Civil Engineering - Surveying - Discussion

Here h= 1.5, it is not correct because here 30 m is hypotenuse (i.e. equals to L , which is also used in slope correction formula ) but why use as a base or horizontal for finding h? Please explain me.

Let me clearly explain,
tanθ = height/ base.
=> tanθ = 1/20,
=>θ = 2.86.

Now actual slop correction = l(1-cosθ).
=> 30(1-cos2.86) = 0.0373m.
=> 3.73 cm.

But 30 M is along the gradient, not horizontal distance.

So, the value of h won't be exactly 1.5 M.

Thank you all for explaining the solution.

Let me clearly explain this for those who are confused.

The formula for slope correction is;
C=h^2/2L.

Now since grade is given is 1 in 20.

That means for 20 m horizontal movement there will be a movement of 1m vertical..therefore accordingly for 30 m it will30/20 =1.5m.

Now, put this into the formula and you will get 3.75 as answer.

The exact formula is = (L- √(L^2-h^2)) Now in words
Where L=length and h =meaning is same as above,
L=30m And h=1.5m..

Thanks for the explanation.

Slope means= just an inclined line. always 1st value indicates vertical distance & 2nd indicates horizontal distance...So h=1.5 how?

see h=(V/H.), where V=vertical distance H=Horizontal distance.

So, first Given(1 V/20 H)=(1/20)=0.05 this value in meter (we calculated slope length for 1 meter but given(30m), h=height between two endpoints of the slope.

We have to calculate for 30 m) so just multiply =0.05*30=1.5 m Now, don't use (h^2/2L) formula it's approx.

By using the formula, we get 0.03752346684 in meter now in option all value in cm.

So just convert in cm = 3.75234668427 answer.

h=height between two endpoints of the slope, hope this concept is clear.

h^2/2L.

h= 1.5 (because of 1:20, 1/20 * 30 = 1.5),
h= 2.25/60,
h= 0.375m,
h= 0.375 * 100,
h= 3.75cm.

How are they getting 1.5*2=2.25? Please explain.