Tire diameter effect on RWHP
Le Mans Master
Joined: Jul 2005
Posts: 5,328
Likes: 405
What Ranger has done with his data logging is extreemly valuable, even for us road racers.
The challange many ( basicly me as I am math challanged - and I come from a family of engineers
) is what is called the moment of inertia. http://en.wikipedia.org/wiki/Moment_of_inertia
"The Moment of Inertia of an object about a given axis describes how difficult it is to change< its angular motion about that axis. For example, consider two discs of the same mass, one large and one small in radius. Assuming that there is uniform thickness and mass distribution, the larger radius disc requires more effort to accelerate it (i.e. change its angular motion) because its mass is effectively distributed further from its axis of rotation. Conversely, the smaller radius disc takes less effort to accelerate it because its mass is distributed closer to its axis of rotation. Quantitatively, the larger disc has a larger moment of inertia, whereas the smaller disc has a smaller moment of inertia."
First you need to know the weight and dimensions of the rim. Next, weigh the rim with the tire on it and however much air you usually keep in the tire.
Now for the equations:
I = 1/2 MR^2 (the "^2" denotes "to the second power" aka "squared")
This is the equation for a solid shere which you would use on the rim. Obviously, the rim isn't perfectly uniform,
but this will give you a close enough estimation to be able to compare two different wheel/tire combos. M stands for the weight (do it in pounds) of just the rim and
R (half the diameter of the rim) is the radius (do it in feet).
I = 1/2 M (R1^2 + R2^2)
This would be for the tire. You know the mass because it is the weight of the wheel and tire combo minus the weight of the wheel.
R1 is the inner radius (which equals half the diameter of the rim) and R2 is half the diameter of the entire tire.
Add the two I's together and you get the moment of inertia (resistance to rotation) in pound feet. This is for one rim and tire which needs to be compared
to the other type of rim and tire combination. Obviously, multiplying by 4 for each combo gives their total values.
http://ls1tech.com/forums/showthread...3&page=3&pp=20
http://en.wikipedia.org/wiki/Moment_of_inertia
and In classical Mechanics
F = m * a
Where
F = force
m = mass ( weight )
a = accleration
Or when we change that forumula to a = F/m
If we reduce the Mass of the rotating object ( the wheel ) and keep the same force, there is faster acceleration.
In addition we need to look at Rotational wt ( unsprong weight)
and redcing wt in general
http://www.thirdgen.org/sheddingpounds
"A vehicle's total weight is the sum of all of its parts and affects its ability to accelerate, brake and corner. Reducing the total weight will enhance the vehicle's performance because less weight needs to be controlled and therefore, less energy is required. Unsprung weight is the weight under the springs which moves up and down as the vehicle rides over uneven roads and leans in the corners. Reducing unsprung weight allows the springs and shock absorbers to be more effective in controlling the suspension's movement. Additionally, a vehicle's rotational weight includes all parts that spin including everything in the vehicle's driveline from the engine's crankshaft to its wheels and tires. This affects the energy required to change speed as the vehicle accelerates and brakes. As you would guess, reducing the weight of any of these rotating components will enhance the vehicle's performance because less energy will be required to increase or decrease their speed."
Source: http://www.tirerack.com/wheels-techpage-1/108.shtml
When all the caculations are done, saving 1 lb of rotation (unsprong wt ) is about the same as removne 3.5 lbs of normal wt off the car.
Other good caculations:
Engine tuners http://wahiduddin.net/calc/calc_hp_dp.htm
Physicis or Racing http://www.miata.net/sport/Physics/
Physics of Racing " the Drive Wheel" http://www.miata.net/sport/Physics/Part26.html
Phew Sorry for the long thread ( and I only had two cups of coffee this morning
The challange many ( basicly me as I am math challanged - and I come from a family of engineers
) is what is called the moment of inertia. http://en.wikipedia.org/wiki/Moment_of_inertia"The Moment of Inertia of an object about a given axis describes how difficult it is to change< its angular motion about that axis. For example, consider two discs of the same mass, one large and one small in radius. Assuming that there is uniform thickness and mass distribution, the larger radius disc requires more effort to accelerate it (i.e. change its angular motion) because its mass is effectively distributed further from its axis of rotation. Conversely, the smaller radius disc takes less effort to accelerate it because its mass is distributed closer to its axis of rotation. Quantitatively, the larger disc has a larger moment of inertia, whereas the smaller disc has a smaller moment of inertia."
First you need to know the weight and dimensions of the rim. Next, weigh the rim with the tire on it and however much air you usually keep in the tire.
Now for the equations:
I = 1/2 MR^2 (the "^2" denotes "to the second power" aka "squared")
This is the equation for a solid shere which you would use on the rim. Obviously, the rim isn't perfectly uniform,
but this will give you a close enough estimation to be able to compare two different wheel/tire combos. M stands for the weight (do it in pounds) of just the rim and
R (half the diameter of the rim) is the radius (do it in feet).
I = 1/2 M (R1^2 + R2^2)
This would be for the tire. You know the mass because it is the weight of the wheel and tire combo minus the weight of the wheel.
R1 is the inner radius (which equals half the diameter of the rim) and R2 is half the diameter of the entire tire.
Add the two I's together and you get the moment of inertia (resistance to rotation) in pound feet. This is for one rim and tire which needs to be compared
to the other type of rim and tire combination. Obviously, multiplying by 4 for each combo gives their total values.
http://ls1tech.com/forums/showthread...3&page=3&pp=20
http://en.wikipedia.org/wiki/Moment_of_inertia
and In classical Mechanics
F = m * a
Where
F = force
m = mass ( weight )
a = accleration
Or when we change that forumula to a = F/m
If we reduce the Mass of the rotating object ( the wheel ) and keep the same force, there is faster acceleration.
In addition we need to look at Rotational wt ( unsprong weight)
and redcing wt in general
http://www.thirdgen.org/sheddingpounds
"A vehicle's total weight is the sum of all of its parts and affects its ability to accelerate, brake and corner. Reducing the total weight will enhance the vehicle's performance because less weight needs to be controlled and therefore, less energy is required. Unsprung weight is the weight under the springs which moves up and down as the vehicle rides over uneven roads and leans in the corners. Reducing unsprung weight allows the springs and shock absorbers to be more effective in controlling the suspension's movement. Additionally, a vehicle's rotational weight includes all parts that spin including everything in the vehicle's driveline from the engine's crankshaft to its wheels and tires. This affects the energy required to change speed as the vehicle accelerates and brakes. As you would guess, reducing the weight of any of these rotating components will enhance the vehicle's performance because less energy will be required to increase or decrease their speed."
Source: http://www.tirerack.com/wheels-techpage-1/108.shtml
When all the caculations are done, saving 1 lb of rotation (unsprong wt ) is about the same as removne 3.5 lbs of normal wt off the car.
Other good caculations:
Engine tuners http://wahiduddin.net/calc/calc_hp_dp.htm
Physicis or Racing http://www.miata.net/sport/Physics/
Physics of Racing " the Drive Wheel" http://www.miata.net/sport/Physics/Part26.html
Phew Sorry for the long thread ( and I only had two cups of coffee this morning
Wheels/tires do not perform "work", they store kinetic energy (KE). (I noticed in your link you referred to I in terms of LB-FT when it should be LB-FT^2.) Work is done ON the wheel/tire by the engine to increase their KE with the amount of work done (KE added) given by the KE (after) minus the KE (before). KE=1/2Iw^2 (where w is the angular velocity) from this link: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html
The rate at which work is being done on the wheel/tire is equal to the rate at which the KE is increasing. HP is the rate at which work is being done so we're left with finding the KE added and dividing by the time it took to add the KE to the wheel/tire to find the average HP required to accelerate it from one RPM to another. (This is from my old Physics book which I tried to scan in but my scanner stopped working.)
We'll use some realistic numbers for the C6Z06:
Tires- 36 LBs @ 24" average diameter
Wheels- 20 LBs @ 15" average diameter
(I used average diameter (which is my guesstimate or SWAG...Scientific Wild Azz Guess) to mean the diameter at which half the weight of the wheel or tire is outside that point and the other half is inside that point. Since the majority of the wheel/tire weight is at the outermost point, we can't really use the average of the ID and OD as that would give us a large error given the radius is a "squared" term. Using average diameter makes the wheel or tire a hoop so we would use the following formula from the link above: I=MR^2.)
I for the tire is I=(16.33 KG)(.3048m)^2=1.517 kg-m^2
I for the wheel is I=(9.07 KG)(.1905m)^2=.3292 kg-m^2
Tire+wheel I=1.517+.3292=1.846 kg-m^2
We'll use a video from the following link:
http://www.livernoismotorsports.com
There's one of a stock LMB C6Z06 with the engine running from ~1700 RPM to ~6900 RPM during the run which took 18 seconds total.
1700 RPM @ the crank=(1700/3.42/60)(2)(pi)=52.05 rad/sec at the wheel.
KE (before)=1/2(1.846)(52.05)^2=2501 Joules
6900 RPM @ the crank=(6900/3.42/60)(2)(pi)=211.3 rad/sec at the wheel.
KE (after)=1/2(1.846)(211.3)^2=41201 Joules
KE (after)-KE (before)=41201-2501=38700 Joules
1 Joule=.738 ft-lb
38700(.738)=28560.6 ft-lb which is the total work done to increase the KE of the wheel/tire when the engine went from 1700 to 6900 RPM.
To get the rate at which work was done, divide by the 18 sec dyno run giving: 28560.6 ft-lb/18 sec= 1586.7 ft-lb/sec
1 HP=550 ft-lb/sec, so we have 1586.7/550=2.885 HP for one tire/wheel or 5.77 HP for both tires/wheels.
Again, tires/wheels do not perform "work" thus you can't use the HP=T*RPM/5252 formula. Work is performed on the tire/wheel requiring the solution for HP be performed by determining the rate the KE is changing. For small changes in weight and diameter, you won't get much diffference in the calculations and you certainly won't feel it in the seat of your pants. Also, as your acceleration rates go down (as you go up in the gears), the time to add KE increases which reduces the HP required to accelerate the wheel/tire until you reach top end where HP going to inertia is zero and there is no HP loss to inertia. As Ranger said, other than possibly 60' times, you won't be able to detect small changes in wheel/tire weights/diameters with a stop watch...not to say there isn't any, but it's not much.
Just to add a note as to the low HP number above, remember that was an 18 second run from 1700-6900 RPM in 4th gear...make that a 3.5 second run in first gear from 1700-6900 RPM (a really good 0-60 MPH time), and the HP number goes up significantly (to about 30 HP) and now you can see where the 60' time in a drag race is affected by decreasing inertia...as long as you can get traction.
Last edited by glass slipper; Apr 17, 2007 at 07:33 PM.
Le Mans Master
Joined: Sep 2004
Posts: 5,707
Likes: 4
From: Los Angeles
Yes, In the end that rotational mass is certainly worse than the same mass anywhere else on the car. You have to remember you're not talking about unsprung mass, as that would also include all your brake hardware, ball joints, spindles and half of the A-arms. That has more to do with the handling than acceleration. Rotational mass is what you're getting into here. Ranger's excellent data information should show these differences in the long term but as I said and Ranger confirmed, even up to 50 lbs. would be hard to detect on any single 1/4 mile pass because there are so many other minor details that could disguise the small gains/losses of wheel diameter.
You and I go out on the track with passengers/students that probably average about 200-250lbs. or more and we can feel the difference but without a scientific procedure of data aquisition like Ranger, we really don't know how much it really affects us. We don't think twice about putting our in-car cameras, floor mats, hardbars, wallets
and other non essential stuff but it does all add up. As you know, the few horsepower difference gained or lost in wheel mass is easily overshadowed by a driver.
Le Mans Master
Joined: Jul 2005
Posts: 5,328
Likes: 405
A heavier wheel and tire combo did show up on a dyno. A C6 went from 507 RWHP down to 499 RWHP with only a tire change. It was from a wheel/tire combo of 19" PSC's to an 18" combo of C6 motorsports wheels and Nitto 555RII tires. The latter (heavier) wheel/tire combo made 8 RWHP LESS HP on the same dyno with the same car under the same conditions. 
Last edited by glass slipper; Apr 17, 2007 at 12:40 PM.
Thread Starter
Burning Brakes
Joined: Feb 2005
Posts: 1,113
Likes: 8
From: Whitehall PA
When I asked this question, I had a feeling some old, or young, physics majors would chime in. I certainly got more than I expected. If I am ever asked about the performance effects of wheel diameter my answer will be "not much". Then I will reference this thread. Impressive!
