Civil Engineering - Surveying - Discussion

Discussion Forum : Surveying - Section 1 (Q.No. 38)
38.
Offsets are measured with an accuracy of 1 in 40. If the point on the paper from both sources of error (due to angular and measurement errors) is not to exceed 0.05 cm on a scale of 1 cm = 20 m, the maximum length of offset should be limited to
14.14
28.28 m
200 m
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
22 comments Page 2 of 3.

JAY PANCHAL said:   8 years ago
Why 1.414?

Anvesh said:   1 decade ago
How it will come?

Mohamed Elbesy said:   8 years ago
I think 200 is the correct answer.

1/40 accuracy on the map means 50 cm on the ground according to scale.
This means the offset is rounded to the nearest 50 cm and this applies only on 200 m.

Naga raju Gowd said:   9 years ago
How answer is 28.28?

I am not getting this. Please explain me.

Akhilesh yadav said:   9 years ago
The displacement due to angular error =Lsin.
and it should be equal to displacement due to Linear error = L/r.
Corresponding displacement on paper=√2(L/r)(1/S)= √2(Lsin-/S).
Corresponding if a limit of accuracy in plotting is 0.05cm then,
√2(L/r)(1/S)=.05,
L= (.05x40x20)/√2.
= 28.28.
Where S=1 in accuracy.
r=1 in scale.

Satish said:   9 years ago
Due to angular & linear sin45 = 0.7 then it is 1.414.

Abhay said:   10 years ago
The center-to-center dimension for a 45-degree bend is equal to the desired size of the offset times the cosecant 1.414.

A cosecant is used to determine the distance between the centers of the two bends used to make an offset. A 45-degree angle has a cosecant of 1.414.

Ihab said:   10 years ago
Why 1.414?

Darshika said:   10 years ago
L = 0.05*s*r/1.414 = 28.28 m.

Ganpati Patel said:   10 years ago
L = s*r*error/1.414.

= 20*40*0.05/1.414 = 28.28 m.


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