A System of Useful Formulae Adapted to the Practical Operations of Locating And

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= 7. 7820505 C. = 00 41 37" -46 C s 1720 Q3/ 40" -26 F K 70 14 42" -27 1800 00 00" -00 = 1780 45 17" = 1720 05 24" -65 = 60 39 52" -35 211730 20 07" -65 = J (1800 C 6 ) = 860 40 03"- S 6 =860 40 03" -82 r 1 + J h = 502 075 feet C 6 =60 39 52" -35 ch 6 = 58-233 feet co. Ar. Sin. = 0-0007349 log. = 2-7007686 sin. = 9-0646679 log. = 1-7661714 the distance from the mouth of the switch on the outside of the turnout to the frog angle in the main track.
TURNOUT FROM INSIDE OF CURVE. 99 (45) Having thu
...s completed our investigation of the problem direct, we will now examine it reversed, by supposing the radius of the main track given, as before, viz. , r = 5729 597; h = 47 feet ; d = -416 feet ; and the frog angle Fr = 7 14 42". 27 ; the switch angle, Sw = 1 08 12"; to find the radius of the turnout = /, and the position of the frog.
Draw the lines ee, cc representing the outer and inner rail of the main track, and the dotted line //, corresponding to the switching of the outer rail ; then, draw the radius F C ; then, from F draw the line F C indefinitely, making an angle with F C = the frog angle ; then, from F draw the line F C", making an angle with F C = the switch angle -f- the frog angle, and equal in length to the radius of the dotted line = C G; then, with a radius = C F and with C" as a centre, draw the angular dotted line at S, and this dotted line will intersect //at the mouth of the switch.


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