Question

Two roads meet at an angle of 127 degree 30 min. Calculate the necessary data for setting out a curve of 15 chains radius to connect the two straight portions of the road (a) if it is intended to set out the curve by chain and offsets only, (b) if a theodolite is available. Explain carefully how you would, in both cases, set out the curve in the field.

Answer 1

Answer:

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Answer 2

Answer:

Angle of intersection = 127°30’

Deflection angle (Δ) = 180° - 127°30’ = 52°30’

Length of chain = 20 meter.

Radius of curve = 15 chains x 20 = 300 m.

Tangent Length T = R tan(Δ/2) = 300 × tan(52°30'/2) = 147.94m

Radial offset method:

$O_{x}=\sqrt{R^{2}-x^{2} }-R$

Assuming interval as 20m.

The first half of the curve is set from point of the curve ($T_{1}$)

$O_{20}=\sqrt{300^{2}-20^{2} }-300 = 0.67m$

$O_{40}=\sqrt{300^{2}-40^{2} }-300 = 2.66m$

$O_{60}=\sqrt{300^{2}-60^{2} }-300 = 5.94m$

$O_{80}=\sqrt{300^{2}-80^{2} }-300 = 10.48m$

$O_{100}=\sqrt{300^{2}-100^{2} }-300 = 16.23m$

$O_{120}=\sqrt{300^{2}-120^{2} }-300 = 23.11m$

$O_{140}=\sqrt{300^{2}-140^{2} }-300 = 31.06m$

$O_{147.94}=\sqrt{300^{2}-147.94^{2} }-300 = 34.49m$

The second half of the curve may be set from point of tangency ($T_{2}$).

Perpendicular offset method:

$O_{x}=R-\sqrt{R^{2}-x^{2} }$

The first half of the curve is set from point of the curve ($T_{1}$)

$O_{20}=300-\sqrt{300^{2}-20^{2} }=0.67m$

$O_{40}=300-\sqrt{300^{2}-40^{2} }=2.68m$

$O_{60}=300-\sqrt{300^{2}-60^{2} }=6.06m$

$O_{80}=300-\sqrt{300^{2}-80^{2} }=10.86m$

$O_{100}=300-\sqrt{300^{2}-100^{2} }=17.16m$

$O_{120}=300-\sqrt{300^{2}-120^{2} }=25.05m$

$O_{140}=300-\sqrt{300^{2}-140^{2} }=34.67m$

$O_{147.94}=300-\sqrt{300^{2}-147.94^{2} }=39.01m$

The distance ‘x’ from $T_{1}$ for locating the apex point.

x = R sin(Δ/2) = 300 × sin(52°30'/2) = 132.69m

$O_{132.96}=300-\sqrt{300^{2}-132.96^{2} }=30.94m$

The second half of the curve may be set from point of tangency ($T_{2}$).

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