Tyre Data

[Chris Boon]

Hey all,
I have heard that Goodyear was going to/has made a tyre specific for Formula SAE and that some data was available. However, I dont seem to be able to find anything on their website or on the internet in general. I was wondering whether anybody knew where it was?

Also I read in a couple of previous discussions a couple of years back people had Goodyear data sent to them for initial use with ADAMS. Does anyone still have any of this data?

Cheers,
Chris

Suspension, UQ Racing

[Chris Boon]

Hey all,
I have heard that Goodyear was going to/has made a tyre specific for Formula SAE and that some data was available. However, I dont seem to be able to find anything on their website or on the internet in general. I was wondering whether anybody knew where it was?

Also I read in a couple of previous discussions a couple of years back people had Goodyear data sent to them for initial use with ADAMS. Does anyone still have any of this data?

Cheers,
Chris

Suspension, UQ Racing

[kwancho]

I assume you've looked at the website, but on it, they do have tire data for their old tires from 1999. The older tires are also in the FSAE Tire Testing Consortium

[Ché]

Chris not sure if you're looking for more than this but I think what you are looking for might be on their website...

http://www.racegoodyear.com/sae.html

or you can try the Formula SAE Tire Databook (PDF) (http://www.racegoodyear.com/pdf/sae/toc.pdf)

hope that helps...if you want more detailed infor then you should purchase the Tire Testing Consortium data, try searching this forum for TTC or tire test data we got the data about a year ago if I recall

[Erich Ohlde]

I don't believe they have data for their new compound (The one that 4 of the top 5 places in Autocross at FSAE west were running). But I'm sure they will have it soon. Try emailing the goodyear guys and see what they say, there should be an email link on the http://www.racegoodyear.com/sae.html page

[Paul Garcia]

The Goodyear data on their website is inconsistent. Reality check what you get. We run some 20x7x13's. I was using their 20x6.5x13 tire data until we get the TTC data and found Lateral Force Coefficients of 3.

That is a tire I want to get my hands on.

[Alan]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Paul Garcia:
The Goodyear data on their website is inconsistent. Reality check what you get. We run some 20x7x13's. I was using their 20x6.5x13 tire data until we get the TTC data and found Lateral Force Coefficients of 3.

That is a tire I want to get my hands on. </div></BLOCKQUOTE>

Its not uncommon for a tire by itself, on a Force and Moment machine to generate these kinds of numbers. The reason why your car can't do 3 Gs is because of that whole load transfer thing. Just as a quick exercise do a lateral load transfer calculation for your car at 3 Gs and see what your inside tire loads are.

[Paul Garcia]

Alan

Ok, load transfer seems like a valid consideration...only if I have to that is! But maybe what I need to do is figure out how to still corner with both my inside wheels in the air if I'm going to be able to generate grip 3 times the weight of the car on the outside! IN all seriousness, I'm confused. I'm quoting Milliken when he shows typical tire data and his statement that typical Grand Prix tire data will show peaks at up to a 1.8 lateral force coefficient (Fy/Fz).

Now, when I use the 99' data for a 20x8.0x13 I'm getting coefficients of around 1.7-2.0, very reasonable. Its that 20x6.5 data that gives me coefficients of 3.0.

However, I openly admit I don't do this for a living! I was trying to use the data to create a rough "Stability Diagram," also from Milliken. I wonder how useful that would even be since I am producing it for an instantaneous weight transfer "snap-shot." All in a day's worth of learning. Any thoughts?

[Kerry]

What you need to keep in mind is that tire data is gathered in a very controlled laboratory environment on a belt, not asphalt. These are probably the reasons your tire data shows friction ratios as high as 2 or 3 - even if you did get your hands on these tires you would be hard pressed to get your car to pull more than ~1.8-2.0 g's with pure mechanical grip.

If the limiting factor was really weight transfer, wouldn't we all be rolling our cars?

[Buckingham]

You also need to keep in mind that you know nothing about the actual piece of lab equipment the data was collected on. If the piece of equipment was designed to handle loads representative of a fully loaded SUV (lets say 4000 lbs on a single corner under dynamic conditions). How do you know the machine will collect good data on an FSAE tire? (lets say 400 lbs on a single corner under dynamic conditions). Generally, you never want to operate below 10% of full range of a piece of test equipment if you can help it.

Also, you should never exterpolate tire data for two reasons. First, it is a non-linear device. Second, research how these models are made, when the person processed the data into a model they chose an operating envelope to which they matched the model to the data. Do you know whether you are interpolating or exterpolating the tire data?

[Paul Garcia]

I don't know what Goodyear did to extrapolate the data but I am doing neither. I am simply picking points off their Lateral Force curves and dividing that force by the load on the tire. Doing so gives coefficients of up to 3, which according to Milliken is twice what we would expect.

When I use data from the same website for a slightly different size tire (8 instead of 6.5" wide) I get much more reasonable coefficients of 1.7. Apparently, there may be a number of reasons why the other data is inaccurate.

[Jersey Tom]

Upwards of 2.5 doesn't surprise me. Milliken references GP car tires. Much harder than ours. FSAE tires are literally as soft and grippy as you can get. On a heavier car though theyd get destroyed.. way too much load.

Think about it. Your whole car corners at 1.8G lets say. Thats Fy = 1170lb, the most of which is being generated by your outside front tire.. which has say Fz = 310lb. To generate even 65% of the total cornering force, thats friction coefficient of 2.4

[Kerry]

Why is most of the cornering force being generated by the outside front tire? Does your car have an oversteer problem or are you still on the brakes? Something else? I don't think one tire generating 65% of the total grip is an accurate guess. I'm just guessing too, but I'd say a reasonable approximation would be that the portion of the cornering force coming from one tire is about equal to the portion of the car's weight on that tire. If your car was truly balanced your sway bars, etc. would give the outside tires approx. equal load mid-corner. Yes, the bulk of the cornering force would come from the outside tires, but only because there is more load on them, not because they have a higher coef. of friction. The inside tires would actually (usually, based on tire data/Claude Rouelle seminar) have a higher coefficient of friction due to the load sensitivity of tire (this is why sway bars work, no?). If your car is pulling 1.8 g's, maybe the outsides are operating at a hair less than 1.8 and the insides at bit more.

Does Milliken reference these GP tires as what they do on course or what the tire data says? 2.5 still sounds high to me. Our tires are very soft and grippy and would get destroyed on a heavy car, but if we put their tires on our car we would slide across the parking lot like a hockey puck. My guess is that at operating temperature our tires have very similar characteristics/grip. The tire compound is chosen so that the weight of the car maintains the tire in the correct temperature range - not to hot (our tire on their car) and not too cold (their tire on our car).

[Jersey Tom]

Two words, weight transfer.

Front inside tire might only be loaded to 15-20 lb on a hard turn. Outside to 300. Fronts I imagine will be at a much higher SA than the rear. Thus the majority of the force comes from the front outside tire.

[Kerry]

Hmmm. I agree that the inside tire could have as little as 15-20 lb on it in the corner. I question why the front would have a much higher slip angle. Lets assume your car is working pretty well (if you're fronts have higher slip angle it's not because your front end is sliding out). In a road (passenger) car in a normal corner (.2 g's? ref. RCVD) your fronts would almost definately be contributing to the bulk of the corner, have a higher SA, etc., but in a road car you try to get the car neutral to increase the cornering potential. In other words, if your front and rear tires are not both working near their peak slip angle, aren't you wasting cornering power? I realize that for us with our amateur drivers we need to dial in a bit of understeer to make the car more stable/predictable, but I think we probably see more similar front/rear slip angles than a road car. I question whether or not we would be able to corner at 1.8 g's if we didn't have similar forces coming from the front and rear tires (because I question the ability of one tire to have a coef. of friction as high as 2.5 on the track).

A while back we were looking at data for speed and lateral acceleration to determine actual turn radius and comparing that with the steer angle to try to figure out the understeer gradient. Our drivers were fastest with a little bit of understeer because the closer we got to neutral the more they were spinning or almost spinning and slowing down to "catch" themselves. While the car was still pushing, it was far less than when the car was set up so that it felt like driving a road car. It almost felt like rotating around your belly button vs. rotating around the rear axle. It is probably important to note that this was on a skidpad, not an autocross type course, so maybe in the transient nature of our competition it is possible that our fronts have a lot more slip angle than the rears (or maybe our shocks weren't right, or our drivers are slowing us down).

Tom, I've heard several theories for why tire data says you can get 3 g's but you can never actually get that on the track, but I've never heard yours. It's an interesting one, but I'm not quite biting yet. I'll keep thinking about it...

[Ben C-M]

Given that for a balanced car total front grip = total rear grip (at steady state), I'd assume that the grip provided by the outside tires is similar. This will be changed by ARB stiffnesses so there may be more load transfer at one end of the car (to offset the difference in SA perhaps), but the grip would still be similar.

[Jersey Tom]

Assuming your car is balanced pretty well, cg height and track front and rear is the same..

Same weight transfer front and rear axle. Say tire loads are

FL 300
FR 25
RL 300
RR 25

Not only does it make sense that the front outside is at a higher SA than the outside rear because the front is heavily steered while the rear SA is just from the body slip angle.. but it must be for other reasons. For the car to yaw and rotate in the turn there must be yaw moment generated by a difference in force from the axles. Front must be greater than rear to go thru the turn, and everything else being equal more cornering force must come from higher SA.

Like I said, for the car to corner at 1.8G.. if most of that (say 65%) is taken up by the front axle and say 90% of the load is on the outside tire.. it must be at 2.3 friction coefficient.

[Kerry]

The fact that the front tire is steered does not mean that it has a higher slip angle. Look some equations for slip angle given the car's velocity vector and yaw rate. It is entirely possible to have equal front and rear slip angles. By definition, a car with higher slip angles at one end of the car is not balanced.

Also, don't discount the aligning torques produced by tires. All four wheels contribute to yaw with those moments as well as with forces. I know that summing forces and moments for a car with equal loads and on the outer tires will result in a turn in the expected direction - whether this is exactly how it happens on the track is a different story, but I still think each tire will have a coef. of friction just slightly different from the total lateral acceleration of the car (given a well set-up non-aero car).

[Jersey Tom]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">I still think each tire will have a coef. of friction just slightly different from the total lateral acceleration of the car </div></BLOCKQUOTE>

Can't be. Crazy amounts of weight transfer and load sensitivity. Id have to run the numbers but should be substantially different.

[Ben C-M]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Jersey Tom:
For the car to yaw and rotate in the turn there must be yaw moment generated by a difference in force from the axles. </div></BLOCKQUOTE>

Jersey, the reason cars yaw is the centripital force around a corner. One of the many definitions of under/oversteer in a steady state corner is too much grip on the rear/front of the car, respectively. Look at the effect of understeer on a car - the front grip is less then the rear grip and it turns around a larger radius turn then desired, but it still is yawing.

This changes when the car is on turn entry or exit, and the grip at each end of the car will be different.

[Kerry]

I'm not quite sure that centripital force is the reason cars yaw... you do need some kind of moment generating force and moment arm, don't you?

Maybe someone who has a working simulation can comment on the kinds of slip angle the are seeing? Or maybe someone who has created one can comment on the terms that end up in the yaw acceleration equation? I am thinking it's not as simple as I originally thought... maybe coriolis forces in there, too? I'll look at some of my papers and report back.

Tom, I was just looking at some of the data from the TTC for one of the Goodyear tires. It does appear that I may have been mistaken about the amount of variation in the coef. of friction from wheel to wheel. It does confirm that the inside tires work more efficiently, which leads me to believe that for the average grip (car's acceleration) to be 1.8 g's, the outer tires need to have a coef. of friction LESS than 1.8 g's. Since the large majority of the force is comming from the outside tires, it may be very very close to 1.8, but still less. That, when averaged with the inside tires (contributing very little), which may have coef. of friction of more than 2 with only 25 lb on each tire, results in the 1.8 g's that the whole car sees. Does this make any sense to anyone? It still seems very possible to me that outside front and rear tires are at similar slip angles and contributing similar amounts of grip, though.

[Buckingham]

The sum of moments is equal to the polar moment of inertia times the angular acceleration of the vehicle. When a car is negotiating a skidpad at steady state velocity, the angular velocity is equal to one full revolution divided by the lap time. This value is steady state, so the angular acceleration must be equal to zero.

Moments are required to change yaw velocity, but not to maintain yaw velocity (in an ideal steady-state case).

[Paul Garcia]

I am definately liking the reasoning that Kerry and Donavan are providing, that in a SS turn of a neutral car there is no yaw acceleration or yawing moments. I.E The front and rear yaw moment are exactly balanced.

So Kerry, I don't have the TTC data but you are saying that they are HIGHLY load sensitive? Which is why we see a degrade in what you are calling coefficients of friction? Correct me if I'm wrong but isn't that the wrong "phrase?" Should we be talking about "lateral force coefficients" because its not just sliding friction but also (mostly?) deflection in the side-wall and not contact patch sliding? I don't mean to argue semantics...

[Jersey Tom]

Good point. My bad.

[Kerry]

Yes, I believe I am misusing "coef. of friction" here. "Friction Ratio," maybe?