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 2 of 3.
Mohit Kumar said:
9 years ago
Explain the question properly.
Mk said:
9 years ago
Sir please explain this question in detail.
Learner said:
8 years ago
I can't understand it. Please explain it.
Gurkamal Singh said:
8 years ago
Centrifugal ratio for road=1/4,
Centrifugal ratio for railway=1/8,
Centrifugal ratio(c.r.)=V^2/Rg where R is the radius and g is acc. due to gravity............Eq. 1,
and also V^2/R=Acc.rate x t where t is the time..........Eq. 2,
V^2 is directly proportional to C.R. since R and g is constant.R is constant since same radius is given for road and railway.
Therefore V is directly proportional to (C.R.)^(1/2),
Now from Eq. 2.
t=V^2/acc. rate*R ..........Eq. 3.
And we also know that L=V*t where L is length v is vel. and t is time.
Putting the value of t from Eq. 3 in above equation we will get,
L=V^3/acc. rate*R
Now L is directly proportional to V^3 since acc. rate and radius are constant for both road and railway.
And V is directly proportional to C.R.^(1/2),
hence L is directly proportional to C.R.^(3/2),
Now L1(road)=k*(1/4)^(3/2).
L2(railway)=k*(1/8)^(3/2).
Dividing L1 by L2
We get L1/L2=(sq. root of 8)=2*(sq. root of 2)=2*1.414=2.828.
Centrifugal ratio for railway=1/8,
Centrifugal ratio(c.r.)=V^2/Rg where R is the radius and g is acc. due to gravity............Eq. 1,
and also V^2/R=Acc.rate x t where t is the time..........Eq. 2,
V^2 is directly proportional to C.R. since R and g is constant.R is constant since same radius is given for road and railway.
Therefore V is directly proportional to (C.R.)^(1/2),
Now from Eq. 2.
t=V^2/acc. rate*R ..........Eq. 3.
And we also know that L=V*t where L is length v is vel. and t is time.
Putting the value of t from Eq. 3 in above equation we will get,
L=V^3/acc. rate*R
Now L is directly proportional to V^3 since acc. rate and radius are constant for both road and railway.
And V is directly proportional to C.R.^(1/2),
hence L is directly proportional to C.R.^(3/2),
Now L1(road)=k*(1/4)^(3/2).
L2(railway)=k*(1/8)^(3/2).
Dividing L1 by L2
We get L1/L2=(sq. root of 8)=2*(sq. root of 2)=2*1.414=2.828.
Asma said:
8 years ago
Please explain it clearly.
JAY PANCHAL said:
8 years ago
Explain clearly?
Ranjeet kumar said:
8 years ago
Well said, Thanks @Gurkamal.
Vijay reddy said:
8 years ago
Please explain it clearly.
Sahil prajapati said:
8 years ago
Well said. Thanks for the explanation.
Muhammad said:
7 years ago
Thanks @Gurkamal Singh.
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