I can't believe this hasn't been discussed (unless I'm using the wrong search terms, or I'm too dumb to see the simple answer). Anyways, for auto-x, I've basically heard/gone under the impression that you'd want the smallest rolling diameter possible. Best acceleration of the start, which matters in auto-x. So, I went with 205/50/15 Azenis (like the rest of the world ).

I'm contemplating a brake upgrade because of my latest venture in HPDEs, which is a neat upgrade dipping into the Audi parts bin. It snags the A8 12.3" rotors, and the caliper carriers off the TT, and keeping the stock calipers...voila, it fits on my Passat. It should give me a substantial amount of heat capacity in my underbraked car (3200 lbs w/ 11.3" front rotors, 9.4" rears), and a little more brake torque up front. Hrm, I'm getting off topic...

So, with the 12.3" rotors, there's no way that will fit under my 15" rims, so I'm forced to bump up to 16, which is fine by me (I'll hopefully be able to snag BBS VZ's @ $135 a piece in 16x7.5"). Now, here comes the question (took me long enough). The two sizes that fit under my fenders would be the 215/40/16 or the 225/50/16 (using Hankook Z211 sizes). Substantially different rolling circumferences. Using numbers from a tire size calculator, the sizes are 71.5" circumference, versus 78.1", respectively (22.8" diameter versus a 24.9"). 886 revolutions/mile versus 811. Although better acceleration is yielded with the 215/40 tire, its also has 8.4% more rotational speed at a given vehicle speed. So if I'm doing 100 mph on the 225/50/16 tire, it'd be the equivalent of doing 108 mph on the 215/40/16 tire. The problem I see here is that the brakes have to haul the car down an additional 8.4% in speed, right? Also, wouldn't you have less brake torque due to the smaller wheel diameter?

I'm in a world of confusion. I think I need a couple beers or something...

 

All depends on how big your caliper is? BBS built me some 15 X 7" wheels, EP maximum size to go over my light weight Wilwood callipers, 12.2" rotors, just over 14.2" of inside wheel clearance at the hub mountig surface. No one else would build them except BBS, 0" outer 7" inner 108 mm back spacing to carry 23 X 9 X 15" Goodyear slicks. If you want a copy of the drawings let me know.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by stealthx32 &raquo;</TD></TR><TR><TD CLASS="quote">Although better acceleration is yielded with the 215/40 tire, its also has 8.4% more rotational speed at a given vehicle speed. So if I'm doing 100 mph on the 225/50/16 tire, it'd be the equivalent of doing 108 mph on the 215/40/16 tire. The problem I see here is that the brakes have to haul the car down an additional 8.4% in speed, right? Also, wouldn't you have less brake torque due to the smaller wheel diameter?

I'm in a world of confusion. I think I need a couple beers or something...</TD></TR></TABLE>

100mph with a big tire is 100mph with a small tire. It's just that the smaller tire has to make more revolutions at the same given speed. Only Mr. Speedometer would get confused.

As far as brake torque... Moving the caliper outward generates more brake torque at the hub due to larger "leverage". But, as far as i understand it, the actual brake force actually acts something like this: From the brake caliper inward to the hub. Then outward towards the ground and finally to the contact patch along the ground opposite to the direction of travel. The wheel/tire actually acts as a force reducer much in the same way it does under acceleration. So a smaller wheel would in theory put down MORE brake torque to the ground in the same way it puts down more engine acceleration torque. Brake and Engine torque are essentially the same force, just directed in opposite directions.

 

What's funny is that most people over-think autocross. It's about being able to pilot a well balanced car with precision through whatever is thrown at you. It's about overall lightness of the car, the peak and consitent stick you can generate though the tires, and to a lesser degree how fast you can accelerate or brake the car longitudinally.

The most important thing in autocross is the transitional characteristics of the car; so if you have car that you can really slam through a course and do it smoothly and quickly; it's a fast car. Brakes mean almost nothing (assuming they aren't broken). And rolling diameter of the tires is a tool to helping you get the gearing you need....smaller is often better but sometimes not. Depends on where you top out 2nd gear sometimes. In the end, you want to have a package that meets with the best collective needs of a particular situation.

In order of importance, make the car:
not push in slow corners (this is deadly if it happens)
be super fast in transitions
be as well balanced in mid-rpm 2nd gear corners as possible
stop and accelerate well

And then most importantly learn what the car can do and make it do it. There is literally "seconds" out there for novices once they learn how to anticipate what the car will do. Instead of driving past a cone and then turning the steering wheel, you can learn how to enter with more speed, turn the steering wheel earlier, get the car turning in early (but not necessarily changing direction), and then having half your turn done when you pass the cone.

I can't think of anything that describes the difference between an advanced autocrosser vs a novice autocrosser than that actually. And once you've learned how to drive the crap out of the car, then you will understand that things like rolling diameter and incremental brake upgrades are only marinally important in the overall equation.... we are speaking of autocrossing don't forget. Road racing is completely different.

So, is wheel diameter important? Yes. But it's all within the context of many other things; so it's hard to generalize it more than to say it plays into gearing... just like you said. Is smaller or larger diameter better? Depends.

Cheers!
Chris Shenefield
RedShift Motorsports, Inc.
http://www.redshiftmotorsports.com
2002 STS National Champion STS

 

Well said, Chris.

Here's some questions to answer your question. What is the rolling circumference of your current tire? Are you happy with the acceleration characteristics? If so than the answer to your question is to get a tire as close in circumference to yours as possible. I think the 10mm size difference between the 215 and the 225 is not too significant.

But before you do any of this, have you maximized your current brake system? Have you explored track only pads? Was your brake fluid changed to a high performance fluid before your track event? Did you bleed the system between sessions? Can you add ducting to better cool your brakes? (this is a very cheap mod that can be hugely helpful)

Regards,
Alan

 

your purpose is hpde, not autocross, right? just want to make sure i am reading you right.. anyway..
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">So if I'm doing 100 mph on the 225/50/16 tire, it'd be the equivalent of doing 108 mph on the 215/40/16 tire. The problem I see here is that the brakes have to haul the car down an additional 8.4% in speed, right? Also, wouldn't you have less brake torque due to the smaller wheel diameter? </TD></TR></TABLE>
Brakes are an energy dissipation mechanism. The translational kinetic energy of the main chassis would be 1/2mv^2 ... this value doesnt change between the larger and smaller wheels, except for the mass of the wheels (including tires), which probably is not what we are really worried about with this question, I think.

The rotational kinetic energy can be described as 1/2*rotational moment of inertia*(rotational speed)^2 .. the flywheel, drive shafts, tires-- all rotating components whose rotational speed is related to the mph have this contribution to the energy the brakes must dissipate.. The smaller wheels will probably have a smaller rotational moment of inertia, but they will have a higher rotational speed... since the rotational speed value is squared, this is probably the more dominant term at higher speeds..

So, it seems the smaller wheels will probably have higher rotational energy that needs to be dissipated by the brakes. BUT, I expect this increase in rotational energy to not be a terribly significant term, compared to the translational kinetic energy term for speed ranges and tire diameter ranges likely to be discussed. I am too lazy to look up terms and values to run the calculations, though..

As to the brake torque, the torque exerted by the caliper is dependent on the pad's distance from the center of the wheel (among other things, obviously), and if that doesn't change with wheel size, I don't see how brake torque will change.

A decreased radius of the tire's contact patch with regard to the axis of rotation means that the road can exert less torque on the wheel to resist the braking force. So if braking force is constant between the two overall wheel radii, then the smaller wheel radius setup will have more apparent braking power, it seems.. (meaning the calipers have to exert less force to lock up the tires).

hmm.. note, I have been refering to the overal radius as mentioned in the original post, not to rim size.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">As to the brake torque, the torque exerted by the caliper is dependent on the pad's distance from the center of the wheel (among other things, obviously), and if that doesn't change with wheel size, I don't see how brake torque will change.</TD></TR></TABLE>

He did say that he also swapped Audi TT calipers, which I'm assuming have a larger caliper piston surface area (either bigger piston or more of them). So the clamping force of the caliper is larger and so is the braking torque. I don't know why you'd do the caliper if they were the same...unless it's necessary to get the rotors to fit...

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">A decreased radius of the tire's contact patch with regard to the axis of rotation means that the road can exert less torque on the wheel to resist the braking force. So if braking force is constant between the two overall wheel radii, then the smaller wheel radius setup will have more apparent braking power, it seems.. (meaning the calipers have to exert less force to lock up the tires).</TD></TR></TABLE>

You're correct...and here's a story...

When I was doing physics at my previous college, I had a project in advanced mechanics/dynamics where we were allowed to do "any experiment" involving mechanics. So I determined the change in braking force (at the tire; braking torque is at the rotor/hub) after I changed tires: changing the coefficient of friction. So I measured/spec'd every piece from the pedal to the pad (and the rest of it) to determine the forces and torques for a given pedal pressure (physicsts don't bother with anything but ideal situations ). Anyway, the braking force of the ground on the tire is in the opposite direction of the braking torque from the caliper on the rotor. Tangential friction forces induces forward rotation, instead of the relative reverse rotation the brakes are inducing. I made a sweet *** diagram of a bunch of forces on different parts but it's at school...

 

Here's an easy explanation. You accelerate up to some speed by putting energy into the car. So you come to a corner and step on the brakes - they must remove energy to slow you down. It makes no difference what size you wheels, tires, or brakes are, it is exactly the same amount of energy regardless.

This ignores things like brakes getting too hot to work, which is a separate issue.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by kb58 &raquo;</TD></TR><TR><TD CLASS="quote">Here's an easy explanation. You accelerate up to some speed by putting energy into the car. So you come to a corner and step on the brakes - they must remove energy to slow you down. It makes no difference what size you wheels, tires, or brakes are, it is exactly the same amount of energy regardless.</TD></TR></TABLE>

That's true except that the smaller wheel and tire set are traveling at a different rotational speed than the larger wheel/tire set so the rotational kinetic energy is different and does make a difference.

 

Chris, thanks for clearing up that up, but along with furthering my skill, I'd like to have all the help I can get (cuz I'll need it ).

How much did they run you?

But there is more rotational speed in the drivetrain @ 100 mph in a small tire, meaning more inertia to overcome w/ the brake system.

I'm happy with it, but who could want less acceleration?

I threw on some Carbotech Panther XPs, and had some fresh (one week old) Valvoline SynPower DOT4. I didn't bleed between session, but pedal travel did remain constant as far as I could tell. I was on street tires though, and I'm sure I could have overcome them quite easily (although I never quite had the ***** to try). Ducting is next on my list, but the point of the discussion was to find my overall best possible combination. Or at least try.

TT caliper carriers, not actual calipers. My understanding was that although clamping pressure is the same as what I have now, I would have more torque due to the larger diameter rotor...but this isn't overly important in the optimum rolling diameter discussion. I shoulda stuck to one question per thread.

True, but you could argue that the same amount of energy is dissipated on the street hauling your car down from 100 mph in a .5 mile distance. I was more interested in what would be the difference at the threshold.

And what'd you find out? If the wheel diameter was consistent and only tire compound was changed, I'd expect the braking force to be the same unless you exceed the maximum grip of the tire, no? If you're only asking the tire to generate X amount of braking force (as a result of Y amount of brake torque generated by a constant Z lbs. of pedal pressure), my thinking would be that it would be consistent amount of force.

Ah, this probably explains it the most. Although...more energy dissipated just means more heat, right? But if I assume my braking system is maximized to the point where it has the ability to lock the stickiest tire I throw at it and has the heat/cooling capacity to last me a full 30 minute HPDE session...the extra rotational speed of the smaller tire would cause me to have longer braking distances, right? If the tire has X mu as its coefficient of friction, and assuming I can hold it at its threshold for the entire length of the braking zone and dynamic load on the tire is the same due to weight shifting, it would be in fact longer? Or would it be about the same because although there's more revolutions per a given distance, each revolution is shorter...?

I guess its time to do real calculations and find the lesser of two evils. I'm beginning to like kb58's answer...

 

It's been a long time since I have cracked open my physics books, but I would say the wheel diameter will effect the feel of the brakes.

Take an extreme example, like a traditional 10-speed bicycle braking system. The shoes are clamping on the rim of the wheel, and you can generate plenty of braking force to lock up the wheel. now, take the same size shoe and put it on a "rotor" that is only 4" in diameter. It will require a lot more "pedal force" to lock up the wheels. If you keep those same rotors, and decrease the wheel diameter down to 6", you will be able to lock them up again.

With a smaller wheel, you will need less pedal pressure to exert the same amount of braking force at the contact patch.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by stealthx32 &raquo;</TD></TR><TR><TD CLASS="quote">

But there is more rotational speed in the drivetrain @ 100 mph in a small tire, meaning more inertia to overcome w/ the brake system.

</TD></TR></TABLE>

Ok maybe, but it is also offset by the lighter wheel/tire mass and smaller diameter. But anyways, driveline inertia is insignificant compared to the inertia of 3,000lbs moving at 100mph. Not convinced? Try the following experiment at your own risk (std disclaimer). Like those dudes in the appearance forum, secure the car on jackstands with the front wheels off the ground, rev it up to 100mph and a little tap on the brakes will bring the whole driveline to a halt within a blink of an eye almost... Driveline inertia is a relative joke compared to the available braking force...

 

well that this point in the theory discussion, technically, we should all put the clutch in during hard stops, because then you're taking away the kinetic engergy of the rotating engine from the equation. Less weight to slow down...

 

Changes in the tire compound were far more significant than changes to pad compound. The braking force is also dependent upon the tire friction (not braking torque). Azenis have a higher static coeff of friction than Potenza RE92's so the brake distances were shorter with them ("I know from experience, dude." [/Chris Farley]).

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">Ah, this probably explains it the most. Although...more energy dissipated just means more heat, right? But if I assume my braking system is maximized to the point where it has the ability to lock the stickiest tire I throw at it and has the heat/cooling capacity to last me a full 30 minute HPDE session...the extra rotational speed of the smaller tire would cause me to have longer braking distances, right? If the tire has X mu as its coefficient of friction, and assuming I can hold it at its threshold for the entire length of the braking zone and dynamic load on the tire is the same due to weight shifting, it would be in fact longer? Or would it be about the same because although there's more revolutions per a given distance, each revolution is shorter...?</TD></TR></TABLE>

The rotational speed of both the big wheel and the small wheel go to zero. The smaller wheel dissipated kinetic energy at a faster rate than the larger wheel: at a certain point, the [difference in rotational speed^2] &lt; [difference in radius (and mass)]. The kinetic energy reduction of the small wheel surapsses that of the large wheel. Also, the torque from the pad on the rotor is the same for both, but the torque from the road on the tire is smaller for the small wheel (it tries to make the wheel go forward). Newton*meters (torque and work) is also known as a joule (energy). Speaking in terms of work, the distance at which the force is applied is smaller (circumfrence of wheel is smaller) so the work is smaller. More work equates to more energy, so the less work that has to be done, the less energy. Just another way to think about it. Edit: That is for one cycle, however. You can determine the difference in work of each setup by calculating the work done per cycle and then multiply by the number of cycles (rotations). This is assuming the force is constant throughout: the weight on the tires changes as the car speed changes.

This is all assuming the tires haven't locked up. Once the tires lock, the stopping distance is dependent only on the tires so long as the brake keeps em locked.


Modified by GSpeedR at 10:03 AM 10/20/2003

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">What about engine braking? Above certain engine speed, piston assembly inertia is large relative to piston forces on crank, and the crank tries to keep the wheels moving. I thought I read somewhere that the Viper produces something like 180hp worth of engine braking at 5000rpm, which is more than my motor puts out.

</TD></TR></TABLE>

Engine braking isn't as powerful as your brakes. Therefore, you still have to slow down the engine faster than it would by itself, and that makes your brakes absorb more energy. It seems quite simple to me. Clutch in, given no traction/brake overheating issues would mean faster stopping times and distances.

Yet...I don't think I've EVER seen any racer do this. Must be a good reason Re-read Chris Shenefield's post.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by PseudoRealityX &raquo;</TD></TR><TR><TD CLASS="quote">Engine braking isn't as powerful as your brakes. Therefore, you still have to slow down the engine faster than it would by itself, and that makes your brakes absorb more energy. It seems quite simple to me. Clutch in, given no traction/brake overheating issues would mean faster stopping times and distances.

Yet...I don't think I've EVER seen any racer do this. Must be a good reason Re-read Chris Shenefield's post.</TD></TR></TABLE>

You're saying you've never seen a racer brake with clutch-in or the opposite?

I agree that the engine decelerates much slower than the brakes and the engine does add additional rotational inertia. The car does feel like it's slowing faster in gear (high rpm), and it certainly feels more stable under braking... So downshifts during braking are only to maximize accelerate the car during corner exit?

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">

You're saying you've never seen a racer brake with clutch-in or the opposite? </TD></TR></TABLE>

never seen one with clutch in

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">I agree that the engine decelerates much slower than the brakes and the engine does add additional rotational inertia. The car does feel like it's slowing faster in gear (high rpm), and it certainly feels more stable under braking... So downshifts during braking are only to maximize accelerate the car during corner exit?</TD></TR></TABLE>

looking at it from a physics standpoint, yes....it's to make sure you're in the right gear when you decide to stop braking, AND, as a saftey concern, so you CAN use the throttle if you need to avoid something.

of course....the whole deal is that any of us can overload the tires anyway while still using the engine there.....so it's not going to gain you any performance advantage unless you have really crappy brakes, and then....you shouldn't be on a track.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">But if I assume my braking system is maximized to the point where it has the ability to lock the stickiest tire I throw at it and has the heat/cooling capacity to last me a full 30 minute HPDE session...the extra rotational speed of the smaller tire would cause me to have longer braking distances, right?</TD></TR></TABLE>
If you are at threshold of tire grip all the way down, and the total translational kinetic energy of the car is the same for both instances, both stop the same. The brake pads (not the tire friction) have to dissipate the rotational kinetic energy, so they may be dissipating more heat with the smaller diameter, but if you have the excess cooling and clamping capacity to hold the tire at it's peak grip, that won't change the braking distance.

(disclaimer, I have a headache right now, so I may be missing something obvious..)