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 1 of 3.
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)
Sameer sopori said:
7 years ago
@Williams.
For an angle of deflection in RADIAN then use =S/2R.
For an angle of deflection in DEGREE then use =(S/2R)* (180/π),
For an angle of deflection in MINUTES then use = (S/2R)*(180/π)* 60.
For an angle of deflection in RADIAN then use =S/2R.
For an angle of deflection in DEGREE then use =(S/2R)* (180/π),
For an angle of deflection in MINUTES then use = (S/2R)*(180/π)* 60.
(1)
Ved prakash said:
8 years ago
Actually the angular method of curve setting ,by Rankine's deflection method the deflection angle b/w chord and point of tangency is =1720C/R i, where c is the first sub chord and R is the radius.
Shrikant said:
8 years ago
Angle of deflection= (360/4π)* S/R.
This gives answer 28.64 S/R.
which is in degrees,, soo one degree is 60 min.
Thus, 28.64 S/R degrees in radians is answer E) 1718.9 S/R.
This gives answer 28.64 S/R.
which is in degrees,, soo one degree is 60 min.
Thus, 28.64 S/R degrees in radians is answer E) 1718.9 S/R.
(1)
Josh said:
10 years ago
I don't think this answer (E) is correct.
It is only correct if the sub chord, S, was actually the sub arc length. Otherwise, the real answer should be sin^-1 (S/2R)*60.
It is only correct if the sub chord, S, was actually the sub arc length. Otherwise, the real answer should be sin^-1 (S/2R)*60.
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)
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)
Yamanoor said:
9 years ago
Angle of deflection=S/2R radians.
= 180 * S/2 * R degree,
= 180 * 60 * S/2 R* minutes,
= 1718.9S/R.
= 180 * S/2 * R degree,
= 180 * 60 * S/2 R* minutes,
= 1718.9S/R.
Karyeija Felex said:
4 years ago
For all practical purposes sub chord is equal to sub arc length, so E is correct.
(4)
Gman said:
9 years ago
The angle of deflection = ((180/π) * s * 60)/(2 * R)°.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers