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Curve No. 1 SE

The document provides calculations for the TRL and maximum speed on a curve with a radius of 1200 m and curvature of 1.458°. The maximum sectional speed is determined to be 50 KMPH, with cant deficiency and excess values calculated to be within permissible limits. Additionally, the desirable transition length is established at 5.6 m, while the provided transition curve length is 90 m, confirming compliance with safety standards.

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Dy CE Con-2 Rourkela
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23 views1 page

Curve No. 1 SE

The document provides calculations for the TRL and maximum speed on a curve with a radius of 1200 m and curvature of 1.458°. The maximum sectional speed is determined to be 50 KMPH, with cant deficiency and excess values calculated to be within permissible limits. Additionally, the desirable transition length is established at 5.6 m, while the provided transition curve length is 90 m, confirming compliance with safety standards.

Uploaded by

Dy CE Con-2 Rourkela
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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TRL & Maximum Speed Calculation for 1.458° Curveature 1200 m.

Radius
Curve No. 1 Maximum sectional Speed = 50 KMPH
Limiting Value of Ca = 165 mm Gr.A,B,C
R= 1200 m ( 1.458 °) Limiting Value of Cd = 75 mm
Proposed Cant = 20 mm
A) GV
2
1750 x 50 ^2
Ca + Cd = = = 28.7 mm.
127R 127 x 1200

Cant deficiency Cd calculated = 28.7 - 20 = 9 mm. < 75mm. O.K.

B) Cant Excess Consideration


Considering speed of Goods Train = 50 Km. Per Hour
2
GV 1750 x 50 ^2
Ca + Cd = = = 28.7 mm.
127R 127 x 1200.000
Cant Excess = 20 - 28.7 = -8.7 mm. < 75 mm. O.K.

C) Maximum sectional Speed based on Ca & Cd (assume above)

Vmax = 0.27 x 1200.000 x( 20 + 9 )


= 50 Kmph Say 50 Kmph (Max. Permissible Speed)
Desirable Transition Length = Maximum of the following: -

(i) L = 0.0056 Ca x Vmax = 0.0056 x 20 x 50 = 5.6 m.


(ii) L = 0.0056 Cd x Vmax = 0.0056 x 9 x 50 = 2.4 m.
(iii) L = 0.72 x Ca = 0.720 x 20 = 14.4 m.
Desirable Transition Curve Length (Maxm. from the Equn. (i) - (iii) = 5.6 m
Length of Transition Curve Provided as = 90 m hence OK
Cant Gradient = (Ca / Trl. X 1000)= 1 in 4500 > 1 in 720 , hence OK

PROP. SE = 20 mm.
PROP.TRL = 90 Mtr.
PERMISSIBLE SPEED = 50 KMPH

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