“Earlier we noted that the kinetic energy of a vehicle was, in part, dependent on the weight of the vehicle; the heavier the vehicle, the more energy it will have at a given speed. It might seem logical, therefore, that a heavier vehicle may require more distance to skid to a stop than a similar, but lighter vehicle. Contrary to this line of thought, the increased friction generated by a heavier vehicle in a skid, directly compensates for the fact that the heavier vehicle initially had more energy. A heavier vehicle may indeed be more difficult to “lock-up” than a lighter vehicle, but once in a skid, the heavy and light vehicles will require the same distance to stop from the same initial speed. For this reason, vehicle weight is not included in the skid-to-stop velocity formula.
V= / 255 m S
V is velocity (km/h)
m is the friction coefficient
S is skidmark length (metres)”
m is low maybe .1 but assumed constant for comparisons. V then controls S. The engineer says, more mass equals more friction but it all nets to no effect on stopping distance. Loss of m means all attempts to change velocity and/or direction are impaired. Low m means all command inputs to the vehicle must be softened. Assuming m of .7 for dry pavement and .1 for icy pavement, control inputs are 1/7th as much as on ice. Which reminds me of the statement, Don’t ask the truck to do what it cannot do. The truck can still do 65 on a straight. It can only slow or turn 1/7th as well. Its stability is probably only 1/7th of normal. Kind of nuts we even drive at 65.