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How does Car weight impact braking?

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Old 06-15-2005, 05:56 PM
  #16  
macnewma
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Thanks for the explanation guys. I understand what they were saying about the tires.

Is there a reason why it has such a dramatic effect on the limits of cornering but what appears to be a less dramatic effect on the limits of braking distance? Both appear to be affected by the same physical properties.
Old 06-15-2005, 06:01 PM
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M758
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Cornering uses a static friction model and braking uses a rolling tire.

When cornering the tire as not (or little rolling motion) in the direction of cornering. Underbraking most of the friction is of the rolling type. The slopes of the curves are not same under those two types of fiction models.

Do remember in Colorchanages case slinding vs rolling friction is probably more accurate measure of stopping. Just imagine a spinning out of control on asphalt run off area. Things can get quite complex in gravel, grass or dirt.
Old 06-15-2005, 06:03 PM
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Larry Herman
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I was at a Tech Talk where former Penske Race Engineer Don *** used physics and charts to explain EXACTLY why as you increase the weight on a tire, the lateral grip increases disproportionately LESS. Don't even try to ask me to explain this. He showed it though, for both braking and lateral acceleration. It may not be logical or intuitive, but it is true.

He also used physics to show that if you can reduce the weight transfer (by lowering the car, or increasing the track) the cornering ability will go up due to the increased overall level of grip. Don is just thinking on a whole 'nother level.
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Old 06-15-2005, 06:24 PM
  #19  
Greg A
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Originally Posted by renvagn
See the latest issue of Pano for a great article on breaking systems. "Improved breaking is achieved with bigger rotors, tire patch and less weight. Hope this helps.
The article does NOT say that bigger rotors improve braking.

Originally Posted by Bob Gagnon and Bill Gregory, June, 2005 Pano, pg. 47
The only ways to make your Porsche stop faster are to lower the car's center of gravity, fit higher coefficient of friction, large contact patch tires, lower the car's weight and make sure the brake bias is optimized, not fitting larger brakes
Greg A
Old 06-15-2005, 06:42 PM
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ColorChange
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Adrial: I dispute the accuracy of the 13-1 graph. Give me some time and I'll put together actual tire data. My data shows a much more linear curve for cf performance as a function of load. Also, you have to understand that even adding 1,000 lbs to a car ... a huge difference, only adds 250 lbs per tire statically, and maybe 350 under heavy braking in a typical car. So, don't think 1 tires feels the full 1,000 lbs.


In the real world, a heavier car you would use larger tires. That is why a 4,000 lb Mercedes stops in about the same time as the 1,900 Elise.

Technically speaking, if you really did add 1,000 lbs to the passenger seat, the car would take longer to stop. BUT not a LOT longer.
Old 06-15-2005, 06:48 PM
  #21  
Geo
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The old adage about being able to lock up the tires meaning you have enough brakes is misleading and sad to say, one of the people who have put this into the mind of many is none other than the late great Carroll Smith.

I could rig a braking system that would use SS blocks for pads and put enough pressure behind it, you could lock it up, but modulation would be hell. I doubt the braking distances would be anything any of us would want.
Old 06-15-2005, 07:09 PM
  #22  
macnewma
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Geo, was that in reply to my post stating "enough force to reach threshold braking"? If so, I agree completely. I was only interested in differences of braking distance, not modulation, fade, pedal force, pedal feel, etc. Those other attributes are all vital and separate bad systems from good systems.

I just didn't want the argument that my Pinto, when loaded with 10,000 lbs could no longer brake as well as when at stock weight. That was irrelavent to the issue of coefficient of friction as we were then discussing whether the actual braking system was adequate for the greater weight.
Old 06-15-2005, 07:25 PM
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This is an excellent article that describes the relationship between weight and friction, and it is a linear relationship..
http://www.school-for-champions.com/...ctioncoeff.htm

A quote from the experimrnt where you pull a book resting on the table with a scale, and repeat with another book on top .

"For example, measure the weight of a book. Then use the scale to measure the force required to start the book sliding along a table. From these two measurements, you can determine the static coefficient of friction between the book and the table. You can verify that the friction equation is true by adding a second book and repeating the measurement. The force required to pull two books should be twice as much as for one book." (not 1.8 X, it's 2X so the relationship is linear)

This description sure sound like the weight is directly proportinal to friction.



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Old 06-15-2005, 07:29 PM
  #24  
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Wow. Suddenly everyone went insane. Mass is very important in stopping distance. I'll type slow so everyone can understand.

KE=mass(velocity^2). When KE is 0, the car is stopped. So the problem is getting as quickly as possible from KE>0 to KE=0.

Brakes convert KE into heat. The maximum rate at which they can do this is limited by the tires and nothing else (if the car is decent).

Colorchange argues that the tires total capacity is unchanged if all else is equal other than mass. This is the mistake. TIRE GRIP DOES NOT GO UP PROPORTIONALLY WITH NORMAL FORCE! Not even close really. (normal force is basically the weight on the tire for a non-downforce car).

If tire grip did go up proportionally with normal force, all the things we know to be true about vehicle dynamics would be false. Cars would basically be untunable, since all suspension settings untimately are juggling the normal force on each tire.

Chris Cervelli
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Old 06-15-2005, 07:30 PM
  #25  
Rich Sandor
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This description sure sound like the weight is directly proportinal to friction.
Only when the tyres are locked up, though. The coeffecient of friction is different when the wheels are turning...it stays constant with decelleration if the wheels are still turning. (whereas it drops with vehicle speed when locked up.)

I'm pretty sure that if the Cf changes, than that should change the ratio of weight to friction.
Old 06-15-2005, 08:55 PM
  #26  
Brian Morris
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The non-linear relationship between vertical force on a tire and the lateral force it generates is called "tire load sensitivity". It's first introduced on page 26 of Milliken. As Chris C. said all suspension/car engineering, and driving, is based on this concept.

--Brian Morris
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Old 06-15-2005, 10:26 PM
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Dennis K
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TIRE GRIP DOES NOT GO UP PROPORTIONALLY WITH NORMAL FORCE!
This can't be emphasized enough. It's the foundation of suspension tuning.
Old 06-15-2005, 10:36 PM
  #28  
Sanjeevan
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Great, I accept , TIRE GRIP does not go up proportionally with normal force,....I have read 0 books on suspension compared to about 4 books on track driving, to rectify this I just placed an order for Fred Puhn's and Don alexanders books on car setup....
Old 06-16-2005, 12:08 AM
  #29  
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BTW, Everyone there is an excellent article in this months Pano - Braking system basics.
Old 06-16-2005, 08:08 AM
  #30  
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Originally Posted by Premier Motorsp
Wow. Suddenly everyone went insane. Mass is very important in stopping distance. I'll type slow so everyone can understand.

KE=mass(velocity^2). When KE is 0, the car is stopped. So the problem is getting as quickly as possible from KE>0 to KE=0.

Brakes convert KE into heat. The maximum rate at which they can do this is limited by the tires and nothing else (if the car is decent).

Colorchange argues that the tires total capacity is unchanged if all else is equal other than mass. This is the mistake. TIRE GRIP DOES NOT GO UP PROPORTIONALLY WITH NORMAL FORCE! Not even close really. (normal force is basically the weight on the tire for a non-downforce car).

If tire grip did go up proportionally with normal force, all the things we know to be true about vehicle dynamics would be false. Cars would basically be untunable, since all suspension settings untimately are juggling the normal force on each tire.

Chris Cervelli
Premier Motorsports
KE=1/2 M V^2


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