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
2.828
3.828
1.828
0.828.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
26 comments Page 1 of 3.

A.J UET, LHR said:   2 years ago
In question, it is asking "length of transition curve of road (L1) to railways (L2) ".

The simplified formula is.

L1/L2 = ( (C.Rfor road) ^3/2) /( (C.R for railway) ^3/2).

Put values in the above formula.

L1/L2 = ( (1/4) ^3/2)/( (1/8) ^3/2) ) = 2.828.
(5)

Asad said:   4 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 the 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 the 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=(√8)=2*(√2) = 2*1.414 = 2.828.
(3)

Sk prabhakar said:   4 years ago
For roads. -L=12.8√R.
For railways- L=4.526√R.
(1)

Dheeraj kataria said:   4 years ago
Please explain in a short and simple way. So that each student can understand clearly the actual problem.

Hirdesh Gupta said:   4 years ago
C.R.=v^2/Rg ---> eqn1
From this formula, v is proportional to under root of C.R.
L=V^3/CR ---> eqn2.

Now from enq 1 we can say that.
L1=(C.R.1)^(3/2),
L2=(C.R.2)^(3/2),
C.R.1=1/4,
C.R.2=1/8,
THEN L1/L2=(1/4)^(3/2)÷(1/8)^(3/2)=2^(3/2)=2.828.
(2)

Pintu Kumar said:   5 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 the 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 the 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=( √8)=2*(√2) = 2*1.414 = 2.828.
(1)

Muhammad said:   6 years ago
Thanks @Gurkamal Singh.

Sahil prajapati said:   6 years ago
Well said. Thanks for the explanation.

Vijay reddy said:   6 years ago
Please explain it clearly.

Ranjeet kumar said:   7 years ago
Well said, Thanks @Gurkamal.


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