Does anyone know approximately how much distance is needed
12-22-2001, 10:07 AM
#12
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Need one more detail ...
... exactly how steep (in degrees) was that downhill grade?
Assuming perhaps a 0.5 sec lag before you're hard on the brakes, and a 0.85g max deceleration with snow tires, I can calculate the minimum theoretical stopping distance. The effect of braking downhill is important and can be calulated if you know the slope.
Assuming perhaps a 0.5 sec lag before you're hard on the brakes, and a 0.85g max deceleration with snow tires, I can calculate the minimum theoretical stopping distance. The effect of braking downhill is important and can be calulated if you know the slope.
12-22-2001, 10:22 AM
#13
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Ok assuming quite a few things but doing a rough calc
Let's go over the assumptions.
1) we are using a static friction calculation (i.e. your wheels never locked up and slid).
2) According to C&D (Oct 1st issue) 70-0 braking distance in feet is 164 ft
3) Conditions of vehicle/road/tires/etc. is similar enough to C&D test that we can use their test to do a baseline to get the Coeficient of friction for your tires to the road.
v = vehicle intial velocity (m/s)
u = coeficient of friction
m = mass of vehicle (kg)
g = 9.8 m/s/s
d = distance traveled (m)
So the work (friction) to stop the vehicle would be -umgd. Using the work energy principle we can then say
-umgd = -(mv^2)/2
Therefor d = (v^2)/2ug
This in itself is not very usefull, since we do no know u. But from the C&D test we can solve backwards from u, giving us u ~ 1.
Note -- This implies stopping distance independent of vehicle mass. I said quick, rough estimate off top of my head :-).
(Putting u as 1, v = 70mph (31.2928 m/s) you get a stopping distance of 163.915 ft (49.9612 m))
So for a v = 132 mph (59.01 m/s) you get a stopping distance of 582.87 ft (177.66 m)
This again is very rough, probably very wrong as well, please correct me if I am wrong ( I fully expect it :-) ).
1) we are using a static friction calculation (i.e. your wheels never locked up and slid).
2) According to C&D (Oct 1st issue) 70-0 braking distance in feet is 164 ft
3) Conditions of vehicle/road/tires/etc. is similar enough to C&D test that we can use their test to do a baseline to get the Coeficient of friction for your tires to the road.
v = vehicle intial velocity (m/s)
u = coeficient of friction
m = mass of vehicle (kg)
g = 9.8 m/s/s
d = distance traveled (m)
So the work (friction) to stop the vehicle would be -umgd. Using the work energy principle we can then say
-umgd = -(mv^2)/2
Therefor d = (v^2)/2ug
This in itself is not very usefull, since we do no know u. But from the C&D test we can solve backwards from u, giving us u ~ 1.
Note -- This implies stopping distance independent of vehicle mass. I said quick, rough estimate off top of my head :-).
(Putting u as 1, v = 70mph (31.2928 m/s) you get a stopping distance of 163.915 ft (49.9612 m))
So for a v = 132 mph (59.01 m/s) you get a stopping distance of 582.87 ft (177.66 m)
This again is very rough, probably very wrong as well, please correct me if I am wrong ( I fully expect it :-) ).
12-22-2001, 11:44 AM
#16
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Ok, thats in ideal road conditions I am sure. So the road this happened on had a pretty steep
grade, I would assume that could add at least 100feet to stopping distance. Do you use AIM or anything? We need to talk :-)
12-22-2001, 12:45 PM
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Not a rough calc, ...
... and that's also what I was going to do.
Since he's braking downhill, the force of gravity is not perpendicular to the road, but a little trigonometry can make that correction.
A coefficient of friction of 1.0 is probably generous for snow tires, 0.85 - 0.90 should be closer. Using 0.90, the 583 ft increases to 648 ft.
Also, depending on the circumstance, the stopping distance may need to consider the time it takes to get on the brakes. At 132 mph, he's going 194 ft/sec. If there's a 1/2 sec lag, 97 ft should be added to the "stopping distance".
Now all we need is the angle of that hill ...
Since he's braking downhill, the force of gravity is not perpendicular to the road, but a little trigonometry can make that correction.
A coefficient of friction of 1.0 is probably generous for snow tires, 0.85 - 0.90 should be closer. Using 0.90, the 583 ft increases to 648 ft.
Also, depending on the circumstance, the stopping distance may need to consider the time it takes to get on the brakes. At 132 mph, he's going 194 ft/sec. If there's a 1/2 sec lag, 97 ft should be added to the "stopping distance".
Now all we need is the angle of that hill ...
12-22-2001, 02:18 PM
#20
Substantial...probably not...
...the very heavy S4 front rotor is designed more to be able to withstand the massive heat build up (more mass takes longer to heat) from a single high speed stop than it is designed to withstand repeated stops at lesser MPH like lighter weight rotors on the aftermarket kits are designed to (being lighter, they cool off faster between stops).