I was discussing tire growth with a friend earlier today and I said every time I calculate it I get around 5% growth. Since the question comes up here from time to time I thought I would share the math.
This is for the 33.5" MT 3183, although the Hoosier 33.5" 18700 is pretty much identical. I've had multiple sets of 3183s that roll out between 104.5 and 107 and the finish line RPM seems to be the same regardless of the difference sitting on the starting line.
For the run I looked at today my dragster went 7.463 at 178.75 mph on the time slip. The data logger recorded 7075 driveshaft RPM at the finish line. Car has a 4.11 rear gear. Here's the math:
7075/ 4.11111111111 = 1720.95 tire RPM
*33.5(pi) = 181118.12 inches per minute
/12 = 15093.18 feet per minute
/5280 = 2.8586 miles per minute
*60 = 171.51 calculated miles per hour
178.75-171.51 = 7.24 mph faster on the time slip
7.24/171.51 = .0422 or 4.2% growth
But, Calculus shows that the instantaneous velocity at 1320’ is about 2 mph faster than what the time slip shows for my car since it is accelerating through the speed trap. So using a top speed of 180.75 results in a growth of (180.75-171.51)/171.51 = .054 or 5.4% growth.
I've done similar math on other cars/conventional slicks and always end up at about 5%. I'd be curious how the big and little bubba compare as I understand they grow less.
The speed trap starts at the mph cone which is 66ft before the finish line on a 1/4 mile track.
You know trap speed is just a manipulation of this: distance = Vavg x T → Vavg = distance/T.
You already calculated tire speed at the finish line.
Now calculate the the tire speed at 66ft from the finish line from your driveshaft data. This point is about .2517 seconds prior to your ET. (T = 66feet/Vavg)
What's the calc say for tire growth at that point?
How do they compare?
The extra credit formula is how to relate the the tire sidewall stiffness, its radius of gyration and some relationship for the tire growth in the middle of the "tread" to wheel speed to calculate an instantaneous rolling radius due to centripetal force which then is plugged into vehicle speed calcs or the double integration of your accelerometer data to calculate distance.
There are easier ways to do this since you have data collected at a constant recording rate.
Assuming your recording driveshaft speed at 100Hz or better, you pick off the data points and do math or add math channels in to do it for you. In your example, though, using the "calculus" finish line speed delta and dividing it by the calculated Vtirediameter is not quite correct.This message has been edited. Last edited by: Rick!,
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