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 1 of 3.
Asad 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.
(3)
Pintu Kumar said:
6 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.
(2)
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.
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.
A.J UET, LHR said:
4 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.
(8)
Hirdesh Gupta said:
6 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.
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)
Dheeraj kataria said:
6 years ago
Please explain in a short and simple way. So that each student can understand clearly the actual problem.
Sk prabhakar said:
5 years ago
For roads. -L=12.8√R.
For railways- L=4.526√R.
For railways- L=4.526√R.
(2)
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.
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