. American engineer and railroad journal . dius of the main track = A,the radius of the turnout — A, (he standard gauge = g,and the narrow gauge = g. In the triangle ./ /> C we have the side ./ C = A* + A^ — tr, the side A J> = A + i g, and the side li C =A? -/?. , . (/?-/?) + (A- + i^ - g) + (A- + ig)Let J = ^ « then, from trigonometry, we haveSin. i BAC :/i -(^ + ig-g)] [s-{/i + ig)] (A + ) C^+) («4) metre = ft.—to find the frog angle D A E = /? . / C,fig. 12 : _(573 69-38o)4.(573-69+2-625—3-28i)+()s— ^ _ s — {R-ykg-g) = ^Mo 02148438 s — (/?+ ii g) = 1
. American engineer and railroad journal . dius of the main track = A,the radius of the turnout — A, (he standard gauge = g,and the narrow gauge = g. In the triangle ./ /> C we have the side ./ C = A* + A^ — tr, the side A J> = A + i g, and the side li C =A? -/?. , . (/?-/?) + (A- + i^ - g) + (A- + ig)Let J = ^ « then, from trigonometry, we haveSin. i BAC :/i -(^ + ig-g)] [s-{/i + ig)] (A + ) C^+) («4) metre = ft.—to find the frog angle D A E = /? . / C,fig. 12 : _(573 69-38o)4.(573-69+2-625—3-28i)+()s— ^ _ s — {R-ykg-g) = ^Mo 02148438 s — (/?+ ii g) = R + ig - g = 573 034 ar. comp. A + h g = ar. comp 7 4172267 Extracting sq root 2) i .5 ^ C = 2° lo 19 sin. whence the angle I> A C is 4 20 38 —the answer find the frog angle D .1 E = B A C, fig. n, ofIhe third or double pointed frog, given the radius of themain track = A, the radius of the tournout = A, thestandard gauge - g, and the narrow gauge = Example No. i .• Given the radius of the main trackA = 573 69, the radius ot the turnout A^ = 380. the stand-ard gauge ;, = 4 ft. 8A in. = the narrow gauge j, =3 ft. o in.—to find the frog angle D A E = B A C, fig. 12 : _(573-69-3So) +(573-69 +2-354-3-ooo) + (3So + 2-354)_.,, ,,,J— —574-544 Here in the triangle A /> Cv/e have the side A C = Ri ■, the side A B = R + ig — •/, and the side /> CR - R. As before, let s = Ihe i sum of the three sides, or (R - R) + {R - ig) + jR + ig -g)
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Keywords: ., bookcentury1800, bookdecade1890, booksubjectrailroadengineering
