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$$d = \dfrac{V_1^2-V_2^2}{2g\left(\dfrac{a}{g}\pm G\right)}$$

I am wondering if the car travel on the road of a sag curve or crest curve, there is two gradients of the curve. (G1, G2). Which gradient value should be used in the formula ? The one with larger gradient?

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Recall that tires have a no slip condition when they interface with the road and the car is rolling. One gradient will cause the car to rock about an axis aligned with the direction of travel (passenger - driver) and the other gradient will cause the car to dive upon stopping (front - back). The gradient that is transverse to the direction of travel (i.e. the gradient that causes rocking not diving) will not be a factor unless the dynamic forces are large enough to exceed the breaking friction of the tire/road interface. Depending on how sophisticated your model is, this is also a function of speed and ambient temperature as the breaking friction coefficient is going to depend on the tire temperature (and road temperature). Assuming normal road conditions and normal speeds (not race track calculations) the only gradient that would contribute to stopping distance in a meaningful way is the gradient that that is aligned with the direction of travel.

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There are 4 grades on each Vertical curve. G1 up and down G2 up and down

The most critical grade is the bigger of the two while going downwards (-ve).

If you think about it, trying to stop a car going down hill is harder than a car going uphill at the same speed.

ex: if a curve goes up 2.71% then goes down 3.50% then the value of the G term in the denominator is + (- 3.50/100)

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