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 1 of 3.

Anvesh said:   1 decade ago
How it will come?

Subhajit said:   1 decade ago
s*r/(40*1.414).

= (20*40)/56.56.

= 14.14 answer.

Rajesh said:   10 years ago
Mention the formula clearly to calculate.

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

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

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

Ihab said:   10 years ago
Why 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.

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

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.

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

I am not getting this. Please explain me.


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