The crossover between two parallel tracks with an intermediate straight length can be designed by applying any one of two methods.
Coles design is a simple layout. In this case, two parallel tracks at a distance D from each other are connected by a crossover with a small length of the straight portion of the track lying between the two theoretical noses of the crossing. The straight portion of the track (ST) can be calculated using the formula
where G is the gauge of the track and N is the number of the crossing. The overall length (OL) of the crossover from the tangent point of one track to the tangent point of the other track is found by adding the lengths of the curve leads of the two turnouts and the length of the straight portion in between the two TNC (Fig. 15.8).
Overall length = OL of one turnout + ST + OL of other turnout
Example 15.1 A 1 in 8.5 crossover exists between two BG parallel tracks with their centres 5 m apart. Find the length of the straight track and the overall length of the crossover. Use Coles method. Given D = 5 m, N = 8.5, G =1.676 m.
(i) ST = (D - G)N - G^l + N2
= (5 - 1.676)8.5 -1.676^1 + N2 = 13.91 m
(ii) OL = ST + 4GN
= 13.91 + 4 x 1.676 x 8.5 = 70.89 m
Example 15.2 A crossover is laid between two BG straight tracks placed at a distance of 5 m c/c. Calculate the (i) overall length, (ii) radius of the curved lead,
(iii) lead distance. Heel divergence of 1 in 12 crossing = 133 mm.
Solution The crossing number is equal to 12 and the intermediate portion is straight.
(i) ST = (D - G) N - Gjl + N2
where D = 5 m, G = 1.676 m, and N = 12,
ST = (5 - 1.676) 12 - 1.676 yjl + 144 = 39.88 - 20.18 = 19.69 m
(ii) OL = ST + 4GN
= 19.69 + 4 x 1.676 x 12 = 100. 13 m
(iii) Radius of the turnout curve (R):
R = 1.5G + 2GN2
= 1.5 x 1.676 + 2 x 1.676 x 12 x 12 = 485 m
(iv) Lead of crossing (L) = curve lead - switch lead
= 2GN -V2Rd - d2
= 2 x 1.676 x 12 -<J2 x 485 x 0.133 - (0.133)2 = 40.2 - 11.4 = 28.8 m
In IRS design, the distance from the TNC measured along the straight track is given by the formula
ST = (D - G - G seca) cota (15.8)
ST = Dcota - Gcota/2 (15.9)
where ST is the distance from TNC to TNC along the straight track, D is the distance from centre to centre of two tracks, G is the gauge, and a is the angle of crossing.
Similarly, the distance from TNC to TNC along the crossover is given by the formula (Fig. 15.8)
CF = (D - G - G seca) coseca + G tana (15.10)
where CF is the distance from TNC to TNC along the crossover, D is the distance from centre to centre of two tracks, G is the gauge, and a is the angle of crossing.
Example 15.3 A 1 in 12 crossover of IRS type is laid between two BG parallel tracks with their centres 5 m apart. Calculate ST and the distance from TNC to TNC along the crossover.
G = 1.676 m, N = 12, D = 5.0 m, a = 4° 45' 49"
(i) ST = D cota - G cota/2
= 5 x 12 - 1.676 x 24.04 = 19.7 m
(ii) CF = (D - G - G sec a) cosec a + G tan a
= (5.0 - 1.676 - 1.682) x 12.04 + 1.676 x (1/12)
= 19.91 m