance PS, fig. 8, being called the length of the 
tangent, the length of the tangent of the above catenary is 

2z cosh^ - / sinh - . 
2 cj c 

The length of PQ may be called the length of the normal, 


and for the catenary it is c cosh^ - or y^/c. 


8. Find the length of an arc of the catenary 

c - — 
y— (e" -{-e « ). The rule is given in Art. 38. Fig. 62 

shows the shape of the curve, being the origin, the distance 
A being c. The point P has for its co-ordinates x and y. 



Now -f-= ^(e'' —e ^). Squaring this and adding to 1 
and extracting the square root gives us ^(e^ + e ^). The 
integral of this is ^ (e ^ — e ^ ) which is the length of the 
arc AP, as it is when x = 0. We may write it, s = c sinh w/c. 

\. R 





Fig. 62. 

9. Find the area of the catenary between OA and 
SP, fig. 62. 

Area =1 -^ (e^ ■{■ e *" ) dw^ 



c2 ros 1 _«~| 
Or the area up to any ordinate at x is & sinh x\c. 

or ^ (e '^ -e " ). 

The Catenary i/ = -{e^"^ + e~^"') revoh^es about the axis of x^ find 
the area of the hour-glass-shaped surface generated. See Art. 48. 

Area = - \ {e^ "^ -\- e~="''f . dx 




IC ^ ff-'lx IC' 


between the ordinates at x = x and .^'=0. 

It is curious that the forms of some volcanoes are as if 
their own sections obeyed the compound interest law like an 
inverted pump rod. The radii of the top and base of such a 


volcano being a and h respectively and the vertical height h, 
find the volume. See Art. 46. Taking the axis of the 
volcano as the axis of x, the curve ij = be^^ revolving round this 

axis will produce the outline of the mountain if c = -r log ^ . 
The volume is ttJ 6- . e"'"" . da)=^ ^- e-'^^ 

= 2^(e--l). 


Now a — be^^, so that our answer is ^ {a^ — 6-). 

100. Harmonic Functions. 

Students ought to liavc already plotted sine curves like 
y = asm{bx + e)...(l)on squared paper and to have figured out 
for themselves the signification of a, b and e. It ought to be 
unnecessary here to speak of them. Draw the curve again. 
Why is it sometimes called a cosine curve ? [Suppose e to 

be ^ or 90°.] Note that however great a; may be, the sine of 

{bx-\-e) can never exceed 1 and never be less than — 1. The 

student knows of course that sin = 0, sin j (or 45°) = '707, 

sin ^ (or 90°) = 1, sin -J (or 135°) = '707, sin tt (or 180°) = 0, 

sin ^ (or 225°) = - '707, sin '-^ (or 270°) = - 1, sin ^~ 

(or 315°) = --707, sin 27r (or 360') = again and, thereafter, 
sin ^ = sin (^ — 27r). Even these numbers ought almost to 
be enough to let the wavy nature of the curve be seen. 
Now as a sine can never exceed 1, the greatest and least 
values of y are a and — a. Hence a is called the amplitude 
of the curve or of the function. 

When A" = 0, y — a sin e. This gives us the signification 
of e. Another way of putting this is to say that when ba; 

was = — e or A" = — J , y was 0. When x indicates time or 

when bx is the angle passed through by a crank or an 
eccentric, e gets several names ; Valve-motion engineers call 



it the advance of the valve ; Electrical engineers call it the 
lead or (if it is negative) the lag. 

Observe that when bx — 27r we have everything exactly 
the same as when x was 0, so that we are in the habit of 

calling -J the periodic value of (i\ 

Besides the method given in Art. 9, I advise the student 
to draw the curve by the following method. A little know- 
ledge of elementary trigonometry will show that it must 
give exactly the same result. It is just what is done in 
drawing the elevation of a spiral line (as of a screw thread) 
in the drawing office. Draw a straight line OM. Describe 

Fig. 63. 

a circle about with a as radius. Set off the angle BOG 
equal to e. Divide the circumference of the circle into any 
mimber of equal parts numbering the points of division 
0, 1, % 3, &c. We may call the points 16, 17, 18, &c., or 
32, 33, 34, &c., when we have gone once, twice or more times 
round. Set off any equal distances from B towards M on 
the straight line, and number the points 0, 1, 2, &c. Now 
project vertically and horizontally and so get points on the 
curve. The distance BM represents to some scale or other 
the periodic value of x or 27r/6. 

If OG is imagined to be a crank rotating uniformly 
against the hands of a watch in the vertical plane of the 
paper, y in (1) means the distance of G above OM, hx means 
the angle that OG makes at any time with the position OM, 
and if x means time, then h is the angular velocity of the 
crank and 27r/6 means the time of one revolution of the crank 


or the periodic time of the motion, y is the displacement at 

any instant, from its mid position, of a slider worked vertically 

from G by an infinitely long connecting rod. 

A simple harmonic motion may be defined as one which 

is represented by s = a sin (bt + e), where s is the distance 

from a mid position, a is the amplitude, e the lead or lag or 

advance, and 6 is 27r/T or ^irf where T is the periodic time or 

/is the frequency. Or it may be defined as the motion of a 

point rotating uniformly in a circle, projected upon a diameter 

of the circle (much as we see the orbits of Jupiter's satellites, 

edge on to us), or the motion of a slider worked from a 

uniformly rotating crank-pin by means of an infinitely long 

connecting rod. And it will be seen later, that it is the sort 

of motion which a body gets when the force acting upon it is 

proportional to the distance of the body from a position of 

equilibrium, as in the up and down motion of a mass hanging 

at the end of a spring, or the bob of a pendulum when its 

swings are small. It is the simplest kind of vibrational 

motion of bodies. Many pairs of quantities are connected by 

such a sine law, as well as space and time, and we discuss 

simple harmonic motion less, I think, for its own sake, than 

because it is analogous to so many other phenomena. Now 

let it be well remembered although not yet proved that if 

y = a sin (bx + c) then -?i = ab cos (bx + c) 

and ly . dx = -- cos(bx + c). 

101. When c = and t = 1 and a = 1 ; that is when 

y — ^mx (1), 

let us find the differential coefficient. 

As before, let a; be increased to x-\-hx and find y + By, 

y-^hy — sin {x + hx) (2). 

Subtract (1) from (2) and we find 

hy = sin {x + hx) — sin x, 
or 2 cos (x + \Bx) sin \hx. (See Art. 3.) 

Hence ^ = cos (a? + JS^) ^^^^ (3). 

a sin (Kv -{• c). 175 

It is easy to see by drawing a small angle a and recollect- 
ing what sin a and a are, to find the value of sin a -r a as a 
gets smaller and smaller. Thus in the figure, let POA be 
the angle. The arc PA 
divided by OP is a, the 
angle in radians. And 
the perpendicular PB 

divided by OP is the pj g^ 

sine of the angle. Hence 

= -pr-r and it is evident that this is more and more 

a PA 

nearly 1 as a gets smaller and smaller. In fact we may 

take the ratios of a, sin a and tan a to one another to 

be 1, more and more nearly^ as a gets smaller and 

smaller. If we look upon ^Bx as a in the above expression, 

we see that in the limit, (3) becomes 

dy r • * 

-r^ = cos X if y = sin x.* 

* The proof of the more general case is of exactly the same kind. 
Here it is : 

i/ = actln(bx + c), 
y + Sy = a sin {b{x + Sx) + c}, 

Sy = 2a COB {bx+c + ^b, Sx) sin (^ 6 . dx). See Art. 3. 

Sx ^ ^ ^ ^b.Sx 

Now make 5a; smaller and smaller and we have 

— = ab cos (bx 4. c) and hence I a cos (bx + c) . dx = - sin (bx + c). 

Again, to take another case : — 

If ysa cos (bx + e), this is the same as 

y = asin( 6x+e + - J=a 8in(6aT + c), say. 

Hence -=^ = a6 cos (6a; + c) 

dx ^ 

= a&C08 ( 6a: + e + - j 
^ = — ab sin (bx + e). 
Hence / a sin (bx + c) . dx =— - cos (bx + c)« 



And hence 

I cos A' 


sm a% 

102. Now it is not enough to prove a thing like this, it 
must be known. Therefore the student ought to take a book of 
Mathematical Tables and illustrate it. It is unfortunate that 
such books are arranged either for the use of very ignorant 
or else for very learned persons and so it is not quite easy to 
convert radians into degrees or vice versa. Do not forget 
that in sin x or cos x we mean x to be in radians. Make out 

such a lit 

tie bit ot tab 

le as this, w 

hich is taken 

L at ranc 


Angle in 


or angle in 

y = sina; 


















If it is remembered that Bi/ -^ Bx in each case is really the 

average value of -^^ for one degree or ^01745 of a radian, it 

will be seen why it is not exactly equal to the cosine of x. 
Has the student looked for himself, to see if "7547 is really 
nearly equal to cos 41° ? 

103. It is easy to show in exactly the same way that if 

dy F 

y = cos X, ji- = - sin x and Isin x . dx = - cos x. The — 

sign is troublesome to remember. Here is an illustration : 

Angle in 


or angle in 

i/ = C08a; 










- -3584 

sin d. 


Notice that y diminishes as x increases. Notice that 
sin 21° or sin (-3665) = '3584. 

104. Here is another illustration of the fact that the 
differential coefficient of sin x is cos x. Let AOP, fig. 65, be 
6. Let AOQhe -h SO. Let PQ be a short arc drawn with 
as centre. Let OP = 0Q= 1. PR is perpendicular to QB. 

Fig, 65. 

Then AP ^y = sin e, BQ = sm(0 -^ Be) = i/ + By, QP = B0 
and EQ = 8y. Now the length of the arc PQ becomes 
more and more neai-ly the length of a straight line between 
P and Q as B6 is made smaller and smaller. 

Thus Yyp 01^ ]Si is more and more nearly equal to cos PQR 
or cos 0. 

imit*' -J2J = cos d if y = sin 0. 

In the limit 


Similarly if z = cos = OA, Bz = — BA = — RP and 
dz RP . >, 

d9=-QP = """^- 

Illustrations like these are however of most value when a 
student invents them for himself. Any way of making the 
fundamental ideas familiar to oneself is valuable. But it is 
a great mistake for the author of a book to give too many 
illustrations. He is apt to give prominence to those illustra- 
tions which he himself discovered and which were therefore 
invaluable in his own education. 

105. Observe that ii ^=A sin ajc 4- B cos ax; 

<^% and ^=«V 

Compare this with the fact that if y = e«*, ;7-^=aV) ;7^=^V ^^ 




the higher applications of Mathematics to Engineering this resemblance 
and difference between the two functions ^ and sin ax become im- 
portant. Note that if i stands for s] -\ so that i^= -\^ {*=1, &c. 

Then if y=e^ 1^~ ~^V> ;7:^=^"V j^^^ ^^ ^i*^ *^® s^^^® function. 
Comi)are Art. 99. 

106. Exercise. Men >yho have proved Demoivre's theorem 
in Trigonometry (the proof is easy ; the proofs of all mathe- 
matical rules which are of use to the engineer are easy ; 
difficult proofs are only useful in academic exercise work) say 
that for all algebraic purposes, co^ax=\{e^^-\-e~^^)^ and 

sin CW7 = ^ (e^*** — e~^^). If this is so, prove our fundamental 


107. Example. A plane electric circuit of area A sq. 
cm. closed on itself, can rotate with uniform angular velocity 
about an axis which is at right angles to the field, in a uni- 
form magnetic field H. H is supposed given in c.G.s. units; 
measuring the angle ^ as the angle passed through from the 
position when there is maximum induction HA through the 
circuit ; in the position 6, the induction through the circuit is 
evidently A.H. cos 6. If the angle 6 has been turned 
through in the time t with the angular velocity q radians 
per second, then 6 = qt. So that the induction /= AH cos qt. 
The rate of increase of this per second is —AqH ^in qt, and 
this is the electromotive force in each turn of wire. If there 
are n turns, the total voltage is — nAqH^m qt in C.G.S. units; 
if we want it in commercial units the voltage is 

- nAqH 10"8 sin qt volts, 

being a simple harmonic function of the time. Note that the 
term voltage is now being employed for the line integral of 
electromotive force even when the volt is not the unit used. 

Example. The coil of an alternator passes through a 
field such that the induction through the coil is 

/= ^0 + A^ sin {6 + fi) + Ar sin {rO + Cr), 

where 6 is the angle passed through by the coil. If q is the 


relative angular velocity of the coil and field, 6 = qt. If there 
are n turns of wire on the coil, then the voltage is w -7- , or 

nq {Ai cos (qt + ej) + ^,.r cos (rqt + e^)]. 

So we see that irregularities of ?• times the frequency in 
the field are relatively multiplied or magnified in the 

electromotive force. 

108. In Bifilar Suspension, if W is the weight of 
the suspended mass, a and b the distances between the 
threads below and above, h the vertical height of the threads; 
if the difference in vertical component of tension is n times 
the total weight W, and 6 is the angle turned through in 
azimuth, the momental resistance offered to further turning is 

{{l-n')Wjsmd (1). 

Note that to make the arrangement more ' sensitive ' it is 
only necessary to let more of the weight be carried by one of 
the threads than the other. 

The momental resistance offered to turning by a body 
which is 6 fi-om its position of equilibrium, is often propor- 
tional to sin 6. Thus if W is the weight of a compound 
pendulum and OG is the distance from the point of support 
to its centre of gravity, W . OG . sin 6 is the moment with 
which the body tends to return to its position of equilibrium. 
If M is the magnetic moment of a magnet displaced 
from equilibrium in a field of strength H, then HM sin 6 is 
the moment with which it tends to return to its position of 
equilibrium. A body constrained by the torsion of a wire or 
a strip has a return moment proportional to 0. When 
angular changes are small we often treat sin as if it were 
equal to 6. Sometimes a body may have various kinds of 
constraint at the same time. Thus the needle of a quadrant 
electrometer has bifilar suspension, and there is also an elec- 
trical constraint introduced by bad design and construction 
which may perhaps be like a6 + bO^. If the threads are stiff, 
their own torsional stiffness introduces a term proportional to 
6 which we did not include in (1). Sometimes the constraint 
is introduced by connecting a little magnetic needle rigidly 



with the electrometer needle, and this introduces a term 
proportional to sin 6, In some instruments where the 
moving body is soft iron the constraint is nearly propor- 
tional to sin 26, Now if the resisting moment is M 
and a body is turned through the angle hO, the work 
done is M. Bd, Hence the work done in turning a 
body from the position 6^ to the position ^.^, where S.^ is 

greater than ^i, is / M .dd. 

. 6, 

Example. The momental resistance offered by a body 
to turning is asin^ where 6 is the angle turned through, 
what work is done in turning the body from 6^ to 6.J 

Answer, / a . sin 6 .d6 = — a (cos ^2 — cos d^=^a (cos 6^ — cos 6.). 
J e, 

Example. The resistance of a body to turning is partly 
a constant torque a due to friction, partly a term b6-{-c&-, 
partly a term esinO; what is the work done in turning from 
^ = to any angle ? 

M the torque = a-{-bd + cd" + e sin =f(0) say, 

V the work done 

= aO + ibO' + icd'-\-e(l- cos 6) = F(e) say. 

This is called the potential energy of the body in the 
position 6. 

-^-j where 

/ is the body's moment of inertia about its axis. When a 

body is at 6 its total energy E is \I i~\ + F {$). 

If the total energy remains constant and a body in the 
position 6 is moving in the direction in which 6 increases, 
and no force acts upon it except its constraint, it will con- 
tinue to move to the position 6^ such that 

So that when the form oi F {6) is known, 61 can be calculated, 
if we know the kinetic energy at 6. 


Thus let M=e sin 0, so that F(0) = e(l- cos 0), then 
1/ (jY + e (1 - cos (9) = e (1 - cos e,), 

from which ^i the extreme swing can be calculated. 

Exercise. Show that if the righting moment of a ship 
is proportional to sin 4)6 where is the heeling angle, and 
if a wind whose momental effect would maintain a steady 
inclination of llf degrees suddenly sends the ship from rest 
at ^ = and remains acting, and if we may neglect friction, 
the ship will heel beyond 33J degrees and will go right over. 
Discuss the effect of friction. 

A body is in the extreme position 0^, what will be 
its kinetic energy when passing through the position 
of equilibrium ? Answer, F (Oi). 

Thus let M^hO + ce^-^e sin 6. 

Calculate a, the angular velocity at = 0, if its extreme 
swing is 45°. Here 

6, = ^ and F(e) = ^6^ -f ^c6' + e(l- cos 6), 

from which we may calculate a. 

Problem. Suppose we desire to have the potential 
energy following the law 

V=F(e)= a6^ + hO^ + ce'""^ + h sin 26, 

and we wish to know the necessary law of constraint, we see 

at once that as Jllf = -y^ , 

M = f al9'' -f 3&^ + mc€'''^ + 2/i cos W. 

Problem. A body in the position 6^, moving with the 
angular velocity a, in the direction of increasing 6, has a 
momental impulse m in the direction of increasing 6 
suddenly given to it ; how far will it swing ? 

The moment of momentum was /a, it is now loi + m and 

if a' is its new angular velocity ot' = a + y . 


The body is then in the position 0^ with the kinetic 

energy |/ ( a + y ) and the potential energy F{Oq), and the 

sum of these equated to F{6^ enables 6i to be calculated. 

The student will easily see that the general equation 
of angular motion of the constrained body is 

fid) may include a term involving friction. 

109. Every one of the following exercises must be worked 
carefully by students ; the answers are of great practical use 
but more particularly to Electrical Engineers. In working 
them out it is necessary to recollect the trigonometrical 

cos 2(9= 2 cos2^ - 1 = 1-2 sin2^, 
2 sin ^ . cos </) = sin (^ + </>) + sin {6 - ^), 
2 cos 6 . cos <fi = cos (^ + </)) + cos {$ — (j)), 
2 sin ^ . sin <^ = cos (0 — (f)) — cos (6 + <^). 

Such exercises are not merely valuable in illustrating the 
calculus; they give an acquaintance with trigonometrical 
expressions which is of great genei-al importance to the 

The average value of /(.r) from ii' = ooi to .T = a?.^ is 

evidently the area I f(x) . dx divided by .r., — a'i. 

Every exercise from 6 to 20 and also 23 ought to be 
illustrated graphically by students. Good hand sketches of 
the curves whose ordinates are multiplied together and of the 
resulting curves will give sufficiently accurate illustrations. 

X sin 2ax 

sin- (Lxdx 

o r . , , cos (a-\-h)a; cos (a — h)x 

3. sm ax , cos bx . dx = ~ ~ ^ rf- 

J 2(a + 6) 2(a-6) 

2 4a 

X . sin 2ax 


f . . , , sm(a~b)x sin (a + h) x 

k sm ax . sin bx . d^• = —^ r-r ^, , . (— 

; 2 (a — 6) 2 (« + b) 

5. I cos a^ . cos bx.cix — — ^, ~ h 

sin {a + b) x sin (a — b) x 
2(a + 6) "^ ~27a^;^6)~ 


6. The area of a sine curve for a whole period is 0. 

r2.1T r2jr "I 

I sinfl7.fZic=— cosiT = — (1 — 1) = 0. 

7. Find the area of the positive part of a sine curve, 
that is 

I sin ^ . c^a; = — cos x = — (— 1 — 1) = 2. 

Since the length of base of this part of the curve is tt, 


the average height of it is - . Its greatest height, or ampli- 
tude, is 1. 

8. The area of y — a-\-b sin x from to 27r is 27ra and 
the average height of the curve is a. 

9. Find the average value of sin- x from a; = to ^ = 27r. 
As cos 2^' = 1 - 2 sin^ x, sin^ ^ = | (1 - cos 2x). The integral 

of this is |.2J - J sin 2x, and putting in the limits, the area is 
(^27r-isin47r-04-isinO) = 7r. The average height is 
the area -j- 27r, and hence it is ^. 

10. The average value of cos- x from ^; = to iz? = 27r is \. 

In the following exercises s and r are supposed to be 
whole numbers and unequal : 

11. The average value of a niri^ {sqt ■\- e) from ^=0 to 
^ = T is g , s being a whole number and q = ^irlT. T is the 

periodic time. 

12. The average value of aQ,o^^isqt-\-e) from i = to 

13. I cos sqt . sin sqt ,dt=^0. 



14. I sin sqt . sin rqt ,dt — 0. 



15. I cos sqt . cos rqt .dt — 0. 


16. I sin sqt . cos rqt .dt = 0. 

17. The average value of sin^ sqt from to ^T is ^. 

18. The average value of cos^ sqt from to ^T is J. 

19. I sm 55^ . sin rqt . a^ = 0. 


20. I cos 55^ . cos rqt ,dt = 0. 

21. Find /sin a; . sin (x + e) . dx. 

Here, sin {x-\- e) = sin a? cos e -\- cos a; . sin e. 

Hence we must integrate sin'* x . cos e-^-smx . cos a; . sin e, 

a? sin 2a; 


sin'' X .dx = 

(sin X . cos X .dx = \ (sin 2a? . (^a; = — J cos 2a;, 

and hence our integral is 

fx sin 2a;\ , ^ 

( ^ J— J cos e — % cos 2a7 . sin e. 

22. Prove that (sin qt . sin (^-^ + e) . f/^ 

/^ sin2gA 1 ^ , . 

23. Prove that the average value of sin qt . sin {qt ± e) or 
of sin (qt + a) sin (5'^ + a ± e) for the whole periodic time T 

(if q = y) ^s i cos e. 


This becomes evident when we notice (calling qt-\- a = (p), 

sin (ft . sin ((f) ±e) = sin (ft (sin (j) cos e ± cos (f) . sin e) 

= sin^ <l> . cos e ± sin (f> . cos </> . sin e. 

Now the average value of sin- (/> for a whole period is |^, 
and the average value of sin <^ . cos <^ is 0. 


By making a = ^ in the above we see that the average 

value of cos qt . cos {qt ± e) is ^ cos e, or the averages value 
for a whole period of the product of two sine flinctions 
of the time, of the same period, each of amplitude 1, is 
half the cosine of the angular lag of either behind the 

24. Referring to Art. 106, take 

cosa(9 = J(e'^ + e"'^), 

or take e^^ = cos ad -f i sin aO, 

e"^ = cos ad — i sin aO, 

and find L^ cos ad . dO. 

This becomes \j{e^''-''^'-Ve^'-''^')dd 

(b+ai)0 . 1 ib~ai)e) 

e + V . e y 

b — ai 

* (6 + ai 

1 W f 1 aid 1 

i^ "^r-; — ^-^ +1 — 

6 ~ ai 

and on substituting the above values it becomes 

I e cos a^ . 
Similarly we have 

e cos a0.dd= - — ^ e^ (b cos a^ + a sin a^) . . .(1). 

fe^« sin adds = -^^-^ e^« (b sin a(9 - a cos a(9) (2). 


110. Notes on Harmonic Functions. In the fol- 
lowing collection of notes the student will find a certain 
amount of repetition of statements already made. 

111. A function ^ = tt sin qt is analogous to the straight 
line motion of a slider driven from a crank of length a 
(rotating with the angular velocity q radians per second) 
by an infinitely long connecting rod. x is the distance of the 
slider from the middle of its path at the time t At the zero 
of time, ^ = and the crank is at right angles to its position 

of dead point, q = 27rf = -=- , if T is the periodic time, or 

if/ is the frequency or number of revolutions of the crank 
per second, taking 1 second as the unit of time. 

112. A function x = a sin {qt -}- e) is j ust the same, except 
that the crank is the angle € radians (one radian is 57 '2957 
degrees) in advance of the former position; that is, at 
time the slider is the distance a sin e past its mid-position*. 

* The student is here again referred to § 10, and it is assumed that he 
lias drawn a curve to represent 

x=6c""'sin {qt + i) (1). 

Imagine a crank to rotate uniformly with the angular velocity q, and to 
drive a slider, but imagine the crank to get shorter as time goes on, its 

length at any time being fl€~ . 

Another way of thinking of this motion is: — 

Imagine a point P to move with constant aiujnlar velocity round O, 

keeping in the equiangular spiral path APBCDEF; the motion in question 
is the motion of P projected upon the straight line MON and what we have 
called the logarithmic decrement is tt cot a if a is the angle of the spiral. 



113. A fiuictiou X = a sin (ryi + e) + o! sin {(jt + e') is the 
same as X = A sin (5'^ + ^); that is, the sum of two crank 
motions can be given by a single crank of proper 
length and proper advance. Show on a drawing the 
positions of the first two when ^ = 0, that is, set off 

YOF = e and OF = a, 
YOQ = e' and OQ = a/. 

Complete the parallelogram OPRQ and draw the diagonal 
OR, then the single crank OR = A, with angle of advance 
YOR = E, would give to a slider the sum of the motions 
which OP and OQ would separately give. The geometric 

proof of this is very easy. 

Imagine the slider to have 

a vertical motion. Draw 

OQ, OR and OP in their 

relative positions at any 

time, then project P, R 

and Q upon OX. The 

crank OP would cause 

the slider to be OP' above 

its mid-position at this 

instant, the crank OQ 

would cause the slider to 

be OQ' above its mid-po- 

^^g- ^'^' sition,the crank OR would 

cause the slider to be OR' above its mid-position at the same 

instant; observe that OR' is always equal to the algebraic 

sum of OP' and OQ. 

We may put it thus: — "The s.H.M. which the crank OP 
would give, + the s.H.M. which OQ would give, is equal to 
the S.H.M. which OR would give." Similarly "the s.H.M. 
which OR would give, - the s.H.M. which OP would give, is 
equal to the s.H.M. which OQ would give." We sometimes 
say: — the crank OR is the sum of the two cranks OP and OQ. 
Cranks are added therefore and subtracted just like vectors. 

that is, the constant acute angle which OP everywhere makes with the curve, 
or Tfcot a = aT/2 and g = 27r/r, so that cot a = alq. If fig. 66 is to agree with 
fig. 67 in all respects NM being vertical and P is the position at time 0, then 
€= angle iVOP -7r/2. 


114. These propositions are of great importance in dealing 
with valve motions and other mechanisms. They are of so 
much importance to electrical engineers, that many practical 
men say, "let the crank OP represent the current." They 
mean, " there is a current which alters with time according 
to the law (7= a sin (5^ 4- e), its magnitude is analogous to 
the displacement of a slider worked vertically by the crank 
OP whose length is a and whose angular velocity is q and 
OP is its position when t = 0." t 

115. Inasmuch as the function w = a cos qt is just the 

same as asinf^^ + ^j, it represents the motion due to a 

crank of length a whose angle of advance is 90°. At any 
time t the velocity of a slider whose motion is 

{V = a sin (qt + e), 

is v^aq cos (qt 4- f) = -^ or ^r 

= aq sin 

(,* + e + |). 

that is, it can be represented by the actual position at any 

instant of a slider worked by a crank of length representing 

aq, this new crank being 90° in advance of the old one. 

cl?x dv 
The acceleration or -^^ or -7- or v is shown at any instant 

by a crank of length aq^ placed 90° in advance of the v 
crank, or 180° in advance of the x crank, for 

Accel. = - aq^ sin {qt + e) 

= aq^ sin {qt-\-e + ir). 

The characteristic property of S.H. motion is that, numerically, 
the acceleration is q^ or 47r-/- times the displacement, /being 
the frequency. 

If anything follows the law a sin {qt + e), it is analogous 
to the motion of a slider, and we often say that it is repre- 
sented by the crank OP ; its rate of increase with time 
is analogous to the velocity of the slider, and we say that it 
is represented by a crank of length aq placed 90° in advance 
of the first. In fact, on a S.H. function, the operator djdt 
multiplies by q and gives an advance of a right angle. 



116. Sometimes instead of stating that a function is 
A sin (qt + e) we state that it is a sin qt + b cos qt. 

Evidently this is the same statement, if a? + h^ = A^ and if 

tan e = -. 


It is easy to prove this 
trigonometrically, and gra- 
phically in fig. 68. Let 

The crank OP is the sum 
of 08 and OQ, and tan e or 

tan FOP = -. 

Fig. 68. 

117. We have already in Art. 100 indicated an easy 
graphical method of drawing the curve 

x = asm (qt + e), 

where a; and t are the ordinate and abscissa. 

Much information is to be gained by drawing the two of 
the same periodic time, 

a; = a sin (qt + e) and x — a sin (qt + e), 

and adding their ordinates together. This will illustrate 113. 

118. If the voltage in an Electric Circuit is V volts, the 
current G amperes, the resistance R ohms, the self-induction 
L Henries, then if t is time in seconds, 




Now if = Co sin qt, 

T- = G^q cos qt, • 

so that Y= RGq sin qt + LG^q . cos qt, 

and by Art. 116 this is 

. V=Go^/W+LY^^^(qi + ^) (2); 


nR^ 4- L^q^ is called the impedance ; 

tan e = -^* = — — ^ , if/ is frequency ; 

6 is lag of current behind voltage. 
Hence again if V = VoSin qt (3), 

then C = ;7=^Bin(qt-tan-^)l (4). 

Notice that if V is given as in (3) the complete answer 
for G includes an evanescent term due to the starting 
conditions see Arts. 98, 147, but (4) assumes that the simple 
harmonic V has been established for a long time. In practical 
electrical working, a small fraction of a second is long enough 
to destroy the evanescent term. 

119. We may wiite the characteristic property of a simple 
harmonic motion as 

S+<»'*=o W- 

(compare Arts. 26 and 108) and if (1) is given us we know 
that it means 

x = a8mqt+hQ0>iqt or x =^ A ^m {qt + e) (2), 

where A and e, or a and h are any arbitrary constants. 

Example. A body whose weight is W lb. has a simple 
harmonic motion of amplitude a feet (that is, the stroke is 
2a feet) and has a frequency / per second, what forces give 
to the mass this motion ? 

If X feet is the displacement of the body from mid- 
position at any instant, we may take the motion to be 

x= a sin qt or a sin 2irf. t, 

and the numerical value of the acceleration at any instant is 
^ir^px and the force drawing the body to its mid -position is 
in pounds ^ir^f'^xW-^ 322, as mass in engineer's units is weight 
in pounds in London -r- 32 "2, and force is acceleration x mass. 


120. If the connecting rod of a steam or gas engine were 
long enough, and we take W to be the weight of piston and 
rod, the above is nearly the force which must be exerted 
by the cross-head when the atmosphere is admitted to both 
sides of the piston. Observe that it is when a? is and is 
proportional to o), being greatest at the ends of the stroke. 
Make a diagram showing how much this force is at every 
point of the stroke, and carefully note that it is always act- 
ing towards the middle point. 

Now if the student has the indicator diagrams of an 
engine (both sides of piston), he can first draw a diagram show- 
ing at every point of the stroke the force of the steam on 
the piston, and he can combine this with the above diagram 
to show the actual force on the cross-head. Note that steam 
pressure is so much per square inch, whereas the other is the 
total force. If the student carries out this work by himself 
it is ten times better than having it explained. 

Since the acceleration is proportional to the square of the 
frequency, vibrations of engines are much more serious than 
they used to be, when speeds were slower. 

121. As we have been considering the motion of the piston 
of a steam engine on the assumption that the connecting rod is 
infinitely long, we shall now study the effect of shortness of 
connecting rod. 

In Art. 11, we found 6" the distance of the piston from the 
end of its stroke when the crank made an angle 6 with its dead 
point. Now let x be the distance of the piston to the right 
of the middle of its stroke in fig. 3, so that our x is the old s 
minus r, where r is the length of the crank. 

Let the crank go round uniformly at q radians per second. 

Again, let t be the time since the crank was at right 

angles to its dead point position, so that 0— - = qt, and we 
find ^ 


x z= - r cos -^ I \l - \/ 1 -^ sin2 oi , 
x= r iiin qt + I \i — \/ 1 — j~ cos^ qt [• 


Using the approximation that Vl — a = 1 — ^a if a is 
small enough we have 

X =^ r sm qt -\- ^. cos- qt 

But we know that 2 cos- qt-'l= cos 2qL (See Art. 109.) 

Hence x = r sin qt + ^i ^^os (2qt) - — (1). 

We see that there is a fundamental simple harmonic motion, 
and its octave of much smaller amplitude. 

Find -jr and also -f- . This latter is 
at dt^ 

acceleration = — rq- sin qt y cos '2.qt. 

It will be seen that the relative importance of the octave 
term is four times as great in the acceleration as it was in 
the actual motion. We may, if we please, write 6 again for 

2^ + 2 and get 



— - = r^' cos 6 + ~ cos 




When ^ = 0, the acceleration is rif- H — ? . 
When B — 90^ the acceleration is — - 


When = 180°, the acceleration is — rq^ + —j- , 

(q is 27rf, where / is the frequency or number of revolutions 
of the crank per second). 

If three points be plotted showing displacement x and 
acceleration at these places, it is not difficult by drawing a 
curve through the three points to get a sufficiently accurate 
idea of the whole diagram. Perhaps, as to a point near the 
middle, it might be better to notice that when the angle 
OPQ is 90°, as P is moving uniformly and the rate of change 
of the angle Q is zero, there is no acceleration of Q just then. 
This position of Q is easily found by construction. 


The most important things to recollect are (1) that 
accelerations, and therefore the forces necessary to cause 
motion, are four times as great if the frequency is doubled, 
and nine times as great if the frequency is trebled ; (2) that 
the relative importance of an overtone in the motion is 
greatly exaggerated in the acceleration. 

122. Take any particular form of link motion or radial 

valve gear and show that the motion of the valve is always 
very nearly {t being time from beginning of piston stroke or 
qt being angle p?^sed through by crank from a dead point), 

x = a^ sin {qt + 61)4-02 sin {^qt + 63) (1). 

(There is a very simple method of obtaining the terms Oj 
and a^ by inspection of the gear.) When the overtone is 
neglected, a^ is the half travel of the valve and ei is the angle 
of advance. In a great number of radial valve gears we find 
that 6-2 = 90°. The best way of studying the effect produced 
by the octave or overtone is to draw the curve for each term 
of (1) on paper by the method of Fig. 63, and then to add 
the ordinates together. If we subtract the outside lap L 
from cc it is easy to see where the point of cut-off is, and how 
much earlier and quicker the cut-off is on account of this 
octave or kick in the motion of the valve. 

In an example take «i = 1, e^ = 40°, a^ = '2, e^ — 90"^. 

The practical engineer will notice that although the 
octave is good for one end of the cylinder it is not good for the 
other, so that it is not advisable to have it too great. We 
may utilize this fact in obtaining more admission in the up 
stroke of modern vertical engines ; we may cause it to correct 
the inequality due to shortness of connecting rod. 

Links and rods never give an important overtone of 
frequency 3 to 1. It is always 2 to 1 . 

In Sir F. Bramwell's gear the motion of the valve is, 
by the agency of spur wheels, caused to be 

a? = «! sin {qt -f Cj) + Oo sin {^qt -t- e,) (2). 

Draw a curve showing this motion when 

a, = 1-15 inch, e^ = 47", a._ = '435, e, = 62°. 

P. 13 


If the outside lap is 1 inch and there is no inside lap, find 
the positions of the main crank when cut-off, release, cushion- 
ing and admission occur. Show that this gear and any gear 
giving an overtone with an odd number of times the funda- 
mental frequency, acts in the same way on both ends of the 

123. If a? = Oi cos (^1^ + €i) -H 02 cos {q4 ■\- e^ (3), 

where q^^^irfi and q^—'^irf^, 

there being two frequencies; this is not equivalent to one 
S. H. motion. Suppose a^ to be the greater. The graphical 
method of study is best. We have two cranks of lengths a^ 
and tta rotating with different angular velocities, so that the 
effect is as if we had a crank A rotating with the average 
angular velocity of a^, but alternating between the lengths 
tti + tta and ai — tta ; always nearer a^s> position than Og's ; in 
fact, oscillating on the two sides of Oi's position. If q^ is 
nearly the same as q^ we have the interesting effect like 
beats in music*. 

Thus tones of pitches 100 and 101 produce 1 beat per 
second. The analogous beats are very visible on an in- 
candescent lamp when two alternating dynamo-electrical 
machines are about to be coupled up together. Again, tides 
of the sea^ except in long channels and bays, follow nearly 
the s. H. law ; a^ is produced by the moon and a^ by the sun 
if ai=2'la2, so that the height of a spring tide is to the 
height of a neap tide as 3*1 to 1*1. The times oi full are 
times of lunar full. The actual tide phase never differs more 
than 0*95 lunar hour from lunar tide ; 095 lunar hour = 098 

124. A Periodic Function of the time is one which 
becomes the same in every particular (its actual value, its 
rate of increase, &c.) after a time T. This T is called the 

* Analytically. Take cos (^vf^t + eg) =co8 {^irf-J, - 2t (/^ -/2)f + e^}, 
therefore a; = r cos (2t/i« + 6) , 

where r^ = a-^ + a^ + 2a^a^ cos {27r (/^ -/g) t + Cj - Cg} , 

and the value of tan 6 is easily vrritten out. 


periodic time and its reciprocal is called f the frequency. 
Algebraically the definition of a periodic function is 

where n is any positive or negative integer. 

125. Fourier's Theorem can be proved to be true. It 
states that any periodic function whose coraplete period is T 
(and g is ^tt\T or 27r/) is really equivalent to the sum of a 
constant term and certain sine functions of the time 

fif) = ^0 + ^1 sin {qt + E^ + A^ sin {%qt + E^ + 

^3 sin (3^'^ + £'3) 4- &:c (1). 

In the same way, the note of any organ pipe or fiddle 
string or other musical instrument consists of a fundamental 
tone and its overtones. (1) is really the same as 

/(^) = ^0 + «i sin 5^ + 61 cos qt + ag sin ^qt -f 6., cos Iqt + &c., 
if ar + &r = A^ and tan JS^^ = — , &c. t 

126. A varying magnetic field in the direction x follows 
the law X = a sin qt where t is time. Another in the 
direction y, which is at right angles to x, follows the law 

Y—a cos qt. 
At any instant the resultant field is 

R = VZmT^ = a = a constant 
making with y the angle 6, where tan = F/X, or ^ = qt 

Hence the effect produced is that we have a con- 
stant field R rotating with angular velocity q. 

When the fields are 

X = Oi sin (qt + Cj) and F = aa sin (qt + e^), 

it is better to follow a graphical method of study. The 
resultant field is represented in amount and direction by the 
radius vector of an ellipse, describing equal areas in equal 

Let OX and OY, fig. 69, be the two directions mentioned. 
Let OAi in the direction OX = a^ . With 0-4 1 as radius describe 




a circle. Let YOO be the angle ei. Divide the circle into 
many equal parts starting at and naming the points of 
division 0, 1, 2, 3, &c. Draw lines from these points parallel 
to OY, Let 0A» in the direction OF be ag. Describe a 
circle with OA2 as radius. Set off the angle X'OO'as e^ and 

divide this circle at 0', 1, 2, 3, 4, &c., into the same number 
of parts as before. Let lines be drawn from these points 
parallel to OX, and where each meets the corresponding line 
from the other circle we have a point whose radius vector 
at any instant represents, in direction and magnitude, the 
resultant magnetic field. 

If OX and OY are not at right angles to one another, 
the above instructions have still to be followed. 

If we divide the circle OA2 into only half the number of 
parts of OJ-i we have the combination of X = Oj sin (qt + fi) 
and F= ttg sin (2qt 4- 62). 

If we wish to see the combination X = any periodic 
function and Y any other periodic function, let the curve 



from Mo to N^ show F, M^N^ being the whole periodic time ; and 
let the curve from M^ to N^ show X, the vertical distance M^Ni 

Fig. 70. 

being the whole periodic time. If Pi and Pj are points on the 
two curves at identical times, let the horizontal line from P^ 
meet the vertical line from Pj in P. Then at that instant 
OP represents the resultant field in direction and 

Carry out this construction carefully. It has a bearing 
on all sorts of problems besides problems on rotating mag- 
netic fields. 

127. The area of a sine curve for a whole period or for 
any number of whole periods is zero. This will be evident if 
one draws the curve. By actual calculation; let s be an 

integer and g= -— -, 

•2. 1 r^ ' 
sin sqt.dt = cos sat 

because cos5 -^ Tor cos s2it = 1 and cos = 1. 



cos j 


Again, I 


cos sqt .dt = — 


■^. 1 
sm sqt 

.0 J 

1 / . 27r ^ 



because sin s~^ T= sin s27r = and sin 0=0. 


128. If the ordinates of two sine curves be multiplied 
together to obtain the ordinate of a new curve : the area of 
it is for any period which is a multiple of each of their 
periods. Thus if s and r are any integers 


sin sqt . cos rqt . dt = (1), 


sin sqt . sin rqt. dt = (2), 


I cos sqt . cos rqt . dt = . .(3). 

These ought to be tried carefully. '1st as Exercises in 
Integration. 2nd Graphically. The student cannot spend 
too much time on looking at these propositions from many 
points of view. He ought to see very clearly why the 
answers are 0. The functions in (1) and (2) and (3) really 
split up into single sine functions and the integral of each 
such function is 0. Thus 

2 sin sqt . cos rqt = sin (s + r) ot 4- sin {s — r) qt, 

and by Art. 127, each of these has an area 0. 

The physical importance of the proposition is enormous. 
Now if 5 = r the statements (2) and (3) are untrue, but (1) 
continues true. For 


/•T rT 

I sin^ sqt . dt = / cos^ sqt . dt = ^T 

.'o Jo 

whereas (1) becomes the integral of ^ sin 2sqt which is 0. (4) 
ought to be worked at graphically as well as by mere inte- 
gration. Recollecting the trigonometrical fact that 

cos2^ = 2cos2^-l or 1 - 2 sin^ ^, 

and therefore that 

cos^ qt=^ cos 2qt + i, sin^ g'^ = J — J cos 2qt, 

the integration is easy and the student ought to use this 
method as well as the graphical method. 

129. To illustrate the work graphically. Let 0C\ fig. 71, 
be T. Taking s = 2, the curve OPQRSC represents sin sqt 



Its maximum and minimum heights are 1. Now note that 
sin^^^^ is always + and it is shown in OP'Q^R^Sfi. It 
fluctuates between and 1 and its average height is \ or 

Fig. 71. 

the area of the whole curve from to (7 is ^T. The fact 
that the average value of sinsqt. x sinrqt is 0, but 
that the average value of sin sqt x sin sqt is \, is one of 
the most important in practical engineering "work. 

130. Illustration in Electricity. An electric dynamo- 
meter has two coils ; one fixed, through which, let us suppose, 
a current G flows ; the other moveable, with a current c. At 
any instant the resultant force or couple is proportional to 
Go and enables us to measure Gc. But if G and c vary 
rapidly we get the average value of Gc. Prof. Ayrton and 
the author have carried out the following beautifully illus- 
trative experiment. They sent a current through the fixed 
coil which was approximately, G=Gq sin 27rft. This was 
supplied by an alternating dynamo machine. Through the 
other coil they sent a current, c = Co sin 2'nf't whose frequency 
could be increased or diminished. It was very interesting to 
note (to the average practical engineer it was uncanny, 
unbelievable almost) that although great currents were 
passing through the two coils, there was no average force — 
in fact there was no reading as one calls it in the laboratory. 
Suppose / was 100 per second, /' was gradually increased 
from say 10 to 20, to 30 to 40 to 49. Possibly about 49 to 


51 a vague and uncertain sort of action of one coil un the 
other became visible, a thing not to be measured, but asf 
increased the action ceased. No action whatever as f 
became 60, 70, 80, 90, 97, 98, 99, but as / approached 100 
there was no doubt whatever of the large average force ; 
a reading could be taken and it represented according to the 
usual scale of the instrument ^CqCo; when/' increased beyond 
100 the force suddenly ceased and remained steadily until 
/' became 200 when there was a small force to be measured ; 
again it ceased suddenly until /' became 300, and so on. We 
know that if C and c had been true sine functions there 
would have been absolutely no force except when the 
frequencies were exactly equal. In truth, however, the 
octaves and higher harmonics were present and so there were 
slight actions when /and/' were as 2 to 1 or 1 : 2 or 1 : 3, &c. 
This is an extremely important illustration for all electrical 
engineers who have to deal with alternating currents of 

131. Exercise in Integration. C and c being alternating 
currents of electricity. When G = Co sin qt and c = cv sin(qt ± e) 
and these two currents flow through the two coils of an 
electro-dynamometer, the instrument records ^ CoCo . cos e as 
this is the average value of the product Cc. 

When G and c are the same, that is, when the same 
current G = G^ sin (qt + e) passes through both coils, the 
instrument records the average value of G^ . dt, or 

^j G,'8m'{qt + e).(lt (1), 

which we know to be ^Gq^ The square root of any such 
reading is usually called the effective current, so that 

~r^ Gq is what is known as the effective value of C^o sin qt 


Effective current is defined as the square root of 

mean square of the current. Thus when an electrical 

engineer speaks of an alternating current of 100 amperes he 

means that the effective current is 100 amperes or that 

G= 141*4 sin {qt + a). Or the voltage 1000 means 

t; = 1414 sin (</^ + /3). 



Exercise. What is the effective value of 

tto 4- ^1 sin {qt + ti) + A.2 sin {^qt + 63) + &;c. ? 

Notice that only the squares of terms have an average 
value, the integral of any other product being during a 
complete period. Answer: \/ a^ -^ ^{A-^ -{-A.^ -{■ kc). 

Observe the small importance of small overtones. 

If V = — (sin qt + J sin ^qt + i sin aqt &c.). 


we shall see from Art. 135 that this is the Fourier expression 

for what is shown in the curve (fig. 72) the distance OM being 

called Vq and the distance OQ being the periodic time T, where 

q= rp } s-i^d V is measured upwards from the line OQ. 


Fig. 72. 

The effective « = ^^ Vl + * + ^V + ,V + &<=. 


Again in fig. 73, where PM = Vq and OQ = T, 


(sin qt — ^ sin Sqt + ^ sin 5qt — &c.) . . .(4). 

Fig. 73. 

The effective ?; = —-";. VI +-J^ + ^ + &c (5). 

Again note the small importance of everything except 
the fundamental term. 


Exercise. If C=Go-{-A^smqt-\- B^ cos qt 

4- ^2 sin 2qt + B^ cos 2qt + <fec ((3), 

and if 

c = Co + «! sin qt 4- 6i cos g'< + aa sin 2^'^ 4- h^ cos 2^^ + &c. . . .(7). 
Average Cc = G^c^ + J (^itti 4- BJ)^ 4- ^./ta + BJ)^ + &c.). . .(8). 
It will be seen that there are no terms like A.]),^ or A.^^, 

132. Let AB and BG be parts of an electric circuit. In 

yj^^ AB let the resistance be R 

^- --^- -■-- „„ ->, ^j^^ j^^ there be no self-in- 

^•vsAAAAAAAAA/Tfnrinnpnnr-*^ duction. In BG let the re- 

^ ^ T,- ^ i ' ^ sistance be r and let there be 

^'^' ^^- self-induction Z. If 0= Co sin ^i 

is the current passing. Let Vab &c. represent the voltage 

between the points A and B, &c. Let F^ mean the effective 

voltage between A and B. 

Vab = -^C'o sin qt, 

Vbc ■■= Go Vr« 4- ly sin ^^^ + tan-^ -^') , see Art. 118, 

Vac = Co \/(i24-r)2 4-?Y sin (^« 4- tan"^ ^^-) , 

F..= -^i2C„, F..= i^^^V^+>' 

F^o=-^(i2 + r)Ooyi4- ^'^' 

Observe that V^c is alTvays less than Vj^^ 4- Vbc^ or 
the effective voltage between A and C is always less 
than the sum of the effective voltages between A and 
B and between B and C. 

Thus take Co =141*4, jR=l, ?'=1, lq=l, and illustrate 
a fact that sometimes puzzles electrical engineers. 

133. Rule for developing any arbitrary function 
in a Fourier Series. 

-|*The function may be represented as in fig. 75, P^ repre- 
sents the value of y at the time t which is represented by 



OE, 00 represents the whole periodic time T. At G the 
curve is about to repeat itself. (Instead of using the letter 







' E > 


Fig. 75. 

t we may use x or any other. We have functions which are 
periodic with respect to space for example.) Assume that y 
can be developed as 

y^a^^-ai sin qt + hi cos qt -}- a.^ sin 2qt + h.^ cos 2qt 

+ as sin 3^^ + 63 cos ^qt 4- &c. . . .(1), 



It is evident from the results given in Art. 127 that a^ 
is the average height of the curve, or the average 
value of y. This can be found as one finds the average 
height of an indicator diagram. Carry a planimeter point 
from to FPHGCO, and divide the whole area thus found by 
OC. If we have not drawn the curve ; if we have been given 
say 36 equidistant values of y, add up and divide by 36. 
The reason is this; the area of the whole curve, or the 
integral of y between the limits and T, is a^T, because 
the integi'al of any other term such as Wi sin qt or 63 cos Sqt 
is 0. In fact 

rT rT 

sin sqt .dt or I cos sqt . dt is 0, 

J .'0 

if s is an integer. 

tti is twice the average height of the curve which results 
from multiplying the ordi nates y by the corresponding 


ordinate of sin^^; for, multiply (1) all across by sinqt, 
and integrate from to T, and we have by Art. 128 

rT rr 

I y.sinqt.dt==0-\-aij ain' qt,dt -^-O -\-0 + &i}c,=^aiT...{l); 

Jo Jo 

dividing by T gives the average value, and twice this is 
evidently Oi. Similarly 


y.coaqt,dt-=^b/r (2). 


In fact, by the principles of Ai't. 128, a.^ and bs are twice 
the average values of y sin sqt and y . cos sqt^ or 

2 f^ 
^^»=^t/ y. sin sqt, dt 

i .' 


134. In the Electrician newspaper of Feb. 5th, 1892, the 
author gave clear instructions for carrying out this process 
numerically Avhen 36 numbers are given as equidistant 
values of y. 

In the same paper of June 28th, 1895, the author de- 
scribed a graphical method of finding the coefficients. 
The graphical method is particularly recommended for de- 
veloping any arbitrary function. 

Students who refer to the original paper will notice 
that the abscissae are very quickly obtained and the curves 

In this particular case we consider the original curve 
showing y and time, to be wrapped round a circular cylinder 
whose circumference is the periodic time. The curve is pro- 
jected upon a diametral plane passing through ^ = 0. Twice 
the area of the projection divided by the circumference of 
the cylinder is di. Projected upon a plane at right angles to 
the first, we get bi in the same way. When the curve is 
wrapped round s times instead of once, and projected on 
the two diametral planes, twice the areas of each of the 



two projections divided by s times the circumference of the 
cylinder give cig and hg* 

Prof Hem'ici's Analyzers, described in the Proceedings 
of the Physical Society, give the coefficients rapidly and 
accurately. The method of Mr Wedmore, published in the 
Journal of the Institution of Electrical Engineers, March 1896, 
seems to me very rapid when a column of numbers is given as 
equidistant values of y, 

135. When a periodic function is graphically represented 
by straight lines like fig. 72 or fig. 73 we may obtain the 
development by direct integration. Thus in fig. 76, the 
Electrician's Make and Break Curve: 


P Q 

Fig. 76. 

y=OA, or 2^0 say, from t- to t = OP =^iT; 

y = from t = iT or OP, to t==T or OQ. 


Evidently ao = Vo, 5' = -^; 

2 ri^ 

/ b 


2 fi^ 2 r*^ 

as = 7w I 2i'o . sin sqt . dt, h = jp\ 2vq cos sqt . dt. 

4i;o T [■ 



"^=-T-25;^Lr'- r^ 

, 4t;o T V^T 27r 1 

* The method is based upon this, that 

a, = ^\y.m.nsqt.dt = -—^\y.d (cos sqt) = - — / 1/ • ^ (cos sqt). 

Drawing a complete curve of which y (at the time t) is the ordinate and 
COB sqt is the abscissa, we see that its area as taken by a Planimeter 
divided by sir gives a^ . This graphical method of working is made use of 
in developing arbitrary functions in series of other normal forms than sines 
and cosines, such as Spherical Zonal Harmonics and Bessels. 

By the above method 

' ^'=hfy' 

d (sin sqt). 



2^0 , ... 2vo 
aii= (cos SIT — cos 0) = 

/ if « ii 
\- 2 if s ij 

if « is even\ 
IS odd / 

= — if .9 is odd, 


ha=—^ (sin sir - sin 0) = 0. 


Hence the function shown in fig. 76 becomes 
y = Vf,+ -- (sin at + \ sin Zqt + J sin ^qt + &c.) (1). 


Fig. 77. 

If the origin is half-way between and A (fig. 76), as in 
fig. 77, so that instead of what the electricians call a make 
and break we have Vq constant for half a period, then — Vq 
for the next half period, that is, reversals of y every half 
period, we merely subtract v^, then 

y = — ^ (sin qt-\-l sin ^qt + i sin oqt + &c.) . . .(2). 

Let the origin be half-way between and P, fig. 70; the 
^ of (1) being put equal to a new t-\-\T, 

sin sqt where s is odd, becomes sin sq {t -f \T), 

sin s -^ (t-\-^T) or sin Uqti-s'^], 

where 5=1, 5, 9, 13 &c. this becomes cos sqt, 

„ 5 = 3, 7, 11, 15 &c. „ „ — cos sqty 

and consequently with the origin at a point half-Avay between 
and P, 


y = vo + — (cos qt-^ cos Sqt + i cos Sqt - 1 cos Iqt + &c.). 


136. To represent a periodic function of x for all values 
of X it is necessary to have series of terms each of Tvhich 
is itself a periodic function. The Fourier series is the 
simplest of these. 

137. If the values of y, a function of x, be given for all 
values of x between a; = and x — c\y can be expanded in a 
series of sines only or a series of cosines only. Here 

we regard the given part as only half of a complete periodic 
function and we are not concerned with what the series 
represents when x is less than or greater than c. In 
the previous case y was completely represented for all 
values of the variable. 

I. Assume y = ai sin qx + a^ sin 2qx + &c. where q = tt/c. 
Multiply by ^insqx and integrate between the limits 
and c. It will be found that all the terms disappear except 

as sin'^ sqx . dx which is Ja«c, so that a^ is twice the average 

value oiy,^ixisqx. 

Thus let y be a constant m, then 


2 /•' 

C /a 

, 2m p 

m sm sqx .dx= cos sox 

esq Lo ^ _ 

= (cos 57r — 1 ) = — II s IS odd, 

csq^ ' SIT 

= if s is even. 
Hence m=^ — (sin qx-\-\ sin Zqx + \ sin hqx + &c.)*. 

II. Assume y = 6o + &i cos 5-5; + &2 cos 2^-^ + &c. Here 60 
is evidently the mean value of y from a? = to a; = c. In the 

* Exercise. Develope y—mx from a;=0 to .r=c in a series of sines. 

mx — a^ sin qx + a^ sin ^qx + Ac, where g- = - , 

«8 IS ~ I mx . sm sg^a; . «« = -g^i" ^^^ ^^^ ~ *5^ • ^^^ -""/^ • 

For this integral refer to (70) page 365. 


2mc f IT 1 . 27r 1 . Stt „ \ 

'= 1 sin-a;--8in — a;+-sin — x- &c. j. 

TT \ c 2 c 6 c J 


same way as before we can prove that hg is twice the average 
vakie of t/ cos sqx*. 

138. In Art. 118 we gave the equation for an electric 
circuit. The evanescent term comes in as before but we shall 
neglect it. Observe that if V is not a simple sine function of 
t, but a complicated periodic function, each term of it gives 
rise to a term in the current, of the same period. Thus if 

F=Fo + SF,8in(5^^ + 6'«)'^ (1), 

R \IB? + X V^2 V ^ R ] ^ ^ 

If Lq is very large compared with R we may take 

^■^i-^Isq''''^^''^^^''^ •••^^^• 

Thus, taking the make and break curve for V, fig. 76, 

F=F„+ — \Hmqt-\■\sm^t^&c.) (4), 


V 2VT 
^=^- ^«^ (^os qt-hi cos 3^^ + ^ cos 5qt + &c.) . . .(5), 


which is shown by the curve of fig. 73, being at j . 

139. When electric power is supplied to a house or 
contrivance, the power in watts is the average value of CV 
where G is current in amperes and V the voltage. 

* Thus let y=ma? between x=0 and x = c. Evidently bQ=^mc, and we 

^ , tmr 4wt / 1 „ 1m „ \ 

find y~~o ( cos9a; + -co8 3gx + 2^cos5gar + <S:c. J . 

There are many other normal forms in which an arbitrary function of 
X may be developed. Again, even of sines or cosines there are other forma 
than those given above. For example, if we wish generally to develope y a 
function of x between and c as y = Sa^ sin a„^x by the Fourier method, the 

essential principle of which is I sin a,j.T . sina^-r . rfa;=0, where m and n 

are different ; we must have o„ and o-, roots of — ^.— - =a. In the ordinary 
Fourier series s is oo . 

i — =: COS e — tan^ — 

v/r2 4- pQ"^ V rj 


Let F= Fo sin qt and 0—0^ sin (g'^ — e). 

Then P = \GqVq cos ef, or half the product of the ampli- 
tudes multiplied by the cosine of the lag. When the power 
is measured by passing G through one coil of a dynamometer 
and allowing F to send a current c through the other coil, if 
this coil's resistance is r and self-induction I 

c= ,^' - sinfg^-tan-^^) (6). 

What is really measured therefore is the average value 
of Gc, or 


Usually in these special instruments, large non-inductive 
resistances are included in the fine wire circuit and we may 
take it that Iq is so small in comparison with r that its square 
may be neglected. If so, then 

cos le — tan~^~) 
apparent power V rJ 

true power cos e 

Observe that tan~^ — is a very small angle, call it a, 

apparent power cos e cos a + sin e sin a 

-E-f = = cos a -H sm a . tan e. 

true power cos e 

Now cos a is practically 1, and sin a is small, and at first 
sight it might seem that we might take the answer as 
nearly 1. 

But if e is nearly 90° its tangent may be exceedingly 
large and the apparent power may be much greater 
than the true po'virer. 

It is seldom however that e approaches 90° unless in coils 
of great diameter with no iron present, and precautions taken 
to avoid eddy currents. Even when giving power to a 
choking coil or unloaded transformer, the effect of hysteresis 
is to cause e not to exceed 74°. 

140. True Power Meter. Let EG and GD be coils 
wound together as the fixed part of a dynamometer, and let 

P. 14 





<#^c _.^ 

DB be the moveable coil. The current C + o passes from E 

to G. Part of it c goes along the 
non-inductive resistance GF which. 
has a resistance R. The part G 
flows from G to D and D to B and 
through the house or contrivance. 
The instantaneous value of Rc.C 
is the instantaneous power. 

The coils EG and GD are care- 
fully adjusted so that when c = 
and the currents are continuous 
currents, there shall be no deflec- 
tion of the moveable coil DB. Hence the combined action 
of C + c in EG and of C in GD upon C in DB is force 
or torque proportional to cG, and hence the reading of the 
instrument is proportional to the power. With varying 
currents also there will be no deflection if there is no metal 
near capable of forming induced currents. 

Fig. 78. 


o ^ o 



141. The student ought to get accustomed to translating 
into ordinary language such a statement as 
(1) of Art. 119. Having done so, consider a 
mass of W lb.* hanging from a spring whose 
stiffness is such that a force of 1 lb. elongates 
it h feet. If there is vibration ; when W is at 
the level CO, fig. 79, a? feet below (we imagine 
it moving downwards) its position of equili- 
brium 00, the force urging it to the position 
of equilibrium is x-^h pounds, and as the 

moving mass is — (neglect the mass of the 

spring itself or consider one-third of it as being added to the 
moving body), 

W . X 

— X the acceleration = r . 
9 h 

The acceleration = -t4t • The acceleration is then pro- 

* The name W lb. is the weight of a certain quantity of stuff ; the inertia 
of it in Engineers' units is W-i-S2'2. 

Fig. 79. 


portional to x, and our ^ stands for g-- in (1) of Art. 119, and 

(2) shows the law connecting a) and t 

Notice carefully that the + sign in (1) is correct. The 
body is moving downwards and x is increasing, so that doo/dt 

is positive. But -j— is negative, the body getting slower in 

its motion as a) increases. 

142. Imagine the body to be retarded by a force which 

is proportional to its velocity, or b -r. • Observe that this acts 

as J acts, that is upwards, towards the position of equi- 


Hence we may write 

Wd^x dx X 
Jdi^^^di-^h'^^ (^>- 

We shall presently see what law now connects x and t in 
this damped vibration. 

143. Suppose that in the last exercise, when the body is 
displaced x feet downwards, its point of support B is also y feet 
below its old position. The spring is really only elongated 

by the amount x—y, and the restoring force is —7-^ • Con- 
sequently (1) ought to be 

Wd'x dx x_y 

J dt'^^ drh'h ^^^' 

Now imagine that the motion y is given as a function of 
the time, and we are asked to find a; as a function of the time. 
y gives rise to what we call a forced vibration. If 2/ = 
we have the natural vibrations only. 

We give this, not for the purpose of solving it just now, 
although it is not difficult, but for the purpose of familiar- 
izing the student with differential equations and inducing 
him to translate them into ordinary language. 



144. Notice that if the angular distance of a rigid body 

from its position of equilibrium is 0, if / is its moment of 

inertia about an axis through the centre of gravity, if H6 is 

the sum of the moments of the forces of control about the 

same axis, and if jP vr is the moment of frictional forces 

which are proportional to velocity, 

4'+^!+^^=^^' <'^)' 

if B' is the forced angular displacement of the case to which 
the springs or other controlling devices are attached. 

145. The following is a specially good example. Referring 
back to Example 1 of Art. 98, we hud CR, the voltage in 
the circuit, connecting the coatings of the condenser. If we 
take into account self-induction L in this circuit, then the 
voltage V is ♦ 

^C' + ii = '' w- 

We may even go further and say that if there is an 
alternator in the circuit, whose electromotive force is e at any 
instant (e, if a constant electromotive force would oppose G 
as shown in the figure) 

RG + L^ = v-e (5). 

But we saw that the current G =^ — K-jj (6). 

Using this value of G in (5) we get 

LK^ + RK^+v = e (7). 

Kow imagine that e is given as a function of the time and 
we are asked to find t; as a function of the time. 

e gives rise to what we call a forced vibratory current 

in the system. If e = we have the natural vibrations only 
of the system. Having v, (6) gives us (7. 


146. If (7) is compared with (2) or (3) we see at once 
the analogy between a vibrating mechanical system 
and an electrical one. 

They may be put 

W d^x ..dx X y ^f , . - 

^S+^S+^=i'E^-*'^-i (^)- 

The mass — corresponds with self-induction L. 

The friction per foot per second &, corresponds with the 
resistance R. 

The displacement x, corresponds with voltage v, or to be 
seemingly more accurate, v is Q the electric displacement 
divided by K. 

The want of stiffness of the spring h con-esponds with 
capacity of condenser K. 

The forced displacement y corresponds with the forced 
E.M.F. of an alternator.! 

147. The complete solution of (8) or (9), that is, the 
expression of x or i; as a function of t, will be found to 
include: — 

(1) The solution if i/ or e were 0. 

This is the natural vibration of the system, which dies 
away at a rate which depends upon the mechanical friction 
in the one case and the electrical friction or resistance in the 
other case. We shall take up, later, the study of this vibra- 
tion. It ought to be evident without explanation, that if y 
or e is 0, we have a statement of what occurs when the 
system is left to itself 

(2) The solution which gives the forced vibrations 

The sum of these two is evidently the complete answer.! 

148. Forced Vibration. As the Mechanical and 
Electrical cases are analogous, let us study that one about 


which it is most easy to make a mental picture, the mechanical 
case. We shall in the first place assume no friction and neglect 
the natural vibrations, which are however only negligible 
when there is some friction. Then (8) becomes 

dt?^wi.'—my <i*^>- 

Let y = a sin qt be a motion given to the point of support 
of the spiral spring which carries W ; y may be any compli- 
cated periodic function, we consider one term of it. 

We know tha t if y were 0, the natural vibration would 
be a; = 6 sin f i 4/ ^-jj + ^^^ ) , where h and m might have any 

values whatsoever. It is simpler to use n"^ for glWh as 
we have to extract its square root, n is 27r times the 
frequency of the natural vibrations of W. We had better 
write the equation as 

— -f i^^x = n^y = n-a sin qt (11). 

Now try if there is a solution, 00 = A r,in qt + B cos qt If so, 

since ,- = — Aq- sin qt — Bq^ cos qt ; equating the coefficients 

of sinqt and also those of cosqt, — Aq- -[■ n^A = n^a, so that 

A =— — ^ , and — Bq^ + ii^B = 0, so that B = unless n = q, 
n^ — q^ ^ * 

We see that we have the solution 

^^n^Zrjf^'^'i^ (12)- 

This shows that there is a forced vibration of W which is 
synchronous with the motion of the point of support ; its 

amplitude being , times that of the point of support. 

Now take a few numbers to illustrate this answer. Let a = 1, 

let " be great or small. Thus ^ = -X means that the forced 
n ^ n ^^ 

frequency is one tenth of the natural frequency. 




Amplitude of j 




W 's motion j 


JF's motion 


101 1 














- 4-762 


1026 ; 


- 0-8 




- 0-333 




- 0042 




- 0010 

Note that when the forced frequency is a small fraction 
of the natural frequency, the forced vibration of TT is a 
faithful copy of the motion of the point of support B ; the 
spring and W move like a rigid body. When the forced is 
increased in frequency the motion of TT is a faithflil magni- 
fication of B's motion. As the forced gets nearly equal to the 
natural, the motion of W is an enormous magnification 
of B's motion. There is always some friction and hence the 
amplitude of the, vibration cannot become infinite. When 
the forced frequency is greater than the natural, W is always 
a half-period behind B, being at the top of its path when 
B is at the bottom. When the forced is many times the 
natural, the motion of W gets to be very small ; it is nearly 
at rest. 

Men who design Earthquake recorders try to find a 
steady point which does not move when everything else is 
moving. For up and down motion, observe that in the last 
case just mentioned, W is like a steady point. 

When the forced and natural frequencies are nearly equal, 
we have the state of things which gives rise to resonance 
in acoustic instruments ; which causes us to fear for suspen- 
sion bridges or rolling ships. We could easily give twenty 
interesting examples of important ways in which the above 
principle enters into engineering problems. The student 
may now work out the electrical analogue for himself and 
study Hertz' vibrations. 

149. Steam engine Indicator vibration. The 

motion of the pencil is to faithfully record the force of 


the steam on the piston at every instant ; this means that 
the natural vibrations of the instrument shall be very quickly 
destroyed by friction. Any friction as of solids on solids will 
cause errors. Indeed it is easy to see that solid friction 
causes diagrams to be always larger than they ought to be. 
Practically we find that if the natural frequency of the 
instrument is about 20 times that of the engine, the diagram 
shows few ripples due to the natural vibrations of the indi- 
cator. If the natural frequency is only 10 times that of the 
engine, the diagram is so ' upset ' as to be useless. 

The frequency of a mass — at the end of a spring whose 


yieldingness is hy see Art. 141, is ^ ^gj Wh, neglecting friction. 

We shall consider friction in Art. 1 GO. What is the frequency 

of a mechanism like what we have in an indicator, controlled 

by a spring ? Answer : If at any point of the indicator 

mechanism there is a mass - , and if the displacement of 

this point is .9, when the displacement of the end of the spring 

(really the piston, in any ordinary indicator) is 1 ; imagine 

w w 

that instead of — we have a mass s^ — at the end of 


the spring. Thus the frequency is ^ a/ ^ 

To illustrate this, take the case shown in fig. 80 ; GAB is 
a massless lever, hinged at 0, with 
S- the weight W at B. The massless 

e spring is applied at A. 

I When A is displaced downwards 

from equilibrium through the dis- 
tance x, the extra pull in the spring 



is -r . The angular displacement of 

^ig- 80. the lever, clockwise, is j-j . Mo- 


ment of Inertia x angular acceleration, is numerically equal 

to moment of force. The Moment of Inertia is — OB^, 



The angular acceleration is yr-j , where x stands for -^ , so 

that -0i?^;^ + 7. 0^ = 0, 
g OA h 

^+^.^.^ = 0. 

And yya ^^ what we called 5, so that s^W takes the place 

of our old W when W was hung directly fronri the spring. 

150. Vibration Indicator. Fig. 81 shows an in- 
strument which has been used for indicating quick vertical 
vibration of the ground. 


c o a o 

Fig. 81. 

The mass GPQ is supported at F by a knife edge, or by 
friction wheels. The centre of gravity G is in a horizontal 
line with P and Q. Let FG=a, GQ=:h, PQ = a+b=l 
The vertical spring AR and thread BQ support the body at 
Q. As a matter of fact AR is an Ayrton-Perry spring, which 
shows by the rotation of the pointer R, the relative motion of 
A and Q; let us neglect its inertia now, and consider that 
the pointer faithfully records relative motion of A and Q. 
It would shorten the work to only consider the forces at P 
and Q in excess of what they are when in equilibrium, but for 
clearness we shall take the total forces. 

When a body gets motion in any direction parallel to 
the plane of the paper, we get one equation by stating 
that the resultant force is equal (numerically) 
to the mass multiplied by the linear acceleration 
of the centre of gravity in the direction of the 
resultant force. We get another equation by stating 


that the resultant moment of force about an axis 
at right angles to the paper through the centre of 
gravity is equal to the angular acceleration, multi- 
plied by moment of inertia about this axis through 
the centre of gravity. I shall use x, x and ic to mean 

fj nf* Cm IT 

displacement, velocity and acceleration, or x, -j- and -v- . 

Let P and A get a displacement x^ downward. Let Q 
be displaced x downward. Let the pull in the spring be 
Q = Qq-\- c{x — Xi) where c is a known constant (c is the 
reciprocal of the h used in Art. 141). Let W be the weight 
of the body. Then if Pq and Qo be the upward forces at the 
points marked P and Q when in the position of equilibrium, 

Q„(a + Z>)= Wa and Po+Qo= W. 

J, bW ■ aW 
Hence ^o = ^^^, Qo = ^-7-^ (1), 

Q=Qo + c{x-x,). 

Now G is displaced downwards =- Xi H ? x, so that 

^ a+6 a+b 

W-P-Q='^{b-£, + aB]^^ (2). 

The body has an angular displacement clockwise about its 
centre of mass, of the am< 
moment of inertia about G 

X -— X 

centre of mass, of the amount r« So that if / is its 


--Qb + Pa = ^^^^^(x-x,) (3). 

Hence (2) and (3) give us, if M stands for — , and if 

/ = Mk^ where k is the radius of gyration about G, 


If ki is the radius of gyration about P, we find that (4) 
simplifies to 

I [^ 
if n stands for tta/ ir?= ^tt x natural frequency, and e^ stands 

for 1 — r-2 • Call X — Xihy the letter y because it is really y 

that an observer will note, if the framework and room and 
observer have the motion x^. Then 2^ y = x — XiOv x = y-\-Xx 

y + Xi-\- n^ (y + ^'i) = ^Xi 4- n^Xi . 

So that y + n^y = (^-l)xj_ (5), 

or y + w'2/ + p^i = ^ (6)- 

Let Xj^ = A sin qt. 

We are neglecting friction for ease in understanding our 
results, and yet we are assuming that there is enough 
friction to destroy the natural vibration of the body. 

We find that if we assume y = a sin qt, then 
_ al ([- . 

That is, the apparent motion y (and this is what the 
pointer of an Ayrton- Perry spring will show ; or a light 
mirror may be used to throw a spot of light upon a screen), 

is T-« -r^ — ^ times the actual motion of the framework and 
A-j* n^ — q^ 

room and observer. If q is large compared with n, for 

example if q is always more than five times n, we may take 

it that the apparent motion is y-^ times the real motion and 

is independent of frequency. Hence any periodic motion 
whatever (whose periodic time is less say than Jth of the 
periodic time of the apparatus) will be faithfUlly indi- 

Note that if al ~ Tc^ so that Q is what is called the point 
of percussion, Q is a motionless or * steady ' point. But in 
practice, the instrument is very much like what is shown in 


the figure, and Q is by no means a steady point. Apparatus of 
the same kind may be used for East and West and also for 
North and South motions. 

151. Any equation containing ■— or -^ or any other 

differential coefficients is said to be a " Difierential 
Equation/' It will be found that differential equations 
contain laws in their most general form. 

Thus if o) is linear space and t time, the statement -f^ = 

d^x . ^^ 

means that ,- , (the acceleration), does not alter. It is the 

most general expression of uniformly accelerated motion. 

When we integrate and get -7^= ci, we have introduced 

the more definite statement that the constant accelei-ation is 
known to be a. When we integrate again and get 


^ = at + h, 

we are more definite still, for we say that h is the velocity 
when ^ = 0. 

When we integrate again and get 
X = ^af -i-bt-^-c, 
we state that x = c when ^ = 0. 

Later on, it will be better seen, that many of our great 
general laws are wrapped up in a simple looking expression 
in the shape of a differential equation, and it is of enormous 
importance that when the student sees such an equation he 
should translate it into ordinary language. 

152. An equation like 

g-^S-«2-^l-^=^ «. 

if Py Q, R, S and X are functions of x only, or constants, is 
said to be a linear differential equation. 

Most of our work in mechanical and electrical engineering 
leads to linear equations in which P, Q, &c., are all constant 
with the exception of X. Thus note (8) and (9) of Art. 146. 


Later, we shall see that in certain cases we can find the 
complete solution of (1) when X is ; that is, that the solu- 
tion found will include every possible answer. Now suppose 
this to be y =/(*'). We shall see that it will include four 

arbitrary constants, because ^^ is the highest differential 

coefficient in (1), and we shall prove that if, when X is not 0, 
we can guess at one solution, and we call it y= F (os), then 

j,=f{x) + F(x) 

is a solution of (1). We shall find in Chap. III. that this is 
the complete solution of (1). 

In the remainder of this chapter we shall only consider 
P, Q, &c., as constant ; let us say 

da^^^d^^^d^^^dx^^y-^ ^^^' 

where A, B, G, E are constants and Xis a function o{ a;. 
We often write (2) in the form 

153. Taking the very simplest equation like (3). Let 

J-«y = o • W' 

it is obvious (see Art. 97) that 

y = Me^'' (5) 

is the solution, where M is any constant whatsoever. 

154. Now taking ^-«'.y = (6), 

we see by actual trial that 

2/ = if €«* + i\r6-«^ (7) 

is the solution, where M and N are any constants whatsoever. 

But if we take ^^n^y = (8), 


we see that as the a of (6) is like ni in (8) if i means V— 1, 

y =: Me''^ -\- Ne-""^ (9) 

is the sohition of (8). If we try whether this is the case, by 
difFerentiat.Ton, assuming that i behaves like a real quantity 
and of course i^ = — 1, i^ = — i, i* = l, {^ = i, &c., we find that 
it is so. But what meaning are we to attach to such an 
answer as (9) ? By guessing and probably also through re- 
collection of curious analogies such as we describe in Art. 106, 
and by trial, we find that this is the complete solution also, 

y — Mi sin tix + N^ cos nx (10). 

As (10) and (9) are both complete solutions (Ai-t. 152) be- 
cause they both contain two arbitrary constants which may be 
unreal or not, we always consider an answer like (9) to be the 
same as (10), and the student will find it an excellent exercise 
to convert the form (10) into the form (9) by the exponential 
forms of sinew; and coscu?, Art. 106, recollecting that the 
arbitrary constants may be real or unreal. Besides, it is im- 
portant for the engineer to make a practical use of those 
quantities which the mathematicians have called unreal, 

155. Going back now to the more general form (3) when 
X = O, we try if 2/ = Me'^ is a solution, and we see that it is 
so if 

, m^ + Am^ -\- Bni" -{■ Gm-\- E = ^ (1). 

This is usually called the auxiliary equation. Find the 
four roots of it, that is, the four values of m which satisfy it, 
and if these are called 7?ii, "nfi^y Wg, m^, we have 

as the complete solution of (3) when X = 0; M^, &c., being 
any arbitrary constants whatsoever. 

156. Thus to solve 

if we assume y = e*"*, we find that m must satisfy 
?M^ + oiit^ -f om^ — om — 6 = 0. 


By guessing we find that m = l is a root; dividing by 
m — 1 and again guessing, we find that m= — l is a root ; 
again dividing by m + 1 we are left with a quadratic expres- 
sion, and we soon see that m= — 2 and m = — 3 are the 
remaining roots. Hence 

is the complete solution, il/i, i/s) &c., being any constants 

157. Now an equation like (1) may have an unreal root like 
7n + 7ii, where t is written for J— 1, and if so, we know from 
algebra, that these unreal roots go in pairs ; when there is 
one like m + ni there is another like ni — ni. The corre- 
sponding answers for y are 

y — M € ('"~"^' ^ 4- N 6 <"*+*"^ * 
or e"*^ {i¥i€-^'^ + iVi€+"'*), 

and we see from (10) that this may be written 

y = e"** {31 sin nx + N cos nx], 
where M and iV" are any constants whatsoever. 

158. Suppose that two roots m of the auxiliary equation, 
happen to be equal, there is no use in writing 

because this only amounts to (M^ + M,^) e*"^ or ife**^ where M 
is an arbitrary constant, whereas the general answer must 
have two arbitrary constants. In this case we adopt an 
artifice ; we assume that the two roots are m and 7n + h and 
we imagine h to get smaller and smaller without limit : 

= e"'^ (Ml -H if^e^^^), 
but by Art. 97, e^ = 1 + /^A- + ^ + j^3 4- &c., 

therefore y = e"^^ (m^ + M^ + MJiw + M^ ^ + ^c) - 

Now let i/2^ be called N and imagine h to get smaller 


and smaller, and M2 to get larger and larger, so that MJi may 
be of any required value we please, say N, and also 

as h gets smaller and smaller without limit we find 

If this reasoning does not satisfy the reader, he is to 
remember that we can test our answer and we always find it 
to be correct. 

159. It is in this way that we are led to the fol- 
lowing general rule for the solving of a linear differential 
equation with constant coefficients. Let the equation be 

fl + A^^+Bf;^, + S^. +G^^ + % = 0...(1). 

Form the auxiliary equation 

m** + ^m^-i + ^m'*-^ + &c. + Gm + iT = 0. 

The complete value of y will be expressed by a series of 
terms : — For each real distinct value of m, call it a, , there will 
exist a term Mie*»'*'; for each pair of imaginary values a^ ± ySgi, 
a term 

6»*^ (ilfa sin ^^ + iVs cos ^.^) ; 

each of the coefficients M^y M^, N2 being an arbitrary 
constant if the corresponding root occurs only once, but a 
polynomial of the r— 1th degree with arbitrary constant 
coefficients if the root occur r times. 

Ea^rcue. g+ 12^ + 66 f|+ 206 g 
aa:^ dx* dx^ do(? 

+ 345^ + 234y = 0. 

Forming the auxiliary equation, I find by guessing and 
trying, that the five roots are 

- 3, - 3, - 2, - 2 + 3i, - 2 - 3i. 

Consequently the answer is 

y = (ifi + N^x) €-^ + M^e-"^ + €-^ (i/g sin Sx + N, cos 3^). 


Exercise. 1. Integrate ^t^ - 4 ^ 4- 3y = 0. 
Answer : y= A^+ Bg^. 

% lutegrate g-loJ+34y = 0. 

Answer : y = e^ {A sin ^x + B cos 3^j. 

3. Integrate g + 6| + 9, = 0. 
Answer : y = (^ + 5i») e~^. 

4 Integrate g-l^g + esg- lo6|+ 169, = 0. 

Here vv' - 12m^ + 627?t2 - Ib^m + 169 = 0, and this will be 
found to be a perfect square. The roots of the auxiliary 
equation will be found to be 

3 + 2i, 8 + 2'i, 3 - 1i, 3 - 2i 
Hence the solution is 

y = 6^* {(^1 + B^x) sin 2*- + (^2 + BrO)) cos 2^j. 

We shall now take an example which has an important 
physical meaning. 

Natural Vibrations. Example. 

160. We had in Art. 146, a mechanical system vibrating 
with one degree of fi-eedom, and we saw that it was analogous 
with the surging going on in an Electric system consisting of a 
condenser, and a coil with resistance and self-induction. We 
neglected the friction in the mechanical, and the resistance 
in the electric problem. We shall now study their natural 
vibrations, and we choose the mechanical problem as before. 
If a weight of TTlb. hung at the end of a spring which 
elongates x feet for a force of x -^Ji lb., is resisted in its 
motion by friction equal to 6 x velocity, then wc had (8) of 
Art. 146, or IL^ h^ --0 

d^x hq dx xq ^ ,^. 

-rfF + F-rf*+m = *^ (!>• 

i\ 15 


Let ^^ be called 2/ and let -^ = ?i= ; (1) becomes 

5+2/.J + n'. = (2). 

Forming the auxiliary equation we find the roots to be 

We have different kinds of answers depending upon the 
values of / and n. We must be given sufficient information 
about the motion to be able to calculate the arbitrary con- 
stants. I will assume that when t ia the body is at w = 
and is moving with the velocity Vq. 

I. Let / be greater than n, and let the roots be — a 
and - 13. 

IL Let /be equal to n, the roots are —/and — / 
IIL Let /be less than n, and let the roots be — a ± bi. 
IV. Let/= 0, the roots are ± ni. 
Then according to our rule of Art. 159, 
In Case I, our answer is 


^ = 0, we can calculate A and B and so find x exactly in 

terms of ^ ; 

In Case II, our answer is 

In Case III, our answer is 

X = €-«* {A sin U + B cos U] ; 
In Case IV, our answer is 

X — A^mnt'\- B cos nt. 

161. We had better take a numerical example and we assure 

the student that he need not grudge any time spent upon it 

and others like it. Let n = 3 and take various values of /. 

For the purpose of comparison we shall in all cases let a? = 

"*" dx 

when ^ = 0; and ^. = 20 feet per second, when i = 0. 

and if we are told that x = when < = and -y; = Vo when 


Case IV. Let/= 0, then a;== A sin nt + 5 cos nt, 
= AxO + Bxl, so that 5 = 0, 



= nA cos ?«^ — iiB sin n^, 

20 = SA, so that ^ = ^-. 
Plot therefore *•= 6*667 sin 3^. 

This is shown in curve 4, fig. 82. It is of course the 
ordinary curve of sines: undamped S.H. motion. 

Fig. 82. 

Case III. Lety = "S. The auxiliary equation gives 

771 = - -3 ± \/-09 - 9 = - -3 ± 2-985t. 

Here a = '3 and 6 = 2-985 in 

x^e-^'^lAsinU-tBcosbt} (1). 

You may not be able to differentiate a product yet, although 
we gave the rule in Art. 90. We give many exercises in 
Chap. III. and we shall here assume that 


— = - ae-"* (A sin ht + B cos bt) 

+ 6e-««(^cos6e-5sin60 (2). 



Put .£=0 when t = and -r: = 20 when t = 0. Then 


B = from (1) and 

and hence os = 6-7€-*«* sin 2-985^.* 

This is shown in curve 3 of fig. 82. Notice that the period 
has altered because of friction. 

Case II. Let/= 8. The roots of the auxiliary equation 
are ?/i = — 3 and — 3, equal roots. Hence 

^• = (^+^0^"'' (!)• 

Here again we have to differentiate a product and 

~ = B€-''-S{A+Bt)e-'^' (2). 

Putting in a; = when ^ = and -17 = 20 when ^ = 0, 
^ = from (1) and J5 = 20 from (2). 

Hence o) = 20^ . e-'K 

This is shown in curve 2 of fig. 82. 

Case I. Let f=o. The roots of the auxiliary equation 
are — 9 and -- 1, 

0! = Ae-''^ + Be-\ 


Putting in the initial conditions we have 

= A-hB, 20^-9A-B. 

Hence J. = - 21, 5 = 2^, 

«<'=2i (€-*-€-»«)• 

This is shown in curve 1 of fig. 82. 

Students ought to take these initial conditions 

OS =10 when t^O and -77 = when t = 0. 


This would represent the case of a body let go at time 
or, in the electrical case, a charged condenser begins to be 
discharged at time 0. 

Notice that if we differentiate (1), Art. 160, all across we 

have f using v for -j-) , 

d^v hq dv a ^ 
— ^ — P— ?) = 

dt^^ Wdt^ Wh 

We have therefore exactly the same law for velocity or 
acceleration that we have for x itself. 

Again, in the electrical case os K -r: represents current, 

if we differentiate all across we find exactly the same 
law for current as for voltage. Of course differences are 
produced in the solutions of the equations by the initial 

162. When the right-hand side of such a linear differential 
equation as (2) Art. 152 is not zero and our solution will give the 
forced motion of a system as well as the natural vibrations, it 
is worth while to consider the problem from a point of view 
which will be illustrated in the following simple example. 

To solve (11) Art. 148, which is 

-J- -\- ii^x — n^a sin qt (1), 

the equation of motion of a system with one degree of 
freedom and without friction. 

Differentiate twice and we find 

d*w ^d^x , , . . 

Hence from (1), ^+(712 + 52)^+^^^ = (2). 

To solve (2), the auxiliary equation is 

m^ + (ii' + 50m2 + g'2|i2^0 (3), 

and we know that + ni are two roots and + qi are the other 
two roots. Hence we have the complete solution 

«? = ^ sin nt + B cos nt -^-G sinqt-^ D cos qt (4). 



© @ © 

Now it was by differentiating (1) that we introduced 
the possibility of having the two extra arbitrary constants C 
and D, and evidently by inserting (4) in the original equation, 
we shall find the proper values of C and D, as they are really 
not arbiti-ary. It will be noticed that by differentiating 
(1) and obtaining (2) we made the system more complex, 
gave it another degree of freedom, or rather we made it 
part of a larger system, a system whose natural vibrations 
are given in (4). When we let a mass vibrate at the end of 
a spring, it is to be remembered that the centre of gravity of 
the mass and the frame which supports it and the room, 
remain unaltered. Hence vibrations occur in the supporting 
frame, and there is friction tending to still the vibrations. 
If there is another mass also vibrating, this effect may be 
lessened. For example in fig. 83, if M vibrates at the end of 

the strip MA, clamped in 
the vice A, any motion 
of M to the right must be 
accompanied by motion of 
A and the support, to 
the left. But if we have 
two masses Mi and il/g (as 
in a tuning fork), moving 
in opposite directions at 
each instant there need be 
no motion of the supports, 
consequently the system 
MjMo vibrates as if there 
were less friction, and this 
principle is utilized in 
tuning forks. Should a 
motion be started, different 
from this, it will quickly 
become like this, as any part of the motion which 
necessitates a motion of the centre of gravity of the 
supports, is very quickly damped out of existence. 
The makers of steam engines and the persons who use 
them in cities where vibration of the ground is objected to, 
find it important to take matters like this into account. 

163. If y is a known function of oc, we are instructed by 

Fig. 83. 


(3), Art. 152, to perform a complicated operation upon it. 
Sometimes we use such a symbol as 

{e'-\-Ae'-{-Bd^+ce+E)y = x, 

to mean exactly the same thing ; 6y meaning that we differ- 
entiate y with regard to x, 6^y meaning that we differentiate 
y twice, and so on. 

6, 6^, &c., are symbols of operation easy enough to under- 
stand. We need hardly say that 6'^y does not mean that 
there is a quantity 6 which is squared and multiplied upon 
y: it is merely a convenient way of saying that y is to be 
differentiated twice. 66y would mean the same thing. 
On this same system, what does (6 -\-a)y mean ? It means 

^+ay. What does (6'' + A0 + B)y mean? It means 

(Py dy 

~T^ -}- -4 1^ -h By. (6-{-a)y instructs us to differentiate y and 

add a times y, for a is a mere multiplier although is not so, 
and yet, note that (6 -\-a)y= 6y + ay. 

In fact we find that 6 enters into these operational 
expressions as if it were an algebraic quantity, although it is 
not one. 

If u and V are functions of x we know that 

e {u + v)= 6u + Ov. 

This is what is called the distributive law. 

Again, if a is a constant, 6au = a6u, or the operation 6a is 
equivalent to the operation aO. This is called the com- 
mutative law. 

Again 6^6^ — 6^+^.^ ^\^\^ jg ^j^^ index law. When these 
three laws are satisfied we know that 6 will enter into 
ordinary algebraic expressions as if it were a quantity. 6 
follows all these laws when combined with constants ; but note 

that if u and v are functions of x, vdu meaning v -j- , is a 

very different thing from 6 , uv. When we are confining our 


attention to linear operations we are not likely to make 

Thus operate with 6-{-b upon (0 -f a) y. Now 

{e-\-a)y=-ey-\-ay or ^ + ay. 
Operating with 6 ■\-h means "diflFerentiate (this gives us 
/7/S"^^^) ^^^ ^^^ ^ times -^- + ay" Consequently it gives 

We see, therefore, that the double operation 

{e + h){d + a) 

gives the same result as 


In this and other ways it is easy to show that although 6 ia 
a symbol of operation and not a quantity, yet it enters into 
combinations as if it were an algebraic quantity, so long as all 
the quantities a, h, &c. are constants. Note also that 

is the same as {0 + h){d -\- a). 

The student ought to practise and see that this is so and 
get familiar with this way of writing. He will find that it 
saves an enormous amount of unnecessary trouble. Thus 
compare such expressions as 

with jaa^-^ + (a^ +ah)e + h^} y, 

164. Suppose that Dy is used as a symbol for some curious 
operation to be performed upon y, and we say that Dy = X ; 
does this not mean that if we only knew how to reverse the 


operation, and we indicate the reverse operation by D~^ or p. , 
then y = I)~^ X or yy ? We evidently mean that if we operate 
with D upon D~^X, we annul the effect of the D~^ operation. 
Now if -^ + ay = X , or l-j- -\- a] y — X , or {6 ■{■ a) y =X , let us 
indicate the reverse operation b}^ 

y-^^x-^^y^^^i^+^rx (1), 

X X ,^^ 

or , or ^ (2). 


Keeping to the last of these; at present ^ is a mere 

symbol for an inverse operation, but v-\-a 

y=-^a (^> 

submits to the usual rules of multiplication, because (3) is 
the same as (6 + a)y= X (4); 

and yet (4) is derived from (3) as if by the multiplication of 
both sides of the equation by {6 + a), 

Again,take^ + (a + 6)^+a62/ = Z (5), 

or {e^+{a-{-h)e + ab]y=^X (6), 

or (e + a)(d + h)y = X (7). 

Here the direct operation 6 + a performed upon (0 -{•h)y 
gives us X ; hence by the above definition 

(^+6)y=^„ (8). 

and repeating, we have 


~(e-hh)\d-\-aj "^^^' 


But it is consistent with our way of writing inverse operations 
to write (6) as 

^^e'-^(a+b)e+ab ^^^^' 

and so we see that there is nothing inconsistent in our treating 
the 6 -{-h and ^ + a of (9) as if ^ were an algebraic quantity. 

165. We know now that the inverse operation 

{^+(a+6)(9 + a61-^ (1), 

may be effected in two steps ; first operate with (6 -\- h)~^ and 
then operate with (6 + a)~\ 

Here is a most interesting question. We know that if 6 
were really an algebraic quantity, 

1^1 1 ^ .(2). 

^ + (a + 6) ^ + a6 b-a\d-\-a O-^b, 
And it is important to know if the operation 

irrs(5n"ff+b) ^^^' 

is exactly the inverse of ^ + (a + b) d + ab?...(4). Our 

only test is this ; it is so, if the direct operation (4) com- 
pletely annuls (3). Apply (3) to X and now apply (4) to 

the result; if we apply (4) to ^ X, we evidently obtain 

dX tf + ct 1 

(d-{-b)X or -T- + 6X ; if we apply (4) to j^v X we evi- 
dently obtain (^ + a) X or -^ + aX, and 


b — a 

We see therefore that (3) is the inverse of (4), and that we have 
the right to split up an inverse operation like the left-hand 
side of (2) into partial operations like the right-hand side of 
(2). We have already had a number of illustrations of this 
when the operand was 0. For it is obvious that if a^, oui, &c. 
are the roots of the auxiliary equation of Art. 159, it really 
means that 

e^'^A e>'-' + J5l9"-=^ + ka. ■\- GO ■\- H 
splits up into the factors {6 — ttj) {6 — a^), &c, 

f + 6X-(f + «X)^=X. 


Observe that if -J^ = X, or 6y = X, or y — -^, or y = 6~^X, 

the inverse operation ^^ simply means that X is to be inte- 
grated. Again, 0~^ means integrate twice, and so on*. 

* Suppose in our operations we ever meet with the symbols 6^ or d~^ 

or 6"^ &c., what interpretations are we to put upon them ? It is not very 
necessary to consider them now. Whatever interpretations we may put 
upon them must be consistent with everything we have ah"eady done. For 

example 6^ will be the same as 60^ and d~ -^ will be the same as d^ d~^ 

or 6~^6^. We have to recollect that all this work is integration and we use 
symbols to help us to find answers ; we are employing a scientific method of 
guessing, and our great test of the legitimacy of a method is to try if our 
answer is right; this can always be done. Most of the functions on 
which we shall be operating are either of the shape ^e"* or B sin bx or sums 
of such functions. Observe that 


if n is an integer either positive or negative. There is therefore, a likelihood 
that it will help in the solution of problems to assume that 

or that e^A6*^ = Aa^6^ (1). 

Again 0B sin bx = Bb cos bx = Bb sin. (bx + ^\ , 

e^B sin bx= - Bb^ sin 6a; = Bb^ sin (&a; + tt) , 

and ^ 0^B sin bx=Bb'' Bin (bx + n'^j (2). 

Evidently this is true when n is a positive or negative integer ; assume it 
true when n is a positive or negative fraction, so that 

623 sin bx = Bb2 sin f bx + ^^ j 


There are certain other useful functions as well as e*"* and si7i bx such 

that we are able to give a meaning to the effect of operating with 6^ upon 
them. It will, for example, be found, if we pursue our subject, that we shall 
make use of a function which is for all negative values of x and which is a 
constant a for all positive values of x. It will be found that if this function 
is called /(.r) then 

0if(x)=a4-x-^ (4), 

and the meaning of d^ or ^^ or ^~i or 9~^ &c. is easily obtained by 
differentiation or integration. The Mnemonic for this, we need not call it 

proof or reason, is d^x'^ =~=^~ x"^'^. Let w=i, m=0 and we have 
*^ \m-n ^ 

286 for engineers. 

166. Electrical Problems. Circuit with resistance 
R and self-induction L, 

let -^ be indicated by 0, then 


V=(R + Le)G or = 

R + Ld' 

In fact in all our algebraic work wo treat R + "LO as if it 
were a resistance. 

Condenser of capacity K farads. Let V volts, be voltage 
between coatings. Let G be current in amperes into the 
condenser, that is, the rate at which Q, its charge in coulombs, 

is increasing. Or ^ = 7^ = "7^ (-^^^ ^^' ^^ ^ ^ usually 

assumed to be constant, (7 = TT—r- . 


The conductance of a condenser is KO, therefore 

Hence the current into a condenser is as if the condenser 
had a resistance — ^. 

Fig. 84. 

Circuit with resistance, self-induction and capacity, fig. 84. 
All problems are worked out as if we had a total resistance 

R + I-^ + xFZJ (!)• 


dhx^=-. — ^x 2. But |- i has no meaning. Give it a meaning by assuming 

|_IL2 ~ 

that whatjs true of integers, is true of all numbers, and use gamma function 
of I or ! ^ which is Jir instead of l~^' It is found that the solutions 
effected by means of this are correct. 


167. In any network of conductors we can say exactly 
what is the actual resistance (for steady currents) between 
any point A and another point B if we know all the resist- 
ances 7\, r^, &c. of all the branches. Now if each of these 
branches has self-induction ^i, &c. and capacity Ki, &c. 

what we have to do is to substitute r^ + l^d -f t^-t, instead of 

Ti in the mathematical expressions, and we have the resistance 
right for currents that are not steady. 

How are we to understand our results ? However com- 
plicated an operation we may be led to, when cleared of 
fractions, &c. it simplifies to this; that an operation like 

a-\-b0-\- cB' 4- dO' + ed'+fd' + &c 

a' + b'd + c'0'' + d'e' + e'e'~tf'e'-{-&}c. ^ ^' 

has to be performed upon some voltage which is a function 
of the time. On some functions of the time which we have 
studied we know the sort of answer which we shall obtain. 
Thus notice that if we perform (a -}- bd -\- &}c.) upon e** we 

{a + b2 + col' + da^ + ea.^ +/a^ + &c.) e*^ (2). 

Consequently the complicated operation (1) comes to be a 
mere midtiplication by A and division by A', where A is the 
number a -f 6a + ca^ + &c. and A' is the number 

a + Vol + c V -|- &c. 

Again, if wc operate upon in sin {nt + e), observe that 

6"- would give — mn^ sin {nt + e). 


^ „ „ +mn^mn{nt-\-e\ 

and so on ; 


would give inn cos {nt + e). 

6^ „ „ — mu^ cos {nt + e). 

6^ „ „ + mn^ cos {nt + e). 

And hence the complicated operation (1) produces the same 
effect as -^ — ^^ , where 

p — a — cn^-hen* — &c., q=^b — dn- -^fn^ — &c. 
a = a' - c'n2 -\-e'n^ - &c., ^ = b' - dn" +fn' - &c. 


Observe Art. 118, that p + qO operating upon m sin (tit 4- e) 
multiplies the amplitude by Vp^ + q^ii^ and causes an advance 

of tan~^— . The student ought to try this again for himself, 

although he has already done it in another way. Show that 

(p + qO) sin nt = V^T^ sin (nt + tan-^ 2!!:^ . 

Similarly, the invei-se operation l/(a + ^6) divides the 

amplitude by Va" + yS^/i^ ^j^^j produces a lag of tan~^ — , and 



a labour-saving rule of enormous importance. 

168. In all this we are thinking only of the forced 
vibrations of a system. We have already noticed that 
when we have an equation like (1) or (2) Art. 152, the 
solution consists of two parts, say y —f(x)-{- F{x)\ where 
f{x) is the answer if X of (2) is 0, the natural action of 
the system left to itself, and F {x) is the forced action. 
If in (2) we indicate the operation 

{ d' . d^ ^ d'' ^ d „\ , ^ 

then D (y) = X gives us 

Where I)-^{0) gives /(^) and B-^(X) gives F(x). 

Thus if^+ay^O, or fj- + a]y = 0, or (^ + a)y = 0, we 

know Art. 97, that y = ^.e-**^. 



so that if ^■-\-ciy'= X, the complete solution is 

Hence we see that x is not nothing^ but is Ae' 

ELECTRICAL EXAM^lS^^^V , ^^ 239 

"We are now studying this latter paiX't^the forced part, only. 
In most practical engineering problems the exponential terms 
rapidly disappear. 

169. Thus in an electric circuit where V=(R + Ld) C, if 
V = Vq sin qt, 

we have already found the forced value of G, 

^_ Fosin^ 

and according to our new rule, or according to Art. 118, this 

0= . ZV s^^ f^^-tan-^^) ....(1). 

But besides this term we have one 


and according to the above rule (Art. 168) this gives a term 

A,e ^' (2). 

Or we may get this term as in Art. 97, 

iJ(7+X^ = 0,or ^=-j;0. 

This is the compound interest law and give's us the answer (2), 
and the sum of (2) and (1) is the complete answer. If we 
know the value of G when ^ = 0, we can find the value of the 
constant Ai ; (2) is obviously an evanescent term. 

Thus again, suppose V to be constant = Vq, 


R + ie- 


It is evident that G = ~ is the forced current, for if we 
Y ^ 

operate on G=-~ with R + LOwe obtain Fo, and the e vanes- 


cent current is always the same with the same R and L 
whatever Fmay be, namely Ai€~l^, 

The complete answer is then 



G^A,e-L^+^^ (2). 

Let, for example, C = when t — Oy then 
V V 

and (2) becomes (7= — " (l - e"!') (3). 

The student ought to take Fo=100, jK=1, L=1 and 
show how C increases. We have had this law before. 

170.. Example. A condenser of capacity K and a 
non-inductive resistance r in parallel; voltage V at 
their terminals, fig. 85. The two currents are c— V/r, 

C=KOV, and their sum is C+c=v(^^ + Ke\ or v(^-'~^^\ 


so that the two in parallel act like a resistance .. ,T^n * 

o g^c> 


Fig. 85. 

r '' 


C + c = — Vl + r'^K'^n'', sin ini + tan-^ rKn\ 

c = — sin nt, C = V^n sin ( n^ 4- ^ ) • 


171. A circuit with resistance, self-induction and 
capacity (fig. 86) has the alternating voltage V= Fosin z*^ 
established at its ends ; what is the current ? 

Answer, G= -—, and by Art. 167 

R + Ld + ^^ 

r- ^'^'^ ^^ V * 

l + RK.e + LK.e^ {l-LK7i') + RKe 

C = : Sin nt +K — tan ^ - — ^ rr » • 

^{l-LKnJ + R^KHi-' V 2 l^-LKnV 

The earnest student will take numbers and find out by 
much numerical trial what this means. If he were only to 
work this one example, he would discover that he now has a 
weapon to solve a problem in a few lines which some writers 
solve in a great many pages, using the most involved mathe- 
matical expressions, very troublesome, if not impossible, to 
follow in their physical meaning. Here the physical mean- 
ing of every step will soon become easy to understand. 

' Numerical Exercise. Take Fo = 1414 volts, K=l micro- 
farad or 10-«, R = 100 ohms and n = 1000, and we find the 
following effects produced by altering L. We give the 
following table and the curves in fig. 87 : 

Fig. 86. 

A BCD shows how the current increases slowly at first from 
A where Z = as X is increased, and then it increases more 
rapidly, reaching a maximum when Z= 1 Henry and diminish- 
ing again exactly in the way in which it increased. EFG 
shows the lead which at X = 1 changes rather rapidly to a 
lag. The maximum current (when LKn^=l) is the same as 
if we had no condenser and no self-induction, as if we had a 
mere non-inductive resistance R. It is interesting to note 
in the electric analogue of Art. 160 that this LKn^=l is the 
relation which would hold between L, K and n (neglecting the 

P. 16 



small resistance term) if the condenser were sending surging 
currents through the circuit R, L, connecting its two coatings. 

/y, ill 
















Lead of 



ciuTent, in 

current, in 

current, in 























2425 1 




1-961 1 




1-644 1 




1-414 ! 




1-240 ! 


















Lead of 

current, in 


















— 1 











8 ^ 


i ^ 





iu 3 














-2 '4 

100 o 

•6 -8 1-0 1-2 1-4 1-6 1-8 2-0 22 2-4 2-6 2-8 S'O 



Fig. 87. 

Experimenting with numbers as we have done in this example 
is much cheaper and much more conclusive in preliminary 
work on a new problem, than experimenting with alternators, 
coils and condensers. 


172. Even if a transformer has its secondary open there 
is power being wasted in hysteresis and eddy currents, and the 
effect is not very different from what we should have if there 
was no such internal loss, but if there was a small load on. 
Assume, however, no load. Find the effect of a con- 
denser shunt in supplying the ''Idle Current." 

The current to an unloaded transformer, consists of the 
fundamental term of the same 
frequency as the primary voltage, 
and other terms of three and five 
times the frequency, manufactured 
by the iron in a curious way. 
With these "other terms" the con- 
denser has nothing to do ; it cannot 
disguise them in any way ; the total 
current always contains them. We shall not speak of them, 
as they may be imagined added on, and this saves trouble, 
for if the fundamental term only is considered we may 
imagine the permeability constant ; that is, that the primary 
circuit of an unloaded transformer has simply a constant 

In fact between the ends of a coil (fig. 88) which has 
resistance R and self-induction L, place a condenser of capacity 
K. Let the voltage between the terminals, be F= Fo sin nt. 
Let G be the instantaneous current through the coil and 
let c be the current through the condenser, then + c is the 
current supplied to the system. 

Fo sin nt 







Fig. 88. 

Now G: 


and c = Fo sin ??^ -7- -^^ , or c = Kn Fo cos nt, 


_ 1 + RKd -H LKd' ^_ l-LK n' + RK.e y 
R + LO R + Ld 

by our rule of Art. 167. 

It is quite easy to write out by Art. 167 the full value of 




+ c, but as we are not concerned now with the lag or lead, 
we shall only state the amplitude. It is evidently 

"V R' + JJn^ 

and the effective value of (7 + c (what an ammeter would 
give as the measure of the cuiTent), is this divided by \/2. 
Observe that C + c is least when 

(Note that if L is in Henries and n = 27r x frequency (so 
that in practice n = about GOO), K is in farads. Now even a 
condenser of J microfarad or J x 10"" farad costs a number of 
pounds sterling. We have known an unpractical man to 
suggest the practical use of a condenser that would have cost 
millions of pounds sterling.) 

When this is the case, the effective current C + c, is 
R/JR^ + L'^n- times the effective value of G. 

The student ought to take a numerical case. Thus in an 
actual Hedgehog Transformer we have found iJ = 24 ohms, 
L = 6'23 Henries 7i = 509, corresponding to a frequency o^ 
about 81*1 per second. The effective voltage, or Fo-rV2 
is 2400 volts. In fig. 89 we show the effective current cal- 
culated for various values of K. The current curve ABCD is 
a hyperbola which is undistinguishable except just at the 






















































•* -5 -6 -7 -8 -9 10 M 


•2 1-3 1-4 \-h 

Fig. 89. 

vertex, from two straight lines. The total current is a 
minimum when K — Lj(R? -^ LSi^)^ in this case 618 micro- 


farad ; and the effect of the condenser has been to diminish 
the total cun-ent in the ratio of the resistance to the im- 
pedance. It is interesting on the curve to note how the 
great lag changes very suddenly into a great lead. 

173. If currents are steady and if points A and £ are 
connected by parallel resistances 7\, 7\, r^y if V is the 
voltage between A and B, and if the three currents are 
Ci, C2, C3, and if the whole current is (7; then 
_F _7 _F 

In fact the three parallel conductors act like a conductance 

1 1 IN 

- + -■ + - 

Also if G is known, then 



i\ r^ Vs 

Now let there be a self-induction I and a condenser of 
capacity k in each branch, and we have exactly the same 
instantaneous formulae if, for any value of r, we insert 

The algebraic expressions are unwieldy, and hence nu- 
merical examples ought to be taken up by students. 

174. Two circuits in parallel. They have resistances 
Vi and ?-2 and self-inductions l^ 

and I2 . how does a total current ASL^i^i&SliSiSi 

C divide itself between them ? 


If the current were a con- 
tinuous current, Cj (fig. 90) in 
the branch 7\ would be " p- ' qq 


TT V • r.+ LO ,^ 

Jtieiice it IS now Ci = -. rz~n ^. 

If a= Oo sin nt, then by Art. 167 

'^'"''"Vcn + ny + i/i + Lyrr 

sin «^ + tan^ — tan^ -^^ 

In the last case suppose that for some instrumental 
purpose we wish to use a branch part of C, but with a 

lead. We arrange that tan~^ tan~^ — ^^ shall be 

7*2 ?•! + ?•, 

equal to the required lead, and we use the current in the 

branch r^ for our purpose. 

175. Condenser annulling effects of self-induction. 
When the voltage between points A and B follows 
any law w^hatever^ and we wish the current flowing into 


A and out at B to be exactly -^ , whatever V may be, and 

when we have already between A and B a coil of resistance 
R and self-induction L, show how to arrange a condenser 
shunt to effect our object. 

Connect A and 5 by a circuit containing a resistance ?•, 
self-induction I and condenser of capacity K, as in fig. 91. 

The total current is evidently 

V V 

r 1 k '^ I ^ . 

M-hLd^ ^j. 1 ' 
r^ld + j^^ 


R L or, bringing all to a common 

Fig. 91. denominator and arranging 

terms, it is 

1 + e (vK + RK) -^e^-{lK + LK) y 

R+e (RrK -\-L)+&' {RIK -I- LrK) + LIKO' 

Observe that as V may be any function whatsoever of the 
time Ave camiot simplify this operator as we did those of 


Art. 167. Now we wish the effect of the operation to be the 
same as p F! Equating and clearing of fractions we see that 

R + e (BrK+ B?K) + &" (IilK+ RLK) 

must be identical with 

jR + (9 {RrK + X) + (9^ {RIK + LrK) + LIKO'. 

As V may be any function whatsoever of the time, the 
operations are not equivalent unless LIK = ; that is, l—O) 
so there must be no self-induction in the condenser circuit, 

RrK + R^K = RrK + L ; that is, i(^= 4 ; 

RIK + RLK = RIK + LrK ; that is, R = r; 

so the resistance in the condenser circuit must be equal to 
that in the other. 

In fact we must shunt the circuit R + L^ by a 

1 Ii 

condenser circuit R + ==3 where K = =^ . 

176. If in the last case F= Fq sin nt, the operator may 
be simplified into 

R-K(Rl + rL) n^ + (9 {RrK + L - LlKn'} ' 

, . R-K(Rl-\- rL) n^ RrK + L- LlKn' 

^^^'* l-K{l-\-L)n' ~ K(r + R) 

then although the adjustment alters when frequency alters, 
we have for a fixed value of n the current flowing in at A and 
out at B proportional to V and without any lag. If R — r 

the current is equal to ^ . 

177. To explain why the effective voltage is some- 
times less between the mains at a place D, fig. 92, than 
at a place B further away from the generator. This 
is usually due to a distributed capacity (about J microfarad 
per mile is usual) in the mains. We may consider a dis- 
tributed capacity later ; at present assume one condenser of 


capacity K between the mains at B. Let the non-inductive 

resistance, say of lamps, between 

000000/ ? *^® mains at B be r. Let the 

jn resistance and self-induction of 

Kfrnlr the mains between D and B be 

If R and L. Let v be the voltage 

— X ^^ ^» ^^^ ^ the current from D 

Fig. 92. to B. 

The current into the condenser is v -^ ifn^^ ^^^'• 


The current through ?- is - , so that 

O=(if0 + i). (1). 

The cZro/) of voltage between D and 5 is 

{R + X^) or (jK + LO) (kB + ^) t;, 

or 1^ 4- (rK + ^) (9 + LKe\ V. 

Now if V = t;o sin nt, the fZrop is 

The voltage at D is the drop plus v, or 

1 b A-f ir7 square of effective voltage at D 
square of effective voltage at B 

= {l + ^-LKn^\{RK + ^^'n\ 

and there are values of the constants for which this is less 
than 1. As a numerical example take 

r= 10, R = '\, K=lx 10-^ w = 1000, 

and let L change from to 05, "01, '02, '03 &c. 


The student will find no difficulty in considering this 
problem when r + ^^ is used instead of r in (1) ; that is, when 
not merely lamps are being fed beyond B, but also coils 
having self-induction. 

Most general case of Two Coils. 

178. Let there be a coil, fig. 93, with electromotive force 
E, resistance i2, self-induction L, 
capacity K\ and another with e, r, RLC m ric 
I, h Let the mutual induction f § § I 

be m. Em § fe |o« 

Using then R for R -\- Ld ■\- -^^, I | 

I r for r -{- 
the equations are 

and r for r -1-/(9 4--^^^. ^^^- '^^• 

E = RG + m6c,] 
e = meC + rc j ^^' 

Notice hoAV important it is for a student not to trouble 
himself about the signs of and c &c. until he obtains his 

From these we find 

Re — mOE ,^. 

'=w^.^^ ^^>' 

^ rE-mOe ,„. 

^^Rr-m^e' ^^^- 

We can now substitute for i2, r, E and e their values and 
obtain the currents. 

Observe that E may be a voltage established at the 
terminals of part of a circuit, and then R is only between 
these terminals. 

The following exercises are examples of this general case. 

There are a gi-eat. many other examples in which mutual 
induction comes in. 

179. Let two circuits (fig. 94), with self-inductions, 


be in parallel^ with mutual induction m between 


1 r c 
Fig. 94. 

(1) At their terminals let v = Vo sin nt ; (2) of last exercise 

becomes c = fs iTTv. v- Or, chanerinof R into R-\-L6 and r 

into r 4- 16, 

R + {L-m)e 

^""[Rr- (LI - m'') n^) + (Zr + IR) 6 ^* 

(2) How does a current A sin nt divide itself between two 

such circuits ? Since -p^ = ^ we can find at once -^ 

G r-mO G+c 

A ^ A R + {L-m)e 

and >. -— . Answer : c = Tr, x ^Vr r—7i — r^ operatmsf 

G + c {R + r)-{-{L+l-2m)0 ^ ^ 

on -4 sin nt. 

We think it is hardly necessary to work such examples 
out more fully for students, as, to complete the answers they 
have the rule in Art. 167. 

180. In the above example, imagine each of the 
circuits to have also a mutual induction with the com- 
pound circuit. We shall use new letters as shown in fig. 95. 





viznnnfg W ^ c+Co=C 


Fig. 95. 

If V is the potential difference between the ends of the 
two circuits which are in parallel. Using r, /jl and m to stand 
for r 4- W, fiO and mO, 

V = ?-iCi + fiCi + nh (ci + Co). 


Hence the equations are 

V = (7\ + 71h) Ci + (/X + nil) Ca, 

V = (ft + m^) Ci + (r., + mg) c.,, 

^. _ (r, + m,)-(/^ + mO ^ 

' (ri + mi)(r2 + 7?i2)-(/A + mi)(/A + m.2) *"^ ^' 

with a similar expression for Cg. 

Also a total current C divides itself in the following way 

^ 7-1+ ro — 2//- ^ '^' 

If we write these out in full, we have exceedingly pretty 
problems to study, and our study might perhaps be helped by 
taking numerical values for some of the quantities. If we care 

to introduce condensers, we need only write ^ + ^^ + r^ with 

proper affixes, instead of each ?•; fi becomes fiO and m 
becomes mO. 

To what extent may we make some of the ma negative? 
I have not considered this fully, but some student ought to 
try various values and afterwards verify his results with 
actual coils. Taking (2) without condensers 

_ f\+ (h+ fn-2 — 1^ — '^^h) p 

181. Rotating Field. Current passes through a coil 
wound on a non-conducting bobbin ; the same current 
passes through a coil wound on a conducting bobbin. The 
coils are at right angles and have no mutual induction ; 
find the nature of the fields which are at right angles at the 
centre of the two bobbins. Let the numbers of turns be rii 
and 71.2 • Instead of a conducting bobbin imagine a coil closed 
on itself of resistance 7\, and ng turns and current c. For 
simplicity, suppose all three mean radii the same, and the 
coils 7i2 and % well intermingled. One field F^ is propor- 
tional to iiiG per square centimetre, call it iiiC The other F, 
is proportional, or let us say equal to,??2(7 + ^gC per square cm. 
Take the total induction / through each of the coils as 


proportional to the intensity of field at its centre, say b times. 
Then for the third coil, we have 

= nc + risdl or = rgC + bn-iO {n.C -\- n^c), 


so that — c = 

and hence F., = v..G 


\-^bn^^e n + bn^^e' 
If then G= Cq sin qty 
Fi = 71jGq sin qt, 

■^■2= 2* = — , sin 7^-tan-i q): 

Art. 126, shows the nature of the rotating field. We 
can assure the student that he may obtain an excellent 
j-otating field in this way. 

It is evident that bn-i really means the self-induction of the 

bn^ . . 

third coil, and — - means its time constant. A coil of one 

turn, — that is, a conducting bobbin, will have a greater time 
constant than any coil of more than one turn wound in the 
same volume. It is evident that if the bobbin is made large 
enough in dimensions, we can for a given frequency have an 
almost uniform and uniformly rotating field by making 

n^^b— q=nj. 

This is one of a great number of examples which we 
might give to illustrate the usefulness of our sign of opera- 
tion 6. 

182. In Art. 1 78 let E= V the primary voltage of a trans- 
former, the primary circuit having internal resistance B and 
self-induction L ; let the secondary have no independent 
E.M.F. in it; let its internal resistance be 7\ and self-induction 
I and let it have an outside non-inductive resistance p, of 
lamps. Let the voltage at the secondary terminals be i^ = cp. 



Then in (1), (2) and (3) Art. 178, let F^V, e=0; instead of R 
use R-\-Ld. Instead of r use r + Id which is really i\-{- p + W, 


(7 = 

Br + (Rl -trL)e + {LI - m"") ^ 


Rr + (Rl + rL)e+(Ll-m^)d' "" 
Note that the second equation of (1) Art. 178 is 

0==meG + (r + W)c (2)* 

I. From (2)*, if C = Co6**, c = 

r + la 

— c 

r + la 

If r is small compared with la 

— c 



I ' 












Fi^. 96. 

II. If C = Cq sin qt, again using (2)* 

Wr^-\'Pq'' V 2 rJ 



effective c 
effective G 

V^ + £' 

Except when the load on the secondary is less than it 
ever is usually in practice, r is insignificant compared with 
Iq (a practical example ought to be tried to test this) and we 
may take 

C=Gq sin qt. 

c= Gg-j- sin (qt — tt), 






It may become important in some application to re- 
member that the ratio of the instantaneous values of — c and 
G is that of 

r sin qt + Iq cos qt to mq cos qt, 

and this sometimes is oo . t 

Returning to (1). Let LI — m^ (this is the condition 
called no magnetic leakage) and let Rr be negligible. In 
any practical case, Ri' is found to be negligible even when r 
is so great as to be several times the resistance of only one 
lamp, t 

'f'^- -^ = B^Z W' 

SO that — c is a faithful copy of F as a function of the time. 
C is so also. 

If N and n are the numbei*s of windings of the two coils 
on the same iron, 

m : L : l^Nn : N^ : 71^ (5), 

so that — c = ^ (6); 

that is, the secondary current is the same as if the trans- 
formed voltage ( — 7^ ^ ) acted in the secondary circuit, but 
as if an extra resistance were introduced which I call the 
transformed primary resistance IR-^). 

If the volumes of the two coils were equal, and if the 

volumes of their insulations were equal, R ^^ would be equal 

to Vy the internal resistance of the secondary. Assume it 
so and then 

-^=2;tt7 ^'^^ 


also /)c or v = — (8). 

As Vi is usually small compared with p, 

and — * is called the drop in the secondary voltage 
due to load. 

^2 IP 

As — = P, the power given to lamps ; - = -^ and the 

fractional drop is — ^ P and is proportional to the Power, or 

to the number of lamps which are in circuit. 

183. The above results may be obtained in another way. 
Let / be the induction, and let it be the same in both 
coils. Here again we assume no magnetic leakage, 

V^RC-^NOI (1), 

O^rc + ndl (2). 

Multiplying each equation by its N or n and dividing by its 
^ or r and adding 

NV . (m n\ ^^ 

-^ = ^ + U+-)^^ (•^)' 

where A — NC + nc, and is called the current turns. 

Now when we know the nature of the magnetic circuit, 
that is, the nature of the iron and its section, a square centi- 
metres, and the average length X centimetres of the magnetic 
circuit, we know the relationship between A and /. I 
have gone carefully into this matter and find that whatever 
be the nature of the periodic law for A, so long as the 
frequency and sizes of iron &c. are what they usually are in 
practice, the term A is utterly insignificant in (3). Reject- 
ing it we find 

'^ SZhk '^^--^^^ '"' "^'"^ ""^^' •••^*^- 


Thus, in a certain 15 00- watt transformer, J? =27 ohms, 
N = 460 turns, internal part of r = '067 ohms, ii = 24 turns, 
effective V is 2000 volts or F= 2828 sin qt where q = 600 say, 
a= 360, \ = 31. When there is no load r = x ; on full load 
r = nearly 7 ohms. 

We have called R -^^ the transformed resistance of the 

. / 24 \2 

primary. It is in this case 27 ( .^^1 or '073 ohms. 

If the primary and secondary volumes of copperthad been 
equal, no doubt this would have been more nearly identical 
with '067, the internal resistance of the secondary. 

^r,.— or — '- is the fractional drop in / from what it is at 

no load. When at full load r* = 7 ohms the fractional drop is 
greatest, and it is only 1 per cent, in this case. Because of 
its smallness we took a fractional increase of the denominator 
as the same fractional diminution of the numerator of (4). 

Consider / at its greatest, that is, at no load ; -^V is the 



integral of F or — -^^^ cos 600^. So that the amplitude 

. 2828 ^^^ 
° ^^600x460 • 

Multiply this, the maximum value of / in Webers, by 10* 
to obtain c. a s. units, and divide by a = 360, and we find 
2856 c. G. s. units of induction per sq. cm. in the iron, as the 
maximum in this transformer every cycle. 

61 being ^ F / f 1 + vfa — ) > we have from (2) the same 

value of — re that we had before in (6) of Art. 183. 

184. Returning to (7) of Art. 182. Let us suppose that 
there is magnetic leakage and that Vx is really r^ + VO. 
If one really goes into the matter it will be seen that this 
is what we mean by magnetic leakage. Then we must 
divide by 

p + 2n + 21' By 
instead of /> -f 2ri. In fact our old answer must be divided by 

1+ 2r . 


or neglecting 2?'i as not very important in this connection ; 

our old answer must be divided by 1 -\ 6. This means 

. . P 
that the old amplitude of v must be divided by 


1 H ^ or 1 H ^,— nearly, 

^ P" 

if the leakage is small, and there is a lag produced of the amount 

tan-i — ^ . We must remember that a is 27r/* if f is the 

frequency. We saw that P, the power given to the lamps, is 

inversely proportional to p, so we see that the Aractional 

2r P 
drop due to mere resistances is — \- , the fractional drop 

due to magnetic leakage is la^pF", and the lag due to 

magnetic leakage is an angle of afP radians where a is a 
constant which depends upon the amount of leakage, and / 
is the frequency. 

185. Only one thing need now be commented upon in 
regard to Transformers. If Fis known, it has only to be inte- 
grated and divided by N to get I. Multiply by 10^ and divide 
by the cross-section of the iron in square centimetres, and we 
know how /3, the induction per sq. cm. in the iron, alters with 
the time. The experimentally obtained /3, H curve for the 
iron enables us to find for every value of ^ the corresponding 
value of H,w[\d H multiplied by the length of the magnetic 

cuTCUit in the iron gives the gaussage, or y^ x the ampere 

turns A. Hence the law of variation of A is known, and if 
there is no secondary current, we have the law of the 
primary current in an unloaded transformer or choking 
coil. This last statement is, however, inaccurate, as one 
never has a truly unloaded transformer, even when what is 
usually called the secondary, has an infinite resistance. 

186. Sir W. Grove's Problem ; the effect of a condenser 
in the primary of an induction coil when using alternating 

ABB, fig. 97, is the primary with electromotive force 
E— Ed sin nt, resistance R and self-induction L. BA is h 

P. IT 


condenser of capacity K, and ?' is a non-inductive resistance 

in parallel with the condenser. G 

qq^q'q^j^ -? the current in the primary, has an 

S JTI amplitude Co, say. 

® Kli1j|r rpjjg condenser has the resist- 

*^" It is quite easy to write out 

the value of Cq when r and K have any finite values*. 

But for our problem we suppose r=0 or else r=oo . When 
r = 0, the resistance \^R-\-L6 and the current is EI{R + Z^), 

n^-—^ (1). 

When r=oc , the resistance is 

R + Le+-^^ or j^ , 

or ^^ j^ , by Art. 167, 

A rr.- £Jo'K'n^ Ej . 

\Kn J 

Now (2) is greatei' than (1) if 2KLn^ is greater than 1, 
so that the primary current is increased by a condenser of 

capacity greater than o;-— g. Again, there is a maximum 

current if iT = y— ^ ; in this case the condenser completely 

destroys the self-induction of the primary. 

* When both r and K have finite values, the parallel resistances between 
2S and A, together form a resistance rl{l + rKd), and the whole resistance of 

the circuit for C is jR + L^ + - tptSO that 

1 + rKd 

(1 + rKd)EQ sin lit 

{R + r- LrKii') + {RrK+L) d 

02 = - 

" ~ (E + r - LrKn^f + (RrK-^LYii^ ' 
and the lag of C is easily written. 


187. Alternators in series. Let their e.:m.f. be ei and 
e.2 and let C be the current through both. The powers exerted 
are efi and 020. Now if 

ei=E sin (nt+a) and e.2=Esm (nt — a),t ei +e.j=2j&cos a . sin?^^. 

If I is the self-induction of each machine, r its internal 
resistance, and if 2R is the outside resistance and if Pi and 
P2 are the average powers developed in the two machines, 

^ 2^cosa. sin?i^ Ecosa • / . , , In \ 

^ = -zTTTi — ^ ^tt = ,-' sm nt — tan~^ v, 

2;^ + 2r-f-2P V(P4-r)2 + ^'^'' V i^ + r/ 

= ilf cos a sin {nt — e) say, 

Pi = p/^ cos a . cos (a + e), 

Po = ^Jlfii' cos a . cos (a — e). 

Hence P2 is greater than Pi, and machine 2 is retarded 
whilst machine 1 is accelerated ; hence a increases until 

a = - , and when this is the case, cos a = 0, so that Pi = 0, 

P2 = and the machines neutralize each other, producing no 
current in the circuit. Alternators cannot therefore be 
used in series unless their shafts are fastened together. 

188. As we very often have to deal with circuits in 
parallel we give the following general formula ; if the electro- 
motive forces ^1, ^2 and e^, fig. 98, are constant, 

v^e^ — c^Vi = e.2 - 0.^2 ^e-i-c^n ( 1 ), 

and Ci + C2 + C3 = (2). 

Given the values of e^, e._,, e^ and 7\, i\, r^ we easily find 
the currents, because 

, = (?i + ?. + ^V(i + i + l) (3). 

Fig. 98. 



Now if the es are not constant, we must use ?'i + liO, &c., 
instead of mere resistances. 

189. Alternators in Parallel. Let two alternatoi-s, each 
of resistance r and self-induction I, and with electromotive 
forces, ei = E sin (iit + a), and e^ = E sin (nt — a), be coupled 
up in parallel to a non-inductive circuit of resistance R. 
What average electrical power will each of them create, 
and will they tend to synchronism ? If Ci and e.2 were con- 
stant or if I were 0, then v = ei — CiV = e., — cr = (ci + c^) R. 

And hence 

^^ = 2r^H^+3-^-^ 



Now alter r to ?• 4- 16, because the e's are alternating. 
The student will see that we may write 

62 =61 (a — bO), 

e^ = 62 (a -H hO), 

where a^ -h 6W =1, a = cos 2a, 6/1 = sin 2a. Then 



with a similar expression for Ca in terms of e, except that 6 
is made negative. If we write out (1) by the rule of Art. 167, 
there is some such simplification as this : — 

Let tan <f) = — ^— - — =^ and tan ^^ = ^— ^ 

^ 2Rr + r^ — Pn^ ^ R+r — aR 

tan ylr^ = t,-; d • 

^ R + r-aR 

Then c, = if sin (ni -f a — <^ -}- i|ri), 

C2 = il/ sin (nt — a — <j) -{- yfr^, 

the angles <^, yjry, s^^ being all supposed to be between 
and + 90^ 


The average powers are 

P„_ = ME cos ((f>-^|r.,), 

where M'= ^ .^^.. l^^""^ , ^J' E^. 

Po COS (<l>-ylr^y 

It' 11= oo we see that 

, , In ^ . sin a , , 

tan 6 = — , tan -v/r, = , = — tan vr... 

^ r ^ 1 — cos a ^ " 

Hence P,^ cos(^-^ ) 

P. COS(«/)+>^i) 

In this case it is obvious that Pi is greater than P2. 


The author has not examined the general expression for p- 

with great care, himself, but men who have studied it say 
that it shows P, to be always greater than Po. Students 
would do well to take values for i\ I, R and a and try for 
themselves. If Pi is always greater, it means that the 
leading alternator has more work to do, and it will tend to go 
slower, and the lagging one tends to go more quickly, so that 
there is a tendency to synchronism and hence alternators 
will work in Parallel. 

190. Struts. Consider a strut perfectly prismatic, of 
homogeneous material, its own weight neglected, the resultant 
force F at each end passing through the centre of each end. 
Let ACB, fig. 99, show the centre line of the bent strut. Let 
FQ = yhe the deflection at P where OQ = x. Let OA=OB = I. 
y is supposed everywhere small in comparison with the length 
21 of the strut. 

Fy is the bending moment at P, and ^ is the curva- 
ture there, if E is Young s modulus for the material and / is 


the least moment of inertia of the cross section everywhere, 
P, about a line through the centre of area of the sec- 

tion. Then as in Art. 60 the curvature bemg - t~* 

we have 

II--^^ (1) 

Now if the student tries he will find that, as in 
the many cases where we have had and again shall 
have this equation, (see Art. 119) 

7/ = a cos X 

\lm <^) 

satisfies (1) whatever value a may have. When 
a: = we see that y — a, so that the meaning of a is 
known to us ; it is the deflection of the strut in the 
middle. The student is instructed to follow carefully 
99. the next step in our argument. 

When A" = /, 2/ = 0. Hence 


a cos ^ a/ VTr = (3). 

* Notice that when we choose to call -~ the curvature of a curve, if the 
expression to which we put it equal is essentially positive, we must give such 
a sign to -r^ as will make it also positive. Now if the slope of the curve of 
fig. 99 be studied as we studied the curve of fig. 6, we shall find that 
-r-^ is negative from a; = to a: = O^ , and as y is positive so that ^ is positive, 

«re must use - —-^ on the right-hand side. 

It will be found that the comjilete (see Arts, loi and 159) solution of any 
such equation as (1) which may be written 

ia y = ^ ^^^ nx + B sin nx 

where A and B are arbitrary constants. A and B are chosen to suit the 
particular problem which is being solved. In the present case it is evident 
that, as ?/ = when x = l and also when x= - I, 

0=iA cos 111 + B sin 111, 

= A cos 111 - B sin 7il, so that B ia 0. 


Now how can this be true ? Either a = 0, or the cosine 
is 0. Hence, if hending occurs, so that a has some value, 
the cosine must he 0. Now if the cosine of an angle is the 

angle must be ^ or -^ or -^ , &c. It is easy to see why we 
confine our attention to » *. 

Hence the condition that bending occurs is 

is the load which will produce bending. This is called 
Euler's law of strength. The load given by (4) will produce 
either very little or very much bending equally well. It 
is very easy to extend the theory to struts fixed at both ends 
or fixed at one end and hinged at the other. 

For equilibrium under exceedingly great bending, the 

equation (1) is not correct, as -r~ is not equal to the curva- 
ture when the curvature is great, but for all engineering 
purposes it may be taken as correct. 

191. We may take it that F given by (4), is the load 
which will break a strut if it breaks by bending. If / is 
the compressive stress which will produce rupture and 
A is the area of cross section, the load /J. will break the 
strut by direct crushing, and we must take the smaller 
of the two answers. In fact we see that /A is to be 
taken for short struts or for struts which are artificially*!" 
protected from bending, and (4) is to be taken for long struts. 
Now, even when great care is taken, we find that struts are 
neither quite straight nor homogeneous, nor is it easy to 
load them in the specified manner. Consequently when 
loaded, they deflect with even small loads^ and the}^ break 
with loads less than either /A or that given by (4). 

* This gives the least value of W. The meaning of the other cases is 
that y is assumed to be one or more times between x = and x = l, so that 
the strut has points of inflexion. 

t This casual remark contains the whole theory of struts such as are used 
in the Forth Bridge. 


Curiously enough, however, when struts of the same section 
but of different lengths are tested, their breaking loads 
follow, with a rough approximation to accuracy, some rule as 
to length. Let us assume that as F=fA for short struts, 
and what is given in (4) for long struts, then the formula 

F = -a 


may be taken to be true for struts of all lengths, because it is 
true both for short and for long ones. For if I is gieat 
we may neglect 1 in the denominator, and our (5) is really (4) ; 
again, when / is small, we may regard the denominator as 
only 1 and so we have W=fA. We get in this way an 
empirical formula which is found to be fairly right for all 
struts. To put it in its usual form, let I — Ak^, k being the 
least radius of gyration of the section about a line through 
its centre of gravity, then 




where a is ^//Ett^ or rather / and a are numbers best 
determined from actual experiments on struts. 

If F does not act truly at the centre of each end, but at 
the distance h from it, our end condition is that y = h when 
00=^1. This will be found to explain why struts not perfectly 
ti-uly loaded, break with a load less than what is given in (4). 
Students who wish to pursue the subject are referred to 
pages 464 and 513 of the Engineer for 1886, where initial 
want of straightness of struts is also taken account of 

192. Struts with Lateral Loads. We had better confine our 
attention to a strut with hinged ends. If the lateral loads are such 
that by themselves and the necessary lateral supporting forces, they 
produce a bending moment which we shall call ^ (.r), then (1) Art. 190 

Thus let a strut be uniformly loaded laterally, as by centrifugal force 
or its own weight, and then <j> {a:)~^ w' {l — x)^ if w' is the lateral load 
per unit length. 


We find it slightly more convenient to take <f) (x) = ^Wl cos ^x 

where W is the total lateral load ; this is not a very different law. 

dh/ F , Wl n ^ 

d^^+m^-^^W'''2i''=^ (1)- 

We find here that 

y="-r3 cos- r (2). 

Observe that when i^=0 this gives the shape of the beam. 
The deflexion in the middle is 

.. = -i?^ (3), 

and the greatest bending moment fi is 

M = ^yi+iTf/,or 

-i--^7(^^--) (^)- 

If TF=0 and if fx has any value whatever, the denominator of (4) 
must be 0. Putting it equal to 0, we have Euler's law for the strength 
of struts which are so long that they bend before breaking. If Euler's 
value of F be called U, or U=ETii^/4l^y (4) becomes 

>^=i^''^T&F • '^)- 

If Zc is the greatest distance of a point in the section from the 
neutral line on the compressive side, or if I-^Zc=^Z, the least strength 
modulus of the section, and A is the area of cross section, and if / is 
the maximum compressive stress to which any part of the strut is 
subjected, n F 

Z'^ A- 

Using this expression, if /3 stands for -r (that is Euler's Breaking 

load per square inch of section), and if w stands for -r (the true break- 

ing load per inch of section), then 

(^-?)(^-f)=:^ («)• 

This formula is not difficult to remember. From it lo may be found. 

Example. Every point in an iron or steel coupling rod^ of length 
26 inches, moves aljout a radius of r inches. Its section is rectangular, 



d inches in the plane of the motion and b at right angles to this. We 
may take W=lbdrn^'7-62940y in pounds, where 7i = number of revolu- 
tions per minute. Take it as a strut hinged at both ends, for both 
directions in which it may break. 

1st. For bending in the direction in which there is no centrifugal 

force where/ is ^^^ 

Euler's rule gives 



Now we shall take this as the endlong load which will cause the 
strut to break in the other way of bending also, so as to have it equally 
ready to break both ways. 

2nd. Bending in the direction in which bending is helped by 
centrifugal force. Our w of (6) is the above quantity of (7) divided by 
bdf or taking 



Taking the proof stress / for the steel used, as 20000 lb. per sq. inch 
(remember to keep^ low, because of reversals of stress), and recollecting 

the fact that / in this other direction is r-s^ , we have (6) becoming 


8-4 X 108 1 - 308 







Thus for example, if 6= 1, Z= 30, r= 12, the following depths c? inches, 
ai'e right for the following speeds. It is well to assume d and calculate 
n from (8). 

<^ 1 I 1-5 2 2-5 3 4 6 

0|205 277 327 368 437 545 

Exercise. A round bar of steel, 1 inch in diameter, 8 feet long, or 
Z = 48 inches. Take i^=1500 lb. Show that an endlong load only 
sufficient of itself to produce a stress of 1910 lb. per sq. in., and a 
bending moment which by itself would only produce a stress of 816 lb. 
per sq. inch ; if both act together, produce a stress of 23190 lb. per 
sq. inch. 

For other interesting examples the student is refeiTed to The 
Philosophical Magazine for March, 1892. t 



193. In Chapter I. we dealt only with the differentia- 
tion and integration of ic" and in Chapter II. with €«^ and 
sin aw, and unless one is really intending to make a rather 
complete study of the Calculus, nothing further is needed. 
Our knowledge of those three functions is sufficient for 
nearly every practical engineering purpose. It will be found, 
indeed, that many of the examples given in this chapter might 
have been given in Chapters I. and II. For the differen- 
tiation and integration of functions in general, we should 
have preferred to ask students to read the regular treatises, 
skipping difficult parts in a first reading and afterwards 
returning to these parts when there is the knowledge 
which it is necessary to have before one can understand 
them. If a student has no tutor to mark these difficult parts 
for him, he will find them out for himself by trial. 

By means of a few rules it is easy to become able to 
differentiate any algebraic function of cc, and in spite of our 
wish that students should read the regular treatises we are 
weak enough to give these rules here. They are mainly used 
to enable schoolboys to prepare for examinations and attain 
facility in differentiation. These boys so seldom learn more 
of this wonderful subject, and so rapidly lose the facility in 

question, because they never have learnt really what -^ 

means, that we are apt with beginners to discourage much 
practice in differentiation, and so err, possibly, as much as the 
older teachers, but in another way. If, however, a man sees 
clearly the object of his work, he ought to try to gain this 
facility in differentiation and to retain it. The knack is 
easily learnt, and in working the examples he will, at all 


events, become more expert in manipulating algebraic and 
trigonometric expressions, and such expertness is all-important 
to the practical man. 

In Chapters I. and II. we thought it very important 
that students should graph several illustrations of 

y = cwJ", y = ae^, y = a sin {hcc + c). 

So also they ought to graph any new function which comes 
before them. But we would again warn them that it is better 
to have graphed a few very thoroughly, than to have a hazy 
belief that one has graphed a great number. 

The engineer discovers himself and his own powers in 
the first problem of any kind that he is allowed to work out 
completely by himself. The nature of the problem does not 
matter; what does matter is the thoroughness with which 
he works it out. 

Graph y = tan ax. We assume that the student has 
already graphed y = «€** sin nx. 

194. If y =f(x), so that when a particular value of on is 
chosen, y may be calculated ; let a new value of x be taken, 
x+Bx, this enables us to calculate the corresponding value 


or y + By^f(x-{-Bx). 

Now subtract and divide by Bx, and we find 
By_ f{x-^Bx)--f(x ) 
Bx~ Bx ' ^^^• 

We are here indicating, generally, what we must do with 
any function, and what we have already done with our famous 
three, and we see that our definition of dy/dx is, the 
limiting value reached by (1) as Bx is juade smaller and 
smaller without limit. 

195. It is evident from this definition that the differential 
coefficient of af(x), is a multiplied by the differential co- 
efficient of fix), and it is easy to show that the differential 
coefficient of a sum of functions is equal to the sum of the 
differential coefficients of each. In some of the examples of 
Chapter I. we have assumed this without proof 


We may put the proof in this form : — 

Let y=u + v + w, the sum of three given functions of x. 
Let X become os-{- hx,f\et u become ii + Su, v become v-{-Sv, 
and w become w -f ^iv. It results that if ij becomes y + 5y, then 

8y = Bit -^ hv + Biu, 

- By Bit Bv Bw 

and 5^^ = ^ + ?- + F~ . 

ox ox ox ox 

, . ,, ,. . dy du dv dw 

and m the limit -r^ = ^- + t" + -? • 

ax ax ax ax 

196. Differential Coefficient of a Product of two Func- 

Let y = uv where u and v are functions of x. When x 
becomes x + Bx, let 

y-\-By = {ii + Bi()(v + Bv) — uv •{- ^l . Bv + v . Bu ■{■ Bu . Bv. 

Subtracting we find 

By = a . Bv -^^ v . Bu -{■ Bu . Bv> 

, By B^ Bu Bu ^ 

and f-^U'^-hv K- + ^'Bv. 

ox ox ox ox 

We now imagine Bx, and in consequence (for this is 

always assumed in our work) Bti^ Bv and By to get smaller 

and smaller without limit. Consequently, whatever -r*- may 

be, -r- . Bv must in the limit become 0, and hence 

dy_ dv da 
dx dx dx' 

The student must translate this for himself into ordinary 
language. It is in the same way easy to show, by writyig 
uvto as uv X w, that if t/ = uvw then 

dy dw , du dv 

-r- = wv -7— + viv -,- + wit -J- " 
ax ax ax ax 

Illustrations. If y = 10a;'' then, directly, ^ = lOa^. But 
we may write it y = 5^' x 2x^. 


Our new rule gives 

^ = 00^ (8^) + 20,-^ (15*'0 = ^O*-" + SOaf = 70af. 

The student ought to manufacture other examples for 

197. Differential Coefficient of a Quotient. 

Let ?/ = - when u and v are functions of x. 

Then y + oy= — —^ . 

Subtract and we find 

^ __ w 4- Sw u _v .8u — to.^v 
"^ " v+Bv V v'^+v .Sv ' 

Bu Bv 
hi/ __ 8x Bx 
Bx" v'^ + v .Sv 

Letting Bx get smaller and smaller without limit, v . Bv 
becomes 0, and we have 

du dv 

dy dx dx 

dx v^ 

Here again the student must translate the rule into 

ordinary language, and he must get very well used indeed to 


the idea that it is ^ -v" which comes first : — 


Denominator into differential coefficient of nume- 
rator^ minus numerator into differential coefficient 
of- denominator, divided by denominator squared. 

A few illustrations ought to be manufactured. Thus 

24a;^ . ,i ^ . t dy .^ ^ 
y = -^— IS really Sa?^, and -j- — ¥)oi^. 
ox ax 

By our rule, ■£= ^^ ^—^ = 40^. 

d^^dy dz^ 271 

dx dz ' dx' 

The student ouffht to work a few like y = -^r— - = 5x~^ or 
again y = — ^ = — 1^~^, and verify for himself. 

198. If y is given as a function of z, and ^ is given as a 
function of x, then it is easy to express y as a function of oj. 
Thus if 2/ = 6 log (az^ 4- g) and ^ = c + c?^ + sin ex, then 

2/ = 6 log {a (c-\- dx-\- sin eaj)^ + ^}. 

Now under such circumstances, that is, y =/{z) and 
2 = F (x), if for X we take x + 5^, and so calculate z + S^, and 
with this same z -\-Bz we calculate y + %, then we can say 
that our 8y is in consequence of our Bx, and 

Bx Sz 8x ^ ^' 

This is evidently true because we have taken care that the 
two things written as Bz shall be the same thing. On this 
supposition, that the two things written as Sz remain the 
same however small they become, we see that the rule (1) is 
true even when Sx is made smaller and smaller without 
limit, and as we suppose that Bz also gets smaller and 
smaller without limit, 

dx"dz ' dx ^ ^* 

This is such an enormously important proposition that 
a student ought not to rest satisfied until he sees very 
clearly that it is the case. For we must observe that the 
symbol dz cannot stand by itself ; we know nothing of dz by 
itself; we only know of the complete symbols dy/'dz or dzjdx. 

We are very unwilling to plague a beginner, but it would 
be fatal to his progress to pass over this matter too easily. 
Therefore he ought to illustrate the law by a few examples. 

Thus let y = az^ and z=hx\ As -^ = Saz^, -y- = 2hx, we have 
^ dz dx 

-r ' -T- = Qabz^x or Qab^x^. But by substitution, y = ab^af, 

and if we differentiate directly we get the same answer. A 
student ought to manufacture many examples for himsel£ 



An ingenious student might illustrate (2) by means of three 
curves, one connecting z and x, the other connecting z and y 
and a third produced by measurements from the other two, 
and by means of them show that for any value of x the slope 
of the y, X curve is equal to the product of the slopes of the 
other two. But in truth the method is too complex to be 
instructive. By an extension of our reasoning we see that 

dy _ dy dw du dv 

dx dw ' du ' dv ' dx 

199. It is a much easier matter to prove that 

dv dx 

:t^x-- = i (4), 

dx dy ^ ^ 

by drawing a curve, because it is easy to see that -y- is the 
cotangent of the angle of which -~ is the tangent. 

Otherwise: — if by increasing i» by &c we obtain the 
increment By of y, and if we take this same By, so foimd, we 
ought to be able to find by calculation the very same Bx with 
which we started. Hence 

By Bx ^ _. 

On this proviso, however small Bx may become, (5) is 
true and therefore (4) is true. 

200. To illustrate (2). If a gas engine indicator 

diagram is taken, it is easy to find from it by applying 
Art. 57, a diagram for h, the rate at which the stuff shows 
that it is receiving heat in foot-pounds per unit change of 
volume, on the assumption that it is a perfect gas receiving 
heat from some furnace. (In truth it is its own furnace; 
the heat comes from its own chemical energy.) Just as 


pressure is -j- , the rate at which work is done per unit 

change of volume : so ^ is -=- . Observe that h is in the 


same units as p, and to draw the curve for h it is not necessary 

to pay any attention to the scales for either p or v. They 


may be measured as inches on the diagram. We know of 
no better exercise to bring home to a student the meaning 
of a differential coefficient, than to take the indicator 
diagram, enlarge it greatly, make out a table of many values 

of j> and V, and find approximately ^ for each value of v. 

This is better than by drawing tangents to the curve. Using 

these values, and having found the values of A or -p at 

every place, suppose we want to find the rate per second at 
which the stuff is receiving heat. If t represents time, 

-rr — —r- . -77 , and hence it is only necessary to multiply h 

, dv 

As -7T is represented by the velocity of the piston, and as 

the motion of the piston is, as a first approximation, simple 

harmonic, we describe a semicircle upon the distance on the 

diagram which represents the stroke, and the ordinates of the 

semicircle represent -j- . We have therefore to multiply 

every value of h by the corresponding ordinate of the semi- 
circle, and we obtain, to a scale easily determined, the 

diagram which shows at every instant -^ . 

Havins: seen that -^^ = -^ . -^ and that — = 1^-7-, we 
° dx dv dx dx dy 

shall often treat dx or dy as if it were a real algebraic 

quantity, recollecting however that although dy or dx may 

appear by itself in an expression, it is usually only for 

facility in writing that it so appears ; thus the expression 

M.dx + N.dy = (1), 

may appear, where M and N are functions of x and y ; but 

this really stands for M+N-^ = (2). 

Again, if y = ax^, we may write 

dy = 2ax . dx (3), 

P. 18 


but this only stands for ~~ = 2ax (4). 

Our main reason for doing it is this, that if we wish to 
integrate (3) we have only to write in the symbol /, whereas, 
if we wish to integrate (4) we must describe the process in 
words, and yet the two processes are really the same. We have 
already used dx and dy in this way in Chap. I. 

Mere mathematical illustrations of Art. 198 may be manu- 
factured in plenty. But satisfying food for thought on the 
subject, is not so easy to find. The law is true; it is not 
difficult to prove it ; but the student needs to make the law 
part of his mental machinery, and this needs more than 
academic 'proof.' 

Let us now use these principles. 

201. Let y = log x ; this statement is exactly the same as 

X = e'^ Hence -,- — e^ — x and -f- = - . We used the idea 
ay ax x 

that the integral of x~^ is logd?, in Chap. I., without proof. 

It is the exceptional case of the integration of «". 

202. If the differential coefficient of sin a? is known to 
be cos X, find the differential coefficient of sin ax, 

y = sin ax = sin u\iu = ax, 

dy J du 

-Y- = cos u and -r = a, 

du dx 

xi X ^y dii du 

so that -!i r= -^ . — = cos uxa — a cos ax. 

dx du dx 

Find the differential coefficient of 3/ = cos ax, knowing 
that the differential coefficient of sin x is cos a?, 

y = cos ax = Bm.lax + -^\ — sin u say, where ;7- = «, 

dy dy du / 7r\ . 

^ — ~l '~^ — ^^s u X a = a cos {ax-{--^j — — asinaa. 


203. Let 2/ = log(x+a). 

Assume x-\-a= ti, or y = loer u, then -,- = 1 and -j^ = - , 

dy _dy dii _\ _ 1 
rfa; dii ' dx u x+a' 

204. y = tan x. Treat this as a quotient, y = , 

cos X 

dy __ cos a? . cos a; — sin a; (— sin x) _ 1 
(ia? ~ cos'^ X cos** a^ ' 

The student ought to work this example in a direct 
manner also. 

205. y = cot X. We now have choice of many methods. 

cos X 

Treat this as a quotient, y = -. , 

^ ^ sm X 

dy _ sin x (— sin x) — cos x (cos x) _ 1 

c^a; ~ sin^ x sin^ a? ' 

or we might have treated it in this way, 

y = u~^ if ii = taniJ7, 

ir'' X -T- = — i^~- X 


1 1 

cosec^ ic. 

dx dx cos* fl? 

11 1 

tan^a; cos'^a? sin^a? 

206. Let y = sin aar^, say y = sin t^, and it = aa-^. 
Then -r- = 2aa7, 

and -=^ = cos u, 


so that ~ = cos ^6 X 2ax = 2aa7 cos ax^, 


Let 3/ = 6*""^*, say y = e", and w = a sin a?, so that 

dy „ c?u 

^ = e", J- = a cos a?, 

so that ~- = e^a cos x, or a cos a? . e**'^"* 



207. y — sec x. We may either treat this as a quotient, 
or as follows ; y = (cos ijc)~^ = u~^ if u = cos x. 

du . dy dy du ,/ • x 

-Y- == — sin a?, j^ = -/ . J— = — u-^ (— sm a?) 
ax ax du ax ^ ' 

sin 33 

sec X . tan x. 


208. In Art. 1 1 the equation to the cycloid was given in 
terms of an auxiliary angle <!>; x = a<t> — a sin (f), y = a — a cos (f). 

Find -— and ~ at any point. 

Here ^^^ J±^^^^ 
dx d(j) ' dx d(f> ' d(j) 

. ,,. ,. sin</> 

= a sm d>(a — a cos 0) = = ^~ . 

^'^ ^^ l--cos<^ 

Also '^ = - (^] = — (^A X ^ 

dx^ dx \dx) d<t> \dx) dx 

(1 — cos <f)) cos <f) — sin d) (sin 6) . 

= ^^ ^^ — ^ JLX2 ^ -^ (a - a cos </)) 

(1 — cos <^)'* ^^ 

-1 ^_a 

'"a(l-cos</))2'~ 2/'*- 

209. If a^ + 2/'*=a'* (1), 

If we want -,- in terms of x only we must find y from (1) 

and use it in (2). But for a great many purposes (2) is 
useful as it stands. 

x^ ifi 
In the same way, if — + ^ = 1, 

?f 4.?^^=0 or ^--^^ 

a?'^ V" dx dx~ a'y' 

Agam, if --2-n=l» :r ="~2-- 


Also if u;^ 4- 2/^ = a^j 

If y = i^-^i sin 2a; + yg sin 4.x;, -j- = cos* a). 

If 2/ = i t^^^^ *' + t^^ ^' 1^>~ ^^^*' ^' 

Let // = V^2 + a^ = u^ if m = a-^ + ct^, ^— = 2a;, so that 

— - = 4?4-i X 2x or V 

210. Let y = sin~^ x. In words, y is the angle whose 
sine is x. Hence x = sin y, 


cos y — 's/i— sin'-^ y = Vl — a;^. 



dx ^/l-.a^' 

We have extracted a square root, and our answer may 
be + or — . We must give to -^ the sign of cos y. 

211. Similarly if y = cos~^ x, 

dy 1 

dx Vl-x^' 

212. Let y — tan~^ x, so that x = tan y, 

~ = —= 1 + tan^y = 1 +x\ 

dy cos^ y '^ 

213. Similarly if 2/ = cot"^ x, then -f-= — ^ — -5 ■ 


214. It will be seen that (2) and (4) of Arts. 198 and 
199 give us power to differentiate any ordinary expression, 
and students ought to work many examples. They ought 
to verify the list of integrals given at the end of the book. 
A student ought to keep by him a very complete list of 
integrals. He cannot hope to remember them all. Some- 
times it is advisable to take logarithms of both sides before 
differentiating, as in the following case : 

y = x^. Here logy = ^ log ;r, 

1 dy 1 , 

y ' dx X 

| = .^(l + log,.).t 

215. In the following examples, letters like x, y, z^ v, w, 
0, &c. are used for the variables ; letters like a, b, c, m, n, &c. 
are supposed constant. A student gets too familiar with x 
and y. Let him occasionally change x into ^ or ^ or v, and 
change ?/ also, before beginning to differentiate. Ho ought 
to test the answer of every integral by differentiation. 

List of Fundamental Cases. 

-7- x'' = nx''-\ L'"» . dx = *•»«+! ; 

dx J m 4- 1 

^(log^) = i, jl.dx^logx; 

d , . . [ , 1 . 

^r-(sm mx)= m cos mx, j cos 'tnx .dx— — sm mx ; 
dx J m 

d . . . r 7 1 

-J- ( cos mx)= — m sin mx, \ sm mx ,dx — cos nix ; 

dx^ ^ J m ' 

., . a [ dx 1 

(tan ax) = — ^ — , — - — = ~ tan ax ; 

cos^a./; j cos'' fr.r a 


d , . . a [ dx 1 ^ 

-^ (cot ax) = — -^— — , -^-T— - = cot ax \ 

dx^ ^vo^ax J^m^ax a 



d , . , . 1 { dx 

-j- (sin ^x) = . , , = 

d .. , . 1 f dx 1 ^ , ^ 

Many integrals that at first sight look different are really 
those given above. Even the use of \/~ or ^~ instead of 
the numerical symbol of power or root, disguises a function 
to a beginner. Thus 

1 . 1 _, 

3 ■- IS -X*, 

ci\/x (I 
and its integral is 

1 / x-^+' \ 3 3 

or z:r-X\ 

a V- i + 1/ 2a 

216. In some of the following integrals certain substitu- 
tions are suggested. The student must not be discouraged if 
he cannot see why these are suggested ; these suggestions are 
the outcome of, perhaps, weeks of mental effort by some 
dead and gone mathematician. Indeed, some of them are no 
better than this, that we are told the answer and are merely 
asked to test if it is right by differentiation. 

Just here, in learning the knack of differentiation and 
integi'ation, the student who has a tutor for a few lessons has 
a gi'eat advantage over a student who works by himself from 
a book. Nevertheless the hardworking student who has no 
tutorial help has some advantages ; what he learns he learns 
well and does not forget. The man who walks through 
England has some advantages over the man who only takes 
railway journeys. In learning to bicycle, I think that on 
the whole, it is better to be held on for the first few days ; 
learning the knack of differentiation and integration is not 
unlike learning to bicycle. 

Exercises and Examples. 

1. i/ = xhgx, ^=l+%^'- 


-^ ' dx 2^Jx 

3. y = losf (tan a?), -t-= - — ^r- - 
'^ ^^ ^' dx sin 2a; 

, 1 — tan X du ... . 

4. y — , -T-— — (sin X 4- cos x\ 

•^ sec a; eta; ^ 

1/1 \ ^y 1 

o. y = log(log<.), ^ = ^1^. 

tto; — ^— — 
6. a; = 6«< sin 6^, "^ = ^/«' + ^' • ^* «" (6^ + c), 

where tan c = - . 


We here use the simplification of Art. 116. The student will 
note that by page 235, Q (standing for djdt), operating n times 
upon sin ht^ multiplies its amplitude by 6" and gives a lead 
of n right angles. He now sees that if operates n times 
upon e"* sin ht, it multiplies by {it? 4- 6^)"^^ and produces a 
lead nc. 

Thus ^1 = (a= + I/) €«« sin (6^ + 2c) ; 

and ^^ = (a-^ + 62)^€«' sin {ht + 3c). 

^, fl-e dp 1 

i^ = 2tan-yj-^^, J = -^7j^ 

8. 2/ = log (6^ + 6-0, :£ 

c?i/ €^ — e" 

cZ^ e^ + e-^ ' 

9- y = ^^^, t=t^/^. 

10. y = aa^+hx + c, -^^ = 2aa; + &. 

12. p^cv-''\ ^=-l-37ct;-'^' 




14. \av'''dv^--^v-''\ 


15. Haf + bt + c)dt=: ^at^ + ^bf + ct + g. 

16. I V^ . (^« = \x^ . c?a; = §a-^ 

17. /f i«/r3.<^e=JJ^ = -i. 
19 ! dx 1 r dx 1 1 _, a; ^ 

"^' 2r^ 

V m 


20. I v^oT+l; . dv. Here let a + 1; = 2/ so that dv = c?y, and 
we have jy^ . dy = ly^ = ^ (a + v)\ 

= f Sa f— + 3a^ I a^ /— , 

4 — m 3 — m 2 — m 1— m 

and in this it is easy to substitute t i- a for y. 


22. I '- -T . Let a + 6^ = y so that h .dx = dy, 

J (a + bx)^ 


3. I -7^=r_ .dt=^-\/a^- 1\ evidently. 
. I . Let x — a — y,dx = dy 

25. Since ~i-, = ^(-^ --^-l 

g^—d? zaxx — a x-^ a/ 

J a?-a? 2a ^ ^^ ^^ ^^ 2a ^^ + a 

^. ., , r dx \ , x — a 

r fix 

26. If x- + 2Ax + B has real factors, then - — ^— j ^ 

j x^-{-2Ax-\-B 

is of the form just given. 

But if there are no real factors, then the integral may 
r (2x 

^^ ''"'^^'' j a? + 2Ax + A^ TB^rAi ^"^ if y = *' + ^ and 

a^ = B — A^ we have | ~ ^-„ which is - tan~* - . 
J y^ + a^ a a 

27. I tan X. dx=— I dx. This is our first example 

J J cosx ^ 

of a great class of integrals, where the numerator of a fraction 

is seen to be the differential coefficient of the denominator. 

Lot ?/ := cos X, then dy = — sin x . dx, so that the above integral 

is — I -^ , or — log y, or — log (cos x). 


28. Let/' (x) stand for the differential coefficient oif{x), 
and Ave are asked to find I — \r7-: — ■ Let f{x) = y, then 
f {oc) . dx = dy, so that the integral becomes 


Hence, if the numerator of a fraction is seen to be the differ- 
ential coefficient of the denominator, the answer is 

log (denominator). 

_^ f X . dx If 2bx .dx 1 , , , „. 

31. Reduce I ^ ^ — - to a simpler form. If the 

J Ct "T" OX "T" cx 

numerator were 2cx + b, the integral would come under our 
rule in Ex. 28. Now the numerator can be put in the shape 

n ,^ 7x ^^ 

-(2c^ + 6) + m- — , 

so we may write the integral as 

r 2cx + h J f nh\ C dx 
^j a + bx + cx^^'^'''^V''~2c)j a + bx + cx' 

The latter integral is given in Example 26. 
na f x + b , . f2x .dx C b. dx 

= vUog(a2 + .7'^) + -tan-i-. 

„_ r sin X .dx 1 f—b sin x .dx 1 i . 7 v 

33. — — T = -,- -— , = -rlog(a + £*cos*'). 

Ja + b cos i« b J CI + 6 cos a; 6 ° 


34 14^ n_+jog^-log^^ 

J xAogx J xiogx 

_ /' (I + \ogx)dx Cdx 
J a; log a; J x 

= log (x log x) — log X 
= log X 4- log (log ;r) — log X 
= log (log a;). 
When expressions involve a"* and (a + 6^)**, try substi- 
tuting y = a + bx or y=^ - + b. 


35. Thus f .--^^-.- = -^-Jl_-. 
*17 f <^^' __Jl_lA] (f' + bx 

^^'"^ J (a + 6^+1 " 2^;^ (a + ba^y- 

2m - 1 

2ma J (a-hbx-y 
and so we have a formula of reduction. 

C da 

J (a-hh 


When expressions involve Va 4- bx try y^ = a-{- bx. 

rp,, C x.dx 2 (2a — 6^) / =— 

Thus -— ..^=_-A_^^ — ^s/a + bx. 
J Wa + bx ob^ 

40. I - Vl 4- log ic . dx. Try ?/ = 1 -f logo;. 

Answer : f (1 + log a;)^. 

41. I VT~^=^- Try€^ = i/. Answer: tan-^e*. 

217. Integration by Parts. Since, if u and v are 
functions of x, 

d . ._ dv du 

dx^ ^^ dx dx' 

uv= j u.dv-\- I V . da, 


or lu . dv = uv — I V . du (1). 

We may write (1) as ju '-r . dx=uv — Iv .-j-.doc. 

By means of this formula, the integral I u . dv may be 
made to depend upon j v .du. 

r rl 

42. Thus to find p**. log a?, c?^. Let 2^ = log a? and -t-=^", 

so that V = r . Formula (1) gives us :r loe: x — \ ^ dx, 

a;«+i /, 1 \ 
or log X . 

43. fx . 6" , dx. 

Let it — x;--j- = e^^, so that v==-€^"'; then formula (1) gives 

us \ x.e"^^ .dx = - .re"* I e"* . dr = - a?€"* ^e"* 

J a aj a a^ 

= - €«* (a? ) . 

a \ aJ 

44. I e** . sin bx . dx. Call the answer A. 

Let u = sin 6^, v = - e^^, then formula (1) gives us 

^ _ _ gaa; gj^ hx \ e^^ . COS bx ,dx = - e"^ sin 6^ 5. 

a a J a a 

But similarly 1 e^* . cos 6a; . dx, which we have called B, 
may be converted, if we take u = cos bx and v = - e"*; 

^ _ _ guas cos hx-\- - I 6*** sin bx.dx — - e"* . cos bx-\-~A, 
a a J a a 


Hence A = - e"* sin h.v ( - e"^ cos hx-\- -A] ,so that 

a a \a a ) 

f . , , €"* (a sin bx—b cos bx) 
A = l e^"" sin bx .dx= - -^ ;— -y^ . 

€°^(a cos bx-\-b sin bx) 
€** COS bx . dx = — 

Similarly i^ = I 

a- + b"- 

218. By means of Formulae of Reduction we reduce 
integrals by successive steps to forms which are known to us. 
They are always deduced by the method of integi'ation by 
parts. Thus 

I ar«e"* . dx = - A-«6«^ - - I x''-' . €«^ . dx. 
j a aj 

If then we have to integrate ar*e"*, we make it depend 
upon x^e"^', again using this formula of reduction we make 
^a^ax depend upon ar^e"*, and so on, till we reduce to x^e"^ or t*"*, 
whose integral we know. 

Thus L-^e^ . (^ = o^-^t* - 3 U-^e* dx 

= x^e^-^\x'e'-2 jxe^ . dxl 

= (x^-Sx'-^Qx-(j)6\ 

Some General Exercises. 

45. 2/ = « sin^ bx, -^^ab sin 2bx. 

46. y=^b sin ax^, -~ = bnax^~^ cos ax^. 
^ dx 

47. y = (a + 6a;")»", -^ = nbx^-^m {a + 6a?»*)»*-^ 

48. y = (a + 6^') e'^*, -^ = €'^(b + ac + bcx). 


49. y — a"^, -^=^a^ . log a. 

50. y=\ogaiCy -r-=—\ . 

^ ^ dx xioga 

-- a — tdv a 


52. v=^^a^-t\ — = 

53. u = 

v^ du Sv"" 

(l_V2)f' dv (1-V2)f 

_ ^/a + t dv _ Vg ( V^- Va) 

57. 2/ = log (sin a?), -7^ = cot x. 

58. „ = logya-^sJ=-^,. 

1 /l — cost dy 1 

61. ^ = tan"^ / , -TT = / . 

Jl+P-Jl-t" dt Ji-ti 

62. a; = sec-i^, ^= "^ 

dt tjf-l' 
63. 3/ = sin (log v)y-r = ■ cos (log v). 


^- _l-\-x dy'^\ — 2x — ob^ 

66. « = log (cot t;), -^ = - -^-s" . 
■^ *=* ^ ^ dv sin 2v 

67. 5=e'(l-f»X^=€*(l-3f''-^). 
69. ^ = 

'^- /^-fd-r rf(9~ (€^-1)'^ ■ 

hr-. Ti- X Zl . Zl 4.1, 4. ^^^ COS ^ 

71. If a? = tan 6 + sec ^, prove that -j^ = 

c?^ (1 - sin ey 

d^x 2 
72. If a; = ^ log ^, prove that 3^ = g • 

78. If 2/ = 6~^ cos a?, prove that -^^ + 4?/ = 0. 

.T>. T^ ^ 4.1,4. d*y 24 

74. If2,= ^__, prove that ^ = ^-^-^. 

75. L"»-^ (a 4- &a:") ^/« dx. 

(1) If ^/5' be a positive integer, expand, multiply, and 
integrate each term. 

(2) Assume a + hx^ — y^ ; and if this fails, 

(3) Assume aaj~" +h — y^: this also may fail. 

76. \ af^ (a + x)^ . dx. Let a + x — y"^, then dx—2y. dy, and 
co^y^—a, so that we have 2 {(y*— 2ay^-h a^)y^ ,dy, or 
2 I (3/^ — 2a3/* 4- ay) c?y, which is easy. 


— 2x~^ .dx = 2i/ .dy so that we have 

78. [ — ^^ . Try a'^a?"* + 1 = ^^ ^nd we find 

1 rc^^_ 1 _ 0) 


79. If a? = -4 sin nt +B cos w<, prove that -j-^ + ?i2a; = 0. 

80. If 1* = xy, prove that ^^, = ^ ^ + ^-^^i • 

81. Illustrate the fact that -^ — i- = ^ 7- (see Art. 83) 

ay . ax ax . ay ^ 

in the following cases : 


w = tan~^ - , ^* = sin (aa;** + 63/**), 
w = sin {oi^y), u = a; sin y + y sin a?, 
« = ba^ log ai/, li = log (tan ^ ) , 

82. y = e**^ sin"* 6a;, -^ = e*'* sin"*-^ 6a? (a sin hx + m6 cos hx). 

83. ic = €-«'cos6^, ^ = (a« + 6=')~2 e""' cos(6^ -n^) where 

tan d = - . 

^1 A 
84. 2/ = a;^loga?,^- ^— 



,,.. d^y 2 COS 00 

85. y = log(8m^),J=-^j^. 

86. Uv = Aiof'^y^^ + A^y^^ + &;c., where 

aj + 6i = Oa + 62 = &c. = n, 
V is called a homogeneous function of x and y of n dimensions. 

Show that fl7(-7^)+v(-T-)= ^^- Illustrate this when 1; = — ^ 

and V = V ^* + 3/"- 

87. In general if u =/ (1/ 4- a^) + i" (y — ax), where / and 
-P are any functions whatsoever, prove that 

dx^^'' dy'' 
the differentiation of course being partial. 

88. If v = (x^ + f + ^)-i, prove that g + + g = 0. 

89. If 5 = ae--* sin y8^ satisfies ^ + 2/ j, + w^s = 0, find / 

and n^ in terms of a and /3, or find a and ^ in terms of / 
and n^, 

90. If y = €** is a solution of 

find a. As an example take 

^__2^^-^^ + 2^ = 
da^ da? dx^ dx ' 

and find its solution. 

Answer : y — ae^-\' be~^ + ce^ + e, where a, 6, c, e are any 

constants whatsoever. 

219. To integrate any fraction of the form 
Ax"^ + Bx"^-"- + Cx"^-^ + &c. 

ax"" + 6a?^-i + ca;"-2 + &c. 
where ?^ and n are positive integers. 



If m is greater than or equal to n, divide, and we have a 
quotient together with a remainder. The quotient is at once 
integrable and we have left a fraction of the form (1) in which 
m is less than n. Now the factors of the denominator can 
always be found and the fraction split up into partial 

For every factor of the denominator of the shape x — a 

assume that we have a partial fraction ^ ; for every 

factor of the shape x^ -\- oix-\- ^ assume that we have a partial 

of the shape — 77; if there are n equal factors each of 

them being x— a assume that we have the corresponding 
partial fractions 

Thus for example, suppose we have to deal with a fraction 

which we shall calK^^^ and that F{pii) splits up into factors 

a? — a, X — y8, a?- + ax + h, {x — 7)** ; we write 

/(^__4_ ^ Gx^B E 

F(x) " x-a'^ x-^'^ af + ax-{-b'^ (^-7)" 

+ 7— ^^-T^+&c (2). 

Now multiply by F{x) all across and we can either 
follow certain rules or we can exercise a certain amount of 
mother wit in finding A, B, C, D, E, F, G, &c. 

Notice that as we have an identity, that is, an equation 
which is true for any value of x, it is true if we put a; = a or 
X— ^ or a? = 7 or x'^ + ax-hb—0. Do all these things and 
we find that we have obtained A, B, E, C and D. To 
find G we may have first to differentiate our identity and 
then put ^ = 7 and so on. You will have found it more 
difficult to understand this description than to actually carry 
out the process. 

Having split our given fraction into partials the integra- 
tion is easy. 



^^ a? A B Gx-\-D 

Q1 ^^ _______ _| J 

Hence x'' = ZX^^ -f I) + 5 (^ - 1) (^=^ + 1) + {Cx -{-D)(x- If. 

Let dr^+l = 0, and we have with not much difficulty 
(7= - J. i) = 0. Put a? = 1, and we have ^ = J. To find B, 
make x = 0, and we find B=^. Hence we have to inte- 

11 11 1 X 

2(07-1)2 2^-1 2l+af' 
and the answer is 

-| -^ + ilog(«-l)-Jlog(^= + l). 

When there are r equal quadratic factors, we assume the 

(x' + cuc + ^y {x' + axi-^y-^'^ °* 

It is not difficult to see how all the constants are deter- 
mined. We seldom, however, have complicated cases in our 
practical work. 

92. Integrate ^-^^-^ or -^^^^-^—^^; 

assume it to be equal to 

M F P 
X "^a;+3"^a;-2' 

so that x^-\-x-l^M{x-\-^){x-'2,)-^Nx (a;- 2)+ P^ {x^- 3). 

As this is true for all values of x, put x = and find M, 
put ^ = — 3 and find N, put x—l and find P. Thus we find 
that the given fraction splits up into 

Qx 3a7+3 2a;-2' 
so that the integral is 

Jloga?-i-ilog(a^ + 3) + ilog(^-2). 



= — + 15a; - 6 log (a; - 1) + 41 log {x - 2). 

94. 1^-7^3-6'^ 

/■/^u.^^1 1 32 1 ^243 1 \, 

95 [ ^-^^ 

= J tan-1^4- i log(H-a^')- i log(l 4-^). 

^3-5a;24-3^ + 9 a; + l"^(a;-3)2 ^--3' 
and we find ^ = - 8, A = - 5, ^3== 17 ; 

so that the integral is 

-81og(^4-l)+-^ + l71og(a;~3). 
^^^- /^ + i- 3 = ^ l"g (^' + 3) + i log i^ - !)■ 


107 r (2a; - 5) dx __ 7 ^ + 1 

j(a; + 3)(a; + i> 2(a; + l)'^^^''^^+3* 

108. - = -7=tan^ — 7=—. 


220. Maxima and Minima. If we draw any curve 
with maxima and minima points, and also draw the curve 

showing the value of -~ in the first curve, we notice that ; — 

where y is a maximum, -f- = and -~ is negative; whereas, 

1 • • • ^V /x J d^y • :• T<. • 

^vnere y is a mmimum, t =^ and -^^ is positive, it in 

any practical example we can find no easier way of discrimi- 
nating, Ave use this way. 

Notice, however, that what is here called a maximum, 
value, means that y has gradually increased to that value 
and begins to diminish, y may have many maximum and 


minimum values, the curve beinsr wavy. Notice that -^ 

(Py . . ^^ 

may be and ~- = so that there is neither a maximum 
a 3c 

nor a minimum value, y ceasing to increase and then begin- 
ning" to increase again. See M, fig. 6. 

1. Find the maximum and minimum values of . 

Answer : ^ and — J. 

2. Find the createst value of .— : t-tt-^ r . 

Answer : 

{a + hy 
3. Prove that a sec 6-\-h cosec ^ is a minimum when 


tan 6 

= V5 

4. When is , a maximum ? Answer : x = ^. 

5. When is x'^ {a — xy^ a maximum or minimum ? 

. ma 

Answer : x = , a maximum. 

m + n 

6. Given the angle (7 of a triangle, prove that 
sin^ A + sin^ 5 is a maximum and cos'^ A + cos^ J5 is a mini- 
mum when A=B. 

7. y = asmx-\-h cos x. What are the maximum and 
tninimum values of y ? 

Answer : maximum is y = \/aFTb^, minimum is - Va^ + ¥. 

8. Find the least value of a ton + bcot 6. 

Answer : 2 ^ah. 

9. Find the maximum and minimum values of 

^^^ + 2^ + 11 
i»2 4- 4^ + 10 ' 
Answer : 2 a maximum and f a minimum. 

Students ought to plot the function as a curve on squared 


10. Find the maximum and minimum values of 

a^-x + 1 
x^ + os-l' 

Answer : maximum, — 1. 

11. Find the values of x which make y = ztr — 

^ a;- 10 

a maximum and a minimum. 

Answer : x=A gives a maximum, a;= 16 a minimum. 

12. What value of c will make v a maximum if ?; = - logo? 

Answer : c = e. 

13. If^ = ^^ -^ yt^Nah gives a mmimum value 


1 A sin" 6 ^ IT . . , 

14. X — , ^ , u —- £(ives a maximum value to x. 

1 - cos ^ 3 ° 

15. What value of c will make v a minimum if 

V = q , ? Answer: c = i. 

16. When is 4^— 15a^4- 12a; — 1 a maximum or minimum ? 

Answer : a; = J a maximum ; a; = 2 a minimum. 

17. tan'^ a? tan** (a — a;) is a maximum when 

tan (a — 2a;) = tan a. 


1 ft _ ^^ ^ = 3 a maximum, 
9 + ^v ^ = — 3 a minimum. 

19. Given the vertical angle of a triangle and its area, 
find when its base is a minimum. 

20. The characteristic of a series Dynamo is 

^=T^a W. 


where a is a number proportional to the angular velocity of 
the armature, and a and s depend upon the size of the iron, 
number of turns &c., E is the E.M.F. of the armature in volts 
and (7 the current in amperes. If r is the internal resistance 
of the machine in ohms and i^ is an outside resistance, 
the current 

^'=4^ (2), 

and the power given out by the machine is 

P:=OR (3). 

What value of R will make P a maximum ? 
Here (2) and (1) give --^ -^ = (7. 

So that 1+5(7=-^, C=if-^-lV 

r-\-R' s \r -\-R J 


we have {-^ - iT + 2i2 {-^- \\ {- ,— ^o,.! = 0- 

Rejecting ^— 1 = because it gives (7=0, we have 

— r-15 — 1 = 7 7^ > and from this R may be found if r and 

r-\-R (r + Ry ^ 

a are given. Take a = l'2, 5 = 0'03, r = *05 and illustrate 

with curves. 

21. A man is at sea 4 miles distant from the nearest 
point of a straight shore, and he wishes to get to a place 10 
miles distant from this nearest point, the road lying along 
the shore. He can row and walk. Find at what point he 
ought to land, to get to this place in the minimum time, if he 
rows at 3 miles per hour and walks at 4 miles per hour. 
Assume that he can equally well leave his boat at one place 
as at another. 


Fig. 100, ^C=4, CT=10 . Let him land at D where 
CD = X. Then AD = Vl6+a-^ and D^ = 10 - a;. 

Hence the total time in hours = r 1 -. — . 

3 4 

This is a minimum when ^^(16 +^)~*=J, or J^'=16+a;*, 
or a; = 4*535 miles. 

22. The candle power c of a certain kind of incandescent 
lamp X its probable life I in hours, was found experimentally 
to approximate on the average to 

Iq _ ]^Qll-897— 00764et> 

where v is the potential difference in volts. The watts w 
expended per candle power were found to be 

tt; = 3-7 + 108 O07-07667W 

The price of a lamp being 2^., the lamps being lighted 
for 560 hours per year, and one electrical horse-power (or 
746 watts) costing £2 for this year of 560 hours, find the 
most economical v for these lamps, so that the total cost in 
lamps and power may be a minimum. 

—J- lamps are needed per year, each costing £01. Cost 

. 56 . 

per year is then -y- in pounds, and this is for c candles, so that 

. 56 
cost per year in pounds per candle is y- . Now £1 per year 

means —^ watts, so that the cost per year per candle is 

56 746 ,, 
_x— watts. 

This added to w gives total cost in watts. 


We have Ic and lu as functions of v. Hence 
R« y 74fi 

<j\j /^ -x\j -|^Q_ii.697+o 07545W i g-^ _I_ ]^Q8 007— 07667V 

is to be made a minimum. 

Answer : i; = 101*15 volts. 

221. Sometimes when a particular value is given to ^ a 
function takes an indeterminate form. Thus for example 
in Art. 43, the area of the curve y=mx~^^ between the ordinates 

/* TYl 
mx~^^ . dx was z (6^~"^ — a}~^). 

Now when «,= 1 the area becomes ^ (1 — 1) or ^, and 

this may obviously have any value whatsoever. 

In any such case, say 4r/—'J > '^^ fipd — ^ and F {a) = 0, we 

£ {X) 

proceed as follows. We take a value of x very near to a and 

find the limiting value of our expression as x is made nearer 

and nearer to a in value. Thus \q\} x — a-\- hx. 

Now as hx is made smaller and smaller it is evident 

that f{x + hx) is more and more nearly f{x) + hx . — — . If 

in this we put x=a, f{x) or f(ci) disappears, and conse- 
quently our fraction*— ^ becomes more and more nearly 

The rule then adopted is this : — Differentiate the numerator 
oidy and call it a new numerator; differentiate the denomi- 
nator only and call it a new denominator; now insert the 
critical value of x, and we obtain the critical value of our 
fraction. The process may need repetition. 

loP" X 
Example 1. Find the value of ° when x = l, 

X ~~' L 


First try, and we see that we have 0/0. Now follow the 

above rule, and we have - , and inserting in this a; = 1 we get 

1 as our answer. 

2. rmd ^—„ — ^, r— when a? = c. 

First try a? = c, and we get 0/0. 

Now try ^, _^ , , and again we get 0/0. 

Now repeating our process we get r - 

sc—l 1 

3. Find ^- when x=l. Answer: -. 

a?" - 1 n 

4. Find when x = 0. Answer : log j- . 

5. Try the example referred to above. The area of a 

curve is ^j (6^~" — a^"") = A. If m, b, a are constants, 

what is the value of A when n = 1? Writing it as 

m — = , 


differentiate both numerator and denominator with regard 
to n, and we have, since 


6^~" log b - a^~" log a 

7)1 - , 

and if we insert n = 1 in this, we get 

m (log b — log a) or m log - , 


which is indeed the answer we should have obtained if 
instead of taking our integral p~^ . dx as following the rule 


a;~" . dx — + c, 

— 7l-\-l 


we had remembered that in this special case 
]a;~^ . dx—logiv. 


222 J Glossary and Exercises. 

Asymptote. A straight line which gets closer and closer 
to a curve sls x ov y gets greater and greater without limit. 

Thus y = - Jx^ — a^ is a Hyperbola. Now as x gets greater 

^ a . 

and greater, so that - is less and less important, the equation 

X T 

approaches more and more y=- x, which is the asymptote. 


The test for an asymptote is that -^ has a limiting value 

for points further and further from the origin, and the inter- 

cept of a tangent on the axis of x, x — y -^ , has a limiting 

value, or the intercept on the axis oiy.y — x -j-, has a limit- 
ing value. ^^ 

Point of Inflection. A point where -^^ changes sign. 

Point of Osculation. A point where there are two or 
more equal values of — . 

Cusp. Where two branches of a curve meet at a common 

Coi^Jugate Point. An isolated point, the coordinates 
of which satisfy the equation to the curve. 

Point d' AiTet. A point at which a single* branch of a 
curve suddenly stops. Example, the origin in y = a? log x. 

The Companion to the Cycloid. x-a{\ -cos (/>), 

The Epitrochoid. x={a + h) cos <^ - m6 cos f ^ + 1 J </>, 
y = (a -f- 6) sin — m6 sin ( T- -f 1 J (f>, 


where h is the radius of the rolling circle, a is the radius of 
the fixed circle, and mb = distance of tracing point along 
radius from centre of rolling circle. Make m=l, and this is 
the Epicycloid. 

The H3rpotrochoid. ^ = (a— fe)cos<^+m6cos f ^-1 J<^, 

y= (rt— 6)sin<^— ??i6sin(7- — 1 ]</). 

Make ??i = 1, and we have the Hypocycloid. 

Take a = 46, and obtain a Hypocycloid in the form 
a;* + y* = ai 

Take a = 26, and obtain the Hypocycloid which is a 
straight line. 

In obtaining the Cycloid, Art. 11, let the tracing point be 
anywhere on a radius of the rolling circle or the radius pro- 
duced and obtain a? = a (1 — m cos <^), y = a (<^ + m sin <^). 
If wi > 1, or < 1, we have a prolate or a curtate Cycloid. 

The Lemniscata (a^ + y^y= a^ (jx^ ^y^) becomes in polar 
coordinates r* = a* cos 26, 

and taking successively ^ = 0, ^ = '1, &c., we calculate r and 
graph the curve easily. 

The Spiral of Archimedes, r = ad. 

The Logarithmic or Equiangular Spiral, r = ae*^ 

The Logarithmic Curve. y=a\ogba)-{- c. 

The Conchoid a^ = (a + xy (¥ - x') becomes 
r = a + 6 sec 6. 

The CiBBoid y^ = a^/{2a — x) becomes r = 2a tan 6 . sin d. 

The Cardioide. r = a (1 — cos 0). 

The Hyperbolic Spiral. r6=a. 

The I.ituusis?^2^ = a2. 

The Trisectrix. r = a (2 cos ^ + 1). 

1. In the curve y=—^ ^, show that there are points of 

a ~i~ X 

inflexion where a? is and a VS; the axis of x is an asymptote 
on both sides ; there are points of maxima where x = a and 
— a ; the curve cuts the axis of ac at 45°. 


2. In aj^y = 36^?^ — a^ show that there is a point of inflexion 
where x = b, y= —r - 

3. If y^x = 4a'^ (2a — a?), show that there are two points of 

, „ . , 3a ^ 2a 

inflexion when oo= -zr- , y=^±-j=-. 
^ V3 

4. If 2/^ {x^ — a^) = ^, show that the equations to the 
asymptotes are y=^-\-x and y=- w. 

5. The curve a^ — y^ — a^ cuts the axis of x at right angles 
at a? = a where there is a point of inflexion. 

6. Show that y = a^xjiah + ^) has three points of in- 

7. Prove again the statements of Exercise 2, Art. 99, 
and work the exercises there. 

8. Find the subtangent and subnormal to the curve y = e*** 
Answer : subtangent - , subnormal ae^^^. 

9. Find the subnormal and subtangent to the catenary. 




subtangent = c 

Answer : subnormal = -J6<'— e "l 

c - c 

10. Find the subtangent of the curve 
x^ — Saycc + y* = 0, 

Hence ^ ^ay-j^ 

dx y^ — ax 

Subtangent at point x, y is y-^ = y ^ — . 

(ty cty — ocr 


11. In the curve 2/^ = find the equation to the 

asymptote. Here i/^ = a?^ I — \=ia^ f 1 -\ 1- --- + &c. j 

by division. As x is greater and greater, — &c. get smaller 
and smaller and in the limit (see Art. 3) 



So we have a pair of asymptotes y = x-\-a, y = — x — a. 

Again, the straight line x = a,a> line parallel to the axis of 
y, is also an asymptote, y becoming greater and greater as 
X gets nearer and nearer to a in value. 

12. Find the tangent to y* + a^ = a*. 

Hence at the point x^, y^ the tangent is ~ — ^ = — a/^\ 

13. In the curve 3/ — 2 = (a? — 1) Va; — 2, where is -^ = x ? 
At what angle does the curve cut the axis ? 

This is infinity where x — 2 and then 3/ = 2 ; that is, the 
tangent at (2, 2) is at right angles to the axis of x. 

Where y=0/\t will be found that a? = 3 and -t- = 2. 

14. In the curve y^ — ax^ ■\- a? , find the intercept by the 
tangent on the axis of 3/, that is, find y — cc-^. 

Sy^ -^ = 2ax + Sx\ 


bo tnat we want y — x — ^r— — or -^ r- and 

3i/2 %f 

this will be found to be 

3 \a +xj ' 

15. The length of the subnormal at x, y, is 20^^, what is 
the curve ? 

Here 2/-,— = 2a V. Hence ^y^ = ^a^x* or y = ax^ is the 

equation to the curve, a parabola. The subtangent is ^ y , 


16. Show that the length of the normal to the catenary^ 
is - y\ 

17. Show that y^ — x^-\- 2hx^y = has the two asymptotes 

b , b 


18. Show that the subtangent and subnormal to the cii'cle 
y^==2ax — x^, are — — — and a — x respectively, and to the 

IT 2 ^\^ 2X4.1. 2ax-x' ,b\ . 
ellipse y^ = —Azax — x^) they are and -- (a — x\ 


19. Find the tangent to the cissoid y* = 

2a — X 

Answer : y = j ^^^^^y l {(3a -x)x^- ax\, 

20. What curve has a constant subtangent ? 

dx ^ dy , ^? 

2/ -J- = a or a^ = a -^ , or x= a log y + c or y = Ce^y 

the logarithmic curve. 

21. Show that a? — y^-\- ax^ = has the asymptote 

P. 20 


22. Show that a curve is convex or concave to the 

axis of X as y and -— have the same or opposite signs. See 

Art. 60. 

223. The circle which passes through a point in a curve, 
which has the same slope there as the curve, and which has 
also the same rate of change of slope, is said to be the 
circle of curvature there. If the centre of a circle has 
a and h for its co-ordinates, and if the radius is r, it is easy 
to see that its equation is 

(a;-a)» + (i/-6)» = ?-« (1). 

Differentiating (1) (and dividing by 2) and again dif- 
ferentiating we have 

a.-a + (y-6)J=0 (2), 

and i+(,_,)g+(|J = (3). 

writing p for -~ and q for j^ we have from (3) 

y-^ = — ^ (4); 

using this in (2) we have 

X'-a=^—^p (5). 

Now p and q and x and y at any point of the curve being 
known, we know that these are the same for the circle of 
curvature there, and so a and h can be found and also r. If 
the subject of evolutes were of any interest to engineers, this 
would be the place to speak of finding an equation connecting 
a and 6, for this would be the equation of the evolute of the 
curve. The curve itself would then be called the involute to 
the evolute. Any practical man can work out this matter 
for himself. It is of more interest to find r the radius of 
curvature. Inserting (4) and (5) in (1) we find the curvature 

1 = ^_ (6) 



A better way of putting the matter is this: — A curve 
turns through the angle Bd in the length Bs, and curvature 
is defined as the limiting value of 

hd 1 dO 

B^'^^'r^dS (^)- 

Now tan = -~ = p, say, so that 6 = tan~^ j^. Hence 

dd _ 1 f^ 
ds ~ 1 -i-p- ' ds ' 

Now ^ = ViT^^=-.'^ = -.^. 

dw ^ dp dx dp dx" 

H-~ \-f.-%i[Htr\' 

Exercises. 1. The equation to a curve is 

ic^ - 1500a;2 + 30000^ - 3000000y = 0. 

Show that the denominator of - in (8) is practically 1 
from ;r = to iz; = 100. Find the curvature where x — 0. 

2. In the curve y = ar* — 4a'3 — ISar^ find the curvature 
at the origin. Answer : 36. 

3. Show that the radius of curvature at x=^a,y = Ooi 
the ellipse — + ^r = 1, is — . 

4. Find the radius of curvature where a; = 0, of the 
parabola, y^ = 4aa;. Answer : 7' — 2a. 

5. Find the radius of curvature of y = 6e"^. 

(1 + a^h-6^^)^ 
Answer : q = a^be^"^, p = ci&e«^, r = -'^ ,, „^ , so that 

where a; = 0, r = ^ -j — ^ . 


6. Find the radius of curvature of y = a sin bx. 

. (l + a262cos2 6a?)5 .,,, . 

Answer : /• = ^^ j-—. — , — — . Where x = 0, r=QC, or 

— ab^ sin ox 

curvature is : where 6a; = - , ?■ = r, ♦ 

2 — ao^ 



7. Find the radius of curvature of the catenary 

Answer -. r=.^ . At the vertex where a; = 0, 1/ = c, r = c. 

8. Show that the radius of curvature of 

7/^ {x — 4m) = 7nx {x — Sm), 

at one of the points where y = 0, is -^ , and at the other, ^r- . 

9. Find the equation of the circle of curvature of the 
curve y* = 4m'^ — a^, where x=0, 3/ = 0. 

10. The radius of curvature of Sa^y = x^, is ?' = ^ . . 

11. In the ellipse show that the radius of curvature is 
(a' — e-x"^)^ -^ ah, where e^=\ ^ , e being the eccentricity. 

12. Find the radius of curvature oi xy = a. 

Answer: (a;^ 4- 2/")* -i- 2a. 

13. Find the radius of curvature of the hyperbola 

Answer: (eV — a^)^ -r- ab, where e- = l-\ — . 


14. In the catenary the radius of curvature is equal and 
opposite to the length of the normal. 

15. Find the radius of curvature of the tractrix, the 


equation to which is y -7- = Ja^ — y^. 

224. Let f(x,y,a) = (1) 

be the equation to a family of curves, a being a constant for 
each curve, but called a variable parameter for the family, 


as it is by taking different values for a that one obtains 
different members of the famil}^ Thus 

f(a;,y,a + Ba) = (2) 

is the next member of the family as Ba is made smaller and 
smaller. Now (2) may be written (see (1) Art. 21) 

/(^,2/,a)+Sa.^/(^,2/,a) = (3), 

and the point of intersection of (2)f and (1) is obtainable by 
solving them as simultaneous equations in oo and y; or again, 
if we eliminate a from (1) and 

^J(^.y,a) = (4). 

we obtain a relation which must hold for the values of x and 
y, of the points of ultimate intersection of the curves formed 
by varying a continuously ; this is said to be the equation of 
the envelope of the family of curves (1) and it can be 
proved that it is touched by every curve of the family. 

Example. If by taking various values of a in 

m , 
^ a 

we have a family of straight lines, find the envelope. Here 
f(x, y, a) — is represented by 

y ow? = (1)*, 

and differentiating with regard to a we have 

+ ^■-^^ = (4)*, 

1 a" , m 

or - == — or a^ = _ , 

X m X 

Using this in(l)*we have 

y — ^mx — X A / — = 0, 

or y — 2 \/nix = 0, or y^ = 4>mx, 

a parabola. 



Example. In Ex. 3, Art. 24, if projectiles are all sent out 
with the same velocity F, at different angular elevations a, 
their paths form the family of curves, 

Fsin a 

2/ = 


Fcosa ^^F^cos^a' 
or . y — xa-^ma^{a^'\-\) = 0, 

where a stands for tan a and is a variable parameter, and 

1 9 

Differentiating with regard to a, 

— x-\- 2ma^a = or a = + - — : 


1 ,/ 1 

is the equation to the envelope, or 


+ 1 


-y == — rtiOi^ -f 


This is the equation to a parabola whose vertex is j — or 


— above the poii^t of projection. 

225."^ Polar Co-ordinates. If instead of giving the 
position of a point P in .r and y co-ordinates, we give it in 

terms of the distance OP called 

r, the radius vector, and the angle 

QOP (fig. 101) called (9, so that 

what we used to call x is r cos 

and what we used to call y is 

r sin 6, the equations of some 

curves, such as spirals, become 

simpler. If the co-ordinates of 

P' are r + Sr and 6 + hO, then 

^ig- 101- in the limit PSP' may be looked 

upon as a little right-angled triangle in which PS = r . BO, 

SP' is Sr, PP' or 8.9 = Jr'(8dy + (8ry so that 

dd y ' ^ \de) 


Also the elementary area POP' is in the limit ^r"^ .hd, and 
the area enclosed between a radius vector at ^j and another 

at ^2 is i I r"^ ' dO, so that if ?' can be stated in terms of 6 it 

is easy to find the area of the sector. Also the angle </> 

between the tangent at P and ?' is evidently such that 

tan cj) = PS/P'S or, t -r- — tan <f). This method of dealing 

with curves is interesting to students who are studying 

If r = a*** (the equiangular spiral) 

— = ha^^ log a, and so, ?' -7-; , or ?'-f- -^ = 1/6 log a, 

so that tan <^ is a constant ; that is the curve everywhere 
makes the same angle with the radius vector. 

Let ^ = rcos 6 so that x is always the projection of the 
radius vector on a line, x — a^^ cos 6. Now imagine the radius 
vector to rotate with uniform angular velocity of — ^ radians 
per second starting with ^ = when ^ = 0, so that 6 — — qt, 
then X = a~^9^ cos qt. 

Thus we see that if simple harmonic motion is the pro- 
jection of uniform angular motion in a circle; damped 
simple harmonic motion is the projection of uniform 
angular motion in an equiangular spiral. See Note, Art. 112. 

Ex. 1. Find the area of the curve r = a (1 + cos 0). Draw 

the curve and note that the whole area is I r^ . dO, or ^iral 

Ex.2. Find the area of ?^= a (cos 2^+ sin 2^). Answer: Tral 
Ex. 3. Find the area between the conchoid and two radii 
vectores. Answer : 

b^ (tan e^ - tan d^) + 2ab log (tan (7r/4 - i^a) -f- tan (7r/4 - ^6,)}, 

226. Exercises. 1. Find the area of the surface gener- 
ated by the revolution of the catenary (Art. 38) round the 
axis of y. 

2. Prove that the equation to the cycloid, the vertex 
being the origin, is 

x=:a(0 + sinO) y = a{l- cos 6), 
if (fig. 102) PB = x, PA= y, OCQ = 6, 



Show that when the cycloid revolves about OF it generates 

(0_.2 Q\ 

•~9 — q) > and when it revolves about OX it 






A X 

Fig. 102. 

generates the volume ttW If it revolves about EF it 
generates the volume bir^a^. 

3. Find the length of the curve ^ay^ — 4a^. 
Answer, . = /J^ 1 + f . <te = fa j(l + ^^ - 1 

4. Find the length of the curve y^ = 2ax — x^. 

Answer : s = a vers~^ x/a. 

5. Find the length of the cycloid. See Art. 47. 

Answer : 5 = 8a (1 — cos ^(f>) = 8a — 4 V4a^ — 2ay. 

parabola y — 

^x + Va + X 

6. Find the length of the parabola y — ^4iax, from the 

Answer : 5 = Nax -^ x^ -{■ a log 


7. Show that the whole area of the companion to the 
cycloid is twice that of the generating circle. 

8. Find the area of r = he^l'^ between the radii n and ra, 


using A — \ ^r^ . d6. 
J 0, 

Answer: -r{r.^ — r^). 



9. Show that in the logarithmic curve x — ae'^, 

5 = clog J + Vc^ + a?-^ + C^. 

c + Vc^ + a-^ 

10. Show that in the curve r = a (1 + cos ^), using 

s = 4a sm ^ . 

11. Show that in the curve r = ^e*/*, 

5 = rVl + c2+(7. 

12. Show that in the cycloid, 

ds \l 2a' 
and consequently s = 4 Va^ — ^ay — 2a. 

13. Show that in the curve x^ + 2/* = a*, 

s = ^ciSx^. 

x" f 

14. The ellipse, — + r^ = 1 revolves about the axis of x. 

^ a- 0^ 

Prove that the area of the surface generated is 
where e^=l -. 

x^ 1/^ 

15. Show that the whole area of the curve, -^ + fi = 1 

IS fTrao. 

16. Find the area of the loop of the curve, y=xtJ —33- . 

Answer : 2a=^ ( 1 — -7 j . 


17. Find the whole area of y = a; + Va^ — {jc\ 

Answer: Tral 

18. Find the area of a loop of the curve r^- = a^ coa 20. 

Answer : Jal 

19. Find the area of the ellipse — „ + r;: = 1 ; that is, find 
four times the value of the integral 




20. Find the area of the cycloid in terms of the angle <p 
(Art. 11). 

Answer : a^ (|</) — 2 sin </> + J sin 2<f)) ; and if the limits are 
(f> = and </) = 27r we have the whole area equal to 3 times 
that of the rolling circle. 

227. A body of weight W acted upon by gi-avity, 
moves in a medium in which the resistance = a?^", where 
V is the velocity and a and n are constants. 

9 dt 

What is the velocity when acceleration ceases? Let Vi 
be this terminal velocity, av^^ = W, or our a = Wvr\ 

dt I 1 


so that 

Thus let n= 2, t=^ log ^-^^ , 
2g ^ Vi — v 

^ , at dx 
or v=^Vi tann -- = ^rr • 

V. dt 


If a; is the depth fallen through, 

Vi^. 1 Qt 

X — — losf . cosh ^- . 

228. Our old Example of Art. 24. 

A point moves so that it has no acceleration horizontally 
and its acceleration downward is ^ a constant. Let y be 
measured upwards and x horizontally, then 


"' dt' 





dy dx 
'~di ' It'' 

' dx 



d^y dx 
^'^dx'' ~dt 

= c2 






y-^-lS'^^ + o.c+h... (I), 

which is a parabola. Compare Art. 24. 

If Ave take y— when .r=0, 6=0. Also we see that a is the 
tangent of the angle which the path makes with the hori- 
zontal at d? = and c is the constant horizontal velocity. If a 
projectile has the initial velocity V with the upward inclina- 
tion a, then c = Fcos a, and tan a = a, so that (1) becomes 

1 gx^ 
y= — ^ TF^ — ;; — f- oc tan a. 

229. Exercises on Fourier. 

1. A periodic function of x has the value /(^) = mx, from 
iv — to a? = c where c is the period, suddenly becoming 


and increasing to mc in the same way in the next period. 

Here, see Art. 133, 7 is — , 


mju = cto -\- ttj sin qx + ^1 cos qx + &e. 

+ ttg sin sqx + hg cos sqx + &c. 

tto is ^mc, 

Gg = - I mx . sin sqx . dx, hg=- I mx . cos sqx . dx. 

c J c J 

Answer : 


mx = ^mc (sin qx +i sin 2qx + J sin Sqx -f i sin iiqx -f&c.). 

2. Expand a? in a series of sines and also in a series of 

Answer : x = 2 (sin ii" — J^ sin 2a; + J sin 3a; — &c.) from — tt 
to tt; 

also a* = — (sin a? — ^ sin Sx -f ^ sin 5x — &c.) from to ^ , 



TT 4 

and ''^" = 9 (<^os A' 4-^ cos 3.r -i-^ cos 5a; + &c.). 


3. Prove t = sin x -\- 1 sin 3./; + i sin 5a^ + &c. 


4. Show that 

5. Integrate each of the above expansions. 

230. 1. The radius of gyration of a sphere about a 
diameter being k and the radius a, prove that 

Here, since a;' + 3/^ = a", and the moment of inertia of a 
circular slice of radius y and thickness hx about its centre, is 


Taylor's theorem. 317 

The moment of inertia is 

I iriny'^ .da)Xi/'=l irmy^ . dx = m^Tra*, 
Jo Jo 

and the mass is m^TraK 

2. In a paraboloid of height h and radius of base a, about 
the axis, P = Ja^. 

About the diameter of the base k^ — ^ (a^ 4- h^). 

3. In a triangle of height h, about a line through the 
vertex parallel to the base, k^ = ^h\ 

About a line through centre of triangle parallel to base 

231. Taylor^ s Theorem. 

If a function of « + h, be differentiated with regard to ic, 
h being supposed constant, we get the same answer as if we 
differentiate with regard to h, x being supposed constant. 

This is evident. Call the function f{u) where u = x-\-h. 

Then -y- f(u) = y- fiu) x ^- = -j- fCu) as -y- is 1, and this is 
dx-^ ^ ^^ dir ^ ' dx du-^ ^ ^ dx 

the same as ifrfi}^) because 

Assume that f(x-\-h) may be expanded in a series of 
ascending powers of h. 

f(x + h) = Xo + XJi + X,h'^+XJi'-^&c (1), 

where Xq, X^, X^ Sac. do not contain k 

^^!i^tA) = o + X,+ 2Z^ + 3X3/1^+ &c (2), 


§f<^)=i^+^\h + ^.h' + &c (3). 

dx dx dx ax 

As (2) and (3) are identical 

^ _dXo ldX,_ 1 d'Xo 

'" dx ' ^"^2 rf;^; "1.2 dx' ' 


1 dX^ 1 d'Xo 

' S dx 1.2,2 dx" ' 


A.lso if h = in (1) we find that Xo =f(x). If we indicate 
-r-^f(^) ^y f'{^)y <ihen Taylors Theorem is 

f (X + h) = f (X) + hf (x) + ^ i" (x) 

^' r'(x) + &c (4). 


After having differentiated f{x) twice, if we substitute for x, 
let us call the result /" (0) ; if we imagine substituted for 
X in (4) we have 

/W=/(O) + A/(0) + j^/"(0) 


/"(0) + &c (5). 

Observe that we have no longer anything to do with the 
quantity which we call x. We may if we please use any 
other letter than h in (5) ; let us use the new letter x, and 
(5) becomes 

/(^•)=/(0) + ^/(0) + j^/'(0) 

r'(0) + &c (6); 


which is called Maclaurin's Theorem. 

The proof here given of Taylor's theorem is incomplete, 
as we have used an infinite series without proving it con- 
vergent. More exact proofs will be found in the regular 
treatises. Note that if x is time and s =f(t) means distance 
of a body from some invariable plane in space ; then if at the 
present time, which we shall call to, we know s and the 

velocity and the acceleration and -5-, &c.; that is, if we know 

all the circumstances of the motion absolutely correctly at 
the present time, then we can predict where the body will be 
at any future time, and we can say where the body was at 
any past time. It is a very far-reaching theorem and gives 
food for much speculation. 


232. Exercises on Taylor. 1. Expand (^4-/^)'^ iu 
powers of h. 

Here f{x) = x^, f {x) = nx''-\ f {x) = n {n - V^x'^'S &c, 
and hence 

{x 4- A)" = X'' + n/t^"-i + ^^ Y ~ -^ fe"-2 + &c. 

This is the Binomial Theorem, which is an example of 

2. ExjDand log (.r + h) in powers of A. 
Here /(..) = log (..), /(.;) = 1 


/'(4 = -*-=,/"(.■) = + 2*-', 

1 A/^ 1 /i^ 1 

and hence log (a; + A) = log ^ + A -^ -3 + -^ -5 — &c; 

^ X A X o Qj 

If we put a; = 1 ,we have the useful formula 
log(l+A) = + A-| + ^-&c. 

8. Show that 
sin (x + h) = smx-\- h cos ^ — y— « sin os — ■= — ^— s cos a} + &c. 

4. Show that 
cos (a? + A) = cosa; — Asini»— = — jrcos^+ z — ^— ^ sina? + &c. 

6. What do 3 and 4 become when ^ = ? 

233. Exercises on Maclaurin. 1. Expand sin x in 
powers of x, 

fix) ^ sin X, -/(0) = 0, 

/(^) = cos^, /(0)=1. 

/'(^) = -sina;, /'(0)=0, 

/» = -cos^, r{0)=-l, 

/X^) = sin^j, /v(0)=0, 

&c. /'(0)=1. 

/>io /^>5 /j,-»7 

Hence sin a; = i^* — -r^ + — — -c + &c. 

|3 • 1 5 \7_ 




2. Similarly cos ^ = 1 — 'r^ -I- 77 — 77, + &;c. 

\A ]i E. 

Calculate from the above series the values of the sine and 
cosine of any angle, say 0*2 radians, and compare with what 
is given in books of mathematical tables. 

3. Expand tan~^ w. Another method is adopted. 

The differential coefficient oHair^xisr—, — , , and by actual 

division this is 1 — a;^ + a;* — a.* -{- x* — &c. 
Integrating this, term by term, we find 

tsin-^a; = x- J^ + ^a* - -f a;' + J^ - &c. 
We do not add a constant because when a? = 0, tan~* ic =« 0. 

4. Expand tan (1 — x) directly by Maclaurin. 

5. Show that 

a*= 1 +0; loga+ 1^ (logay + ^ (log ay+ &c 

6. Show that tan a; = a? + 7r4-T^ + &c. 

6 15 

234. Expand e'^, compare with the expansions of sin Q 
and cos Qy and show that 

e-** = cos — i sin 6, 
cos (9 = J (e^'^ + e-'^), 

sin^ = -i(e^'^-e-^). 

Evidently (cos Q ±i sin ^)" = cos nd + % sin nB^ 
which is Demoivre's Theorem. 

In solving cubic equations when there are three real roots, 
we find it necessary to extract the roots of unreal quantities 
by Demoivre. To find the gth root of a + hi where a and 
h are given numerically. First write 

a + ln = r (cos -\-i sin 0)» 


Then r cos = a, r sin. 6 = b, r = Va* + b^, tan 6 = ^- Cal- 


culate r and 6 therefore. 

- / 1 1 \ 

Now the qth roots are, r^ f cos - ^ + 1 sin - ^ J , 

r9 jcos - (27r + (9) + i sin - (27r + 0)] , 

r^ ]cos - (47r + ^) + t sin - (4>7r + ^) [ (Sec. 

We easily see that there are only q, qth. roots. 
Exercise. Find the three cube roots of 8. 
Write it 8 (cos 0-\-i sin 0), 8 (cos 27r + i sin 27r), 
8 (cos 47r + { sin 47r) and proceed as directed. 

235. The expansion of e^^ is 

1+he + ^h'0' + j-|-^A3<9» + &c. 

Now let 6 stand for the operation -7- , and we see that 

h — 

f(w +h)—e^^f(ai) ; or e ^f{x), symbolically represents Taylor's 

236. An equation which connects x, y and the differ- 
ential coefficients is called a Differential Equation. We 
have already solved some of these equations. 

The order is that of the highest differential coefficient. 

The degree is the power of the highest differential coeffi- 
cient. A differential equation is said to be linear, when it 
would be of the first degree, if y (the dependent variable), and 
all the differential coefficients, were regarded as unknown 
quantities. It will be found that if several solutions of a 
linear equation are obtained, their sum is also a solution. 

Given any equation connecting x and y, containing 
constants; by differentiating one or more times we obtain 
sufficient equations to enable" us to eliminate the constants. 

P. 21 


Thus we produce a differential equation. Its primitive 
evidently contains n arbitrary constants if the equation is of 
the nth. order. 

Exercise. Eliminate a and b from 

y^ax^+bx (1), 

Hence (1) becomes 

^'i^-^4x^^y-' (2>- 

If we solve (2) we find y = Aa^ + Bx, where A and B are 
any arbitrary constants. 

237. In the solution of DifiPerential Equations we begin 
with equations of the First Order and the First Degree. 

These are all of the type M^-N-^ —0, where M and N 
are functions of x and y. We usually write this in the shape 
M. dx + N.dy=0. 


1. (a + x) (b + y) dx -i-dy=0 or (a + x) dx -\- r dy — 0. 

Integrating we have the general solution 
a^ + i;z;2 4-log(6 + 2/) = (7, 
where C is an arbitrary constant. 

It is to be noticed here, as in any case when we can 
separate the variables^ the solution is easy. 

Thus if /(^) F{y).dx-^^ {x) .'f{y).dy = Oy^Q have 
f{x).dx ^{y).dy 
<^(.') ^ F{y) -"' 
and this can be at once integrated. 


{l+x)i/.dx-\-(l -y)x.dy={), 

\ogx + x + \ogy-y=G, 
log xy — G+y — x. 

Integrating, we have sin"^ x + sin~^ y — c. 

This may be put in other shapes. Thus taking the sines 
of the two sides of the equation we have 

X Vl-y2 ^ y Vl_^ = (7. 

'^^ (^ " ^ • iTv ^^^^"^^^ (2/ + 2/') dy = (x + ^) (^^. 

Answer : ^y^ ^ j^^s _ |^ ^_ ^^ _j_ constant. 

5. ^±5") = ^. Answer: (1+^) (1+2/^) = c^. 

l-¥y^ ay ^ f \ a f 

6. sin ^ . cos y . c^o; — cos a? . sin y . (ii/ = 0. 

Answer: cos y = c cos ij?. 

7. {f ■\- xy'')dx -^ {cc" - yo(P')dy =^. 

Answer : loff - = c + ^ . 

y ^y 

^' -^. + a/t-^=^- Answer: VlT^^ + VIT? == C. 

238. Sometimes we guess and find a substitution 
which answers our purpose. Thus to solve 

dy _y^ — x 

dx Ixy ' 

. dx 

we try y — Nxv, and we find \-dv=^0, leading to 

log a? + — = c. 

Solve (y - x) Vr+^ ^ = ?i ( 1 + 7/2)1 



239. If M and N are homog^eneous functions of ic 
and y of the same degree : assume y = vx and the equation 
reduces to the form of Art. 237. 

Example 1. ydx + (2 ^fxy — a;) dy — 0. Assume y — vx, 
vx.dx-^ {2x s/v — x)(v.dx + x. dv) = 0, 
(2^^) dx + (2.X-2 VS - x") dv = 0, 
2dx 2 Vti - 1 , ^ 

X \v V^J 

2 log .r 4- 2 log^; + 2v-* = 0, 
log XV + v~* = (?, 

Answer: y = C€ V », 
where c is an arbitrary constant. 

Let y — vx and we find the answer 

x^ — y^ 

+ log((;^-yi)(.^-y)i)-=0. 

Answer: a^ ^'^.Ay ^-A^. ' 

Remember that two answers may really be the same 
although they may seem to be altogether different. 

4. Solve (x^ + Sxy"") dx + (y^ -f 3^y) dy = 0. 

Answer : ^-^ 4- 6.r'y^ + y*=C. 



5. Solve 3^24.(^^4.^)^ = 0. 

Answer : Zxy + x" + 2^^ ^ q^ (2y + xf. 

6. Solve [x — y cos ^ J c?^' 4- a; cos - .dy =0. 

Answer : x = ce * 


7. (y/ — x)dy-\-y. dx = 0. Answer : 1y = ce K 

8. j;c?2/ — y-dx — ^x^ 4- y^ . cZ.:^ = 0. Answer : x^ = c^ + 2cy. 

9. x-\-y~ ^%j. Answer : (./; - y ) e^ " ^ = (7. 


240. Of the form {ax + by-{-c) dx + (a'i^- + b'y + c') (/y = 0. 

Assume x = w + a, y = v + y^, and choose a and /3 so 
that the constant terms disappear. 

Thus if (3^' - 2y + 4>) dx -{■ (2x - y -^l)dy ==0 ', •a.sdx^dw 
and dy = dv, we have 

(Sw-{-S^ji-2v-2^-\-4>)dw + (2w-\-2oL-v~^ + l)dv = 0. 

Now choose a and /9 so that 

3a- 2^ + 4 = and 2a-/34-l = 0, 

or -a4-2 = 0, or a = 2, /3=5. 

Therefore the substitution ought to be x=tu-^2, y = v-\-b, 
and the equation becomes a homogeneous one. 

Exercise. (3y - 1 x -{■ 1) dx -^ {1y - Sx + 3) dy=^0. ^ 'n.'- 
Answer : (y — x + 1)^ (y + x — ly — c. 

r, • dy 2x — y-\-l _ 

Exercise. -f + ^ ^ — ^ = 0. 

ax zy — x—1 

Answer : x^ — xy-{-y'^-\-x —y=c. 

241. Exact Differential Equations are those which 
have been derived by the differentiation of a function of x 
and y, not being afterwards multiplied or divided by any 
function of x and y. Consult Art. 83. 


Mdx 4- Ndy = is an exact differential equation if 
.dM\ (dN\, ,, df{x,y) ,^ d., . 

the primitive being /(^, y)=c. It will be found that 

(x" '-Safy)dx-\-(y^- a-') dy = 0, 
is an exact differential equation. 

Then x'-Sx^y=^ ^-^'^f''^\ 


Integrating therefore, as if y were constant, and adding Y 
an unknown function of y, instead of a constant, 
f{x,y) = la^-x^y^Y. 

Differentiating as if /c were constant, and equating to Ny 
we have 


-J- = y^ and hence Y=^y^ + c. 

Hence ^x" - afy + ^y^ + c = 0, 

where c is any arbitrary constant. 

242. Any equation M . dx-\- N .dy = may be made 
exact by multiplying by some function of x called an Inte- 
grating Factor. See Art. 83. For the finding of such 
factors, students are referred to the standard works on 
differential equations. 

243. Linear equations of the first order. 
These are of the type 

s+^y=« W' 

where P and Q are functions of x. 

The general solution is this. Let IP.c^a* be called X, 

y=e^{/e^.Q.dx+cJ (2), 

where C is an arbitrary constant. 


No proof of this need be given, other than that if the 
vakie .of y is tried, it will be found to satisfy the equation. 
Here is the trial : — 

(2) is the same as ye^ = I e^Qdx-\- G (3). 



Diiferentiating, and recollecting that -y— = P, (3) becomes 


e^ + ye^P=e^Q (4), 

ov -~- + Py — Q, the original equation. 

To obtain the answer (2) from (1), multiply (1) by e^ and 
we get (4) ; integi-ate (4) and we have (3) ; divide by e^ and 
we have (2). 

We have, before, put (1) in the form 

(e + P)y=Q or y^ie + PrQ. 
and now we see the general meaning of the inverse operation 

(d + P)-\ In fact if I P.dwhe called X, 

{d + P)-i Q means, e^^ || e=^ . Q . dx + c| . . .(5). 

Thus if Q is 0, (0 + P)-i = Ce-^. Again, if P is a con- 
stant a, and if Q is 0, then (6 + a)~^ = Ce~"^, where C is an 
arbitrary constant. We had this in Art. 168. 

Again, if Q is also a constant, say n, 

{6 4- a)-' n = e""^ I ( He'*^ . dx -[- C 


= a€-«* + - (6), 


where G is an arbitrary constant. See Art. 169. " 
Again, if Q = e^^, 

((9 + a)-' e^ = e-^'-^ | j €'«+^) "^ . dx ■{■ g\ 




It is easy to show that when a = — h,y = (G-i-a;) €~"* 
li Q — h sin (ex + e), 

(6 + a)-' h sin {ex -\-e) = e-«* j h | €*»* sin {ex + e) . dx + C 

= Ce-«^ + , sin fco; + e - tan-^ -^ (8). 

244. Example. In an electric circuit let the voltage 
at the time t be F, and let (7 be the current, the resistance 
being R and the self-induction L. We have the well-known 

F=ie(7 + X^, 
or ^^^G=\rV. 


dt L L 

--t (1 f-t ) 
and hence C=6 ^ ]j j^^ F . cZ^ 4- constant -4 [ (1). 

Of this we may have many cases. 

1st. Let F at time 0, suddenly change from having been 
a constant Fj, to another constant V^. Put F=F2 therefore 
in the above answer, and we have 

0=r^'UF,e^ + 4| 

V -it 


To determine A we know that = ^ when^=0; there- 

V V V — V 

fore -^^ = ^^ 4- ^ so that A = — -„ — ^ and hence 
±1 li K 

c=Yi-Y^^I-^r^' (2) 


K/ --t\ 
Thus if Fi was 0, ^=^(1--^^) (3)> 

showing how a current rises when a circuit is closed. 

V -^t 
Again if F2 is 0, ^=:^^^ ^^)' 

showing how a current falls when an electromotive force is 

Students ought to plot these values of C with time. 

Take as an example, F^=100 volts in (3), i2 = 1 ohm, 
L = -01 Henry. 

Again take Vi= 100 volts in (4). Compare Art. 169. 

2nd. Let V at time suddenly become 

Vq sin qt, 

a=e ^ 


V f -t 

~ le^ .smqt.dt-\-A 

itfR . 

R \ ^r ^^ f -f sin qt — q cos qt ) 

-^t V 
This becomes C = Ae ^ + , " sin (qt -e).. .(5), 

where tan e = ~ . 


The constant JL, of the evanescent term Ae ^ , depends 
upon the initial conditions ; thus if (7 = when ^ = 0, 

= ^.-^=^ sing, 

or A = VoLql(R' + Df). 

Students ought to plot curves of several examples, taking 
other initial conditions. 


245. Example. A body of mass M, moving with velocity 
V, in a fluid which exerts a resistance to its motion, of the 
amount fv, is acted upon by a force whose amount is F at 
the time t. The equation is 

M%+fv = F. 

Notice that this is exactly the electrical case, if M stands 
for Z,/for R, F for Volts V, v for C; and we have exactly the 
same solutions if we take it that F is constant, or that F 
alters from one constant value Fi to another F.., or that F 
follows a law like Fo sin qt 

This analogy might be made much use of by lecturers on 
electricity. A mechanical model to illustrate how electric 
currents are created or destroyed could easily be made. 

The solutions of Linear Differential Equations with 
constant coefficients have such practical uses in engineering 
calculations that we took up the subject and gave many 
examples in Chap. II. Possibly the student may do well 
now to read Art. 151 over again at this place. 

246. Example, x ^ = ay-^x + \, 

dy a ^ 1 



— . dx=X= — a loff a-. 
X ^ 

Observe that e-'*^^'^'" = x'"', e^^^K'^ = x^. 
Hence y = x^\ L"" ( 1 + -\ dx -\- c\ y 

y = x^ ||(^-« + x-^-^) dx-^c\, 

" \ — a a 
the answer, where C is an arbitrary constant. 



247. There being continuous lubricating liquid between the surfaces 
AB and EF as of a brass and a journal. 
OC=/iQ the nearest distance between them. 
At the distance ^, measured along the arc 
OAj let the thickness be h. Anywhere in. 
the normal line there, representing the 
thickness, let there be a point in the liquid 
at the distance y from the journal, and 
let the velocity of the liquid there be 2t. 
Then if p be the pressure it can be shown 

'^- 541 (^)' 

if /u, is the coefficient of viscosity of the 
lubricant, and Wq the linear velocity of 
the journal, and ic is the velocity of the 
liquid at any place; we have no space for the reasoning from (1) 

G^Uq dh 
h^ duo 

Fig. 103. 

leading to 

d'^p 3 dh dp 
dx^ h ' dx dx 



dp _ 



d4 3 
dx h 

dh 6/iWg ^^^ _ /^ 

dx ' ^ A3 dx'' 

This is of the shape (1) Art. 243, h in terms of x being given. 
Let X= JP . dx= f f . ^ . dx=3\ogh, e^=h\ Hence 

(t> = h 



QjiUq dh 
h^ dx 

dx + & 


A-'(6;.«„A + = ^?+f3. 

The solution depends upon the law of variation of A. The real case 
is most simply approximated to by h = hQ + ax^: using this we find 

p = C' 




If students were to spend a few weeks on this example they might 
be induced to consult the original paper by Prof. 0. Reynolds in the 
Phil. Trans, vol. 177, in which he first explained to engineers the theory 
of lubrication, t 

Numerical Exercise. Let OB = 2-59, OA = 11 -09 centimetres. 
fi = 2-l6, Aoor 0C=0-001135, a- -0000082, W(, = 80cm. per second. 


Calculate C and C\ assuming jo = at B and at A. 

Now calculate the pressure for various values of x and graph it on 

squared paper. The friction per square cm. being /ij- at y=0, the 

total friction F will be found to be 

between the limits A and B. The total load on the bearing is \p.dx 

bearing is / ^ . ( 

between the limits, if AB covers only a small part of the journal, and 
may be calculated easily in any case. 

The bearing is supposed to be infinitely long at right angles to the 
pai)er in fig. 103, but forces are reckoned i)er cm. of length. 

248. Example. Solve 

Writing this in the form (^ - 4^ + 3) 7/ = 2e^, 

Now (e^ - 4^ + 3)- = i (^-^^ - ^ :^ i) • 

Indeed we need not have been so careful about the J as 
it is obvious that the general solution is the sum of the 
two (^— 3)~^ and (^ — 1)~^ each multiplied by an arbitrary 

Anyhow, y = ^__^_^, 

and this by (7) Art. 243 is 

or y=(C^+a')^\ 

249. Equations like ^ 4- Py = Qy", where P and Q, 
as before, are functions of oo only. 

Divide all across by y* and substitute z = y^~'\ and the 
equation becomes linear. 

Example. (1 — w"^) -^ — cci/ = ax\f. 


Substituting z = y~^ we find 

dz xz ax 

dx \—x^ \-x'^' 

Answer : 

2/-1 = (1 _ ^)*{_ a(l -j;2)-^ + 0} = - a + OVr^T^^. 

Exercise. ^~j -^V =y^ log iz*. 

Answer : - = 1 + Oa; + los: a?. 
2/ ^ 


. 1. Given C^Y-ay = 0. 

This is an equation of the first order and second degree. 

Solve for -^ and we find two results, 


-^ — ai/ = 0, so that log y—ax—A^ = 0, 

-^ 4- a^ = 0, so that log y + ax— A2 = 0. 

Hence the solution is 

(log y - ax - A,) (log y + ax-A^)= 0. 

It will be found that each value of y only involves one 
arbitrary constant, although two are shown in the equation. 

2. Given i + (^^Y = ^, 

This is an equation of the first order and third degree. 

Hence 2 = (.^-l)*, 

and y = |(.r-l)HG 


3. Given (^y_7^ + 12 = 0. 

\dxj ax 

This is an equation of the first order and second degree. 


(3/ - 4^ + Ci) (y - S.'Zr + Co) = 0. 

251. Clairaut's equation is of the first order and of any 

y^xp-Vf{p) (1), 

where p is -^ and f{p) is any function of ~ . 
Differentiate with regard to x, and we find 

^■^iMt=' <^)- 

So, either ^ = (3) 

^ + ^fiP) = 0.... (4) 

will satisfy the equation. 

1=0., =0. 

Substituting this in (1) we have 

y = CX+f{c) (5), 

which is the complete solution. 

Eliminating^ now between (1) and (4) we obtain another 
solution which contains no arbitrary constant. Much may 
be said about this Singular Solution as it is called. It is 
the result of eliminating c from the family of curves (5), and 
is, therefore, their Envelope. See Art. 224. 

Example of Clairaut's equation. 

We have the general solution (5) ... y — cxA — , a family 



of straight lines the members of which differ in the values of 
their c. 

Hence y = 2 Jmx or y^ = 4ma;, a parabola which we found 
to be the envelope of the family in Art. 224. 

This curve satisfies the original equation, because in any 

infinitesimal length, the values of x, y and -^ are the same 

for it as for a member of the family of straight lines. 

252. If a differential equation is of the form 

it can be at once solved by successive integration. We 
have had many examples of this in our work already. 

253. Equations of the form -^^=f{y); multiply by 2 ^ 

cix dx 

and integrate and we have 

Extracting the square root, the equation may be solved, 
as the variables are separated. 

Thus let g = ay 

Proceeding as above, 

y -=^dx. 

Integrating we find 

x=^^\og{ay + JaY-^C}-\-C' (1). 


If this equation (1) is put in the shape 
ce'^^ = ay-\- Ja^ + C, 
it becomes c^e^'' - ^ayce*"^ = C, 

^ la 

or ?/=^6«^ + ^e-«^ (2), 

which looks different from (1) but is really the same. (2) is 
what we obtain at once if we solve according to the rule for 
linear equations, Art. 159. 

254. Solve -7^= a -s~, an equation of the third order 
and first degree. 

Let, g = ?, theng = a5, 

9 = 6e«* so that t^ = - «"^ + G, 

^ a? 

or y == Ae"^ •\- Coo+ C\ where A, 0, C are arbitrary constants. 
This also might have been solved by the rule for linear 
equations, Art. 159. 

255. Solv( 

a. dp 

^^=:dx SO that -=log {p+ V«2+ 1) +C", 

or (7e<»-jp = Vp2 + l. 

1 -^ d. 
20 ~dx 

- 1 -- dti 
Squaring, we find p = ^Ce** — ^r^ e " 


Integrating this we have 

X X 

where G and c are arbitrary constants. 

256. General Exercises on Differential Equations. 

(1) {a? + y-) dx + Ja^^^ .dy = 0. 

x X y 

Answer : sm"^ - + — tan"^ ^ = c. 
a a- a 

/^\ dy X 1 

^^ dx^l-\-x^'^ 2x(l+x')' 

. , _ 1 + JU^'] 

1 f ,, l-fVl+^^'l 

(3) _ + ycOS*--^. 

Answer : i/ = sin i» — 1 + Ce" ^^"'^. 

Answer : (2y -x'^-c) (log (a; + y - 1) + ^ - c} = 0. 

(6) J+2^y = 2a^y. 

Answer : y = (Ce^^' + ^a {2x^ + l)}"*. 

(6) 1 = f ^V + ?^ ^ . Answer : y' = 2c^ + c\ 

^ ^ \dxj y dx 

(7) x-^ + y = y^logx. Answer: y == {ex + log x + 1)-\ 

(8) y = 2x^- + y^ C^Y . Answer : y^ = 2c^ + c\ 

(9) Solve (l + 5)c^^-2|c^y=.0. 

1st, after the manner of the Exercise of Art. 241. 

2nd, as a homogeneous equation. 

Answer: af — y"==cx, 

P 22 


(10) Solve — '- f- 1 -^ J c?y = after the manner of 

Art. 241. Answer : x^ — y^ = cif. 

Answer : y = {x-[-\y [^ {x + 1)^ + C). 

(12) (a; + 2/)2 ^ = a\ Answer : y-a tan-^ ^^ = c. 

(13) xy (1 4- a??/2) ^ = 1. Answer : - = 2 - 2/^ + ce-^J/'. 

Answer : (y —a log a; — c) (?/ + a log a; — c) = 0. 
Answer : y = (ao + aio; + aoa?'' + a^) e~^. 

Answer : a? = (ai + dj^) e-^' + /-^r— Tvj • 
Answer : 3/ = (a? + 1)" (e* + c). 

(19) (1 —x^)-f- — xy = axy\ Answer : - = c Jl — a^ - a. 

(1^) yi = sin (^ - 6). Answer : cot |j - ^-^ I = <^ + c. 

Answer: 3/ = C^ie'^^ + e^ Jf Og-^j cos a; + fOg + ^jsinajl . 
(21) Change the independent variable from a; to i in 
(1 — a?) -~ — a? -^ = 0, if a; = sin f, and solve the equation. 


(22) Also in {a- +x'')-j^^-\-2x -^ = 0, ii x=a tan t, and 
solve the equation. % 

«■ '■ '-■• S=-£/©'. 

, „ d's [dt dH „fdHV\ fdty 

*""* ^- -dt= = -\ds-d^-^W)\^\ds)- 

3. Prove as i^ = -^ / tt, so we have also 

ax dt I dt 

d^y _ fdx (Py _ d'x dy\ ^ /dxV 
dx^ ~ [dt df dt^ ' dt) ' \dt) ' 

and find the equivalent expression for -^^ . 

4. If X = e* show that b,sx^ — x-^-~-t: this equals -^ . 

dx dt dt ^ dt 

Also a^^^Jl-^^^-fl-l]^ 

^^^"^ "^ da?~ dP dt~\dt V dt' 

and a^^-(l-2\(^-l]^ 

^"""^ "^da--\dt V\dt ^Jdt' 

5. Change the independent variable from x to t if x = €*, 

dx^ dx 

in x^ ;r 2 + ^' ;/ + ^^^y = ^> ^^^ ^^^^^ *^® equation. 

258. If we try to find by the method of Art. 47 the length of the arc 
of an elhpse, we encounter the second class Elliptic Integral which is 
called E (k, x). It may be evaluated in an infinite series. Its value 
has been calculated for values of k and x and tabulated in Mathemati- 
cal tables. 

When the angle through which a pendulum swings is not small, and 
we try to find the periodic time, we encounter the first class Elliptic 
Integral which is called F (k, x). It can be shown that the integral of 
any algebraic expression involving the square root of a polynomial of 
the third or fourth degree may be made to depend on one or more of 
the three integrals 



rx /i _ z.2«.2 re . 

f 1 \ i"" ^ 

7r(w, h. 6)=\ 




k, which is always positiA e and less than 1, is called the tnodulus. 
n, which is any real number, is called the parameter. 

The change from the x form to the 6 form is effected by the substi- 
tution, :r = sin 6. When the limits of F and E are 1 and in the x case, 

or - and in the B case, the integrals are called complete, and the letters 

K and E m erely are used for them. B is called the amplitude and 
\/\ — ^2sin2^ is called by the name A^. 

If u = F{k, x) = F{ky B\ then in dealing with fimctions which have 
the same k if we use the names 

^ = amw, 

x=Bnu (in words, x is the sine of the amplitude of w), 
*^l — x^=cmi (or Vl — -^ is the cosine of the amplitude of u), 
Vl - k^x^=dnu (or Vl - k'^^^ is the delta of the amplitude of u), 
it is found that 

sn^w-f cn2?< = l, dn'^u+k^.Bn^u=l, -=-(amw)=dnw, &c. 

Also am { — u)=— am . ti, &c. 

., / . \ sn w. en V. dn v + cni/. snv.dn . -M 
Also sn {u±v) = Yx-== r, , 

and similar relations for en {u±v) and dn {u±v). 

Expressions for sn (tc + v)+sn(u- v), &c. follow. Also for 
sn2u, cn2w, dn2u. 

So that there is as complete a set of formulae connecting these 
elliptic functions, as connect the Trigonometrical functions, and there are 
series by means of which tables of them may be calculated. Legendre 
published tables of the first and second class integrals, and as they have 
known relations with those of the third class, special values of these 
and of the various elliptic functions may be worked out. If complete 
tables of them existed, it is possible that these functions might be 
familiar to practical men. 

259. To return to our dififerentiation of IHinctions 
of two or more variables. 


1. If u = z'^ + 7/^ -\- zy and ^ = sin a; and y = e^, 

and hence ^ = (Si/^ + ^) e* + (2^ + y) cos a;. If this is expressed 

all in terms of x we have the same answer that we should 
have had, if we had substituted for y and z in terms of x in 
u originally, and differentiated directly. 

2. If u= a/ — — „ , where v and w are functions of «?, 

V v^ + w- 

nnd -Y . 

3. If sin (xy) = ma?, find -^ . 

4. If %i — sin~^ - , where z and y are functions of x, find 


5. If \i = tan~^ - , show that du, = ^-^^ — -^ . 

y f + 2^ 

260. Exercise. Try if the equation 

dx^~Kdt ^ ■ 

has a solution like v = e*^ sin (qt + yx), and if so, find a and y, 
and make it fit the case in which v=0 when x = oo , and 
v = asm qt where x = 0. We leave out the brackets of 




-7- = «€"* sin (qt + yx) + e*^ 7 cos (qt + ^cij), 

— — = a^e*^ sin (g^ + yx) + a7€*^ cos (qt + 7.<c) 

+ a7e** cos (qt + yx) — 6**7^ sin (qt + 70;). 

Also -^ = ge*^ cos (qt + 7^), 



SO that to satisfy (1) for all values of t and oc 
a"-rf=:0 or a= ±y, 


ay + ay 


As - is not zero, a = + 7 only, 
2a=» = ^, a= + 


Hence we have 

v= A e"'* sin (qt -f cue) -f i^e"** sin (qt — ax) 

where A and B are any constants, and ol= /u -x— or a/ 

if ^ = 27r/2. Now if i; = when x= oc , obviousl}^ A=0. If 
V = a sin qt where x=0, obviously B = a. 
Hence the answer is 


V — ae 

V " sin (27r?^^ - .'?? W 

K ) 


261. Let a point P be moving in a curved path 

^PQ; let^P = a^,j5P = y, 

Cm *ir (ill 

-^ and -~ being the velocities in the directions OX and OY^ 

-j~ and -7^ being the accelerations in the directions OX 

and OY. 

Let OP=r, BOP —6,x = r cos ^, ?/ = r sin ^. The acceler- 
ation or velocity of P in any 
direction is to be obtained just 
as we resolve forces. Thus the 
velocity in the direction r is 


and in the direction PT which 
is at right angles to r, the 
velocity is Pig^ 104 

-t«in^ + ^!cos^ (2). 











Now diiferentiating a; and y, as functions of the variables 
r and 0, since ( ^— j = cos and f -^ J = — ?* sin 6, 

dx dr ^ . ^ dO ,.-,. 

5«=*'=°^^-'-^'°^rfi ^^^' 

dy dr . ^ ^dO ,^. 

Solving (3) and (4) for j- and r -j- , we find 

dr dx ^ dy . ^ 

= ;i- cos ^ +-^ sm d (o), 

(^^ dt dt 

dO dx . ^ a 
dt dt dt 

dO dx . ^ dy ^ ,_. 

From (1) and (2) we see therefore that -j- is the velocity 



in the direction OP and that r -^ is the velocity in the 

direction PT. Some readers may think this obvious. 

Now if we resolve the x and y accelerations in the 
direction of OP and PT, as we did the velocities, and if we 
again differentiate (3) and (4) with regard to t, we find 

d^x d^y 

Acceleration in direction OP = -i- cos ^ + -i^ sin ^ . . .(7), 

d^x d^y 

Acceleration in direction PT— — ^ sin ^ + -j^ cos ^...(8). 


S={£-'©"}-*-('Sf-S) '"•■■<'>. 

And hence, the acceleration in the direction r is (and 
this is not very obvious without our proof), 

dt' '\dt) 



and the acceleration in the direction PT is 

^dtdt^"^ dt^'°' rdtK dt) ^^^^- 

r^ -J- is usually called A. It is evidently twice the area per 
second swept over by the radius vector, and (12) is - -^ . 

262. If the force causing motion is a central Force, 
an attraction in the direction PO, which ia a function of 
r per unit mass of P, say /(r) ; or mf{r) on the mass m at 

P; then (12) is 0, or r^ -^- = constant, or h constant. Hence 

under the influence of a central force, the radius vector 
sweeps out equal areas in equal times. 

Equating mf{r) to the mass multiplied by the accelera- 
tion in the direction PO we have 

/<')"©-!?■ "»> 

But r"^ -^ = h a. constant. As r is a function of 6 

dr _dr dd _dr h 
^ ^\^h _ h (dA^^)h 
df'~\d6'7'' 7^\dd)]r'' 
and we can use these values in (13) to eliminate t. 

If we use - for r, (13) simplifies into 

f(r)=h2u2(^+u) (14). 

If f(r) = a?'~" or a^*", an attraction varying inversely as 
the ?ith power of the distance, -^ + ^^ = I^ '^^~^ — ^w"~^ say. 

Multiplying by 2 -^ and integrating, we have 

(sr— «-?!— <->• 


Thus let the law be that of the inverse square; 
f{r) — ar~^ or ait^ ; (14) becomes 

(Pu a , 

Let lu — iL — h, then 

The solution of this is, 

-m; = ^ cos (^ + B), 
and it may be written 

i,= l = ^{H-ecos(^-a)l (IG). 

This is known as the polar equation to a conic section, 
the focus being the pole. The nature of the conic section 
depends upon the initial conditions. 

(15) enables us, when given the shape of path, to 
find the law of central force which produces it. Thus if a 
particle describes an ellipse under an attraction always 
directed towards the centre, it will be found that the force of 
attraction is proportional to distance. It is easier when given 
this law to find the path. For if the force is proportional 
to PO, the X component of it is proportional to oo, and the y 
component to y. If the accelerations in these directions are 
written down, we find that simple harmonic motions of the 
same period are executed in these two directions and the 
composition of such motions is well known to give an 
elliptic path. If the law of attraction is the inverse cube 

or f{r) = (7?'~^ = au^, (14) becomes -j^- + ?^ = r^ ^^• 
If ^-l = a-, w = ^6*« + 56-*^ 


If l-~ = ^2 u = Asm0 + Bcos0e, 


giving curiously different answers according to the initial 
conditions of the motion. 


263. If a? = r cos 0,y = r sin 6, so that if y is a function 
of x, r must be a function of ^ ; if i* is any function of x and y, 
it is also a function of r and 6. 

Express f -7- ) and f -^ j in terms of the polar co-ordinates 
r and 6. 

\drJ~\dx)dJi^^\d^)d^ ^^^' 

6 being supposed constant, and 

fdu\ _ fdu\ dx /dic\ dy 

\de)~\Tx)de^\d^jlde ^^^' 

r being supposed constant. 

Now -77; if r is constant = — ?- sin 0, 


-M. if r is constant = r cos ^, 

-7- if ^ is constant = cos ^, 

— ^ if ^ is constant = sin ^. 

Treating (-7- J and (-7-) in (1) and (2) as unkno\vn, and 
finding them, we have 

(|) = .„..g).Jeos..(|) (4). 

Notice that in [-,— ], the bracket means that y is supposed 
to be constant in the differentiation. 

In ( -7- j , it is ^ that is supposed to be constant. 

r, ^, <^ CO-ORDINATES. 


In (o) or (4) treat ( i-l or ( -, ) as it is treated, and find 

^i— and Ti^ . However carefully one works, mistakes are 
da^ dy^ *^ 

likely to occur, and this practice is excellent as one must 

think very carefully at every step. Prove that 

d2u d2u 


d^'^dy2"dr^ "^^r dr'*"?d9^' 

1 du 
r dr 

1 d2u 


264. Sometimes instead of x, y and z, we use r, 6, (f) 
co-ordinates for a point in space. Imagine that from the 
centre of the earth (Fig. 105), we have OZ the axis of the 
earth, OX a line at right angles to OZ, the plane ZOX being 
through Greenwich ; OF a line at right angles to the other 
two. The position of a point P is defined by x its distance 

from the plane ZOY, y its distance from the plane ZOX, z its 
distance from the equatorial plane YOX. Let r be OP the 
distance of the point from 0. Let <^ be the west longitude 
or the angle between the planes POZ and XOZ\ or if Q be 
the foot of the perpendicular from P upoii XOY^ the angle 
QOX is <^. Let Q be the co-latitude or the angle POZ. 
Then it is easy for anyone who has done practical geometry 
to see that, drawing the lines in the figure, QfiO is a 


right angle and OR = x, QR — y, also PQO is a right angle, 
a; = rsm6. cos </>, y = r sin 6 .sincf), z = r cos 6. If u is a 
given function of x, y and z, it can be expressed in terms of 
r, 6 and ^, by making substitutions. It is an excellent 
exercise to prove 

du . ^ , du cos 6 . cos <f) cZm sin ci c??^ 

-y- = sin ^ cos </> ;t- H ^ -ja ^n -ji , 

dx ^ dr r do r sm 6 d<f> 

du . ^ . , du cos 6 ,sin<f> du cos <f> du 
dy ^ dr r dd r sm 6 d<f> 

du _ ^ cZ?^ sin ^ du 
dz dr r dO' 

It will be noticed that we easily slip into the habit of 
leaving out the brackets indicating partial differentiation. 

The average student will not have the patience, possibly 
he may not be able to work sufficiently accurately, to prove 

d2u d^ d2u_d2u l^d^ 
33? + dy2 ■*■ dF " dp" "*" r2 d(92 

1 d2u 2du cotg du 

■^r2sin2^'d^ + 7d?"^"J2-d9 ^ ^• 

This relation is of very great practical importance. 

265. The foundation of much practical work consists in 
understanding the equation 

d^u dhc d^u__ldu 
dx^'^df'^'d?~'K~dt ^ ^' 

where t is time. For example, we must solve (1) in Heat 

Conduction Problems if u is temperature, or in case ;t- = 

and u is electric or magnetic potential, or velocity potential, 
in Hydrodynamics. 

(1) is usually written 

^^-4S (^>- 


We see then in (A) the form that V"u takes, in terms of r, 
and </> co-ordinates. 

We know that if u is symmetrical about the axis of z, 
that is, if u is independent of </>, the above expression 

_, dhi 1 d-ii 2 da cot 6 du ,^. 

^'''=d;^-^?w^-^rd^+-^d0 <^>- 

266. Students are asked to work out every step of the 
following long example with great care. The more time 
taken, the better. This example contains all the essential 
part of the theory of Zonal Spherical Harmonics^ so 
very useful in Practical Problems in Heat, Magnetism, 
Electricity, Hydrodynamics and Gravitation. When to is 
independent of </> we sometimes write (2) in the form 

du K id f „du\ 1 d / . ^du 

'hMei^-'U ^^>' 


u being a function of time t, r and 6, The student had 
better see if it is correct according to (3). 

If -^ = 0, show that the equation becomes 

„d^u - du ^ ndu d^u _ ,^. 

,._ + 2r^^ + cot^^^ + ^, = (2). 

Try if there is a solution of the form u = RP where R is 
a function of r only, and P is a function of 6 only, and show 
that we have 

T^d^R 2rdR ,AdP 1 d'P ,^, 

R^^^Rd^=-''''^^p~de-pd¥ (^^- 

Now the left-hand side contains only r and no 0, the 
right-hand side contains only 6 and no r. Consequently each 
of them must be a constant. Let this constant be called C 
and we have 

drR ldR_RG_ 

dr' +r dr r^ ~" ^^^' 

%^.oie%^PG = (5). 


There is no restriction as to the value of 0, and it must 
be the same in (4) and (5), and then the product of the two 
answers is a value of u which will satisfy (2). The solutions 
of many linear Partial Differential Equations are obtained 
in the form of a product in this way. There are numberless 
other solutions but we can make good practical use of these. 

We have then reduced our solution of the Partial Differ- 
ential Equation (1), to the solution of a pair of ordinary 
differential equations (4) and (5). Now a solution of (4) may 
be found by trial to be r*", and when this is the case we have 
a method (see Art. 268) of proving the general solution to be 

ii = ^r»« + 5r-(»«+i' (6), 

where G is m (m + 1); anyhow (6) will be found by trial to 
answer. Using this way of writing G in (5) and letting 
cos = fjb, we find that we have an equation called Legendre's 
Equation, an ordinary linear equation of the 2nd order 

We now find it convenient to restrict m. Let m be a 
positive integer, and try if there is a solution of (7) in the 

P = 1 + Aj/M + A^^ + A^^ + &c. 

Calling it P^ (/it) or P,^ (0), the answers are foimd to be 

Pq (0) = 1, if m is put 0, Pi (0) = ^, if m is put 1, 

Pg (0) = 3^2 _ ^, if ^ is put 2, Ps (6) = f /!« - f /x, if m is put 3, 

P, (0) = ^fi* - ^fi^ + 3, if m is put 4. 

A student will find it a good exercise to work out these 
to Pg. My pupils have worked out tables of values of Po, Pi, 
P2, &c., to F7 for every degree from ^ = to ^=180°. See 
the Proceedings of the Physical Society, London, Nov. 14, 
1890, where clear instructions are given as to the use of 
Zonal Harmonics in solving practical problems. 

We see then that 

(Ar- + j^,)p„(<)) (8) 


is a solution of (1). A practical problem usually consists in 
this : — Find u to satisfy (1) and also to satisfy certain limit- 
ing conditions. In a great number of cases terms like (8) 
have only to be added together to give the complete solution 

In the present book I think that it would be unwise to do 
more in this subject than to set the above very beautiful 
exercise as an example of easy differentiation. 

(8) is usually called The Solid Zonal Harmonic of 
the mth degree^ P^ (6) is called the Surface Zonal 
Harmonic of the mth degree. 

267. In many axial problems, u is a function only of 
time and of r the distance of a point from an axis, and we 
require solutions of (1) which in this case becomes 

d^u \ du ^\ du 

dr^ r dr k di * 

Let us, as before, look for a solution in the form 

u = BT (2), 

where i? is a function of r only and J' is a function of t only. 
(1) becomes 


dr^ r dr k dt' 
Dividing by RT 

Rdr'^r Rdr~ kT dt~ ^"^ ^^^' 
where /^^ is a constant. 


Then -m = - /cfi^dt or log T= — k/jlH + c, or 

T=Ge-'^f^'i (3), 

where G is an arbitrary constant. We must now solve 

d'R IdR ,^ ^ ,,, 

^+-r;^+^^^=^ (^)- 

Let r = - and (4) becomes 

d2R 1 dR . „ ^ ,^^ 

dF + xdJ + ^ = ^ (^>- 


Assume now that there is a solution of (5) of the shape 

we find that A = G = E= G=0 and in fact that 
x^ X^ x^ x^ 

This is an important series first used by Fourier, although 
it has Bessels name. It is called the Zeroth Bessel and 
the symbol Jo (*") is used for it. Tables are published which 
enable us for any value of x to find Jo(x). Thus then 
R = Jo {^r) is a solution of (4), and hence 

u=Ge-'^|'''J,{^lr) (7) 

is a solution of (1). Any solution of (1) needed in a practical 
problem is usually built up of the sum of terms like (7), where 
different values of iju and different values of G are selected to 
suit the given conditions. 

268. In the linear differential equation 

when P and Q are functions of x, if we know a particular 
solution^ say y=^v, we can find the general solution. 

Substitute y = vii, and we get 

"d^+^id-. + ^Vd^-"' (2). 

Calling -p= u\ (2) becomes 

U V 


log u' + log v^+ I P .dx = constant. 


Let I F .dx = X then u' or -^ = A ~ e-^, 
J ax v^ 

u^B + Al^^e-'^.dx (3). 

Thus we lind the general solution 

y = Bv-\- Av I — €~^ .dx .(4), 

where A and B are arbitrary constants. Even if the 
right-hand side is not zero, the above substitution will 
enable the solution to be found, if v is a solution when the 
right-hand side is 0. 

Easy Example. One solution of 

-r^ + a^x = is ?/ = cos ax. 
dx^ "^ 

Find the general solution. 

Here P = so that jP.dx^X^ 0. 

f dx 
Hence y = B cos ax + A cos ax I — -i, — , 

•^ J cos^ax 

and as I — - — = - tan ax, 

J cos^ ax a 

we have as the general solution 

y = B cos ax+G sin ax. 

Exercise. We find by trial that y = x^"' is a solution of 
ic* ^ + 2a; ^ - m (m + 1) y = 0, see Art. 266. 

Show that the general solution is y=Ax^-^ — ;^^. 


Exercise. We find by trial that y — e^^ is a solution of 


-T^ = a^y, show that the general solution is y = Ae^^ + i?e~^*. 

P. 23 



E.rercise. We saw that u — P^n{t^) is a solution of 
Legcndre's equation Art. 2GG, prove that xl =APin(fi)-\-BQmi^) 
is the general solution, where 



Qm(H') or Qui(0) is called the Surface Zonal Harmonic 
of the second kind. 

Exercise. We saw that Jo(x) was a solution of the 
Bessel equation (5), show that the general solution is 
AJq {x) -f BKq {x)y where 

Kq (x) is called the Zeroth Bessel of the second kind. 

269. Conduction of Heat. If material supposed to 
be homogeneous has a plane face AB, 
If at the point P which is at the dis- 
tance X from AB, the temperature is v, 
and we imagine the temperature the same 
at all points in the same plane as P 
parallel to -4^ (that is, we are only con- 
sidering flow of heat at right angles 

to the plane AB), and if t- is the rate of 

rise of temperature per centimetre at P, 

then — A; -7- is the amount of heat flowing 

per second through a square centimetre of 
area like PQ, in the direction of increas- 
Fig. 106. -^g ^ rpj^-g -g ^^^ definition of k, the 

conductivity. We shall imagine k constant. 

k is the heat that flows per second per square centimetre, 

when the temperature gradient is 1. Let us imagine PQ 

exactly a square centimetre in area. Now what is the flow 

across TS, or what is the value of — A; -j- at the new place, 



which is x + hx from the plane ABl Observe that —k-j- 
is a function of x\ call it f{x) for a moment, then the 


space PQTS receives heat f(x) per second, and gives out 
heat f(x + Bx) per second. 

Now f(x-\- Bw) -f{x) = Bx i^^ . 

These expressions are of course absolutely true only 
when Bx is supposed to be smaller and smaller without limit. 

We have then come to the conclusion that —Bx-j- f(x) 
is being added to the space PQTS every second : this is 

— Bx^ri— f^' 1-] or +k.Bx-j—. 
ax \ ax J oar 

But the volume is 1 x Bx^ and if w is the weight per cubic 
centimetre, and if s is the specific heat or the heat required 
to raise unit weight one degree in temperature, then if t is 
time in seconds, 

5j dv 

W .ox .S-Ti 


also measures the rate per second at which the space 
receives heat. Hence 

1 ^ d'^v ^ dv 

tc, ox. -j-- — W.OX.S.-Tly 

da? at 

dH ws dv /,v 

d^^T-dt ^^)- 

This is the fundamental equation in conduction of heat 
problems. Weeks of study would not be thrown away upon 
it. It is in exactly this same way that we arrive at the 
fundamental equations in Electricity and Hydrodynamics. 

If flow were not confined to one direction w^e should have 

the equation (1) of Art. 265. — is often called the diffusivity 

^ ws 

for heat of a material, and is indicated by the Greek letter 
K', wsis, the capacity for heat of unit volume of the material. 

Let us write (1) as 

dh^ __ 1 ^ /.^K 

dx^'Kdi ^"^' 



270. It will be found that there are innumerable solutions 
of this equation, but there is only one that suits a particular 
problem. Let us imagine the average temperature every- 
where to be (it is of no consequence what zero of tempera- 
ture is taken, as only differences enter into our calculations), 
and that 

F= a sin 27r;i^, or a sin ^'^ (3), 

is the law according to which the temperature changes at 

the skin where a; is ; w or ^ means the number of com- 

plete periodic changes per second. Now we have carefully 
examined the cycle of temperature of steam in the clearance 
space of a steam-cylinder, and it follows sufficiently closely 
a simple harmonic law for us to take this as a basis of 
calculation. Take any periodic law one pleases, it consists 
of terms like this, and any complicated case is easily studied. 
Considering the great complexity of the phenomena occurring 
in a steam-cylinder, we think that this idea of simple 
harmonic variation at the surface of the metal, is a good 
enough hypothesis for our guidance. Now we take it that 
although the range of temperature of the actual skin of the 
metal is much less than that of the steam, it is probably 
roughly proportional to it, so we take a to be proportional 
to the range of temperature of the steam. We are not 
now considering the water in the cylinder, on the skin and 
in pockets, as requiring itself to be heated and cooled ; this 
heating and cooling occurs with enormous rapidity, and the 
less there is of such water the better, so it ought to be 
drained away rapidly. But besides this function of the 
water, the layer on the skin acts as creating in the actual 
metal, a range of temperature which approaches that in the 
steam itself, keeping a large. Our n means the number of 
revolutions of t lie engine per second. 

To suit this problem we find the value of v everywhere 
and at all times to be what is given in (2) of Art. 260, 

V = ae 

"V^sin (2',mt-x^^'j (4). 

This is the answer for an infinite mass of material with 
one plane face. It is approximately true in the wall of a thick 



cylinder, if the outside is at temperature 0. If the outside 
is at temperature v and the thickness of the metal is h 

we have only to add a term y ^ to the expression (4). 

This shows the effect of a steam-jacket as far as mere con- 
ductivity is concerned. The steam-jacket diminishes the 
value of a also. Taking (4) as it stands, the result ought 
to be very carefully studied. At any point at the depth x 
there is a simple harmonic rise and fall of temperature every 
revolution of the engine ; but the range gets less and less as 
the depth is greater and greater. Note also that the changes 
lag more as we go deeper. This is exactly the sort of 
thing observed in the buried thermometers at Craigleith 
Quarry, Edinburgh. The changes in temperature were 1st 
of 24 hours period, 2nd of 1 year period ; we give the 
yearly periodic changes, the average results of eighteen 
years* observations. 

Depth in feet 
below surface 

Yearly range 

of temperature 


Time of highest 

3 feet 

6 feet 

12 feet 

24 feet 





August 14 
„ 26 
Sept. 17 
Nov. 7 

Observations at 24 feet below the surface at Calton Hill, 
Edinburgh, showed highest temperature on January 6th. 

Now let us from (4) find the rate at which heat is 

flowing through a square centimetre; that is, calculate —kj 

for any instant ; calling a/ — = a. 


— aae~*^ sin {2irnt — ax) — aae~'^^ cos {27rnt — ax), 

where x = 0, that is, at the skin, it becomes when multiplied 


hy —k; l—k -J-] = + kaa {sin 27mt + cos 27mt] 

= kaa V2 sin (2'Trnt + ^] , by Art. 116. 

This is + for half a revolution when heat is flowing into 
the metal, and it is — for the other half revolution when heat 
is flowing out of the metal. Let us find how much flows 
in ; it will be equal to the amount flowing out. It is really 
the same as 

kaa V2 | sin 27rnt . dt = kaa^/2 . - , where t = - , 
Jo IT n 

= «\/ 


That is, it is inversely proportional to the square root of 
the speed and is proportional to the range of temperature. 

We have here a certain simple exact mathematical result ; 
students must see in what way it can be applied in an 
engineering problem when the phenomena are very compli- 
cated. We may take it as furnishing us with a roughly 
correct notion of what happens. That is, we may take it 
that the latent heat lost by steam in one operation is less 
with steam jacketing, and with drying of the skin ; is pro- 
portional to the range of temperature of the steam, to the 
surface exposed at cut off, and inversely proportional to the 
square root of the speed. Probably what would diminish it 
more than anything else, would be the admixture with the 
steam of some air, or an injection of flaming gas, or some 
vapour less readily condensed than steam. When we use 
many terms of a Fourier development instead of merely 
one, we are led to the result that the heat lost in a steam 
cylinder in one stroke is 

{0,-0,){h + ^^AI>Jn, 

where 6^ is the initial temperature and 6^ the back pressure 
temperature, r the ratio of cut ofi", n the number of revolu- 
tions per minute, A the area of the piston h and c constants 


whose values depend upon the type of engine and the 
arrangements as to drainage and jacketing. 

271. Students will find it convenient to keep by them 
a good list of integrals. It is most important that they 
collect such a list for their own use, but we have always 
found that it gets mislaid unless bound up in some book of 
reference. We therefore print such a list here. Repetition 
was unavoidable. 

Fundamental cases : 

1. Ix"',dx = V x'"+\ 

J m + 1 

2. j - .dx= log X. 

4. la^.dcc = -, a*. 

J log a 

5. /cos mo) .dx= — sin mx. 
J m 

sin mx .dx = cos mx. 


6. /si 

7. I cot x.dx = log (sin x). 

8. I tan X .dx= - log (cos x), 

9. I tan X . sec x.dx = sec x. 
10. I sec^ x.dx — tan x. 

cosec^ X .dx = — cot x. 


„. /, 

, dx 1 . 
12. I = - tan ax. 

cos^ ax a 


13. / -^—z — = — cot ax. 
J sm^ax a 

1 ^ f dx • , ^ 

14. I — . = sin~* - . 
^ V a^ - ar" « 

f dx 1 , X 

J a'^ + x^ a a 


10. / — _ =-sec' 


dx 1 X 

—- zzi SGC~~ ~ 

X ^a?-a? a a 

17*. I cosh ax .dx — - sinh ax. 
J a 

* From 17 to 23 we have used the symbols (called Hyperbolic sines, 
cosines &c. ) 

Binhx=i (c* - e-*), and cosech x— . . , 

cosha' = i(c*+6~*),8echa:= — ^^ , 
^^ ' coah« 

^ , sinh a; e^-1 ^, 1 

tanh x= — i— = -iiT, — - , coth a; = 

cosha: e'^ + l' tanhx' 

Also if 2/ = sinh x, a: = sinh"^ »/. 
It is easy to prove that 

sinh (a + h) = sinh a . cosh h + cosh a . sinh h, 
cosh (a + &) = cosh a cosh h + sinh a . sinh t, 
sinh {a-b) = sinh a cosh & - cosh a sinh Z>, 
cosh (a - ?^) = cosh a cosh t - sinh a sinh t, 
sinh 2a = 2 sinh a . cosh a, 
cosh 2a = cosh^ a + sinh^ a = 2 cosh'^ a - 1 , 
= 2sinh2a + l. 

If we assume that aJ -1, or i as we call it, submits to algebraic rules and 
t2= - 1, t^= -t, 1*^1, 1^ = 1 &c. we can write a + bi as r (cos 6 + i sin ^), where 

r^=a'^+h^, and tan ^=- . It is easy to extract the 7tth root of a + bi: being 

r^ " I cos - + 1 sin - j , and by adding on 27r to 6 as many times as we please, 
we get n, 7ith roots. ■ 

We also find that c** = cos a + i sin a, 

— ia . . 

e = cos a - 1 sm a. 
If z = a + bi-r (cos + i sin ^) = r . e**, 

log 2 = log r + id = ii log {a^ + b"^) + z tan"' - . 


18. I sinh ax.dx = - cosh ax, 
J a 

19. I sech^ ax .dx = - tanh ax. 
J a 

20. I cosech^ ax.dx = — coth ax. 
J a 

21. [ --M^ = sinh-^ - = log [x + V^M^^K 

22. 1 -^^— = cosh-^ - = log {x + ^/^^~^'l • 

23. f,^,==ltanh-f=llog 
J a^ — x- a a za ° 


a —X 

This is indeterminate because tan~i - may have any number of times 2ir 

in it, and indeed the indeterminateness might have been expected as e '"=1. 
Evidently cosh a;= cos ix, 

sinh x= -i sin ix. 
sinh is usually pronounced shin, 
tanh is usually pronounced tank. 

Prove that if if = sinh~ia; or a; = |(e" - e~«), only positive values of u being 
taken, then e" = x + ^/l + x^, and therefore u = sinh"^ x = log {x + >^1 + x"^). 

Similarly cosh-ix = log {x+ Jx- - 1), 
tanh-ia; = ^log^— |, 

sech-ia: = logQ+yi-l), 

cosech-i a; = log (^ + ^^^ + 1) • 
Now compare 

/ -7==5= 8in"^a;, / , = sinh-i rr = log (a; + Vl+^). 

y aJi-x^ j ^1+x^ 

f dx I 

I - .- = cosh-^a; = log {x + ^Jx"^ - 1). 

j-j^,= tan i.r. j j--,= tanh-»a:=ilog ^~^. 


dx 1 , a; — a 

24 ( ^ = -i- log 

25. f^^ = ver«-.^, f- , _... 

dx ,x[ dx i . ,x — a 

— - SIQ" 

27. I \/^Ta* . rfa? = ia; VaM^ + ia' log {a; -f Vi»Ta«}. 

28. I Var' - a^ . da? = Ja; v/^t^ _ ^2 _ |^2 i^g [_^ _^ \/.x"2 - a^}. 

c?a; 1 . , 0. 1 , a 

■ - = — sm~^ - or = - 

Na?- — a? ^ ^ ^ 

29. I — ^'^^ = - - sin"^ - or = - cos~* - . 


a + Vo^T^ 

30. f ,j^-=^iog ; 

J x\/a^±a? a a + V «' ± a^ 

31. |-\/aM^.d'a; = \/anr^-a 

32. I - si a? -a? . rfa; = Va;^ - a^ - a cos"^ - . 

J a? X 

r x.dx , 

r x.dx , 

35. f a? V^To^ . rfa; = J ^{¥Tc^y. 

36. l*a;\/a2-a;2.(Za; = -iV(a2-a^)». 

37. I V2aa; - a?^ . rfa? = ^^ sJ2ax-a^ + ^sin-*^^ 
] 2 2 a 

33 r ^ ^ fx-\ 

i(a; + l)V^-l Va; + X* 


40. I A / , . dx = sin"^ x — s/\— a?. 

J V 1 — « 

^1- \sJ"'-^^'dx=^J{x^a){x^h) 


4- (a — 6) log {sjx 4- a + -s/a; + 6). 

42. I a;"^-i (a + 6a;")9 . c^a?. 

1st. If ^/g' be a positive integer, expand, multiply and 

integrate each term. 

2nd. Assume a + hx^ — y^, and if this fails, 

3rd. Assume ax~'^ ^-h — y^. This also may fail to give 

an easy answer. 

43. I sin~^ X .dx — x sin~^ x-{- Jl— x"^. 

44. I X log x.dx= — (log a? — 1^). 

45. I xe*^"^ dx = - 6«* ( ic j . 

46. I a?"6«^ . c^a; = - aj'^e"^ - - / a;«-i6«* . c^a--. 
J a a] 

Observe this first example of a formula of reduction to 
reduce n by successive steps. 

^^' j^'^^~"^^r:n.^— l■^7;^^ja;--^'^''• 
^ 1 1 r e"^ 

48. I 6**^ log a? . (^a? = - ^'^ log a; I — dx. 

49. rJ^=jiogl±£!^=loglcot(^-f)l. 
j cos a; ^ ^ 1 - sm a; ^ ( V4 2/J 



51- — 7T = / r *^^ 7 ; it a > 6, 

^ Jb—a tan ^ + Jb+a 

log if a < 6. 

Jb^ - «=» /j; ^ ^ /,— - 

Jb — a tan ^ — V^ + fi 

^^ r . , e*''^ (c sin aa? — a cos cw;) 

o2. e*'* sm ax .ax= — ^ r- — ^ . 

J a^ ^ c- 

f _ , ^^ (c Go^ ax ■\- a sin aa;) 
53. I e<^ cos aa; . rfa; = — ^^ :— — r ^ . 

f . „ , cos X . sin"~^ a; w — 1 f . „__ , 

o4. sin** a? . c?a? = + sin"-^ x . dx. 

J n n J 

[ dx _ cos a? 71 — 2 f dx 

J sin** x~ (w — 1) sin**"* a; ri. — 1 j sin""*-^ x ' 

r 6^* . . 

56. I €"* sin" x.dx= ~- sin**~^ x (a sin x — n cos a?) 

J a^ + n^ ^ 

n(n-l) r .^^,^^^^ 
a2 + ?i=' j 

-K,* r . . , sin (7?t — w) •''■ sin(m + 7i)a? 

57*. I sin 7«a; . sm ?ja; . aa; = — ^r-^ r ^ ^ . 

J 2(m-n) 2(m + n) 

-o f 7 sin (?/i — ?i) a; sin(m + w)a? 

58. I cos 7?ia7 . cos 7ix . aa? = -^) f- H — ~ ^- . 

J 2(m-n) 2(m + ?i) 

-^ r . 7 cos(m+?i)a; cos(7ti—n)x 

59. I sin mx . cos ??a; . ax = -^ ^ — -^^ f- . 

; z (??i + n) 2 (m — n) 

60. I sin- na? . dx =:la; — -— sin 2na;. 
j 4fn 

61. I cos^ ?ia; .dx — — sin 2wa; + Jo;. 

* In integrating any of these products 57 to 61 we must recollect the 
following formulae : 

2 sin vix . sin nx = cos (m -n) x- cos (w + 7j) a;. 
2 cos wa; cos 7/a; = cos (m-w)a: + cos (m + n) x. 
2 sin wx cos jja: = sin (m + n) a: + sin (m - n) a;, 
cos 2«x = 2 cos^ ;ia; - 1 = 1 - 2 siu^ nx. 


In the following examples 62 to 67, m and n are supposed 
to be unequal integers. 

rrr or 2n- 

62. I sin 1710) . sin nx .dx= 0. 

Tir or 27r 

63. I COS ??!« . COS iix . dx = 0. 

64. j sin^ nx,dx = - , I cos^ nx. dx = ^ , if /j is an 
Jo -^ Jo ^ 

I si] 



65. I sin mx . cos nx. dx= 

66. / sin ?7ZA' . cos nx. dx=0 if 7?i — n is even. 


67. I sin 771^ . cos nx .dx— „ if vi — n is odd. 

17V — ?r 



68. I sin"* ic . cos x ,dx = 

sin"*+i a; 

^r. r «. • 7 cos'"-^"a; 

69. I cos"* a; . sin a; . ttic = =— 

J m + 1 

Hence any odd power of cos x or sin x may be integrated, 
because we may write it in the form (1 — sin'^a^)"'cos x or 
(1 — cos^ xj^ sin a?, and if we develope we have terms of the 
above shapes. Similarly sin^ x . cos^ x may be integrated 
when either p or q is an odd integer. 

70. I a?"* sin x.dx = — x^ cos x-^-mi x^"'~^ cos x . c?a7. 

71. I ic*" co^ x.dx = x"^ sin x —m j x'^~^ sin a; . c?a;. 

H-^ fsina; , 1 sin a; 1 fcosx , 
* 2, I — — - . dx = = — -— H :, I — — -; dx. 

»_-, fcosx y 1 cos a; 1 Tsina;, 

J a;"* m-la;"*-^ m-lja;^"~^ 

74. [tan** x.dx = (^^^y~' ^ |'(tan x^-' . dx 


75. A'" Sin-l X.dx = = :r I ,- . 

J 71+1 n+lj l-^-a? 

77. — -r = , tan-i ,.==^ , if 4ac > 6^ 

ja-\-bx+ cx^ V4ac - 6" v 4ac - 6^ 

1 , 2cx + h- ^J¥ - Aiac ... ,, 
log . , it 4ac < o-'. 

Vft^ - 4ac 2ca; + 6 + Vt" - 4ac 

2cx + 6 

If X = a + bx + cx^ and 5' = 4<ac — 6^, then 
^^ fc^a; 2cx-{-h 2c fdx 

„ tdx _ 2cx + b ( 1 3c\ ic^ ("da; 
'^- JX^- q [iX'-'^qX)'^ q'jX- 

Q^ r^' . dx bx+2a b [dx 

01 fdx 1 . a^ b [dx 
^^' J^ = 2-J''Sx-Tajx' 

^^ fdx b , X 1 ^(¥ c\ fdx 

83. f__^__ = 2sin-A/^. 
N{x-a){b-x) V b-a 

84. f_=iL==_=-Asin-.y^pp 
Js/{a + bx){oL-px) V6^ V c^^ + 6a 

85. I , = — sec^ I — . 
Jx\lx''-a'' «^ \^/ 


r rig? _ j_ , vv_+^^ 




f dx 

^ \/a-\-hx-\- caf 

= -p log f c^ + 2 + Vc (a^bx + cx'^yj . 

„„ f dx 1 • -1 ^cx-b 

J Na-\-hx — ca? Vc v 4ac + 6^ 

89. i£±g^ may be altered to 

•^ (^ + 2ca?) 2pc-gh J^ 

2c y/a + bx + ex"" 2c V^ 4- hx + c«2 ' 

and so integrated. 

90. Any integral of the form 1 ^ — ^ — T^r^^> where 

P, Q, ii and S are rational integral functions of x, can be 
rationalized by the substitution oi ax-\-b = v^. 

91. Any integral of the form 1 ,_ dx, where U is 

a+bx-\- cx^, can be rationalized, (1) when b^ — 4>ac is positive 
and c negative, by the substitution , ^ = - — L. 

(2) When ¥ — 4iac is positive and c positive by 

Vc"VF _ 2y 

(3) When 6- — 4ac is negative and a positive by 

VcVF _ 1 +y ' 
A/4ac-62 ~ 1-2/' ' 

If ?7 = a 4- 6a; 4- ca;^, g = 4ac — ¥, S = — , 

92 f cZa;_^ 2(2ca;_j-6) 

[_dx__ 2(2cx-{b)\/ U 2S(n-l) f dx 
JU^s/U (2n-l)qU'' "^ 2n-l JU^-i^' 


^. r, (2ca;+6)\/|7 1 [dx 

nr r /- (2c^ + 6)VF/.. 3\ 3 [dx 

ra;rfa; _ J^ _ ^ [ dx 

J W^ s/a "^ I X 2ja) 

1 . _ / bx + 2a \ .. ^„ 
V - a \x V 6^ - 4ac/ 

-2\fU ., . 
= — = if a = 0. 

f dx ^ 'JU b C dx 

J x-^U~~ d^ 2aj^VF" 

272. The solutions of many Physical Problems are given 
in terms of certain well-known definite integrals some of 
which have been tabulated. The study of these is beyond 
the scope of this book. I say a few words about The 
Gamma Function which is defined as 

[ e-^.x^'-Kdx^Tin) (1). 

By parts, I e~^ . a" ,dx=- e'^x^ +n\e-^. a;"-^ . dx. 

Putting these between limits it is easy to prove that 
— e~^x'^ vanishes when x = and a; = x . 

And therefore / e-^x'^ ,dx = nj e~^ . x^~^ .dx .. .(2). 


Hence T (?i4-l) = n r(w) (3), 

so that if n is an integer 

r(n + l) =, &c. n=\n (4). 

Notice that j/i has a meaning only when n is an integer, 
whereas F (ii) is a function of any value of n. 

Tables of the values of F (n) have been calculated, we need 
not here describe how. The proof that 

ra) = V^ (5), 

as given in most books, is very pretty. The result enables 
us through (3) to write out r(f) or F(— f), &;c. 

A very great number of useful definite integrals can be 
evaluated in terms of the Gamma Function. 


Thus 1. [' sin« e,dd=j cos*^ d.dO 
o r «.-./i ^«-lJ r '»'"+' (fo T(m)V(n) 

4. j a;«e-«* dx = a-<'*+^) F (?i + 1 ). 

Jo 2a Vy 2a ^^' 
1. i £ i/»'-> (1 - y)""-' dy = r (I) r (m) - 2r (I + m) . 

2r (£+? + !) 



sinP e cos9 e,d0 = 



The following notes are intended to be rccod in connection with the 
text on the page whose number appears before the note. The exact 
position on the page is indicated by a t. 

Page 3. The ordinary propositions in Geometry ought to be 
illustrated by actual drawing. The best sign of the health of our race 
is shown in this, that for two generations the average British boy has 
been taught Euclid the mind destroying, and he has not deteriorated. 
Euclid's proofs are seemingly logical; advanced students know that 
this is only in appearance. Even if they were logical the Euclidian 
Philosophy ought only to be taught to men who have been Senior 
Wranglei-s. 95 per cent, of the schoolboys, whose lives it makes 
miserable, arc as little capable of taking an interest in abstract reason- 
ing as the other five per cent, are of original thought. 

Page 43. There is a much more accurate method for finding ^ 

described in my book on the Steam Engine. 

Page 48. This rule is not to be used for values of .r greater 
than 16. 

Page 63. In Art. 39 I have given the flue investigations as usually 
given, but prefer the following method, for I have for some time liad 
reason to believe that in a tube of section A, X)erimeter P, if irib. of 
gases flow per second, the loss of heat per second per unit area of tube 
is proportional to 

if t is the absolute temperature of the gases and v the velocity of the 
gases. Now v oc Wt-^A^ so that the loss of heat per second is pro- 
portional to 


Proceeding as before 


or -A^=G.dS, 


or -A\oge-\-c=CS, 

where c and G are constants. As before, this leads to 

c=^log^i, >^=^log|. 

A 6 
. Whole >S'=7j log ^-, so that the efficiency becomes 



Now if it ia a tube of perimeter F and length ^, JS/A becomes PI I A 
or l/m where m is known as the hydraulic mean depth of the flue, so 

This makes the efl&ciency to be independent of the quantity of stuff 
flowing, and within reasonable limits I believe that this is true on the 
assumption of extremely good circulation on the water side. This 
notion, and experiments illustrating it, were pubhshed in 1874 by 
Prof. Osborne Eeynolds before the Manchester Philosophical Society. 

Mr Stanton has recently {Phil. Trans. 1897) published experiments 
which show that we have in this a principle which ought to lead to 
remarkable reductions in the weights of boilers and surface condensers; 
using extremely rapid circulation and fine tubes. 

Page 95. It ought not to be necessary to say here that the 
compressive stress at any point in the section of a beam such as 

AC AC, fig. 47, is -jz, if z is the distance (say JB) of a point on the 

compression side of the neutral line A A from the neutral line. The 
neutral line passes through the centre of the section. If z is negative 
the stress is a tensile stress. The greatest stresses occur where z is 
greatest. Beams of uniform strength are those in which the same 
greatest stress occurs in every section. 

Page 111. Mr George Wilson (Proc. Royal Soc., 1897) describes a 
method of solving the most general problems in continuous beams 
which is simpler than any other. Let there be supports at ^, ^, C, Dj E. 
(1) Imagine no supports except A and E, and find the deflections at 
B, C and D. Now assume only an upward load of any amount at B, 
and find the upward deflections at B, (7, and D. Do the same for C 
and B. These answers enable us to calculate the required upward 
loads at J5, C and B which will just bring these points to their proper 

Page 139. For beginners this is the end of Chap. I. 

Page 146. In all cases then, 

dll=l'.dt + t(^\dv (23*). 

Exercise 1. By means of (23) express K, I, Z, P and V in terms 
of k and write out the most general form of equation (3) in terms of k. 
Show that among many other interesting statements we have what 
Maxwell calls the four Thermodynamical relations, 

U/ \dpJt' WJp \dp)i' \dt) \dv)t' \d(t>), [dvU 
Exercise 2. Prove that in fig. 55, 

Area ABCB=AE. AF, =A U. AJ==AO.AM=AQ. AB, 
and show that these are the above four relations. 



Exercise 3. Show by using (23) with (7), (8), (11) and (14) that for 
any substance 

and that (20) becomes 

(dk\ d^ 

so that k =1-0 -{-t j( ^ j dv, 

where I'q ls a function of temperature only. 

Bividing dR=K.dt+L.(' 
differential, show that we are 

Dividing dH=K .dt+L.dp by t and stating that it is a complete 
led to 


where Kq is a function of temperature only. 

Page 152. Another way is merely to recognize (8) as being the 
same as 

bH=^ +, 

for l — t(~\, and when a pound of stuff of volume s^ receives bH=L 

at constant temperature (or bt = 0) in increasing to the volume 
«2 (so that dv = «2-«i) we have, since dpjdt is independent of v, 

8Zr=Z = + i.^.(52-«i). 

Page 188. Sine functions of the time related to one another by 
linear operators such as a-{-h6-\-c€^-\-etc. + e6~^+fB~^-\-Qic.^ where 

6 means -y- , are represented by and dealt with as vectors in the manner 

here described. Eepresenting an electromotive force and a current in 
this way, the scalar product means Power. Dr Sumpner has shown 
{Proc. Roy. Soc, May 1897) that in many important practical problems 
more complicated kinds of periodic functions may be dealt with by the 
Vector Method. 

Page 190. The symbol tan~i means " an angle whose tangent is." 

Page 195. To understand how we develope a given function in a 
Fourier Series, it is necessary to notice some of the results of Art. 109, 
very important for other reasons ; indeed, I may say, all-important to 
electrical engineers. 

Page 195. Article 126 should be considered as displaced so as to 
come immediately before Art. 141, p. 210. 

Page 202. The problem of Art. 125 is here continued. 


Page 208. The beginner is informed that 2 means "the sum of all 
such terms as may be written out, writing 1 for s, 2 for 5, 3 for 5, and 
so on." 

Page 209. See Ex. 23, p. 184. 

Page 213. The student ought to alter from v to (7 or to ^ in (9) 
as an exercise. 

Page 213. See Art. 152. 

Page Ml. Remember that the effective value of as.\n{nt-\-e) is 

Page 254. Here again a student needs a numerical Example. 

Page 256. After copper read " and their insulations." 

Page 259. Then 

e-^-\-e.^=E [sin {nt + a) + sin (nt — a)] = 2i7 cos a sin w^. 

{See Art. 109.) 

Page 266. All other statements about this subject that I have seen 
are of infantine simplicity, but utterly wrong. 

Page 269. Insert " and in consequence." 

Page 278. In the same manner show that if 

y=«^ ^=«*ioga. 

Page 281. See the eighth fundamental case, Art. 215. 

Page 299. See (1) Note to Art. 21. 

Page 301. Article 225 should be considered as displaced so as to 
precede Art. 222. 

Page 305. See Ex. 8, Art. 99. 

Page 309. Or (3). 

Page 310. Art. 225 should be read before Art. 222. 

Page 331. I have taken an approximate law for h and so greatly 
shortened the work. 

Page 359. Viscosity. All the fluid in one plane layer moves with 
the velocity v\ the fluid in a parallel plane layer at the distance bx 
moves with the velocity v-\-bv in the same direction; the tangential 

force per unit area necessary to maintain the motion is /* ^ or /* ^- , 

where /x is the viscosity. 

Example 1. A circular tube is filled with fluid, the velocity v at 
any point whose distance from the axis is r being parallel to the axis. 
Consider the equilibrium of the stuff contained between the cylindi-ic 
spaces of radii r and r+5r of unit length parallel to the axis. The 


tangential force on the inner surface is 2wrfi. ,~ and on the outer 
surface it is what this becomes when r is changed to r-\-hr or 
^TTfi -J- (i'-f) 5^1 the difference of j^ressure between the ends gives us 

a force - 2wr ~ dr if x is measured parallel to the axis. The mass of 

the stuflF is 27rr . 6r . m if dr is very small and if m is the mass per unit 

volume ; its acceleration is -^ if ^ is time ; and hence, equating force to 

mass X acceleration and dividing by 27r/x8r, 

d / dv\ r dp _r.m dv 
dr \ dr) fi dx~ fi dt ' 

Example 2. Let -^ be constant; say that we have a change of 

pressure P in the length I so that t^ = y • Let a state of steady flow 

have been reached so that -f=Oy then 

dr\ dry fi I 

If r -7- be called u and if i'/^u be called 2a. then -^ + 2ar = 0, so that 
dr '^ dr ' 

du 4- 2ar . dr = 0, or u + ar^ = constant c. 

rj^ + ar^=^c, or ^ + ar=- (1), 




vi-j^ar' + ^-, = C (2). 

Evidently as there is no tangential force where r=0, -f- = thei-e, 

c must be 0. Hence 

v + ^ar^ = C (2). 

Now v=0 where r^r^, the outer radius of the fluid, and hence 
(2) becomes v=^a {r^^-r^). 

The volume of fluid per second passing any section is 

27r / rv .dr=^ - aro* = nrQ*P/8lfx. 

This enables us to calculate the viscosity of a fluid passing through a 
cylindrical tube if we know the rate of flow for a given difference of 


The References are to pages. 

Academic exercises, Chap. iii. 

Acceleration, 24, 30, 188, 220 

Adiabatic law, 92, 148, 167 

Advance, 186 

Air in furnaces, 65 

Air turbine, 128 

Alternating current formulae, 183 — 5, 

189, 209, 239 
Alternator, 178 

Alternators in parallel and series, 259 
Amplitude, 172 

Analogies in beam problems, 108 
— in mechanical and electri- 
cal systems, 213 
Angle between two straight lines, 16 
Angular displacement, 33 
Angular vibrations, 212 
Apparent power, 209 
Approximate calculations, 2 
Arc, length of, 170 
Archimedes, spiral of, 302 
Area, centre of, 85 

— of catenary, 171 

— of curves, 69 

— of parabola, 71 

— of ring, 80 

— of sine curve, 197 

— of surfaces of revolution, 75, 78 
Asymptote, 301 

Atmospheric pressure, 166 

Attraction, 87, 344 

Average value of product of sine 

functions, 185 
Ayrton, 53, 199 

Ballistic effects, 181 

Basin, water in, 130 
Beams, 48 

— fixed at the ends, 100—108 

— of uniform strength, 102 

— shear stress in, 115 

— standard cases, 97 — 99 
Beats in music, 194 

Belt, slipping of, 165 
Bending, 94—121 

— in struts, 262 
Bessels, 205, 352, 354 
Bifilar suspension, 179 
Binomial Theorem, 34 
Boiler, heating surface, 63 
Bramwell's valve gear, 193 
Bridge, suspension, 61 

Calton HUl, 357 

Cardioide, 302 

Capacity of condensers, 162, 236, 240 

Carnot cycle, 145, 152 

Carnot's function, 145 

Catenary, 62, 170 

Central force, 344 

Centre of gravity, 73, 85 

Centrifugal force, 123 

Centrifugal pump, 131 

Chain, hanging, 61 

Change of state, 150 

— of variable, 339 
Characteristic of dynamo, 296 
Circle, 10 

— moment of inertia of, 82 
Circuitation in electricity, 134 
Cissoid, 302 

Cisterns, maximum volume of, 49 



The References are to pages. 

Clairaut's equation, 334 
Clearance in gas engines, 150 

— in pumps, 131 
Commutative law, 231 
Companion to cycloid, 301 
Complete differential, 143—145, 153 
Compound interest law, 161, 164 
Concavity, 306 

Conchoid, 302 

Condensation in steam cylinders, 

54, 153, 358 
Condenser, electric, 162, 212, 236 

— annulling self-induction, 


— with induction coil, 257 
Conductivity of heat, 341, 354 
Conductors, network of, 237 
Cone, 73, 78 

Conjugate point, 301 
Connecting rod, 191 
Constraint, 179 
Continuous beams, 111 
Convexity, 306 

Cooling, Newton's law of, 163 
Co-ordinate geometry, 6 
Co-ordinates, r, 0, <p, 347 

— polar, 310, 342 

Cosines, development in, 207 
Cos-la;, 277 
Cot-la;, 277 
Cot X, 275 
Coupling rod, 265 
Craigleith quarry, 367 
Crank, 173 

Crank and connecting rod, 12, 191 
Curl, 134 
Current, effective, 200—202 

— electric, 33, 168, 189, 239 
Curvature, 96, 120, 169, 306 

— of beams, 96—121 

— of struts, 262 
Curves, 43 

— ' area of, 69 

— lengths of, 77, 312 
Cusp, 301 
Cycloid, 12, 276, 302, 312 

— companion to, 301 
Cylinder, heat conduction in sides of 

steam-engine, 356 

— moment of inertia of, 82, 

Cylinders, strength of thick, 88 
Cylindric body rotating, 90 

Damped vibrations, 10, 211, 225— 

Definite integral, 68 
Deflection of beams, 96, 118 
Demoivre, 320 
Development, Fourier, 202 

— in cosines, 207 
Differential coefficient, 21, 28, 268 

— complete, 143, 153 

— equation, 220-225 

— equations, general exer- 

cises on, 337 

— equations, partial, 341, 


— partial, 139 
Differentiation of function of more 

than one varia- 
ble, 340 

— of product, 269 

— of quotient, 270 
Diffusivity for heat, 355 
Discharge of condenser, 156 
Displacement, angular, 33 
Distributive law, 231 

Drop in transformers, 255, 257 
Dynamics of a particle, 344 
Dynamo, series, 296 


dx dy 

^( = ^.^,155.271 
dx dz dx 

e<^, 10 

e<" sin bx, 10, 285 
e'^cosbx, 286 
Earth, attraction of, 87 
Earthquake recorder, 215 
Economy in electric conductors, 65, 

— — lamps, 294 

— hydraulic mains, 58 
Eddy currents, 209 

Effective current, 200—202 
Efficiency of gas-engine, 150 

— of heat-engine, 41 

— of heating surface, 64 
Elasticity, 93, 141 

Electric alternator, 178 

— circuit, 33, 168, 178, 189, 
208, 212, 236, 239, 247 



The References are to pages. 

Electric condenser, 162, 236, 240—245 

— — and Ruhmkorf 

coil, 257 

— — as shunt, 243, 


— conductor, 168 

— — economy in, 55, 


— current, efifective, 200—202 

— illustration, alternating cur- 

rents, 199 

— lamps, economy in, 294 

— make and break curve, 201, 


— power meter, 209 

— time constant of coil, 60, 160 

— traction, 59 

— transformer, 33, 249, 253 

— vibrations, 156, 212, 213, 225 

— voltage, effective, 202, 247 

— voltaic cells, 51, 52 
Electricity, partial differential equa- 
tions in, 349 

— problems in, 33, 51, 52, 

55, 57—59, 60, 134— 
136, 156, 157, 162, 168, 
225, 236, 239—261 

— gelf-induction in, 135 
Electrodynamometer, 200 
Electromagnetic theory, fundamental 

laws of, 134 
Electromotive force, 134 

— — in moving coil, 

Ellipse, 8, 10, 11, 83, 158, 276 
Ellipsoid of revolution, 76 
Elliptic functions, 339 

— integrals, 339 
Empirical formulae, 17 
Energy, intrinsic, 143 

— kinetic, 31, 180 

— of moving body, 156 

— potential, 180 

— stored in compressed spring, 

Engineer, 1 
Entropy, 143, 152 
Envelopes, 309 
Epicycloid, 302 
Epitrochoid, 301 
Equality of forces, 217 

Equations, differential, 220—225 

— partial differential, 341, 


— solving, 51 
Equiangular spiral, 186 
Euler's law for struts, 265 
Evanescent term, 169, 190, 208, 240 
Exact differential equation, 339 
Exercises, 38 

— general, on differential 

equations, 33/ 

— — on integration 

and differen- 
tiation, 279 

— maxima and minima, 295 

— on curves, 169, 311 

— on integration of sin x and 

cos^, 182 

— on Maclaurin, 319 
Expansion of functions, 317 — 320 

— of gas, 17 
Experiments, 7 
Explosions, 48 

Exponential and trigonometrical for- 
mulae, 177, 185, 190, 
222, 320 

— theorem, 161 
e«, 161 

Factor, integrating, 328 

Factorial fractions, 235 

Factorials and gamma function, 369 

Falling body, 21, 30 

Feedwater missing in steam cylinder, 

Ferranti effect, 247 
Field, magnetic, about straight wire, 

— — rotating, 195, 251 
Flow, maximum, of gas, 128 

— of gas, 54 

— of liquid, hypothetical, 133 
Fluid friction, 167 

— motion, 125 

— pressure, 121 

— whirhng, 123 

— work done by, 66 
Flywheel, 84 

— stopped by friction, 167 
Force, central, 344 

— due to jet, 27 

— — pressure of fluids, 121 

— lines of, 124 



T/te References are to pages. 

Force of blow, 26 

— of gravity, 87 

— unit of, 26 

Forces on moving bodies, 180 
Forms, indeterminate, 299 
Formulae of reduction, 284, 286 
Fourier development, 208 

— exercises on, 315 

— proof, 183, 184, 197, 201 

— rule, 202, 204 

— theorem, 195 
Fractions, partial, 291 
Frequency, 186 
Friction at pivot, 94 

— fluid, 167 

— solid, 168 
Frustrum of cone, 78 
Fuel on voyage, 49 
Function, 8 

— average value of sine, 185 

— Bessel, 352 

— elliptic, 339 

— gamma, 369 

— hyperbolic, 172, 360 

— of more than one variable, 

137, 341 
Fundamental integrals, 278 

— rules on electricity, 134 

Furnaces, air in, 65 

«7, 26 

Gamma functions, 235, 369 

Gas, 38 

— elasticity of, 94 

— engine, 91 

— — diagram, heat in, 272 

— — formulae, 17, 91, 147, 

150, 272 

— flow of, 55, 127 

— perfect, 136, 150 

— work done by, 66, 72 
Gauge notches, 133 
Gaussage, 134 

Gear, valve, 193 

General case of two coils, 249 

— exercises in differential equa- 

tions, 337 

— — in differentiation 

and integration, 

— rule, 271 

— — for differential equa- 

tions, 224 

General rule for operators, 237 
Girders, continuous, 111 

— shear stress in, 117 
Glossary, 301, 302 
Gordon's rule for struts, 264 
Gradient, temperature, 354 
Graph exercises, 8 — 10 
Graphical Fourier development, 204 

— work in beams, 108 
Gravity, 87 

— motion of centre of, 230 
Groves' problem, 257 
Guldinus' theorems, 80 

Guns, 90 

Hammer, 26 
Hanging chain, 61 
Harmonic functions, 172, 186 
Harmonics, zonal, 205, 354 

— — spherical, 349 
Heat conductivity, 354 

— — equations, 341, 


— equations, 138 

— experiments at Edinburgh, 357 

— latent, 41, 43 

— lost in steam cylinders, 356 

— reception in gas-engine, 272 

— specific, 93, 141 
Heating surface, 63 
Hedgehog transformer, 244 
Henrici, Professor, 205 
Henry, 136 

Hertz, 215 

Horse-power and steam, 41 

Hydraulic jet, 26 

— press, strength of, 89 

— transmission of power, 58 
Hydraulics, 133 

Hyperbola, 10, 11 
Hyperbolic functions, 360 

— spiral, 302 
Hypocycloid, 277, 302 
Hypotrochoid, 302 
Hysteresis, 209, 255 

Idle current in transformers, 243 
Imaginaries, 185 

Incandescent lamp, economy in, 298 
Independent variables, change of, 338 

— — more than 

one, 56, 136, 



Ttie References are to payes. 

Indeterminate forms, 299, 300 
— multipliers, 158 

Index law, 231 
Indicator diagram, 53, 67 

— — gas-engine, 91, 27*2 

— vibration, 215 
Induction, 136 

— coil and condenser, 257 

— in transformers, 256 

— mutual, 249 

— self-, 33, 60 

— self-, and capacity, 240 
Inertia, moment of, 81 

— — of cylinder, 81, 


— — of ellipse, 83 

— — of flywheel, 84 

— — of rectangle, 86 

— — of rod, 84 
Illustrations of meaning of differen- 
tiation, 37, 40, 42, 162, 176, 177 

Inflexion, point of, 20, 301 
Instruments, measuring, 179 
Integral, definite, 68 

— double, 68 

— line, 69 

— surface, 69 
Integrals, elliptic, 339 

— list of, 359—369 
Integrating factor, 144, 327 
Integration, 23, 35 

— by parts, 285 

— exercises on, 183 

— of fractions, 282 
Interest law, compound, 161 
Intersection of two straight lines, 16 
Intrinsic energy, 143 
Isothermal expansion, 92 

Joule's equivalent, 43 
Journal, lubrication of, 231 

Kelvin, 161 
Kinetic energy, 31 

Labour-saving rule, 237 

Lag, 209 

Latent heat, 15, 42 

Lateral loads, struts with, 264 

Lamp, incandescent, 298 

Law, adiabatic, 148 

— commutative, 231 

— of flow of heat, Peclet's, 63 

Law of cooling, Newton's, 163 

— compound interest, 161 

— distributive, 231 

— of Entropy, 148, 152 

— Euler's, for struts, 265 

— of expansion of steam, 17 

— of falling bodies, 21 

— Index, 231 

— of loss of heat in steam cylin- 

der, 356 

— of ^ and t, 18 

— of Thermodynamics, first, 142 

— — — second, 146 

— of vibratory systems, 225 — 230 
Lead in branch electric circuit, 247 
Leakage of condenser, 162 
Legendre's equation, 350 
Lemniscata, 302 

Length of arc, 170 

— of curve, 77, 312 
Level surface, 124 
Limit, meaning of, 22 
Line integral, 69, 134 
Linear equations, 220, 326, 353 
Link motions, 14, 193 
Liquid, flow of, 130 
List of Integrals, 359—369 
Lituus, 302 
Loci, 11 
Logarithmic curve, 10, 302 

— decrement, 11 

— function, 40 

— spiral, 302 
Logarithms, 2, 161 
Log a;, 274 
Lubrication of journal 

Maclaurin, 319 
Magnet suspended, 179 
Magnetic field about straight wire, 
134, 195 

— — rotating, 195, 251 

— force, 134 

— leakage, drop due to, 257 
Make and break curve, 201, 205 
Mass, 26 

— energy of moving, 157 

— of body, variable, 76 

— vibrating at end of spring, 156 
Maxima and minima, 20 

— — exercises on, 

46, 47, 60, 



The References are to pages. 

Maximum current from voltaic cells, 

— flow of air, 128 

— parcel by post, 53 

— volume of cistern, 49 

— power from dynamo, 297 
Measuring instruments, 179 
Merchant and squared paper, 6 
Meter, electric power, 209 
Minimum, 20 

Modulus, 340 
Moment of inertia, 80 
Momentum, 26 
Motion, 30, 157 

— angular, 33, 212 

— of fluids, 125 

— of translation, 344 

— in resisting medium, 314 
Multipliers, indeterminate, 158 
Mutual induction, 250 

Natural vibrations, 225, 220 
Negative and positive slope, I'J 
Network of conductors, 237 
Newton's law of cooling, 163 
Normals, 15, 43, 169 
Notches, gauge, 133 
Numerical calculations, 2 

Octave, 192 
Ohm's law, 33 

— — modified, 136, 168, 189, 

208, 236 
Operation, symbols of, 231 
Orifice, flow of gas through, 54 
Oscillation, 123, 156, 190, 210, 211, 

Otto cycle, 149 

Parabola, 8, 11, 27, 31, 61, 71 
Paraboloid of revolution, 74 
Parallel motion, 13 

— alternators in, 261 
Parameter, 340 

~ variable, 308 
Partial differential equations, 341, 

— differentiation, 39, 137, 155, 

269, 341 

— fractions, 224, 234, 291, 294 
Particle, dynamics of, 344 

Parts, integration by, 284 
Peclet's law of flow of heat, 63 

Pendulum, 179 
Percussion, point of, 123 
Perfect gas, 38 

— — thermodynamics of, 147 

— steam-engine, 41 
Periodic functions, 194, 203 

— motions in two directions, 


— time, 186 
Perpendicular lines, equations to, 

Pivot, friction at, 94 
Point, conjugate, 301 

— d'arr^t, 301 

— of inflection, 301 

— of osculation, 301 

— moving in curved path, 342 
Polar expressions, 342 

Pound, unit of force, 26 

Poundal, 26 


Positive and negative slope, 19 

Potential energy, 32 

Power, apparent, 209 

— electric, 208 

— — transmission of, 58 

— meter, electric, 209 

— true, 209 
Press, hydraulic, 90 
Pressure, 136, 273 

— atmospheric, 166 

— fluid, 121 

Primary, transformed resistance of, 

Product, differentiation of, 155, 269 
Projectile, 310, 315 
Pulley, slipping of belt on, 165 
Pump rod, 164 

Quotient, differentiation of, 270 

Radial valve gears, 14, 193 

Badian, 9 

Radius of curvature, 169, 306 

— of gyration, 82 
Radius vector, 342 

Rate of reception of heat, 92 
Ratio of spcific heats, 93, 139, 141 
Ratios, trigonometrical, 182 
Reduction, formulae of, 284, 286 
Recorder, earthquake, 215 
Rectangle, moment of inertia of, 86 
Resistance, electric, 33 



The References are to pages. 

Resistance, leakage, 163 

— operational, 236 
Besistances in parallel, 245 
Resisting medium, motion in, 314 
Resonance, 215 

Resultant of any periodic functions, 

Revolution, surfaces of, 78 

— volume of solids of, 75 
Rigid body, 61 

— — motion of, 217 
Ring, volume and area of, 80 
Rod, moment of inertia of, 85 
Roots of equations, 224, 321 
Rotating field, 195, 251 
Rotation in fluids, 132 
Ruhmkorff coil, 257 
Rule, 4 

r and d co-ordinates, 342 
r, df <t> co-ordinates, 347 

Secant x, 276 
Secohm, 136 

Self-induction, annulled by con- 
denser, 247 

— — of parallel wires, 

Series, alternators in, 257 

— development in, 207 

— dynamo, 296 
Shape of beams, 109 
Shear stress in beams, 115 
Shearing force in beams, 108 
Simple harmonic motion, 173 

— — — damped, 311 
Sina;, 9, 161, 274 

Sine curve, area of, 173 

— functions, 172 
Sines, curve of, 9, 173 

— development in, 207 
Singular solution, 334 
SHpping of belt, 165 
Slope of curve, 15, 19, 70 
Solution of forced vibration equa- 
tions, 214 

— of linear differential equa- 
tions, 229 

Sound, 94 
Specific heat, 139 

— heats, ratio of, 93, 141 

— volume of steam, 42 
Speed, 21 

Spherical Harmonics, 205, 349, 354 

Spin, 132 

Spiral flow of water, 130 

— hyperboUc, 302 

— of Archimedes (or equi- 

angular), 302 

— line, 173 

— logarithmic, 302, 311 
Spring with mass, vibrating, 156 
Springs, 21, 32, 53 

— which bend, 119 
Square root of mean square, 200, 202 
Squared paper, 6, 7 
State, change of, 150 
Steady point, 219 
Steam, 41 

— work per pound of, 53 

— -engines, 41 

— -engine indicator, vibrations 

of, 215 

— — piston, motion of, 191 
Stiffness of beams, 48 

Straight line, 14 

Strains in rotating cylinder, 90 

Stream lines, 126 

Strength of beams, 48 

— of thick cylinders, 88 

— of thin cyhnders, 91 
Struts, 261 

— with lateral loads, 264 
Subnormal, 44, 169 
Substitution, 279 
Subtangent, 44, 169, 305 
Successive integration, 335 
Sum, differentiation of a, 268 
Surface heating, 63 

— integral, 69 

— of revolution, 75, 78 

— level, 124 
Suspension, 179 

— bridge, 61 

Swinging bodies, 179, 182 
Symbols of operation, 233 

— — — simplification 

of, 237 
Synchronism, 215 

Table of Fundamental Forms, 278 
Tangents, 43, 158, 169 
Tana;, 275 
Tan-la:, 277 
Taylor's theorem, 317 
Temperature, 136 



The Refererwes are to pages. 

Temperature, absolate, 145 

— gradient, 354 

— in rocks, <feo. , 357 
Terminal velocity, 314 
Theorem, Binomial, 34 
Theorem of three moments, 111 

— Guldinus', 80 

— Demoivre's, 320 

— Maclaurin's, 318 

— Taylor's, 317 
Thermodynamics, 42, 138, 153 
Thomson, Professor James, 133 
Three moments, 111 

Tides, 194 

Time constant of coil, 60, 160 

t, diagram, 153 

Torque, 33 

Torsion, 179 

Traction, electric, 59 

Transformer, 33 

Transformers, 252, 257 

— condenser shunt, 243 

— idle current of, 243 
Triangle of forces, 61 
Trigonometry, 2 
Trigonometric and Exponential 

Functions, 177, 185, 222, 234 
Trigonometric Formulae, 182 
Trisectrix, 302 
Tuning-forks, 230 
Turbine, air, 128 
Two circuits, 249 

Valve gears, 14, 193 

Variables, independent, 57, 136, 341 

Variable mass of body, 76 

— parameter, 308 
Velocity, 21, 30, 188 

Vibration, 156, 190, 210, 212, 216, 
225, 230, 238 

— (electrical), 156, 212, 213 

— indicator, 217, 219 

— of indicator, 215 
Volcanoes, 172 

Voltage in moving coil, 178 
Voltaic cells, 51, 52 
Volume of cone, 74 

— of ellipsoid of revolution, 76 

— of paraboloid of revolution, 


— of ring, 80 

— of solid of revolution, 76 
Voyage, fuel consumed on, 49 

Water in steam cylinder, 358 
Watt's parallel motion, 13 
Wedmore, 205 
Weight, 26 
Whirhng fluid, 123 
Willans, 54 
Work, 31—32 

— in angular displacement, 35 

— per pound of steam, 31, 53 

— done by expanding fluid, 66, 


— — gases, 149 

Uniform strength in beams, 102, 103 
Uniformly accelerated motion, 29 
Unreal quantities, 3, 177, 185, 222, 

X . e«*, 285 
x-K- 6—160 

Zonal Harmonics, 205, 349, 354 


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This useful .little book is based upon Mr. Oman's " History of 
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upon an original plan, and will be found a valuable help in the study 
of history. 

Seven Roman Statesmen. A detailed Study of the 

Gracchi, Cato, Marius, Sulla, Pompey, Caesar. Illustrated with 
reproductions of Roman Coins from the British Museum. By C. W. 
Oman. About 320 pages. Crown 8vo., cloth, 68. 

England in the Nineteenth Century. By C. W. Oman, 

M.A., Author of "A History of England," etc. With Maps and 

Appendices. Revised and Enlarged Edition. One vol., crown 8vo., 

3s. 6d. 

Morning Post. — "One finds clearness of statement, simplicity of style, general 

soundness of historical judgment, impartiality, as well as a notable spirit of patriotism, 

which loves to dwell on the greatness and glory of our Empire at home and abroad." 

English History for Boys and Girls. By E. S. Symes, 

Author of '* The Story of Lancashire," "The Story of London," etc. 
With numerous Illustrations. One vol., 28. 6d. 

Lessons in Old Testament History. By the Venerable 

A. S. Aglsn, Archdeacon of St. Andrews, formerly Assistant Master at 
Marlborough College. 450 pages, with Maps. Crown 8vo. , cloth, 4s. 6d. 


( 8 ) 

Arnold's School Series. 


Edited by H. O. Arnold -Forstkb, M.P., Author of "The Citizen 
Reader," "This World of Ours," etc. A magnificent Atlas, including 
48 pages of Coloured Maps, several of them double-page, and Pictorial 
Diagrams. With an Introduction on the Construction and Reading of 
Maps by A. J. Hkbbbrtson, Ph.D. 

Among the notable features of this Atlas are : (1) The Specimens of 
Ordnance Surveys and Admiralty Charts ; (2) the lucid Astronomical 
Diagrams ; (3) the beautifully-coloured Physical Maps ; (4) the careful 
selection of names without overcrowding ; (5) the constant presentation of 
uniform scales for comparison ; (6) a Historical Series of Maps illustrating 
the Building of the British Empire ; (7) an excellent Map of Palestine. 

The size of the Atlas is about 12 by 9 inches, and it is issued in the 
following editions : 

Stout paper wrapper, with cloth 

strip at back, Is. 6d. 
Paper boards, 2s. 

Cloth cut flush, 2s. 6d. 

Limp cloth, Ss. 

Cloth gilt, bevelled edges, Ss. 6d. 


By Andrew J. Hkrbkrtson, M.A., F.R.G.S., Assistant Reader in 
Geography in Oxford University, and Alexis E. Frye. With 
sixteen pages of Coloured Maps, about fifty Outline and Photographic 
Relief Maps, and nearly seven hundred magnificent Illustrations. 

Large 4to. (about 12 by 10 inches), 58. 

This is the first attempt in this country to make the illustrations to the 
book as systematic and important as the text itself. 


By Andrew Herbebtson, Ph.D., F.R.G.S., Assistant Reader in 
Geography at the University of Oxford. Fully Illustrated. Cloth, 
4s. 6d. 

Arnold's New Shilling- Geography. 

The World, with special reference to the British Empire. 160 pp. 
Crown 8vo., cloth, Is. 

The Australian Commonwealth. 

Its Geography and History. A reading book for Schools. 144 pages. 
With Illustrations and Coloured Map. Cloth, Is. 

This book lias been wiitten by an Australian, and the illustrations consist of re- 
productions from photographs and from di-awuigs made by an artist who has sjHjnt 
many years in Australia. 

A Historical Geography. 

By the late Dr. Mobbison. New Edition, revised and largely re- 
written by W. L. Cabbik, Headmaster at George Watson's CoUege, 
Edinburgh. Crown 8vo., cloth, Ss. 6d. 

The Shilling Geography. 

By the late Dr. Morbison. New Edition, revised by W. L. Carrik. 
Small crown 8vo., cloth, Is. 


( I ) 

Arnold's School Series. 


GERMAN WITHOUT TEARS. By Mrs. Hugh Bkll. A versfon 

in German of the author's very iwpular "French Without Tears." With the 
original illustrations. Crown 8vo., cloth, 

Part I., 9d. Part II., Is. Part III., Is. 3d. 

Miss Walker, Clergy Daughters' School, Bristol. — " It pleases me greatly. I fully 
expect to find it a great favourite with the junior forms." 

The Head Teacher.—" Children who could not learn German from this book would 
never learn it." 

LESSONS IN GERMAN. A graduated German Course, with 
Exercises and Vocabulary, by L. Innks Lumsdkn, late Warden of University Hall, 
St. Andrews. Crovm 8vo., 8s. 


Teacher of Modem Languages in the High School of Glasgow. Including care- 
fully graded Exercises, Idiomatic Phrases, and VocabuLvry Crown 8vo., cloth, 
Is. 6d. 

ELEINES HAUSTHEATER. Fifteen little Plays in German for 

Children. By Mrs. Hugh Bkll. Crown 8vo., cloth,. 2s. 

GERMAN DRAMATIC SCENES. By 0. Abel Musobave. With 

Notes and Vocabulary. Crown 8vo., cloth, 2b. 6cL 


MORCEAUX CHOISIS. French Prose Extracts. Selected and Edited 
by R. L. A. Du Pontet. M.A., Assistant Master in Winchester College. The 
extracts are classified under the following headings : Narrations, Desmptions, 
Oenre Didactique, Style Oratoire, Biographie, Style Epistolaire, Anecdotique, Comidie. 
Explanatory Notes and Short Accounts of the Authors cited are given. Crown 
8vo., cloth. Is. 6d. 

POEMES CHOISIES. Selected and Edited by R. L. A. Do 
Pontet, M.A. Cloth, Is, Gd, 

LES FRANCAIS EN MANAGE. By Jetta S. Wolff. With 

Illustrations, Crown 8vo,, cloth. Is, 6d. An entirely original book, teaching 
the ordinary conversation of family life in France by a series of bright and 
entertaining scenes. 
Journal dks D^bats, — " Voici un 616gant volume, qui rendra de vdritables services 
aux Anglais appel^s k sojourner en France." 

Athen^um. — "This lively little volume, with its clever illustrations -will form a 
capital reading-book, especially for girls." 

LES FRANCAIS EN VOYAGE. By Jetta S. Wolff. A com- 

panion volume to the preceding, giving a lively account of travelling on the 
continent. The book is cast in conversational form, and introduces all the most 
useful phrases and expressions in idiomatic French. Cleverly Illustrated. Crown 
8vo., cloth. Is. 6d. 


With Illustrations by W. Foster. Cloth, Is. 3d. ' 

FRENCH DRAMATIC SCENES. By C. Abel Mdsgrave. With 

Notes and Vocabulary. Crown Svo., clothe 2s. 


This book is DUE on the last date stamped below. 

Fine schedule: 25 cents on first day overdue 

50 cents on fourth day overdue 
One dollar on seventh day overdue. 


1 '^ 1947 

-'AN 4 195nU 
MAR 2 1 1952 ' 
OCT 1 9 tgl 

JUN 5=p856 


LD 21-100m-12,'46(A2012sl6)4120 


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