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:
22 comments Page 1 of 3.
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.
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.
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.
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.
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.
(10)
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.
∠ = 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)
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.
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.
Gaurav said:
8 years ago
Nowhere is mentioned that linear measurement is equal to angular measurement then, what is the reason to get answer 28.28?
Aashish said:
8 years ago
This formula is applicable maximum errorrs/√2.
28.28 correct answer.
28.28 correct answer.
(2)
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.
40/100 = 0.4,
0.01m/20m = 0.4m/x,
X^2 = 800,
X = 28.28.
(4)
Chandu said:
7 years ago
The maximum length or limiting length=L * S/sin0.
S=scale,
0=45°
S=scale,
0=45°
(1)
Naga raju Gowd said:
9 years ago
How answer is 28.28?
I am not getting this. Please explain me.
I am not getting this. Please explain me.
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