Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 2 (Q.No. 6)
6.
If the rate of gain of radial acceleration is 0.3 m per sec3 and full centrifugal ratio is developed. On the curve the ratio of the length of the transition curve of same radius on road and railway, is
Discussion:
26 comments Page 3 of 3.
Student said:
9 years ago
For road centrifugal ratio = 1/4.
For railway = 1/8.
Centrifugal ratio = v^2/Rg.
R = radius
(v^2/r) = Acceleration rate * t.
t = time
t = L/v.
L = length.
v = velocity.
Here we get L=(v^3)/(acceleration * r).
Acceleration is given and r is same for road and railway.
So, L is directly proportional to v^3 and from centrifugal ratio = v^2/Rg.
v is directly proportional to square root of the centrifugal ratio.
From here ratio of length for road and railway;
L1/L2= (1/2)^3/(1/square root 8)^3.
= 2 * square root 2.
= 2 * 1.414.
= 2.828.
For railway = 1/8.
Centrifugal ratio = v^2/Rg.
R = radius
(v^2/r) = Acceleration rate * t.
t = time
t = L/v.
L = length.
v = velocity.
Here we get L=(v^3)/(acceleration * r).
Acceleration is given and r is same for road and railway.
So, L is directly proportional to v^3 and from centrifugal ratio = v^2/Rg.
v is directly proportional to square root of the centrifugal ratio.
From here ratio of length for road and railway;
L1/L2= (1/2)^3/(1/square root 8)^3.
= 2 * square root 2.
= 2 * 1.414.
= 2.828.
Manas Kumar sahoo said:
10 years ago
Can anybody clarify the concept?
Yousuf Mallick said:
10 years ago
Anybody help?
Chirag said:
10 years ago
Please some one give the solution.
Anand said:
1 decade ago
Can anybody clarify it?
Suraj said:
1 decade ago
Any one have this solution?
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