David M. Pintaric Posted January 20, 2007 Share Posted January 20, 2007 This thread is directed mostly at Greg, but I'd like anyone else's input on this subject. Please note I am not disagreeing with the process just trying to "edumacate" myself with this racing stuff. In the ST/SU rules cars that are above a certain "Mendoza" line of weight (3,200 lbs I think) get to add "ratio numbers" to that car's power-to-weight ratio. Cars below a second line get to take away numbers from that car's power-to-weight ratio (I think I got that right). What, if any, is the scientific/engineered ratonale to this equation? I think I understand the mindset (and again, agree with it until proven otherwise), but am looking for an engineering response. BTW: this process helps me and my car, so thank you. Thank you in advance. Quote Link to comment Share on other sites More sharing options...
claykos Posted January 20, 2007 Share Posted January 20, 2007 I'm not Greg. But I think he is trying to penalize the inherent advantage of a lighter car. IE a lighter car with the same size tires and power/weight as a heavier car is going to have an advantage. It should be able to transition quicker, generate more absolute cornering force and be generally easier on equipment (last the whole race without over cooking tires). Quote Link to comment Share on other sites More sharing options...
David M. Pintaric Posted January 20, 2007 Author Share Posted January 20, 2007 Thanks for your input, Clay. So to further this, let's compare two cars: #1 2,500 lbs with 250 RWHP/250 lb/ft RWT #2 4,000 lbs with 400 RWHP/400 lb/ft RWT Both cars have proportionately equivalent tires and brakes, and aero is the same. The HP and torque to weight ratios are equal. But because of the massive weight of #2 (and I know this is a drastic example, but it works), then #1 with all items being proportionately equal still has a decided advantage, correct? Is there a mathematic formala to prove this? Quote Link to comment Share on other sites More sharing options...
National Staff Greg G. Posted January 20, 2007 National Staff Share Posted January 20, 2007 David, Clay has it right. As well as the three things he listed (transition, cornering, and brake and tire abuse), there is also the ability to brake (assuming the same size brakes on each vehicle), and the potential for the lighter weight car to have a better aero package in general and have less drag (note that I said potential). You are absolutely correct that the lighter weight 2500lb/250hp car should have the advantage over the 4000lb/400hp car. For those that don't see the reasoning for this, imagine a 40,000lb car with 4000hp--might be a little tough to get it to brake or turn like a 2500 lb car. Although, I'm sure a bunch of us have seen the rocket powered truck that tours air shows, and he accelerates just fine. I'm sure a mathmatician, physicist, or engineer can come up with a nice very complicated formula to explain this. It would be less complicated if the two "masses", otherwise known as racecars, were the exact same shape, size, and had proportiate balancing of the extra weight. Once you throw in different shapes and sizes of the cars, then add in different car balance, and different aerodynamic properties, it becomes much more complicated. Now, throw in different tire compounds and brake pad and rotor material, (and their changing properties as they age), and you have a real math problem to solve. Now, through this all, the assumption is that tire size is equal, but more tire doesn't necessarily always mean faster racecar. But, at least for our discussion, it means more grip and better potential cornering, and ability to get the power down to the ground. This would have to be factored into the equation as well. Anyway, the short answer is that NASA doesn't have the mathmatical formula to describe all of this. But, we do know that in real world racing, it is true that given the same size tire and brakes, that the lighter car has the advantage. So, in your 10:1 wt/hp ratio example above, our ST Rules would actually result in these numbers (both on larger than 275 size DOT approved tires): Car #1 2500 lbs, 232 rwhp Car #2 4000 lbs, 437 rwhp Now, is car #1 still faster than Car #2 ? Let's get them out on the track and race and find out Quote Link to comment Share on other sites More sharing options...
National Staff Greg G. Posted January 20, 2007 National Staff Share Posted January 20, 2007 David, I missed something in your example. You wanted proportionately equal size tires and brakes for the two cars. If they are truly proportionate, then they would be much closer to being equal. The only big factor remaining that would make the larger car slower would be the larger frontal area causing more drag. The other factor to look at is getting the tires to be proportionally the same. Let's say the 4000lb car runs on 345's on all four (just for argument sake). The weight proportion between the cars is 2500/4000 = 0.625 (62.5%). Now, take 62.5% of those 345 size tires, and you get 215 size tires. Now, you have a big American Iron type car with 400 hp on 345's, racing against a touring car with 250 hp on 215's. That is a close race from what I've seen in our PT and TT experience. In fact, those 215's probably have killed the deal for the 250 hp car, because it will have a hard time getting all of that 250 hp down to the ground on turn exit. I'm really looking forward to seeing how this all pans out this year. Hopefully, we see a lot of different car types and sizes. I wish that I had a car that could participate, but I'll have my fun just watching these races in '07. Quote Link to comment Share on other sites More sharing options...
claykos Posted January 20, 2007 Share Posted January 20, 2007 Here's a formula for you... Centripetal force (the force that your tires are generating to hold your car on a curved path at constant velocity) is mV^2/R with m=mass v=velocity and r=radius of the turn. If you double the mass of the vehicle you must double the force required of the tires in order to take that same turn at the same speed. Now it's a bit more complicated than that because the force your tires can generate is proportional to the downward force on them (weight of thecar), but that gives you a basic idea. And also, that is why downforce is SO powerful. You are increasing the downward force on the tire as if the car was heavier but you are not increasing the centripetal force required.... Quote Link to comment Share on other sites More sharing options...
chucktoo Posted January 21, 2007 Share Posted January 21, 2007 This a great place to start - a year of racing and more data will allow fine tuning next year of the formulas. Quote Link to comment Share on other sites More sharing options...
davidfarmer Posted January 23, 2007 Share Posted January 23, 2007 (edited) **Edited for being STUPID** Based on my 15years of racing, and my BS engineering degree (that's bachelors, not the other BS in this cas), I think these numbers are great starting point. It will take time to work it out, but it is also going to take time for people to get there cars up/down to exactly the p-w-r they need also. Looking forward to seeing how things pan out. I'm still looking for a (cheap) Convertible 01' Vette, if anybody comes across one! Edited January 23, 2007 by Guest Quote Link to comment Share on other sites More sharing options...
David M. Pintaric Posted January 23, 2007 Author Share Posted January 23, 2007 David, If that formula works for deceleration, then does it also apply in acceleration? The reason I ask is that when racing against T1 Vettes last year I did not notice a significant (or any) acceleration advantage even though my car had a power-to-weigh ratio advantage (Viper 445 RWHP/ 3700 lbs = 8.31; Vette 375 RWHP / 3300 lbs = 8.. Quote Link to comment Share on other sites More sharing options...
claykos Posted January 23, 2007 Share Posted January 23, 2007 David(farmer) KE=.5mv^2. Changing the mass will not yield the difference you stated, that difference would come from changing the velocity. KE is only linearly dependent on mass. Quote Link to comment Share on other sites More sharing options...
TexaST-1 Posted January 23, 2007 Share Posted January 23, 2007 OMG! Just let me drive. There are obviously much smarter people in racing than me! When in doubt light makes right. Quote Link to comment Share on other sites More sharing options...
davidfarmer Posted January 23, 2007 Share Posted January 23, 2007 doh, sorry about that. I'm in Orlando on vacation, on a cell-phone connection, and obviously not thinking as fast as I'm typing. Corrected, it is a linear correlation, but it still stands that the overall abuse a heavier vehicle has to handle is not easily overcome with power alone. Quote Link to comment Share on other sites More sharing options...
Falcon Posted January 23, 2007 Share Posted January 23, 2007 doh, sorry about that. I'm in Orlando on vacation, on a cell-phone connection, and obviously not thinking as fast as I'm typing. Uh Oh! Is that Disney World I hear in the background? Quote Link to comment Share on other sites More sharing options...
chucktoo Posted January 24, 2007 Share Posted January 24, 2007 In CMC they use a more complex formula for torque and power but only over a limited range of values. We have a great start with ST and lets see how 2007 pans out. Charlie Quote Link to comment Share on other sites More sharing options...
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