aes ~ meer } ee 8 Si Tah —— DOCUMENT COLLECTION / cS UNITED STATES | EXPERIMENTAL MODEL BASIN : NAVY YARD, WASHINGTON, D.C. ee OND Mines RESISTANCE, OF “A HEAVY FLEXIBLE F CABLE FOR TOWING A SURFACE FLOAT BEHIND A_ SHIP BY J. G. THEWS AND L. LANDWEBER RECLASSIFiED 4 ) _geennenitMODaL og ae —- * ee ANG so, ; ci ay ae the : 5 EPARTMENT: .~ bie Oe Oe | | ON TED “OF ...THIS “REPORT aT NOT TO “BE, DIVULGED 4 ‘ i TOM IN ANY “PUBLICATION. SINS HE ; JENT ms HON. .DERIMEDe=FROM™ THIS.REPOR™ et “t “ON TO OFFICER OR »GIMILIAN PER @ NEL, THE. “SOURCE “SHOUED NOT BE REVEALED, PCC. Fi »D3 | no. 4/8 | MARCH 1936 REPORT NO. 418 S SEbLcEOO TOEO O NI 1OHM/T8iN ON THE RESISTANCE OF A HEAVY FLEXIBLE CABLE FOR TOWING A SURFACE FLOAT BEHIND A SHIP by J. G. Thews and L. Landweber U.S. Experimental Model Basin Navy Yard, Washington, D.C. March 1936 Report No. 418 NOTATION Weight per unit length of cable Drag per unit length of cable when normal to stream Drag per unit length of cable when parallel to stream Speed of ship Angle of cable to horizontal Ordinate of any point of cable from origin 0 at lowest point Length of cable from origin 0 Tension of cable at any point To Tension of cable at origin 0 wHanwe st yD Ss L Length of cable between points having equal ordinates p Defined by p? = \ 4x? tr eke or f = F/W r = R/W t=) T/ Tos t, = T,/To: T, =T for 9<0 te =Ts/ Tos T, =T for $>0 Z, = — tan 9/2, $<0 Za = tan 9/2, >0 C = 2f/p aa p24, Ze = PZe y= aps Y u,v Defined by Z4 = tan u, Ze. = tanh v THE RESISTANCE OF A HEAVY FLEXIBLE CABLE FOR TOWING A SURFACE FLOAT BEHIND A SHIP Introduction In an attempt to predict the behavior of a float towed behind a ship by a long, heavy, flexible cable, two distinct problems arose, for whose solution a knowledge of the forces in the cable was required. 1. What must be the strength of the cable? To answer this, it was necessary to know the tension in the cable at the ship. 2. How does the cable affect the behavior of the float? For it is clear that the cable has a two-fold influence upon the float. The horizontal component of the cable tension at the float balances the resistance of the fluid to the motion of the float. In addition to this, there is the vertical component of the tension which in effect increases the displacement of the float and consequently affects its resistance and behavior. To answer this second question it is necessary, then to know the horizontal and vertical components of the tension, or, what is the same thing, to know the tension at the float and the angle that the cable makes with the horizontal at that point. Previous work on the form and resistance of a towing cable has furnished no solution of the present problem. Mathematical expressions! and numerical solutions” (1) A.R.McLeod. On the action of wind on flexible cables, with applications to cables towed below aeroplanes and balloon cables. R. and M. 554. Oct. 1918. (2) H.Glauert. The form of a heavy flexible cable used for towing a heavy body below an aeroplane. R. and M. 1592. Feb. 1934. ———— for the form assumed by a cable used to tow a heavy body below an aeroplane have ~ been developed only for the case where the cable is concave downwards. In addition it has been assumed hitherto that the frictional resistance of the fluid is negli— TF a gible compared with the resultant forces of lift and drag acting on the cable, Mathematical expressions and numerical solutions have therefore been derived here suitable to the present problem in which the cable is concave upwards and where the angle of the cable with the horizontal is so small over a considerable length of cable that frictional resistance cannot be neglected. Hence in addition to the physical assumptions used in reference (1), an assumption regarding the magnitude of the frictional resistance must be introduced. General Analysis Consider a flexible cable of weight W per unit length. Let R* be the drag per unit length of the cable when at right angles to a stream of velocity V; F* the drag per unit length when parallel to the stream. When the cable is inclined at an angle @ to the stream, the force per unit length on the cable due to the stream will be assumed to consist of a component R sin?) at right angles to the length of the cable, and a component F along it. These assumptions are approximations to the actual experimental results. Their experimental basis is reviewed in Appendix I. No other physical assumptions will be made in the analysis of the problem. The general shape of the cable is shown in Fig. 2, the origin O being taken at the point Y RSIN70.d5 R SIN?@d5 wdS PIGS ee of zero slope. Associated with each point P (or Q) of the cable there is the ordi- © nate Y, the arc length S measured from 0, and the angle of inclination $ of the cable with the horizontal at the point P (or Q). These variables are related by the equation dY/dS = "sin, @ y's) ce sacceie! arial eal. Ee a CD *See Appendix II The element dS of the cable at P (or Q) is in equilibrium under the action of the system of forces comprising its weight WdS, the force R sin? dS normal to the cable, the force FdS along the cable and the tension at its ends. Denoting the tension of the cable at any point by T and resolving along the cable, we obtain the equation AT/ASl=y Ret Wnai oat eth ete eee aes 5 oy 2 ea (2) which may be combined with (1) to eliminate sin @ and then integrated to give Be gn ee tet te nt eeRo aT chet ye, hee Me po cuits Gt) where Ty is the tension of the cable at the origin. Resolving also at right angles to the cable we obtain Tdg/dS = Wcos 6+ Rsin*d , 90 (at Q)........ (4d) Equations (2) and (4a) may be combined to eliminate S and then integrated to give a relationship between T and ». For, from (2) and (4a) we obtain aT/Tdd = F epee + megs aa » where f= F/W, r= R/W . . (5) Introducing the substitution Z, = — tan $ » -(5) may be integrated (see Appendix III) to give = -1 -1 2 1 + p*®Z42 log ty = 2f BS Ps tan py + reap? 18 Te pF +4 1+ Z4/ tee oe ta SIAC os ee cle ese ee ee 0G) where p? = \[4r? +1 + 2r, and ty = 1,/To, 7, =T for 9<0 We can now express Y in terms of T and Z, by eliminating S between equations (1) and (4a). Thus we obtain iL sin aY/1,40= Goa @ FR Sin™ @ Side Soe we alkenes! oma ee Taya Making the substitution Z, = — tan 6/2, (7) becomes Z 1 t4Z4dZ, st AT, ata ve | 42 woah ee te NG, fo te Me ete we C8) te) Equations (6) and (8) have been derived for the case $<0. Correspondingly, for $>0, we obtain from equations (2) and (4b) 4 2 F + Wsin 6 = f+ sin aT/Tao = Wcos $—Rsin*> cos@?—-rsins °*°*° - (9) Introducing the substitution Z, = tan $/2, (9) may be integrated (see Appendix III) to give aa 4 2 1+ 2,7/p# log ta = -2f Sry47pz tan Za/p + Srp q7pr 108 7 lz + where t, =T2/To , T2=T for 6>0. Also from equations (1) and (4b), we obtain AR aaa ee Dae ete cht aa i: o ing) W cos 6 + R sin“ Making the substitution Zz = tan 9/2, (11) becomes Z Y= * ATo t2Z2dZ et Spake Grade td ete tay 2 (12) W (p + Za \(i/p a Za } fo} We conclude the general analysis by deriving a relation between the tensions and the length of the cable L between points determined by a prescribed ordinate Y. For a given ordinate Y let Ty, S; for ¢< 0 correspond to Tz and Sz for $>0. Then by (3), T, = To + FS, + WY and Ta = le +. FS2 + WY. Hence 1 1 T2 - T, = F(S2 - S,) =Fxbh GUS BEB es) Gio Ge Seber - (13) noting by (1) that S, is always negative. Application to determination of cable resistance. Approximations. In the particular problem with which we shall concern ourselves, the ends of the given cable are points having the same value for the ordinates. It is required to find the tensions at these ends as functions of the angle ) at the rear end of the cable (¢<0); or, what is the same thing, as functions of Z,. But by (13) Ets T2 —T Es t T= Ts T/T o T,/To Foxat, Been Me abiga re 3 4 ee Cal) Hence a procedure for obtaining numerical solutions of T, and Tz as functions of 2, may be outlined as follows: 1. Compute values of t, against Z, from (6) 7 2. Enter these values of t, in (8) to compute Y/To against 2, by graphical or numerical integration. Thus we obtain values of Y/To against ty. 3. Obtain values of tg against Y/To from (10) and (12), in a similar way. 4, Taking values of t, and tz corresponding to the same Y/To, T, can be computed from (14) and Tg from (13). Thus the problem as stated above is formally solved. Yet it is easily seen that the equations (6), (8), (10) and (12), involving as they do the two independent parameters f and p, do not readily lend themselves to the calculation of a general, dimensionless set of solutions for T, and Tz. It will be shown in Appendix IV that a close approximation is obtained by replacing equations (6), (8), (10), and (12) by the following one-parameter set of equations: log t, =-c tan” '2, Sasa Gbitvastaies alco us aus) sete Bea REO) | iT} con —_ —] ct aE pa N ° e e ° °e ° e 5 6 Seven) ca ote (8a) c 1 + 2 log te = 3 lee q=z Aptana carats mati ap i Na, cones A a metal (021) Za J ySud ean ee oe EE ea SAR Oo 2 where c = 2f/p and Z4 = pZ,, Ze = Dla, ¥ = Ko Ye Forms more convenient for calculation are obtained by introducing the sub- stitutions u = tan™! Z4 and v = 3 log i - Then 24 = tan u and zz = tanh v, and the above equations become ne 2 ier wed cell ale nee, Cee ca See eee eee one (6b) r cu \ gualial Vou (ingciate is Ca Bi ct artgertd ae rd anche Sind anette (8b) (0) ej ameter ttl alae A ik dee ptbiegs Cosco i,, tvbed ae (10b) Vv Vo eleter anh wedvetoimen to ea cts im touercmerean «ak CleD) fe) Tables 1 and 2 for computing y are based upon equations (8b) and (12b) respectively. For, expanding eu into a power series, (8b) becomes 6 u ae, y = log sec u-c ) u tan u du + aT 5 Usbanew dus tee Tew COD ) fo) and similarly from (12b) y = log cosh v+c is v tanh v dv + S \s ¥"! tanhiey dvr eee” het els C120) The coefficients of c* and higher powers were not computed because, in general, c will be small. For values of c = .02, .04, .06, .08, .10, .12 and .14, ty, ta and y were computed from (6b), (10b), (8c) and (12c) respectively. Taking values of t, and t2 corresponding to the same y, T,/FL was computed from (14). Fig. 3 is a plot of 1,/FL against tan 6/2. To facilitate the use of Fig. 3 to compute the tensions as functions of $, Table 3 was constructed, giving values of tan 9/2, sin $ and cos $ against ¢. To illustrate the method, the tension T, and its vertical and hori— zontal components are computed against » for a hypothetical cable 1000 feet long, weighing one pound per foot moving at a speed such that F = f = 1 lb. per ft., p = 20, and hence c = 2f/p = .10. The results are shown in table 4. It is seen that the vertical component T, sin $ is nearly constant for values of $ between 60° and 90°. Conclusion We can now formulate the following procedure for answering the questions raised in the introduction. Suppose a given cable towing a given float. (a) Obtain R and F cable-resistance curves against speed either by new tests or by using the data of Wieselsberger for R and of Kempf for F, for a smooth cable (see Appendix II). Recalling that r = R/w and p* = \4r? + 1+ 2r, determine f = F/W and c = 2 f/p as functions of speed. (b) From Fig. 3 and table 3, compute Ty, T, Sin >, T, cos ¢ for speeds and f values corresponding to C = .02, .04, ....,.14, as illustrated in table 4. As in table 4, the additional displacement qT, Sin $ will vary slowly with the drag of the float, T, cos ¢ . Choose a mean value of qT, sin $ in the neighborhood of the esti-— mated drag qT, cos # . Plot these mean values of T, Sin > against speed. (c) Using these mean values of the corrected displacement corresponding to each speed, obtein the resistance curve of the float. This is equivalent to ob— taining T, cos $ and hence also T, for each speed. Ta is then calculable from (13). Let us illustrate the procedure in (a), (b), and (c) with the example from which table 4 was computed. (a') Suppose that the cable-resistance curves have been obtained, as described in (a) above, and that f and c have been plotted against speed. From these curves read value of speed corresponding to c = .10, and also value of f at this speed. Suppose we find for the speed v = 30 ft./sec. and f = 1.00 1b./ft. (b') From Fig. 3 and table 3 compute Ty» T, sing, T, cos $ for c = .10, and hence at 30 ft./sec. with f = 1.00 1b./ft. This yields table 4. Suppose now that although the resistance of the float and its displacement when towed at 30 ft./sec. are unknown, we can estimate that its resistance will lie between 500 and 1000 pounds; For values of qT, cos 6 in this range we see from table 4 that the additional displacement T, sin $ varies from 1726 lbs. to 1815 lbs. Take 1770 lbs. as a mean value of the additional displacement. Supposing that the weight of the float itself is 300 lbs., we have 1770 + 300 = 2070 lbs. for the total dis— placement of the float when towed by the cable at 30 ft./sec. (c') By model tests, or otherwise, find the resistance of the float for a speed of 30 ft./sec. when its displacement is 2070 lbs. Suppose we find that the resistance at this speed and displacement is 690 lbs., i.e. T, cos $ = 690 lbs., whence, from table 4, we have T, sin @ = 1800 lbs. as a corrected value for the additional displacement. The resistance of the float at 30 ft./sec. could then be corrected using 1800 + 300 = 2100 lbs. for the total displacement. Finally, we see again from table 4 that when qT, cos » = 690 lbs., Tz = 2930 lbs. which is the ten— sion in the cable at the ship at this speed. @ OL’ Le 60°21 88°62 1L°6 0z "be Lb°g Lz°6L 6b°L S0°St 40°9 SP" LL 10°S 1S°8 60°r 60°9 Ghee Sk’ 68P°2 GLL°E 6£0°% 092 °% ZE9°L OLS*t 69¢°L 820°L 156° Lz9° 089° eze° 9SP° 902° 082 * 060° zSL° 0£0° 490° 900° 120° 200° 600° 0 £00° 0) a) 6. (e} apa aim at f ap a quoy a f 1e°o 18°€ LS °€ bere 16°% L9°¢ Le °% 20°? BLL "Lb 12$°*L Lert SEL*L 676° 89L° S65 ° rer’ 162° LLL 840° rr0° 610° S00° A ysoo UT 0°S 7 aa 2 6°¢ 9°€ €°€ o°€ ge v2 ata 0° 8°1 9°L wl ra O°L (eo 0 ap amma et [is apa qu [944 ysoo ut = ¢ TIaVL ° te) €S°p 9°26 88° L°8P Br°E G°2e 02°E Sve 96°% 49°61 $9°% OL°rL iard 86°OL zL°2 %2°3 LLL*L 08°S OZE°L 09°e SLO"L LS°2 06L° 96°L S19° 95° SLb° 9¢°L 29° €0°L L92° crs" L6L° 089° LEL® 905° £80° €t* 9¢0° 60€° 1z0° €02° S00° OOL* np nuey nf A Noes ut on ueq =lz n 9$°L GS°L bS "tL €S°L 7s" 0S°t 87°L SPL vel €*L (aa Lek O°L np n ueq n ts} J Hy apnuyzn ti “L TTaVL d=—n vas ut =f € ‘Sid 2b NVLd YO4 JIVdS 002g OS! 00108 09 Ov O€ O¢ SGI o|e 9S We Oe Die S1 TH+ ae senepn i) 12 TIT, Lb Ol” MMe oe: Pe EC WWRULS OL UV INOIONZdYad JT@VD ‘HIONZT LINN Yad Ovud—u |! nyveens OL TaTIWuvd SIGVd SHLONST LINN Yad OvdG =4 ist 34) IWOTd LV “IVLNOZIHOH HIM 31GV> 40 3TONV =6 |: FTES 40 HIONST LINN Had LHOIAM=M “EE IWO14 Iv 31@VD 4O NOISNSL =! a1gVv> JO HLONAT=71 M/d=4 °M/=7 v2+ lave =.d - TON QNV NOISNL Br Suey Be! ereciieo uk 02 SI O01 08 O09 OS Ov OF Jd1V9S Ted/'L OS 10 TABLE 4. p= 20, c=.10, F=1.0 1b/ft, L = 1000 ft., Ta =T,+Fxb p tan 6/2 T, /FxL T, T, sin 4 T, cos 6 0 oo oo 500 oo -874 6.35 6350 554 6320 1.750 3.89 3890 675 3830 2.634 3.18 3180 824 3070 3.526 2.80 2800 958 2630 4,434 2.60 2600 1100 2350 5.358 2.44 2440 1220 2110 6.306 2.32 2320 1330 1900 7.28 2.23 2230 1435 1710 8.28 2.17 2170 1535 1535 9.33 2.10 2100 1610 1350 10.41 2.05 2050 1680 alee UWos)) 2.00 2000 1726 1000 12.75 1.96 1960 1775 828 14.00 1.92 1920 1805 657 15.35 1.88 1880 1815 487 16.79 1.85 1850 1823 321 18.34 4.82 1820 1813 159 20.00 1.80 1800 1800 0 11 Appendix I The physical assumptions we have made above concerning the components of the forces on the cable normal and parallel to it are based on the experiments and data of Relf and Powell (R.and M. No. 307, January, 1917). They measured the lift and drag of an infinite, straight cable, situated in a uniform stream of air, and in— clined at an angle » to the direction of air flow. Let R be the force per unit length acting on the cable when it is normal to the wind. Let F be the force per unit length acting on the cable when it is paral— lel to the wind. Our assumptions then are that for a given speed the components of the force per unit length acting on the cable are R sin? $ normal to the cable and a constant F parallel to the cable. The values of the normal and parallel compo— nents were computed from the paper cited above for the case of a stranded wire cable of diameter 0.388 inches. Relf and Powell made their measurements with a wind speed of 40 ft./sec. The experimental results are compared with the assumed laws in table 5 and fig. 4 taking R = .074 and F = .0023. TABLE 5. Normal Component Tangential Component b Experimental 074 sin*> Experimental F. 0 0 0 -0023 -0023 10° -0027 0022 - 0028 - 0023 20° - 0089 - 0087 -0031 -0023 302 0174 0185 - 0033 -0023 40° -0280 - 0306 - 0033 0023 50° - 0409 - 0434 - 0033 .0023 60° 0544 -0555 - 0023 -0023 70° 0654 0654 .0018 -0023 g0° .0735 .0719 .0010 .0023 - 0023 12 Appendix II A. Force on a cable normal to stream. If the stream velocity is V, the force per unit length will be of the form R aeeDpt gm Re MM ae oe Oe ee a Ce where D is the diameter of the cable, e is the density of the fluid, and k, a non-dimensional coefficient, is a function of the Reynold's number R‘= WA. y is the kinematic viscosity. Table (6) gives values! of k corresponding to values of R’in the range 105< R’< 106, TABLE 6. R x10 610 620-1808 105° 200 350 4:0°55.0° 6.0 7.0 8.0 k 4.10 1220 1620 1.20 1.14 194 166 133.32 032.33 B. Force on a cable tangential to stream. The force per unit length is given by” PSs VAM e + V)meennes ON ee 2 BS where e = 2.62 x10 ft. A depends upon the curvature and roughness of the cylinder. Table (7) gives corresponding values of A and D for a smooth cylinder. TABLE 7. D(inches) 00 20.0 10. 5e 2.5 1.4 0.7 A x 18 Hoy 1.14 1.17 1.21 1.26 1.31 1.36 (1) Data due to Wieselsberger, 1922. (2) Kempf "On the Frictional Resistance of Surfaces of Various Forms" 1924, Proceedings of the 1st International Congress for Applied Mechanics. 13 Appendix III A. We shall now solve the differential equation encountered in eq. (5). We had at/T = —+* Sin? a 4.. Put Z = — tan 6/2. Then cos $6 + r sin“ 22 4 _ 92 tan Ege in ba pe 5 cok be TEs a6 ae Hence f-7iz 2az t= 20/2 +24 an ee! aa Re eee 4 Buca erie f= 1 + ar — (ari 87s) ee a= — 2a Gly ee) where p? = \/4r? +1 + 2r. Hence we have i 1 = 27/£ + 22 ‘ af 1 = 1/p? = 22/p aT/T = — 2f p? = Z*)(1/p" + Z dZ = — re 1/p* + Z = Lope Lipa ghee 2p(p +z) ap(p — 2 jz as may be readily verified. Integrating, we obtain log T/To = — ya [ ea — 1/p?) tan! pZ — 1/f log (1 + p?Z?) + 2 ee log = 2 so a log (1 — Z?/p?) | where Tp is the tension for Z=0. This last is seen to be equivalent to equation (6). B. Treating equation (9) in a similar fashion, we have aT/T = aS eine on Put Z = tan g/2. Then cos 6 — Yr sin*¢ ° ‘ Riu, : Lun eae: Peale = ee a emezaur ei ® = areas) 1 COS 0) Saye 18S ae 14 A + 2% + + Hence aT/T = eae 4rzZ = af 1 — 47? = dZ hit oe Cte 2) But 1 — 4rz? — 24 2 1 + 4r? — (2r + 22)? = (p? + 2?) (1/p? — 22). : uu t+ ase S27) AZ. _ ao 1 = p* + 22/f tg at/T = 2f (raz 1/p? - Z ire | pe +Z u Leen 1s ~+ 3y/E | 2(1 + pZ 21 — pZ -_— 2 —_ sey log T/Ty = EA Fg E aaa tan 1 Z/p + - log (1 + Z2?/p?) + pitd,,iteZ 1 i 2p 198 t= pz ~ —F lee (1 — p*Z*) where To is the tension for Z=0. This last is seen to be equivalent to equation (10). 15 Appendix IV It has been stated that in finding T, as a function of 6, a close approxima— tion to the set of equations " —~1/ 4 2 ii piZ4q2 log t, ~ 2f Baebes tan py + ore appt 8 Te pF Ae + 1/ 1 + Z4/p fT a7 8 TILL @ s8tsel te. oo *e 2 @ «| '« Ot ar: ©) fe) tay oe (6) a -1 2 722/p? log te = - 2f BS ToBs tan” Ze/p + se a7 log 7 aE + + 4/ 1+ pZe Seer gat ee an, ay ae ae ee stems (10) Z Z where {' Gotan y GREE LAS eata Pal Ladd ° 1- 4, p 1 + p°d, 0 1 + Za*/p*)(1 — p*Za (fromed($)) and (112)! ef aol Sag..2. Bel, spale. 2s 3 SVTGRIO 385 (17) is given by the set of equations log t,° = - tan! pz, Sigotelanmaeai ae Sie, Saasier Yel Buren a ey mel eae) Tog ta? 305s tog J BAND). ae ar say F .. 3010a) Z VitacZada = where J oy : Tp? \ tates oa (from *(Sa) “and (W2a)) 2. ens es (18) Assume that p >> 1. Since p® = \| 4r7 +1 + 2r = 4r, this is equivalent to assuming that 2fr >> 1. Compare (6a) with (6). It is easily shown that the term for log t,° is a good approximation to the sum of the first and third terms for log t4. In fact, one finds on calculation that the difference is given by 2f - —— +1 + Z4/p = tan | ee pZ, + f oe a/pt Spee pad - = (tan DRLa Gee bikie Ania sete: RE sian eli) where 0 < Z,