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
573 S/R
573 R/S
171.9 S/R
1718.9 R/S
1718.9 S/R.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
27 comments Page 1 of 3.

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.
(12)

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.
(11)

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.
(6)

Karyeija Felex said:   4 years ago
For all practical purposes sub chord is equal to sub arc length, so E is correct.
(4)

Prem said:   9 years ago
Please describe the question briefly.
(1)

ABHISHEK KUMAR said:   9 years ago
Please describe it clearly.
(1)

Mohammed Salahuddin Tirandaz said:   9 years ago
Not understanding, Please describe it clearly.
(1)

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.
(1)

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.
(1)

Gman said:   9 years ago
The angle of deflection = ((180/π) * s * 60)/(2 * R)°.


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