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

R=1,433, r= 819, B=34° 20', D=94.

Cos. C= [(R+r) cos. B + D] ÷ (R+r).

XLV.

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

R

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.

In this triangle, x= (180 — A)

Also, CO: IO:: sin. x: sin. z.

But COR; IO

2

- a 90°.

(} A+ a).

=

R
cos. A'

..R:

R cos. A

:: sin. x sin. z.

sin. x

Hence sin. z =

cos.

A'

The triangle ICO may then be

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'.

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

R2 (Ra)2 + (R—b)2,

from which equation,

R=a+b+2ab.

a

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

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 O, 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 HN W, between the line of the switch-rail prolonged and a tangent to turnout curve at frog point O, NOP NIWFS. The angle NOL or NLO, between chord and tangent, half the intersection angle HNW

=

=

=(FS). The angle NOB=NOL+LOB. But NOL is seen to be (F S), and NOB = F; then LOB= F(FS) = } (F+ S). The distance

=

NOB-NOL:

=

LO, from toe of switch to point of main frog, = LB÷sin.

LOB (G— D) ÷ sin. † (F+S).

4. Again: the angle LCY

LO =

NLO (F — S); LY=}

(G D) sin. (F + S). LY÷sin. LCY: = LC; i.e., [ (G — D) ÷ sin. † (F + S)] ÷ sin. † (F — S) =R.

5. R may be found otherwise, as follows:

OK

= OC cos. KOCR cos. F; LM = LC cos. CLM R cos. S; LM-OK=LB; i.e., R (cos. S

D).

-

=

cos. F) (G

=

Hence R = (GD) (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.

IPPE or OK; i.e., R cos. F' =† G+ R cos. F. Hence nat. cos. F' nat. cos. F+ († G ÷ R).

8. The angle LIV, by similar reasoning to that used in relation to LOB, is found to be (FS). The distance LI, from toe of switch to point of middle frog, = LV÷sin. LIV (G − D) ÷ sin. Į (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