Civil Engineering - Surveying - Discussion

Thanks for the explanation.

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..

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.

Thank you all for explaining the solution.

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

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

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.

By using of similarity of the triangle.

H = 30÷401 and putting into slope correction formula.

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.

150/ n^2 because 30 metre chain used and now slope current is 150/ 20^2 = 150 ÷ 400 = 0.375 m or 3. 75 cm.

Helpful explanation. Thank you all.