Brake caliper position....
#1330429
11/04/12 09:20 AM
11/04/12 09:20 AM
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70Cuda383
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So, saw this in Monty's thread of the Ferrari... Quote:
Some may know this trivia... in car design it's ideal to position the brake caliper mass closer to the CofG / center of gravity of the vehicle as done on that Ferrari, the calipers are mounted rearward on the front, and forward on the rear.
OK.
Why?
simply for a weight distribution standpoint? or is there some sort of force vector that occurs and when the calipers are on the "inside" it helps add stability?
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Re: Brake caliper position....
[Re: 70Cuda383]
#1330430
11/04/12 09:35 AM
11/04/12 09:35 AM
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Joined: Jun 2006
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jrlegacy23
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mopar
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The calipers on the front of the front wheels would cause more leverage to attempt to raise the rear of the car when braking hard, causing the rear to be less controllable. With them on the back of the front wheels, not so much, and the front will push down into the road more rather than attempt to tear the car off the front wheels (not that it will). On the rear wheels, the calipers in the rear would allow the rear of the car to lift away from the wheels. The calipers on the front of the rear wheels would, in a sense, cause the suspension to push down more, increasing the coefficient of friction on the rear tires and increase the braking efficiency.
On a normal car, this would not be that noticable. But on a supercar (like Ferrari, Lotus, Lambo's) this would affect there finely tuned handling.
On a street car, I personally like the calipers on the rear portion of the rotors so no debris lays on the top of the calipers and gets forced in between the brakes and rotors when the wheels turn.
[color:"#00FF00"]68 Fastback Barracuda with some stuff[/color]
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Re: Brake caliper position....
[Re: jrlegacy23]
#1330432
11/04/12 09:52 AM
11/04/12 09:52 AM
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70Cuda383
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Interesting theory, but I don't see how that happens...if the calipers were mounted on the frame, maybe. but since they're mounted on the axle, I don't see how it can create any up or down leverage on the wheels.
or rather, regardless of caliper location, the application of brakes will tip the nose of the rear diff down because it's a rotational application of force.
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Re: Brake caliper position....
[Re: 70Cuda383]
#1330434
11/04/12 09:56 AM
11/04/12 09:56 AM
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70Cuda383
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Unlike a car such as a Ferrari that most likely has that perfect 50/50 distribution of weight...what about a nose heavy truck that is more likely to have a 60/40 distribution of weight, would it then make sense to have the calipers as far back as possible to try and improve weight distribution? or is it still all about keeping as much mass between the wheels as possible?
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Re: Brake caliper position....
[Re: autoxcuda]
#1330437
11/04/12 10:01 PM
11/04/12 10:01 PM
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70Cuda383
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OK....Explain to me what exactly Polar Moment of Inertia means.
Obviously it means something different than center of gravity...
does it have anything to do with when the brakes are applied, or is it applicable all the time regardless of brake application?
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Re: Brake caliper position....
[Re: 70Cuda383]
#1330438
11/04/12 11:49 PM
11/04/12 11:49 PM
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Polar moment of rotation simply means how hard it is to get a vehicle to rotate (turn), or stop rotating. In a straight line racer it's a null point. For 99.9% of the vehicles on the street worrying about the caliper location it's turd polishing. On a Ferrari that is being designed optimally to take advantage of every handling trick it's a cheap thing to do that brings a benefit with it. If you can do it easily it's a cheap but slight improvement. If you have to jump through hoops to do it, then don't as odds are very good it will never show up as a measurable improvement.
They say there are no such thing as a stupid question. They say there is always the exception that proves the rule. Don't be the exception.
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Re: Brake caliper position....
[Re: 70Cuda383]
#1330439
11/05/12 12:01 AM
11/05/12 12:01 AM
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Joined: Sep 2005
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mopardamo
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Hello,
In this case "Polar" can be thought of as mass at opposite ends of the car. Moment of Inertia can be though of simply as the force needed to move the car in some direction other than its current path. The further away each mass is placed from the other the harder it is to turn the car. So in an extreme example lets say a car has half the mass at the front bumper and half the mass at the rear bumper with a carbon fiber rod tying them together. In this example the fake car will be a real chore to turn sharp left or right. Handling will be extremely poor. The handling will improve as the mass's are brought closer together.
It has nothing to do directly with brakes. Only the mass of the components and where they are placed.
This help any?
Damon
Last edited by mopardamo; 11/05/12 02:34 AM.
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Re: Brake caliper position....
[Re: 70Cuda383]
#1330441
11/05/12 02:17 AM
11/05/12 02:17 AM
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Joined: Jan 2003
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AndyF
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Quote:
OK....Explain to me what exactly Polar Moment of Inertia means.
Obviously it means something different than center of gravity...
does it have anything to do with when the brakes are applied, or is it applicable all the time regardless of brake application?
Take a 10 lb bar bell and a 10 lb bowling ball. See which is easier to make spin.
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Re: Brake caliper position....
[Re: 70Cuda383]
#1330442
11/05/12 08:39 AM
11/05/12 08:39 AM
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jrlegacy23
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Re-read my post, it is very on target. It has nothing to do with braking forces, but more with stopping distances. This is hard to understand without physically showing you. Let me try to explain. Take a tire, try to stop it from rolling by grabbing it in the front. When you clamp on it and do not move, the tire will want to rollover you (like a flip/vault formula). When you clamp on the back, the tire will try lift you. When you do not move, the tire will be forced downward into the pavement more. This action will inadvertently increase the COF (coefficient of friction) of the said surface and decrease stopping distances of a vehicle. Coefficient of friction is the ratio of the amount of force it take to pull a certain weight across a defined surface. The position of the caliper increases this from the tire surface to the pavement surface. In accident reconstruction (in which I specialize in) the ratio is then multiplied by the weight of gravity (32.2 feet per second squared), to determine the drag factor of a driving surface, this identifies the ability of how fast a tire can stop on any given surface. It will not be a dramatic increase, but enough that exotic car makers will use it. Remember a little here and a little there, will add up to a lot overall.
[color:"#00FF00"]68 Fastback Barracuda with some stuff[/color]
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Re: Brake caliper position....
[Re: jrlegacy23]
#1330443
11/05/12 10:11 AM
11/05/12 10:11 AM
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Joined: Oct 2003
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70Cuda383
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Quote:
Re-read my post, it is very on target. It has nothing to do with braking forces, but more with stopping distances. This is hard to understand without physically showing you. Let me try to explain. Take a tire, try to stop it from rolling by grabbing it in the front. When you clamp on it and do not move, the tire will want to rollover you (like a flip/vault formula). When you clamp on the back, the tire will try lift you. When you do not move, the tire will be forced downward into the pavement more. This action will inadvertently increase the COF (coefficient of friction) of the said surface and decrease stopping distances of a vehicle. Coefficient of friction is the ratio of the amount of force it take to pull a certain weight across a defined surface. The position of the caliper increases this from the tire surface to the pavement surface. In accident reconstruction (in which I specialize in) the ratio is then multiplied by the weight of gravity (32.2 feet per second squared), to determine the drag factor of a driving surface, this identifies the ability of how fast a tire can stop on any given surface. It will not be a dramatic increase, but enough that exotic car makers will use it. Remember a little here and a little there, will add up to a lot overall.
It makes sense, but only if you turn "rotational" forces into "linear" forces. but on a rotating axle/wheel combo, the forces are all "rotational"
but what happens on one side of the tire, the opposite is happening on the other side.
Applying the brakes is opposite of applying power. When you add power to the axle, the tires rotate, the entire axle wants to rotate in the opposite direction, you get "axle wrap" the leaf springs flex, and the yoke of the Diff goes up, this is why pinion angles are set the way they are, so that as the axle rotates up, the U-joints "line up"
when you apply the brakes, you get the opposite reaction. the yoke on the diff will rotate down. no matter where the caliper is on the axle, the force from braking will rotate the yoke/nose of the diff down. what that does to tire planting forces will depend on your suspension set up. Some cars "plant" the tires harder when the yoke rotates up. this is why cal-trac bars are so effective, they "lock up" the front segment of the leaf spring, and turn that axle rotational force into a downward push on the axle, which plants the tire harder.
I would think that you would need a "reverse cal trac" (top of leaf springs, on the rear segment instead of under leaf springs, on the front segment) set-up to make the axle plant downward when braking.
Anyway, with the other examples stated, I now understand "polar moment of inertia" I learned it as "rotational inertia" the further from the center of rotation something is, the more effort it takes to rotate it.
It's the same theory that allows a figure skater to enter a spin with their arms out, then pull everything in to "center mass" and their rotational speed increases.
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Re: Brake caliper position....
[Re: astjp2]
#1330445
11/06/12 02:13 PM
11/06/12 02:13 PM
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astjp2
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Quote:
Think of this concept... 2 identical weight balls, one is hollow, the other is solid. Which one will accelerate faster (slow down faster)?
The solid one will. Anyone know why?
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Re: Brake caliper position....
[Re: astjp2]
#1330446
11/06/12 05:00 PM
11/06/12 05:00 PM
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Joined: Dec 2003
Posts: 22,812 Bitopia
jcc
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Quote:
Quote:
Think of this concept... 2 identical weight balls, one is hollow, the other is solid. Which one will accelerate faster (slow down faster)?
The solid one will. Anyone know why?
yes, except you density of the material is not addressed and therefore a variable?
Last edited by jcc; 11/06/12 05:02 PM.
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