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In the figure, draw L K parallel to G V, and passing through centre of first branch.

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Then MK= (R+r) X cos. B.

NL (R + r) X cos. C.
WL=GK.
NL=r+D+GK.
MK=r+GK.

NL - MK= D.
.: (R+ r) X cos. C — (R+ro) X cos. B=D.
.: (R + r) X cos. C = (R+r) X cos. B + D.
.:. Cos. C=(R + r) X cos. B + D : (R+r).

A = (90° — C) — (90° — B)=B — C.

Example.
R=1,433, r = 819, B= 34° 20', D= 94.
Cos. C = (R+r) cos. B+D; (R+r).
R+r=2,252

log. 3.352568
B= 34° 20', cos.

log. 9.916859
(R+ 7) cos. B - 1,860

3.269427
Add D 94
1,954

log. 3.290925
(R+r)

log. 3.352568

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TO PASS A CURVE THROUGH A FIXED POINT,
THE ANGLE OF INTERSECTION BEING GIVEN.

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Given the intersection angle, A, of two tangents, to find the radius, R, of a curve which shall pass through a point, C; the position of said point, with reference to the tangents or the point of intersection, being known.

1. By what data soever point C is located, they may be commuted by simple processes to the form shown in the figure; namely, the ordinate BC and the distance IC to apex. Call the angle BIC a, and complete the triangle ICO.

'180 - A In this triangle, x =

- a= 90° = (A + a).

2 Also, CO:10:: sin. X: sin. 2.

R

R
But Co=R; IO=

..R:

:: sin. X: sin. 2. cos. | A

cos. LA
sin. X
Hence sin. z =

The triangle ICO may then be

cos. * A

(180,- A)

Example.
A = 40°, BC= 32 feet, IB= 80 feet.

Then BC • IB = 32 - 80 0.4 nat. tan. 21° 49'; and IC=BC : nat. sin. 21° 49' = 32 • .372 = 86 feet.

Also, x= 90° (1 A +a)= 90° (41° 49') 48° 11'.

Next, sin. X, 48° 11'

log. 9.872321 Divided by cos. $ A, 200

log. 9.972986 = sin. 2, 127° 31'

log. 9.899335 Or, since the sine of any angle is equal to the sine of its supplement, the supplement in this case, 52° 29', may be taken directly from the logarithmic table, from which supplement deducting x, or 48° 11', the remainder is the angle y = 4° 18'. Finally, IC= 86

log. 1.934498 Multiplied by sin. X, 48° 11'

log. 9.872321 CD

log. 1.806819 And C D divided by sin. y, 4° 18'

log. 8.874938
= CO=R
say, 855 feet

log. 2.931881 2. In the case of a rectangular intersection, the solution is more simple. It is quite plain, from the figure, that –

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R?= (R — a)2 + (R — b)?, from which equation,

R=a +6+12 ab.

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3. Cases of this kind are disposed of with great ease in the field by means of the curve-protractor.

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TO FIND THE RADIUS OF A TURNOUT CURVE, THE FROG

ANGLES, AND THE DISTANCES FROM THE TOE OF SWITCH TO THE FROG POINTS.

1. Draw the figure as in margin, C being the centre of the turnout curve, CK parallel to main track, and OK, IE, LM, perpendicular to it. Call the angle of the frogs at 0, F; that of the intermediate frog at I, 2 F'; the throw of the switch-rail for single turnout, D; its angle with main track, S; the gauge of the track, G; and radius of outer rail, R.

2. Usually the length and throw of switch-rail and the angles of the frogs at O are given. In that case, to find R, F, 3. The angle HNW, between the line of the switch-rail prolonged and a tangent to turnout curve at frog point 0, NOP NII W = F – S. The angle NOL or NLO, between chord and tangent, - half the intersection angle HNW

(F-S). The angle NOB=NOL+LOB. But NOL is seen to be - $ (F S), and NOB = F; then LOB= NOB - NOL=F Ž (F – S)=} (F+S). The distance LO, from toe of switch to point of main frog, = LB : sin. LOB=(G-) = sin. 1 (F +S).

4. Again: the angle LCY N LÔ = (F– S); LY=; LO } (G

D) • sin. $ (F – S). LY • sin. LCY LC; i.e., [1 (G - D) • sin. $ (F + S)] • sin. $ (F – S)

R.
5. R may be found otherwise, as follows:-

OK OC cos. KOC=R cos. F; LM=LC cos. CLM= R cos. S; LM – OK=LB; i.e., R (cos. S — cos. F)=(G – D). Hence R= (G - D) = (nat. cos. S — nat. cos. F).

6. If R be known, to find F. This equation gives nat. cos. F = nat. cos. S - |(GD) R).

7. To find the angle, 2 F', of the middle frog at I.

IE IP + PE or OK; i.e., R cos. F'=fG+ cos. F. Hence nat. cos. F' = nat. cos. F + (1 G + R).

8. The angle LIV, by similar reasoning to that used in relation to LOB, is found to be : } (F' +S). The distance LI, from toe of switch to point of middle frog, =LV • sin. LIV

(I G – D) - sin. (F' +S).
The preceding formulas translate into the following –

RULES FOR FROGS AND SWITCHES.

9. To find the Angle of Switch-Rail with Main Track. Divide its throw, in decimals, by its length: the quotient will be the natural sine of the angle sought. 10. To find the Distance from Toe of Switch to Point of

Main Frog. Subtract the throw of switch-rail from the gauge of track, both in decimals; call the remainder a. Add together the angle of switch-rail with main track and the angle of the main frog; find the natural sine of half this sum, and call it b. Divide a by b: the quotient will be the distance

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