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A vehicle has wheel base of 5.5 m. What is the off tracking negotiation of a curved path with a mean radius of 31.5 m?

(take width of pavement as 3.5 m)


1. 0.48 m
2. 0.96 m
3. 0.17 m
4. 0.087 m

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Best answer
Correct Answer - Option 1 : 0.48 m

Concept:

The off-tracking on the road is given by:

\({E_{off}} = \frac{{{nl^2}}}{{2R}}\)

where

n = number of lanes

L = wheel-base

R = Radius of the curve

Eoff = Off-tracking

Calculations:

Pavement width is 2.5 m, So Number of lanes(n) = 1

\({E_{off}} = \frac{{{1\ \times 5.5^2}}}{{2 \times 31.5}} = 0.48\;m\)

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