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does it mean that higher the all red time, lower will be the reliability because at 2.5 sec all red time ,speed range was 25-65 which shows larger variation as compared to 2 sec all red time
Sir, Thank you for another informative lecture. Just wanted to clear a doubt with you. For the case when the driver decides to stop, don't we need to take the coefficient of longitudinal friction into account?
very good observation. The basic equation is same in both cases. v^2 = 2as, where a is the rate of deceleration and s is the distance required to stop. when we calculate braking distance in geometric design of highways, a is taken = f*g, and the equation becomes like this---- braking distance (s) = v^2/(2gf). value of f is in the range of 0.4 to 0.5 and therefore rate of deceleration becomes 4 to 5 m/s/s, which is very high for cars and heavy trucks. In this equation we assume that the driver will apply 100 percent brakes and the entire frictional force will be mobilized to stop the car, whereas in the calculation of stopping distance during dilemma zone, we assume a comfortable rate of deceleration (a). That is why we take actual value of 'a' rather than f*g. But here also, car will stop utilizing friction force only. I hope you got the answer to your question.
Excellent explanation.
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wow
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does it mean that higher the all red time, lower will be the reliability because at 2.5 sec all red time ,speed range was 25-65 which shows larger variation as compared to 2 sec all red time
to some extent yes. But remember all red time means loss of effective time for traffic movement
Sir,
How to design a roundabout with signal?
No difference
Clearing time will increase due to presence of circle
Sir,
Thank you for another informative lecture.
Just wanted to clear a doubt with you. For the case when the driver decides to stop, don't we need to take the coefficient of longitudinal friction into account?
very good observation. The basic equation is same in both cases. v^2 = 2as, where a is the rate of deceleration and s is the distance required to stop. when we calculate braking distance in geometric design of highways, a is taken = f*g, and the equation becomes like this---- braking distance (s) = v^2/(2gf). value of f is in the range of 0.4 to 0.5 and therefore rate of deceleration becomes 4 to 5 m/s/s, which is very high for cars and heavy trucks. In this equation we assume that the driver will apply 100 percent brakes and the entire frictional force will be mobilized to stop the car, whereas in the calculation of stopping distance during dilemma zone, we assume a comfortable rate of deceleration (a). That is why we take actual value of 'a' rather than f*g. But here also, car will stop utilizing friction force only.
I hope you got the answer to your question.
you are only talking about type 1 dillema zone sir
Yes. Other two are very special and relevant for research only.