Tim thanks for posting this!
Take a look ay Part 1 weight transfer.
His calculations:
"Lf = 1600 + 3200 / 5 = 2240 lbs, Lr = 1600 - 3200 / 5 = 960 lbs
Thus, by braking at one g in our example car, we add 640 pounds of load to the front tyres and take 640 pounds off the rears! This is very pronounced weight transfer."
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My calculations:
Lf = 1600 + 3200 / 5 = 960 lbs, Lr = 1600 - 3200 / 5 = -320 lbs
so 960 + -320 = 640 lbs?????
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So on my car:
Lf = 1080 + 2400 / 5 = 696 lbs, Lr = 1320 - 2400 / 5 = -216 lbs
so 696 + -216 = 480 lbs of weight transfer in my coupe if it had the same wheelbase and CG as the example which it does not.
That means my weights would then be 1560 front and 840 rear = 2400 total
Lf now is 1560 = .65% front
Lr now is 840 = .35% rear
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So on a Mustang that these brakes were designed for:
a 1998 SVT Cobra has 1950 Front and 1425 rear = 3375 total
Lf = 1950 + 3375/5 = 1065, Lr = 1425 - 3375 / 5 = -390
so 1065 + -390 = 675 lbs of transfer
Lf now is 2625 lbs = .78% front
Lr now is 750 lbs = .22% rear
Now there are some things to consider. Our cars are more stiff and have a much lower CG than a stock Mustang. You must also take into account the larger rear tires we use which will have more friction thus more stopping power.
But ignoring all that.
We are using a braking system engineered to be balanced and stop a car that has 4 times the weight on the front tires than the rear. We now take that same braking system and install it on a car that has less than 2 times the weight on the front than the rear. So after all this do we consider the braking on our cars to be as balanced as a mustang????
So now I have taken a braking system that appears to be at least 50% out of balance (using the laws of physics) and reduced the power of the front brakes by 57% via my proportioning valve. I have tested this system and verified that the system is in fact fairly well balanced.
Funny how all this math and physics can verify reality.
