Isn't weight the force causing the inertia? So more weight, more inertia. More inertia, longer stopping distance.
What takes longer to stop. A 4690 lb '59 Cadillac at 60 mph or a 1500 lb Lotus Elan at 60 mph.
Just sayin'. Or, have I missed the point completely?
Weight is irrelevant. The physics for stopping distance rely only on 2 things: Velocity and, with locked tires where you see skid marks, Coefficient of KINETIC friction. gravity is constant. If you have ABS, you have STATIC friction and you'll stop faster. Like this truck:
https://www.youtube.com/watch?v=ridS396W2BY
That truck is fully loaded and 40 tons GCW. It's traveling at 40mph. If that truck didn't have ABS, it would have slid right into the car. Now. how much does that truck weigh? It has 18 tires which give it a high coefficient of grip WITH ABS which gives it static friction..
Stopping a car is work - work is reducing KINETIC energy to zero.
Work = -umgd u is coefficient of Kinetic friction. m is the mass of the car. g is gravity.
mg = weight d=distance v=velocity
-umgd = 1/2mv^2 or one half the mass times the velocity squared
To solve for d you get d= (v^2)/2ug. the Mass of the vehicle disappears. It's irrelevant. It's all about the velocity and the coefficient of Kinetic friction.
Doesn't matter if it's a 4690 lb '59 Cadillac at 60 mph or a 1500 lb Lotus Elan at 60 mph. Just sayin'. The physics say they'll stop the same distance if they're both laying down skid marks. The difference you see on Car and Driver stopping tests is all about the tires. Same car with tires that have a higher u will stop faster. The Caddy isn't going to have the lotus's tires on it so the tires won't melt while stopping which changes the dynamic. The Caddy will have Caddy tires on it so the u is the same. The only difference is if the lotus has newer tires and the Caddy has worn tires, then the u is different and the Caddy will take longer to stop.
I don't care if you have a Bronco with 29" tires or 37" tires. If your brakes will lock up your tires, you'll stop in the same distance as a Lotus that has a similar u (normal street) tire. If it's got racing slicks with a high grip, then yeah, it'll stop faster.
The picture showed the Bronco's skid marks. But I was GENEROUS in my calculations and gave it the minimum stopping distance with a u of .8 as it's tires looked newer. It was probably closer to .6 because of the road conditions so even at 30 MPH it would have taken longer to stop.
This isn't to diss you, it's to educate. It's a common mis-belief that a heavier vehicle will take longer to stop than a lighter vehicle. Sure, if a vehicle is DESIGNED to stop fast, like with 6 caliber Brembos to manage the braking and super fat and sticky tires to take advantage of the brakes and prevent them from skidding, it'll stop faster. But if you have a fully loaded 18 wheeler and a Honda accord with general street tires on them, and they both lock up their tires, they'll stop in exactly the same distance. If you have 6 caliper Brembos on the Honda with street tires, it won't make a bit of difference. They'll both still skid to a stop in the same distance. In fact a Bronco with big tires that has good brakes but can't quite lock up will stop faster than a Bronco with smaller tires and excellent brakes if both drives smash on their brakes.
always sad to see
