Civil Engineering - Surveying - Discussion

Option (A) is correct.

The question is asking about hypotenusal allowance.
I will explain. This is a bit lengthy. But since there is no option for uploading a figure, I have to type this much. So please read carefully.

For surveying the distance we need is the horizontal distance. But for sloping grounds, we can't directly measure the horizontal distance.

So we will measure the sloping distance and apply the correction (by knowing the slope). We are doing it

(i) either by measuring the full sloping distance once and applying the correction at the office later by using trigonometric relations,
(ii) or we will apply corrections at each chain length so that we don't have to do any corrections at the office. Keep in mind this much.

Here we are using the second method.
Suppose we are measuring a single chain distance L from A to B along a slope of 'theta' degree,
AB = L metre.

The TRUE horizontal distance for this AB is Lcos(theta), which will be less than L metre.
So what we do is we will mark a point B' whose TRUE length is L metre, which is ahead of actual measured point B. Then we will start the next chain from the point B' instead of B.
For that, we take Lm as the horizontal distance and find the sloping distance for that.
AB'cos (θ) = L
AB' = L /cos(θ),
AB' = L sec (θ).

Hypotenusal allowance (HA) = the distance BB' we have to mark ahead of B so that the measured length along the slope is the true horizontal distance.
HA = AB - AB'
= L.sec(θ) - L,
= L [sec (θ) - 1].

l is slope length and D is the horizontal distance and θ angle of slope.

Now the error is;
L-D.
L-Lcos(θ)
L(1-cos(θ)) this is error always positive.
Correction = - error.
= - [L (1-cos(θ)].
= L[cos(θ) -1].

This is my answer.

Slope correction is=L(cos0-1).

But here the question is not about slope correction it is asking about hypotension allowance i,e the correction required to be applied for the given chain length for a given slope.

So, in indirect chaining hypotension allowance is=L(sec0-1).

L(secθ-1) is correct because ;
SLOP . L(1- cos0).

1-cos0 ------ gives eq 1 eg,

Now, cos0 = 1/sce0.
= 1-1/sec0.
= sec0-1/sec0 ------ eq 2.

COMPARE eq 1, eq2.

= sec0(1-cos0)= sec0-1
= solve LHS

sec0(1-1/sec0)
We get sec0 -1.

We know that formula for determining.

L(1-cosθ) we also know that cosθ=1/secθ
Then
L(1-1/secθ)
L(secθ-1)
Here L=30.

Hence;
30(secθ-1).

So this is the Right answer.

Actual slope correction is always -ve therefore, Correction = -[L(1-cosX)].
=(LcosX-L).
=L(cosX-1) ; now L is 30m chain.
So the answer comes out to be;
=30(cosX-1) m.

We know that slope correction is L(1-cosθ) and this is only correct..If you take sec then where that denominator will go then? Please anyone explain me.

@All
Correction along with hz line = L(cosθ-1).
Correction along with slope= L(1-cosθ) or L(secθ-1).

L'=30m, l=L' cosθ.
So, (L'- l)
= L' - L' cosθ
= L'(1 - cosθ)
= 30(1-cosθ) it's answer C.

Hypotensual correction: L(secθ-1).
Slope correction: L(cosθ-1).

Thank you!