And 6° 52'

1°18'

5° 34' amount of curvature between opening of rail and point of frog. By the first method, when the distance between tracks feet we have 13 x 800- 102 feet nearly for distance from origin

of curve to point of reversion.

13

But if the point of reversion be made at the point of frog, the distance between nearest rails of tracks being 7 feet, we have 6: 7 :: 800: 933 3= radius of curve with which to leave frog, and 6: 7: 98: 1143-distance from frog to end of turnout.

Or making the movable rail tangent, and its opening 5 inches, angle of opening being 1° 18', the point of reversion being made at frog, to find the angle of frog, we have:

99274 cosine 6° 55' nearly the same as before.

Suppose the turnout is on a curve running in the same direction, say a 2, with a radius of 2865 feet. Now an 800 feet radius gives a 7° 10' curve, and 7° 10' 2° =5° 10' relative departure from main track. But the radius of a 5° 10' 1109 feet; then

=

=

1/2 x 1109 x 6x -2-115.3. distance from origin of curve to "point of frog.

Therefore to make a turnout from a 2° curve and running the same way would require e 115 feet. It tail rolerent reqqe lliw d

If it were required to keep the distance the same as on a straight line, it would be necessary to make the 7° 10′ curve a 9° 10'

curve of 625 feet radius.

If the 2° curve run in the opposite direction of the turnout, and the radius was 800 feet, then the convergence will be 7° 10' +2° 9° 10′ curve, and the radius of a 9° 10' curve being 625 feet, we

=

have:

x = √2 × 625 × 6 = 7500 86.6distance from origin of curve to point of frog.

When the main track is a curve, and it is required to get on to a side track running parallel thereto.

Note. In treating of turnouts. When the main and side track are curves, the movable rail is considered a part of the curve used for turnout, according to method ist.

Let E M be the main track on a curve of 2865 feet radius. It is proposed with a turnout from E, with a curve of 800 feet radius, to fall upon the side track B N, distant 13 feet from the main track, and running parallel thereto. Now 2865 feet radius denotes a 2° curve, and 800 feet radius is a 7° 10' curve. Therefore the diver

gence of the curve E F from the curve E M is equal to (7° 10' — 2°) =5° 10' curve; and the radius of a 5° 10' curve being 1109 feet, the divergence of the curve EF from the curve EM is equal to that of a curve of 1109 feet radius

By similar reasoning, the convergence of the curve FB towards being parallel with EM is 9° 10' per hundred feet, which may be expressed by a radius of 625 feet from tangent. Then we have 1109+6251734: 1109 :: 13: 8:31 distance of point of reversion from main track. Now since x = 12 R. G, we have by substituting ✔ x 1109 x 8:31 135.7 distance from origin of curve to point of reversion, radius used being 800 feet. The radius of relative curvature being expressed in the formula, we have the proportion 1109: 625 :: 135-7: 76.56 distance from reversion to

2d track.

Suppose it be required to put the side track on the opposite side, then we have 1734: 625 :: 13: 468 distance of point of rever. sion from side track. Then we have the formula √2 × 625 × 4·68 76.48 distance from origin of curve to point of reversion. Then 625: 1109 :: 76 48: 135·7: = distance from point of reversion to side track.

ON RUNNING CURVES BY OFFSETS, OR WITHOUT THE USE OF AN INSTRUMENT FOR MEASURING ANGLES.

FIG. 20.

From a tangent EA let it be required to run a curve ABCD, having for its radius O C. To do this we have only to find HC and its half MCG B.

Suppose the chords A B, BC, CD are equal in length, being 100 feet each. The chords, and consequently the arcs, being equal, the angle H B C is twice the angle GA B. But G A B is measured by half the arc A B = BC, consequently the angle HBC is measured by the whole are BC. But the angle BOC is also measured by the arc B C, consequently the angles HBC and BOC are equal. Now triangle BO Cis isosceles, and B.H being equal to BC triangle HBC is isosceles also; conséquently the two triangles are similar, and we have the proportion:

HC: BC: BC: B O, consequently HC =

BC2
BO'

or HC=

The square of the uniform length of chord divided by radius will give the linear deflection from chord produced to curve, or half of this will give the deflection from tangent produced to curve.

Suppose A0 = 2865 feet, the radius of a 2° curve, then we have

HC=

10000
2865

=3·49 or 3.5 feet nearly; and G B=} of 3-5—1-75.

Since the angle GAB=1° the deflection for 1° per hundred

Suppose we run the curve around to a point which we will call station 10, or 1000 feet from beginning. The point Q, which is less than 100 feet distant from station 10, say 50 feet, being at station 10 4 50.

Suppose this a 2° curve compounded at station 10+50 to a 3°; eurve of 1910 feet radius. Now the instrument setting on station' 10 with a backsight on station 9, the instrumental deflection to 10 +50, 150 feet, will be 1°30'. Now since 1° per 100 feet is 1-75,' that of 1° 30' will be 2:62 feet But the last chord being but 50 feet, or half of a hundred, the deflection will be half of 2-62 hence we have the following rule:

131;

Multiply together half the curvature in degrees: instrumental deflection between the backsight and point required, the length of the last chord and 175, and the product is the distance from chord produced to point required.

Case 3d.

Suppose the curve from 10+50 to station 11 is a 3° curve of 1910 feet radius. Now the deflection from chord to tangent, from station 10 to station 10 + 50, is 0° 30′, and the deflection from tangent to chord between 10 + 50 and 11 is 0° 45', therefore the entire deflection: = 30'+45′ = 1° 15'. Now 1° 15 in a hundred =1·75 × 1 = 2:18, and for 50 feet will be 109 feet. Find station 12 by Case 2d, for 150 feet) x 1·75 = 3·93 station 12 on curve.

thus 21° (= instrumental deflection deflection from chord produced to

Continue the curve around as at first, observing to measure from curve to tangent the same deflection as from tangent to curve, or half the usual chord deflection; the tangent point being supposed, a full station. If not a full station, ascertain the tangent point by Case 2d, and the next full station on tangent by Case 3d.

Having produced two tangents to an intersection at C, ît îs required to connect them with a eurre of given length. Fia. 21.

When the angle made by tangents is not greater than 15° the distance from vertex to the two ends of the curve will not differ materially from half the length of the curve.

Fig.21.

B

Suppose the tangent DC produced 100 feet to E, measure C X = 100 feet, measure E X. Now suppose it is 21 feet.

Now the deflection of 1° for 100 feet is 175, and vature.

21

6

21 1.75

= 12° eur

Suppose it is required to divide the curve into 6 stations. Then

35, the deflection for 2° in 100 feet. Hence it is a 2° eurve.

Or 12° divided by 6 stations gives a 2° curve also. The defleetion being 175 from tangent to curve.

=

Between two fixed points to supply the intermediate points by ordinates from the chord. FIG. 22.

By what has been previously demonstrated, the middle ordinate

4 to 4 will be expressed by

4 x 4
2 R

At 3 the deflection from tan