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210 in the solution of similar problems. The method of computation is as fol- lows: w == Weight per unit of line. Length from lowest point to (4%, y). = 2] = Length between supports at oe elevation. S = Span between supports a equal elevation. H = Horizontal pull (same unit as W1). Y = Pull at x, y in direction of line (same unit as wh). The equation of the catenary is-- (1) 5 pe) \ Sy oa (2) (@) I=Z} oC) 2° @ | =a sinn (2) Differentiating (1) gives-- (3) Gay] e() --e~@) | =sim (2) But from the figure-- dy lw, "7 a ax A whence-- 1 lw ' A -- = --, and a= -- ae Wy Also. from figure-- H=P cos sin | '(2)- Bocce sin. 2 EXAMPLE, Let-- : 2 = 1000 ft. P= 10,000: Ib. ih == 2 1p per ft. of line: Then H=P cos sin "&) = soe? =10,000 cos sin ae 9,949 87 pounds. H 9,949, a Wir 2 andcosh. (:) = 1.005038. Whence-- x eo ee 01003353; ' a a s x of. oo eS (;)- @ 401403) X 2 = 2 2a a = 2 X 4,974.93 X 1.003353 = 998.324 ft, L -- S$ = 1,000 -- 998.324. = 1.676 ft. fips fist [ cosn ()-] = 4,974.93 x 0.05038 = 25.064 ft. (eo) H = 19000 Dip ==°35 ft of which 15. ft, are in air, and 20 4€ gre in water; w, = 6.25 lb. per ft, in air. length THe MarRINE. REVIEW = 625 xX 4/5 = 5 Ib. per ft. in water. j if 19,000 -- in water = WA The length of line submerged may == 3,500. be found on Catenary Chart No. 1 by' H locating the point a (dip = 20 and -- Wt = 3,800) and interpolating on the line og drawn through the point a. More accurate results would be obtained by taking 1/10 ft. for the unit of length, thus:--z, becomes 0.5 lb. per tenth ft, the dip is expressed as-- 20. x 10 = 200. and-- H 19,000 -- = ------ = 38,000. Wi 0.5 H Locating the point a' (dip = 200 and -- U1 = 38,000) and interpolating on the line Ke' drawn through a' the length L is found to be 7,810 tenths ft. or 781 ft. The "length minus span' is found on fatenary Chart No, 2 point a: or. @ £0 be 1.3/7 ft, Now 781 ft. of line in water is about equivalent in weight to-- 4 781 X 2 == 625 ft. in air, and 625 ft. of line-in air with-- A 19,000 -- = ---- = 3,040, or-- 19,000 ---- = 30,400, 0.625 would have a dip (see Catenary Chart 160 No.1, pomt 0b) of ---- == 16 ft. 10 The length minus span (Catenary Chart Noe 2, point. 0) is 1.1 ft, The total give-- length out of water to 16 + 15 = 31 ft. dip is (Chart No. 1, pomt-<) 871 ft, The corresponding length. minus span is (Chart No. 2, pomt ¢) 295. ft The length out of water for the conditions described will be 871 -- 625 = 246 ft. The length minus span for this line will be 2.95 -- ll = P 185 ft: =e 3.070; Wy P = 3,070 X 6.25 = 19,200, (point c', Charts No. 3 and No. 4). The total length of the line in and above the water will be 781 +. 2446 == 1,027 ft. The total length minus span = Ae So. eer g,22. ft, If we had assumed the whole line to be out of water with-- 1.37 167,000 -- 79,000 -- --- en ee 3,040, and-a dip.of 35 ft. L would have been 925 and length minus span = 3.55 ft. If we had assumed the whole tine to be submerged, 7. e., HH . jo. Wi then for a dip of 35 ft, L = 1034 and length minus span = 3.15 ft. If supports are: chosen at unequal elevation. the length of the line is the mean between that for both supports at the lower elevation, and that for both supports at the. upper elevation, not that for both supports at the mean elevation. : 'ele | 3,800, (cl) When H = 391% tons or 79,000 lb., Ne 79,000 -- in air --= ---- = 12,640 Wh 6.25 Dip = about 10%4 ft. (Chart No. 1, point d). = 03: ft (Chart Length minus span No, 2,.point d.) ; P and H are so nearly equal -that they may be used here interchangeably. Stretch due to added pull is then about 79,000: -- 19,000 a x 0.0058 xX 1,027 160,000 ee 225 Et: (See Note e.) (2) -- When H = 83% tons or 167,000 lb. Ht 167,000 -- in air = == 26/00 Wh me O2s ; Dip is about. 5. ft. (Chart "Noo point e). Length minus span about 0.075 (Chart No: 2; point. e¢). Stretch due to added pull is-- <x 0.0058 x 1,027 160,000 = G.2/9 ft. (¢3) When H = 26,600, H 26,600 = i water = ----- = 53,20 Wy 0.5 3 Phe dip for a length of 662 ft & {0} it. A 26,600 -- Un: an = == 42.600. W 0.625 The dip for 364 ft. (182 ft. in each end of a length of 893 ft.) is 15 ft. of the 23.2 ft. for the total dip of the eos ft. fencth.. in = air, Whence the dip for-- S64 {tino ai is.:, 15.0 f 662 ft. in water is..10.3 ft. And _ total dip tor 1,026 ft, in air:.and Water i8.«.:. 25.0 tt. The stretch due to the added load is