See. 45.14 



FRICTION-RESISTANCE CALCULATIONS 



111 



drag of a flat plate of equal area. This takes the 

 form 



1^ = 1 + 0.54 ^ C, 



(45.viii) 



where 



Rp is the friction drag of the convex surface 



Rfo is the friction drag of an equal area of flat plate 



Li is the length of the convex surface 



Si is the area of the convex surface 



Cf is taken for the whole body or ship. 



The sides are vertical at the WL, hence the girth 

 angle from the WL on one side to the WL on the 

 other is 180 deg. 



Taking Li = L and Si = S, in other words con- 

 sidering the whole ship form, L\/Si may be 

 expected to vary from about 10 for a fine form to 

 4 for a full form. For the ABC ship designed 

 in Part 4 it is 5.8. 



Considering only the region of sharp, convex, 

 transverse curvature on a body or ship, the value 

 of Ll/Si for that region may be considerably 

 higher than the values given; in fact, several 

 times as great. If two I'elatively sharp bulges are 

 involved, each having a girth angle of 90 deg, as 

 where the bottom joins the two sides, the bulges 

 and the angles may be combined to form a smaller, 

 separate body with a girth angle of 180 deg. 

 Eq. (45.viii) may then be used as given. There is 

 a question as to whether to use a value of the 

 specific resistance coefficient Cp for the ship 

 length L or for only the bulge length Li . Since 

 the boundary-layer thickness is determined largely 

 by the ship length L it seems reasonable to use 

 the same Cp as is picked from Table 45.d on 

 page 102 for the ship as a whole. 



Taking as an example a modern Great Lakes 

 ore carrier about 635 ft long [MESR, Jul 1952, 

 p. 65], the bilge radius is 3 ft and the length L 

 of uniform bulge is 331 ft on each side. Ll is then 

 109,561 ft^ and Si is (3)Tr(331), or 3,119.6 ft^ 

 whence Ll/Si = 35.12. The service (sea) speed 

 is 14.15 kt, whence R„ for fresh water from 

 Table 45.a is about 1,242 million and Cp from 

 Table 45.d is 1.491(10"'). Then 



Rf 

 Rp 



= 1 + 0.54(35. 12)(1.491)(10"') 



= 1 + 0.0283 



whence A^ of Eq. (22.iv), in Sec. 22.15 of Volume I, 

 is 0.0283, or just less than 3 per cent. The value 

 of A,Cp is 0.0283(1. 491) (10'') = 0.0422(10"'). 



No procedure has as yet been worked out for 

 making allowances for concave transverse curva- 

 ture. This is seldom of great degree, except for 

 the coves of discontinuous-section hulls, the 

 coves along the inside corners of roll-resisting 

 keels, or the coves along the superstructures of 

 some types of submarines. In these cases, the 

 adjacent convex curvatures on the chines and 

 edges largely neutralize the concave curvatures, 

 so that both may be neglected. 



For estimating the effect of longitudinal curva- 

 ture by Horn's method, described in Sec. 22.5 of 

 Volume I, some knowledge or estimate of the 

 sinkage at given speeds, derived from model tests, 

 may be used. Alternatively, ARp may be calcu- 

 lated from the L/B, B/H, and Cp values by the 

 formula given by F. Horn [3rd ICSTS, Berlin, 

 1937, Eq. 2, p. 24], as follows: 



ARp = 0.01 



(11.25 - L/B)- 



[ 



•(0.35 + Cp)(l.3 - ^) 



(45. ix) 



This is a good approximation for normal forms of 

 not extreme proportions and for a range of F„ 

 values from 0.0 to 0.35, T, from 0.0 to 1.18. 



For the ore carrier previously referenced, 

 L/B = 635/70 = 9.07, B/H = 70/24.885 = 

 2.812, and Cp = 0.881, from which ARp (also 

 numerically equalto AiCp) is found to be 



ARp = 0.01 



l_^ 



.25 - 9.07)' 

 5 



+ 



'■■^] 



•(0.35 + 0.881) 1.3 - 



2.812 

 10 



= 0.04326 



For the ABC design, where L/B = 510/73 = 

 6.99, B/H = 73/26 = 2.81, and Cp = 0.62, 

 substitution in Eq. (45.ix) gives 



ARf 



0.0l[^ 



25 - 6.99)- 



+ 2.5 



•(0.35 + 0.62) 1.3 



2.81 

 10/ 



= (0.01)(6.13)(0.97)1.019 = 0.0606 



For the Lucy Ashton, a ship of quite different 

 form [INA, Oct 1953, p. 350ff], having an L/B 



