Civil Engineering - Surveying - Discussion

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Please explain it clearly.

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.

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Can you explain clearly?

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.