# 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 sec

^{3}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, isDiscussion:

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.

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.

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.

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)Ã·(1/8)^(3/2)=2^(3/2)=2.828.

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)Ã·(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.

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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