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
Discussion:
23 comments Page 3 of 3.
Kenneth andres said:
4 years ago
1/100 = 0.01m.
40/100 = 0.4,
0.01m/20m = 0.4m/x,
X^2 = 800,
X = 28.28.
40/100 = 0.4,
0.01m/20m = 0.4m/x,
X^2 = 800,
X = 28.28.
(6)
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.
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.
(11)
Chandhu said:
1 month ago
Given:
Accuracy of offset measurement = 1 in 40
Permissible error on paper = 0.05 cm.
Scale = 1 cm = 20 m.
Step 1: Convert paper error to ground error.
So, the maximum permissible error on the ground = 1 m.
Step 2: Errors involved.
There are two independent errors:
1.Error due to measurement
Linear measurement error≈ L/40.
2.Error due to angular error:
Angular error ≈ L/40 rad for offset measurement.
Step 3: Resultant error.
√(L/40)^2+(L/40)^2 = L√2/40.
This should not exceed 1 m:
Step 4: Solve for L.
L√2/40<=1 m.
Final Answer: Maximum length of offset ≈ 28.8 m.
Accuracy of offset measurement = 1 in 40
Permissible error on paper = 0.05 cm.
Scale = 1 cm = 20 m.
Step 1: Convert paper error to ground error.
So, the maximum permissible error on the ground = 1 m.
Step 2: Errors involved.
There are two independent errors:
1.Error due to measurement
Linear measurement error≈ L/40.
2.Error due to angular error:
Angular error ≈ L/40 rad for offset measurement.
Step 3: Resultant error.
√(L/40)^2+(L/40)^2 = L√2/40.
This should not exceed 1 m:
Step 4: Solve for L.
L√2/40<=1 m.
Final Answer: Maximum length of offset ≈ 28.8 m.
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