Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 1 (Q.No. 7)
7.
If S is the length of a subchord and R is the radius of simple curve, the angle of deflection between its tangent and sub-chord, in minutes, is equal to
Discussion:
27 comments Page 3 of 3.
Nancy said:
7 years ago
@Sameer.
Thankyou for the explanation.
Thankyou for the explanation.
Sai adithya said:
7 years ago
Thanks @Sameer Sopori.
Govarthanan R said:
6 years ago
For an angle of deflection in MINUTES then use = (S/2R)*(180/π)* 60.
= S/R*((180*60)/(2*(22/7)).
=S/R *1718.18,
=1718.18S/R.
= S/R*((180*60)/(2*(22/7)).
=S/R *1718.18,
=1718.18S/R.
(6)
Cemti said:
5 years ago
@Govarthan.
Please explain how did you get the 22/7?
Please explain how did you get the 22/7?
Nadeem said:
5 years ago
Angel of deflection in radian = S/2R.
And in radian =180 * S/2R * °.
And in minute 180 * 60 * S/2R*°.
After simplification =1718.87S/R.
And in radian =180 * S/2R * °.
And in minute 180 * 60 * S/2R*°.
After simplification =1718.87S/R.
(12)
Karyeija Felex said:
4 years ago
For all practical purposes sub chord is equal to sub arc length, so E is correct.
(4)
Shweta said:
2 years ago
The angle of deflection in radian = S/2R.
The angle of deflection in degree = S/2R × 180/π
The angle of deflection in minutes = S/2R × 180/π × 60.
= 1718.18 ×S/R.
The angle of deflection in degree = S/2R × 180/π
The angle of deflection in minutes = S/2R × 180/π × 60.
= 1718.18 ×S/R.
(11)
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