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.

Rinku Kumar meena said:   3 years ago
Actually, the maximum length of offset 1/40 cm it means 0.025mm.

But the above question's maximum length of offset 0 .05cm. It means 0.5 mm or 1/20 cm
So, the maximum length of offset=s/20*r/√2.
S = 20
r = 40.
So, 20/20*40/√2.
= 28.32m.
(10)

Kenneth andres said:   3 years ago
1/100 = 0.01m.
40/100 = 0.4,
0.01m/20m = 0.4m/x,
X^2 = 800,
X = 28.28.
(4)

Abdus samad khan said:   4 years ago
Cosec = 1/sin.
Than sin45 =.707.
Cosec = 1/.707 = 1.414.

Vijender thakur said:   5 years ago
How come 1.414? Please explain the answer in detail.

Shubham P said:   5 years ago
Formula : Limiting error = (L/sin@) * (1/r) * (1/S).

∠ = 45' for max limiting offset.
1/r = 1:40 = accuracy of measurement.
1/S= 1:20 = scale.
Limiting error = 0.05 cm.

L = 20*(2)^0.5.
L= 28.28 m.
(5)

Sachin patil said:   5 years ago
What does 1 in 40 refer? Please explain.
(2)

Chandu said:   7 years ago
The maximum length or limiting length=L * S/sin0.
S=scale,
0=45°
(1)

Er Meghanada (Bhadrak) said:   8 years ago
L = 0.05"40"20*14.14 = 28.28 m.

Aashish said:   8 years ago
This formula is applicable maximum errorrs/√2.

28.28 correct answer.
(2)

JAY PANCHAL said:   8 years ago
Why 1.414?


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