Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 2 (Q.No. 18)
18.
Let angular value of one graduation of a tube of length x be φ seconds and R be the radius of its internal curved surface, then
.Discussion:
9 comments Page 1 of 1.
MJM Gcek said:
3 years ago
Yes, you are correct @Rahendra.
Mesbah Ullah said:
4 years ago
Arc = φ *R......... φ in rad.
φ rad = Arc/R.
φ = Arc/(R*rad) = x/(R*206265).
φ rad = Arc/R.
φ = Arc/(R*rad) = x/(R*206265).
(1)
Shiva said:
4 years ago
Thanks @ Er.Meghanada.
Raju Sarkar said:
4 years ago
Yes, it should be (x/R)206265.
Amjad Ali said:
5 years ago
None of these.
Sensitivity= x*206265/R.
Sensitivity= x*206265/R.
Er Meghanada (Bhadrak) said:
7 years ago
x = Angular value of one graduation of tube of length.
φ = Second.
R = Radius.
D = 206265 sec.
(x/r)*d = (x/r)*206265.
φ = x/206265.
φ = Second.
R = Radius.
D = 206265 sec.
(x/r)*d = (x/r)*206265.
φ = x/206265.
Abhishek Anand said:
8 years ago
Yeah,angle= (arc/radius) radians.
= (d/r) * 206265 sec.
= (206265d/r).
(1rad=206265sec).
= (d/r) * 206265 sec.
= (206265d/r).
(1rad=206265sec).
Rajendra said:
8 years ago
It should be( x/R ) * 206265.
(1)
Bestengineer said:
8 years ago
How? Explain.
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