How does Car weight impact braking?
06-18-2005, 03:10 PM
#91
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I'm seriously thinking of loading up my 968 with as much **** as I can, and just trying it on a strech of dead highway to see what kind of numbers I'd get.
06-18-2005, 09:07 PM
#92
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Formula 1 cars have to be able do decelerate at incredibly high rates. Go find a F1 car that weighs 3500 pounds, then I'll believe that less weight doesn't improve braking.
Its basic physics that even I understand, and I haven't even taken physics in high school yet. An object in motion stays in motion unless another force acts upon it. A stationary object will stay stationary unless another force acts upon it. The less mass an object has, the less force is required to make a moving object stop moving or a stationary object start moving.
Cars accelerate and decelerate better when they aren't sliding. Lets say a 2500 pound car and a 3000 pound car are both going 80 mph. They are both going the same speed, but the 3000 pound car has more momentum, because it has more mass. Since it has more momentum, it will take more force to make it stop. If the amount of force you use to stop the car exceeds the traction thresholds of the tires, you will slide and you won't stop quickly. Adding weight will increase the traction thresholds some, but not enough extra traction to support the extra force needed to keep the vehicle from sliding under heavy decelaration.
Its basic physics that even I understand, and I haven't even taken physics in high school yet. An object in motion stays in motion unless another force acts upon it. A stationary object will stay stationary unless another force acts upon it. The less mass an object has, the less force is required to make a moving object stop moving or a stationary object start moving.
Cars accelerate and decelerate better when they aren't sliding. Lets say a 2500 pound car and a 3000 pound car are both going 80 mph. They are both going the same speed, but the 3000 pound car has more momentum, because it has more mass. Since it has more momentum, it will take more force to make it stop. If the amount of force you use to stop the car exceeds the traction thresholds of the tires, you will slide and you won't stop quickly. Adding weight will increase the traction thresholds some, but not enough extra traction to support the extra force needed to keep the vehicle from sliding under heavy decelaration.
06-19-2005, 01:20 AM
#93
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PT, it's obvious you haven't taken high school physics yet. Maybe you should try to learn from those of us who have taken high school physics, college physics, and practice engineering for a profession.
You are wrong, cars stop the fastest when they are sliding between 10-20% (highest braking force). They accelerate the fastest up to about 50% slip.
Learn more, read more, and comment less until you have enough knowledge to support your views.
You are wrong, cars stop the fastest when they are sliding between 10-20% (highest braking force). They accelerate the fastest up to about 50% slip.
Learn more, read more, and comment less until you have enough knowledge to support your views.
06-19-2005, 02:14 AM
#94
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Here are two graphs that are germaine to the discussion here. Both are from Carroll Smith's book "Tune to Win" page 17. This is a great book BTW.
Below is the graph of decreasing coefficient of friction with load increase. Note this is a racing tire hence the Cf is >1
Below is curve of grip v.s slip, showing that the tire likes 10-20% slip to achieve maximum braking.
Below is the graph of decreasing coefficient of friction with load increase. Note this is a racing tire hence the Cf is >1
Below is curve of grip v.s slip, showing that the tire likes 10-20% slip to achieve maximum braking.
06-19-2005, 09:37 AM
#96
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CupCar: Good data. As you can see from the first graph, cf does drop with increasing load, but not sharply, therefore, all else being equal, adding a few hundred pounds to a car does not make it take lonnger to stop (yes, 1 stop only from reasonable speeds, etc.).
The second graph is in the ballpark of what I am talking about but most of the recent data I have seen shows optimal slip ratio in braking at 10-20%, and optimal slip angle in cornering at 5-7%. We are in the same ballpark.
The second graph is in the ballpark of what I am talking about but most of the recent data I have seen shows optimal slip ratio in braking at 10-20%, and optimal slip angle in cornering at 5-7%. We are in the same ballpark.
06-19-2005, 11:38 AM
#97
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Originally Posted by Cupcar
A nice read is Paul Haney's book The Racing and High Performance Tire.
The Coefficient of Friction is the ratio of the normal force and the lateral force on a contact patch. If a tire had a normal (vertical force) of 100 pounds and supported a lateral force of 100 pounds before sliding then the coefficient of friction would be one.
The equation for Friction Force= Coefficient of Friction X Normal Force.
Rubber is wierd stuff it is not a simple friction material that responds to load as a linear model. Rubber drops it's "coefficient of friction" as the surface loading- pounds per square inch on the contact patch increases. Rubber looses relative grip as it is loaded. This is why sway bar tuning works for example as load is shifted across an axle pair of tires.
Another way to look at it is for each increment of vertical force the tire can support a relatively lesser increment of horizontal force can be supported before the tire slides.
This is also one reason why wings are such an effective device to improve cornering, they provide vertical load with essentially no increase in horizontal (cornering or braking) load. This is why weight is so important in a wing equipped car like an F1 car.
When you push an eraser harder on a table the Total Friction Force is increased through increased Normal Force, however if you looked carefully at the situation and measured the relative increase in lateral force as you pushed the eraser versus the increased normal force you applied you would find that the actual coefficient of friction dropped. The total goes up, the ratio goes down.
This would be the same with the heavier car, all other things being equal in the chassis, the heavier car has to take longer to stop. This would not be the case if rubber were a simple friction material and that is not to say that a heavier car could not be made to stop as fast as the lighter car given larger tires.
The Coefficient of Friction is the ratio of the normal force and the lateral force on a contact patch. If a tire had a normal (vertical force) of 100 pounds and supported a lateral force of 100 pounds before sliding then the coefficient of friction would be one.
The equation for Friction Force= Coefficient of Friction X Normal Force.
Rubber is wierd stuff it is not a simple friction material that responds to load as a linear model. Rubber drops it's "coefficient of friction" as the surface loading- pounds per square inch on the contact patch increases. Rubber looses relative grip as it is loaded. This is why sway bar tuning works for example as load is shifted across an axle pair of tires.
Another way to look at it is for each increment of vertical force the tire can support a relatively lesser increment of horizontal force can be supported before the tire slides.
This is also one reason why wings are such an effective device to improve cornering, they provide vertical load with essentially no increase in horizontal (cornering or braking) load. This is why weight is so important in a wing equipped car like an F1 car.
When you push an eraser harder on a table the Total Friction Force is increased through increased Normal Force, however if you looked carefully at the situation and measured the relative increase in lateral force as you pushed the eraser versus the increased normal force you applied you would find that the actual coefficient of friction dropped. The total goes up, the ratio goes down.
This would be the same with the heavier car, all other things being equal in the chassis, the heavier car has to take longer to stop. This would not be the case if rubber were a simple friction material and that is not to say that a heavier car could not be made to stop as fast as the lighter car given larger tires.
Diminishing CF when download increases is the key factor explaining why lighter cars everything else equal accelerates brakes and corners faster. That's also one of the reasons why you want a low center of gravity on a race car.
06-19-2005, 11:59 AM
#98
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Originally Posted by ColorChange
CupCar: Good data. As you can see from the first graph, cf does drop with increasing load, but not sharply, therefore, all else being equal, adding a few hundred pounds to a car does not make it take lonnger to stop (yes, 1 stop only from reasonable speeds, etc.).
+ When you think about a 300 to 400lbs additional weight and think it would not do much on the scale if you divide the weight on the 4 tyres you also have to keep in mind that the weight transfert when braking (or cornering BTW) means this additional weight has even more impact.
Finally on CupCar' graph the CF drops relatively slowly as download increases. On the graphs I have seen in other books the CF decreases faster.
If somebody has data on modern tyres like some of the R compounds we often use that would be very interesting.
06-19-2005, 01:32 PM
#99
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Vince:Not only do I have modern tire data, I showed it! Please read my post on page 3 of this thread where I show actual data and an 8% drop in cf while adding 440 lbs to the tire (usually over 1,000 to the vehicle). What are you not undertanding? YES the cf drops slightly, NO, it does not significantly affect stopping distance, within limits of course.
06-19-2005, 01:47 PM
#100
Rennlist Member
While it is true that the data are 25 years old, I don't think the principle has changed. Race tires are still at around a 1.5 coefficient of friction without aerodynamic load. Construction has changed, the temperature curves of the rubber are broader and the tires handle better as a result but the same principles apply I think.
Also at 25 years old I still think Carroll Smith's books are among the best out there.
In the example of the 3000 pound Porsche assuming a 60/40 weight distribution going from 3000 to 4000 pounds the corner weights would go from 900 to 1200 in the rear and from 600 to 800 in the front.
Eyeballing the change in Cf on the graph with that corner weight change looks like a significant drop in Cf at each corner to me.
Also at 25 years old I still think Carroll Smith's books are among the best out there.
In the example of the 3000 pound Porsche assuming a 60/40 weight distribution going from 3000 to 4000 pounds the corner weights would go from 900 to 1200 in the rear and from 600 to 800 in the front.
Eyeballing the change in Cf on the graph with that corner weight change looks like a significant drop in Cf at each corner to me.
06-19-2005, 02:08 PM
#101
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I think an 8% change in Cf would equate to over a car length at the braking zone of many corners at many tracks.
I think a car length is pretty significant there!!
I think a car length is pretty significant there!!
06-19-2005, 02:19 PM
#102
Race Director
Originally Posted by Cupcar
While it is true that the data are 25 years old, I don't think the principle has changed. Race tires are still at around a 1.5 coefficient of friction without aerodynamic load. Construction has changed, the temperature curves of the rubber are broader and the tires handle better as a result but the same principles apply I think.
However, IIRC Cf should not be taken as the performance of the tire. Race tires, IIRC, don't preform strictly on Cf. The stickiness of the tires changes the performance vs simply Cf. This is another reason all these physics models being thrown around are probably not accurate enough to describe what is doing on with modern tires.
Keep in mine that today's performance street tires are probably significantly better than the best race tires 25 years ago.
Originally Posted by Cupcar
Also at 25 years old I still think Carroll Smith's books are among the best out there.
06-19-2005, 06:27 PM
#103
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The whole point of this thread is to show that weight is not a critical factor in stopping a car from 60 mph. In the Touareg/Elise example I showed, a 250% weight increase lead to less than a 10% increase in stopping distance. I don't think any rational person can say that weight, in this case, on one realtively low speed stop, that weight is important.
06-19-2005, 09:46 PM
#104
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You have to compare weight as an independent variable.
It is possible to have a heavier vehicle stop faster than a lighter vehicle given the right brake design and tires. Comparing two completely different vehicles does not illustrate the point.
If one only changes the weight in a given chassis and keeps all other variables identical then the heavier car will take longer to stop.
It is possible to have a heavier vehicle stop faster than a lighter vehicle given the right brake design and tires. Comparing two completely different vehicles does not illustrate the point.
If one only changes the weight in a given chassis and keeps all other variables identical then the heavier car will take longer to stop.
06-19-2005, 10:00 PM
#105
Originally Posted by Cupcar
You have to compare weight as an independent variable.
It is possible to have a heavier vehicle stop faster than a lighter vehicle given the right brake design and tires. Comparing two completely different vehicles does not illustrate the point.
If one only changes the weight in a given chassis and keeps all other variables identical then the heavier car will take longer to stop.
It is possible to have a heavier vehicle stop faster than a lighter vehicle given the right brake design and tires. Comparing two completely different vehicles does not illustrate the point.
If one only changes the weight in a given chassis and keeps all other variables identical then the heavier car will take longer to stop.