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To take an example, let us suppose as before that a/7i0 = 1000, so that log (a/h0) = 6'908. By successive approximation we find from (60)
r/a = 8-869,    .................................(61)
so that if a = O27 cm. (as for water),
2r = 4-79cm...............................(62)
The correction to Laplace's formula is here unimportant.
The above is the diameter of tube required to render 7z0 negligible according to the standard adopted.
It may sometimes be convenient to invert the calculation, and deduce the value of h0 from the diameter of the tube (not much less than 4 cm.) and an approximate value of a. For this purpose we may use (60), or preferably (59), taking x =fr for instance. The calculated value of h0 would then be used as a correction. The accompanying small Table may be useful for this purpose.
r/a	-logio('Wffl)	Difference	h0/a
6	1-8275		0-0149
7	2-2319	0-4044	0-0059
8	2-6399	0-4080	0-0023
9	3-0508	0-4109	0-00089
10	3-4639	0-4131	0-00034
We have supposed throughout that the liquid surface is symmetrical about the axis, as happens when the section of the containing tube is circular. It may be worth remarking that without any restriction to symmetry the differential equation of the nearly flat parts of a large surface may be taken to be
dx*    dy*     a? so that z may be expressed by the series
z  A0Io (r/a) + (A1 cos 6 + Bl sin 0) I} (r/a)
+ (Atcos 20 4- 52sin 20) /2 (r/a) + ...,    .... r, 0 denoting the usual polar co-ordinates' in the horizontal plane.]
.(64)hu'HS of  and /y, it in eiim*ly ti'f'?yi;irv i* r'tan the denominator 1 4-a/2r, BO that w may write