Reducing wheel/tire weight - simple data analysis

[eurotic]

So I was bored the other day and the weather was surprisingly nice for where I live for the time of year (10 degrees Celsius in November) so I thought I'd do a very simple test comparing my summer wheel/tire setup to my winter wheel/tire setup so see what sort of quantifiable data I could gather as there is a decent amount of difference in weight. Please understand that this is not an exhaustive test and I was on public roads so I tried to keep it legal.

I'm sharing this to see if anyone might be interested and hopefully I don't get torn apart in the process.

For testing I was in my 2012 S4 with 6MT using my VCDS in turbo mode to just record engine speed. The simple test was to get the car into 2nd gear at idle and then do a WOT pull until the rev limiter which is about 10kph to about 112kph indicated. My summer setup is a set of stock peelers with RE-11's on them weighing 58.4 lbs and my winter setup is a set of VMR 701's with WS-80's that weigh 50.5 lbs which is a weight difference of about 7.9 lbs.

I did 6 runs on each tire with idle around 700 rpm and rev limiter was just over 7200 rpm. I dumped the data into Excel and normalized it to get common reference points (1000rpm, 2000, rpm, ..., 7000 rpm) adjusting the timstamps so that 0 seconds started at 1000 rpm so that I could see how long it took to accelerate to 7000 rpm.

Again, I know this isn't a perfect test but it is a real world test and might show something...and it did I think. Here is the data.

Winter Tires

RPM	Run 1	Run 2	Run 3	Run 4	Run 5	Run 6	Avg.
1000	0.00	0.00	0.00	0.00	0.00	0.00	0.00
2000	1.11	1.09	1.13	1.15	1.12	1.06	1.11
3000	2.04	2.04	2.03	2.06	2.06	2.07	2.05
4000	2.96	2.91	2.88	2.91	2.90	2.93	2.92
5000	3.81	3.75	3.85	3.77	3.78	3.78	3.79
6000	4.74	4.73	4.83	4.81	4.76	4.78	4.78
7000	5.91	5.93	5.92	5.91	5.93	6.02	5.94

Summer Tires

RPM	Run 1	Run 2	Run 3	Run 4	Run 5	Run 6	Avg.	Diff.
1000	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00
2000	1.13	1.12	1.11	1.12	1.19	1.23	1.15	0.04
3000	2.09	2.05	2.06	2.06	2.11	2.19	2.09	0.04
4000	3.01	2.97	3.04	3.01	2.97	3.07	3.01	0.09
5000	3.86	3.92	3.92	3.90	3.91	3.96	3.91	0.12
6000	4.82	4.89	4.85	4.87	4.89	4.93	4.87	0.10
7000	6.03	6.01	6.11	6.13	6.04	6.15	6.08	0.14

You can see that 5 of the 6 runs on the winter tires were in the 5.9 second range with only one run being in the 6.0 second range. The summer tire test was less precise as 3 of the runs were in the 6.0 second range and 3 were in the 6.1 second range. I did a bunch of different combinations of runs that I won't share here (like throwing out the best and worst runs, using just the best runs, using just the worst runs, comparing the best to worst for each, etc). For this post I simply averaged all the results and compared the averages...not the best scientific method but it at least gives us something to start with.

As you can see from the data above it looks like at the end of the ~6 second run the lighter winter tire setup was a little over 0.1 seconds faster. For some reason I was expecting more based on how many people on this forum talk about how dramatic it feels to shed the weight from wheels and tires. However as I thought about it more I was thinking how this might add up in a 1/4 mile run or around a track.

I would love to do a more extensive test but I think that would require entirely different equipment to make it valid. Anyway, there it is, I'd love to hear what anyone has to say especially if anyone else have dared do the same but was too shy to post here. I have my flame suit on for safety at this point.


Shawn

[MrFunk]

Nicely done.
I always notice a seat of pants difference going from summers to winters... real world it's not anything mind boggling but your data certainly shows reduced wheel weight will shave time off your 1/4's etc.

[S4'ed]

I think Car & Driver did this comparison test too...and they reached the same conclusion : heavier wheels/tires decrease acceleration and reduce fuel economy. Basically anything larger than a 17 inch wheel/tire combo results in lower performance, with minimal handling improvement....people want large diameter wheels for the "looks".....

[waxxonMTL]

could also be due to traction vs. cold weather where the summer tires won't grip as much. Why do you keep the performance setup for winter and the heavy setup for summer ? should be the opposite no ? :P

[Race Shooter]

One thing to take into account is if the tires are the same size or not. If you're running two different size tires, the overall rolling diameter and circumference might be slightly different which will affect things pretty greatly as well as its like having smaller gear ratios and therefore increase acceleration.

[SportVier]

Well done. Excellent counter to all the marketing hype around "reducing unsprung weight". And this is a major weight decrease of 8lbs a corner. Despite what the manufacturers and vendors will tell you, dropping a few pounds a corner on a car that weighs 2 tons is just noise.

[chet.rowles]

Well done. Excellent counter to all the marketing hype around "reducing unsprung weight". And this is a major weight decrease of 8lbs a corner. Despite what the manufacturers and vendors will tell you, dropping a few pounds a corner on a car that weighs 2 tons is just noise.

I disagree. .14 second difference is fair, and over a quarter mile the delta would theoretically double. Plus, my current wheels/tyres alone are 11lbs lighter than stock setup, same sizes.
People have paid more $$ for .3 seconds.

Sent from my SAMSUNG-SM-G900A using Tapatalk

[Victor Newman]

Great test, and thanks for sharing. I'd say that's a pretty noticeable difference.

I'd like to know what the moments of inertia are of each setup. Vertical weight gives an indicator of which would perform better within a "normal" range, but the actual inertia info would be great to know.

Weight is telling half the story. 1) So you have to accelerate that mass in the direction the car is going (think of the wheel not actually spinning), but 2) you also have to use power to spin the wheel (inertia).

[Race Shooter]

Well done. Excellent counter to all the marketing hype around "reducing unsprung weight". And this is a major weight decrease of 8lbs a corner. Despite what the manufacturers and vendors will tell you, dropping a few pounds a corner on a car that weighs 2 tons is just noise.

Actually unsprung weight has little to do with acceleration or speed, its more to do with handling, so your attempt at trying to belittle unsprung weight holds no merit and shows you understand little of car dynamics.

Rotational weight has everything to do with this test.

[eurotic]

Hey guys, I'm glad this has generated a bit of a discussion and I'll add in some details that some are asking.

According to Tirerack the revolutions per mile for my summer tire is 801 and for the winter tire it is 803. That said the summer have a lot less tread on them which theoretically would put them closer to being the same revs per mile. That said there are a lot of variables because for sure they are completely different compounds, the summer tire is a 255 and the winter is a 225, the temperature maybe played a factor but it was kind of equally bad for each tire, a little cool for the summer and a little warm for the winter.

More wheel/tire detail:
The summer tires are 255/35R19 that weigh 28 lbs. The winters are 225/45R18 that weigh 25 lbs. So I think that the most significance is that 3 lbs out of the tire as it is the furthest from the center of the hub. And of course the summer wheel is 19" version the 18" winter but I would love to see some sort of test that could really illustrate the energy required to start/stop the extra mass.

Summer setup is only lighter because I have yet to buy new wheels. I haven't bought new wheel partially because of the question of how important is wheel weight. I also am struggling with tire width which plays into who wide of a wheel I should get. The reason I did this test was so that over winter I can stew about it some more and my test helps add a small amount of perspective. Sorry this is about to go on a bit of a tangent...

The width of a wheel, from what my autocross friends say, makes a marked improvement in tire performance. I have read/heard that to get the best performance out of a tire, the wheel should be the same width up to 0.5" wider than the tread width. That means my 9.4" wide RE-11's should ideally go on a 9.5-10" wheel not the 8.5" wheel I have now. But what does that really mean? No one has been able to give me a solid answer but suggest that the tire will be more responsive. But will that translate into lower times at autocross? Will the tires handle heat a little better (i.e. not heat up as much because they will have better sidewall support)?

So my current dilemma, which combo will give me the best results:
All combos using and RE-71R tire in either a 255/35 or 265/35
1 - 9.3" tread on a 9.0" wheel, total weight 46.5 lbs (11.9 lbs lighter than current) - pro: lightest - con: tread wider than wheel
2 - 9.6" tread on a 9.0" wheel, total weight 47.5 lbs (10.9 lbs lighter than current) - pro: wider tire and lighter than other combos - con: tread is even wider than wheel
3 - 9.3" tread on a 9.5" wheel, total weight 48.5 lbs (9.9 lbs lighter than current) - pro: slightly lighter, tread narrower than wheel - con: a little heavier than other combos
4 - 9.6" tread on a 9.5" wheel, total weight 49.5 lbs (8.9 lbs lighter than current) - pro: wider tire - con: tread slightly wider than wheel but it is close, heaviest combination

But I can't answer my question as I have no way to test this as I can't afford to go buy all these combos and do back to back testing, nor have I found anyone that has first hand experience with this.

[bhvrdr]

What was the approx DA for summer vs winter conditions? Id guess thats most of the affect. Ill try and find the article that did controlled testing with several sizes/weights. Made a tiny difference.

EDIT: Here it is. And this was on a little "momentum car" ..

http://www.caranddriver.com/features...d-tires-tested


edit again: i reread the op. Got it. Cool data. Just as an fyi the vcds timestamps are not in seconds though. I do third gear pull comparisons from 2krpm to redline all the time when testing parts and noticed a .6 "unit" difference was really worth about .2 sec.

Mike

[Victor Newman]

So there are three main differences:

1) your winter setup is overall lighter than your summer setup
2) your winter setup is 18" and your summer setup is 19", so the wheel barrel on the winter setup is closer to the center of rotation (distance is squared, mass is not)
3) your winter setup tire is lighter and skinnier than your summer setup, which is farthest from the center of rotation, and would have a larger difference (again, distance is squared, mass is not)

You add all three up, and sure, you've got a difference in vehicles acceleration in a straight line, but you've got likely a bigger benefit in the suspension and steering being able to move that mass more quickly as race shooter is referring to. Like wearing light cleats versus boots.

[S4_Pilot]

Great article here outlining the benefits. I felt a huge difference in handling at least in going to a lightweight wheel especially on the track. My car feels faster as well but not data to back things up.

http://www.focusst.org/forum/attachm...wheel-test.pdf

[phanker]

Some excellent info in that link. Thanks for posting.
Good to confirm my butt dyno doesn't need any recalibration either.

Great article here outlining the benefits. I felt a huge difference in handling at least in going to a lightweight wheel especially on the track. My car feels faster as well but not data to back things up.

http://www.focusst.org/forum/attachm...wheel-test.pdf

[eurotic]

What was the approx DA for summer vs winter conditions? Id guess thats most of the affect. Ill try and find the article that did controlled testing with several sizes/weights. Made a tiny difference.

EDIT: Here it is. And this was on a little "momentum car" ..

http://www.caranddriver.com/features...d-tires-tested


edit again: i reread the op. Got it. Cool data. Just as an fyi the vcds timestamps are not in seconds though. I do third gear pull comparisons from 2krpm to redline all the time when testing parts and noticed a .6 "unit" difference was really worth about .2 sec.

Mike

Hey Mike,

I am not sure what the DA was but I can provide some clarity. I originally tested my winters then changed to summers and did the tests about 45 minutes apart. When I got home and looked at the logs the data wasn't as high resolution as I wanted as I was logging too many parameters and only getting 1 reading per second (even with turbo mode running). So I went back out on my summer set and recorded just RPM to get my 6 runs. Admittedly I couldn't go back out on my winters until 24 hours later when it was very similar conditions.

From the weather station at the airport right near where I was testing here is the data for the days and times I did the test:
November 7 @ 16:00
Temp 8.3 C
Dew Point 4.9 C
Relative humidity 79%
Pressure 99.28 kPa

November 8 @ 16:00
Temp 8.5 C
Dew Point 3.6 C
Relative humidity 71%
Pressure 99.77 kPa

I know I will have to go back and try again in better controlled conditions but hopefully this doesn't completely invalidate the results.

Great article here outlining the benefits. I felt a huge difference in handling at least in going to a lightweight wheel especially on the track. My car feels faster as well but not data to back things up.

http://www.focusst.org/forum/attachm...wheel-test.pdf

See and this is more what I was expecting to see, half a second on a 0-60mph run but maybe the difference is that the S4 has so much more torque. I suspect the reduction of weight might help with traction over uneven surfaces even in our heavy cars.

I wish I could tell a big difference in feel. I think I feel a difference but it could be just placebo.

[db12]

So I was bored the other day and the weather was surprisingly nice for where I live for the time of year (10 degrees Celsius in November) so I thought I'd do a very simple test comparing my summer wheel/tire setup to my winter wheel/tire setup so see what sort of quantifiable data I could gather as there is a decent amount of difference in weight. Please understand that this is not an exhaustive test and I was on public roads so I tried to keep it legal.

I'm sharing this to see if anyone might be interested and hopefully I don't get torn apart in the process.

For testing I was in my 2012 S4 with 6MT using my VCDS in turbo mode to just record engine speed. The simple test was to get the car into 2nd gear at idle and then do a WOT pull until the rev limiter which is about 10kph to about 112kph indicated. My summer setup is a set of stock peelers with RE-11's on them weighing 58.4 lbs and my winter setup is a set of VMR 701's with WS-80's that weigh 50.5 lbs which is a weight difference of about 7.9 lbs.

I did 6 runs on each tire with idle around 700 rpm and rev limiter was just over 7200 rpm. I dumped the data into Excel and normalized it to get common reference points (1000rpm, 2000, rpm, ..., 7000 rpm) adjusting the timstamps so that 0 seconds started at 1000 rpm so that I could see how long it took to accelerate to 7000 rpm.

Again, I know this isn't a perfect test but it is a real world test and might show something...and it did I think. Here is the data.

Winter Tires

RPM	Run 1	Run 2	Run 3	Run 4	Run 5	Run 6	Avg.
1000	0.00	0.00	0.00	0.00	0.00	0.00	0.00
2000	1.11	1.09	1.13	1.15	1.12	1.06	1.11
3000	2.04	2.04	2.03	2.06	2.06	2.07	2.05
4000	2.96	2.91	2.88	2.91	2.90	2.93	2.92
5000	3.81	3.75	3.85	3.77	3.78	3.78	3.79
6000	4.74	4.73	4.83	4.81	4.76	4.78	4.78
7000	5.91	5.93	5.92	5.91	5.93	6.02	5.94

Summer Tires

RPM	Run 1	Run 2	Run 3	Run 4	Run 5	Run 6	Avg.	Diff.
1000	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00
2000	1.13	1.12	1.11	1.12	1.19	1.23	1.15	0.04
3000	2.09	2.05	2.06	2.06	2.11	2.19	2.09	0.04
4000	3.01	2.97	3.04	3.01	2.97	3.07	3.01	0.09
5000	3.86	3.92	3.92	3.90	3.91	3.96	3.91	0.12
6000	4.82	4.89	4.85	4.87	4.89	4.93	4.87	0.10
7000	6.03	6.01	6.11	6.13	6.04	6.15	6.08	0.14

You can see that 5 of the 6 runs on the winter tires were in the 5.9 second range with only one run being in the 6.0 second range. The summer tire test was less precise as 3 of the runs were in the 6.0 second range and 3 were in the 6.1 second range. I did a bunch of different combinations of runs that I won't share here (like throwing out the best and worst runs, using just the best runs, using just the worst runs, comparing the best to worst for each, etc). For this post I simply averaged all the results and compared the averages...not the best scientific method but it at least gives us something to start with.

As you can see from the data above it looks like at the end of the ~6 second run the lighter winter tire setup was a little over 0.1 seconds faster. For some reason I was expecting more based on how many people on this forum talk about how dramatic it feels to shed the weight from wheels and tires. However as I thought about it more I was thinking how this might add up in a 1/4 mile run or around a track.

I would love to do a more extensive test but I think that would require entirely different equipment to make it valid. Anyway, there it is, I'd love to hear what anyone has to say especially if anyone else have dared do the same but was too shy to post here. I have my flame suit on for safety at this point.


Shawn

The real comparison would be to then add a passenger to your car during the tests. The ongoing debate is how unsprung weight equates to total weight etc. Is the 1 lb of unsprung weight really equal to 10 to 20 lbs of total weight etc....

[Victor Newman]

The real comparison would be to then add a passenger to your car during the tests. The ongoing debate is how unsprung weight equates to total weight etc. Is the 1 lb of unsprung weight really equal to 10 to 20 lbs of total weight etc....

There is no way to come up with a number for that, as it's dependent on 1) weight of the wheel, AND 2) inertia of the wheel (where that weight is relative to the center). There is no one number

[mr shickadance]

There is no way to come up with a number for that, as it's dependent on 1) weight of the wheel, AND 2) inertia of the wheel (where that weight is relative to the center). There is no one number

yes and no.

to one side - you could run the same test with 200lbs in the back seat, and 200lbs not in the back seat and compare your results against OP's first test.

the difference would be 200lbs of static weight against 32lbs of rotational mass.

now I get that when you dive into specifics 32lbs does NOT translate 100% as you have the weight placement, etc. but, you certainly have the comparison of how much of a difference static vs rotational.

[db12]

There is no way to come up with a number for that, as it's dependent on 1) weight of the wheel, AND 2) inertia of the wheel (where that weight is relative to the center). There is no one number

Right its not perfect science but by adding overall weight to the vehicle by adding a passenger and see if your test results on say the lighter wheels in this case resulted in equal measurements to the heavier summer tire result would show the comparison

[bhvrdr]

Here are some old posts I used to reference for possibly getting close to calculating unsprung weight loss from wheels....


Article 1

http://www.audiworld.com/tech/wheel13.shtml

Effect of Wheel Mass on Acceleration Eduardo Martinez 2001
What is the effect of a change in wheel/tire mass? Several people have quoted factors of x6 or x8 to determine the equivalent non-rotating mass variation. That is, an increase of 10lbs/wheel could be as detrimental to performance as carrying Arnold Schwarzenegger and Scooby Doo as passengers.

I decided to refresh my Physics and came up with the following conclusion: from the point of view of acceleration, an increase of X in wheel or tire mass is no worse than an increase of 2X in passenger mass. Not 6x, not 8x, just 2x worst case. This is why.

At any given speed/gear combination there is maximum torque T available at the wheels. The torque does two things: (1) opposes the rolling resistance and the aerodynamic drag and (2) accelerates the car. The equation that describes the equilibrium is:

T = I*u + M*r*a + D*r

*: denotes multiplication
T: torque at the wheels
I: total moment of inertia of the rotating parts
u: angular acceleration of the rotating parts
M: total mass of the vehicle
r: external radius of the tires
a: linear acceleration
D: total drag and rolling resistance

The first term, I*u, is the torque used in making the wheels rotate faster. The second, M*r*a, is the torque used to accelerate everything (wheels included) in the direction of movement. The third is the torque used to cancel the drag and rolling resistance.

Another way of writing the equation above is:

T/r = I*u/r + M*a + D

Now the terms are forces. The left side is the force available at the contact patch.

If the tires are not slipping, the angular acceleration and the linear acceleration are related in the following way:

a = u*r

Replacing in the equation and moving things around, we get

T/r - D = (I/r^2 + M)*a

^: denotes exponentiation (r^2 means "r squared")

We can say that the left side is the force available for acceleration. Such force accelerates an "non-rotating equivalent mass" E,

E = I/r^2 + M

Now suppose that we increase the mass of the wheels and tires by an amount X. Both the total mass and the moment of inertia will increase; let's call the new mass M' and the new moment of inertia I'. Obviously,

M' = M + X

What about I'? Well, the moment of inertia of a "punctual mass" [m] (a mass concentrated in a point) at a distance [r] from the axis of rotation is [m*r^2]. That is,
the moment of inertia depends critically on the distance between the mass and the axis of rotation.

In a real wheel+tire combination the mass is distributed in different amounts at different
distances from the center. In order to compute the total moment of inertia we would need to know the mass distribution and use integral calculus. We can do a simpler thing though. We can make the pessimistic assumption that all of the mass increment is
located on the periphery of the tire, that is, at a distance [r] from the center. This assumption is pessimistic because in a real wheel some of the mass will be located closer that [r] and will contribute less to the total momentum (it is not too pessimistic though: most of the mass is located pretty far from the center, if not at the periphery). So now we can compute I',

I' = I + X*r^2

The new "equivalent mass" is,

E' = I'/r^2 + M' = I/r^2 + M + 2*X = E + 2*X

In other words, from the acceleration point of view, the equivalent non-rotating mass increment corresponding to an increment X in rotating mass is - at worst - 2X.

NOTE: After doing some approximations and assumptions about mass distribution in a typical wheel+tire combo, I believe that 1.7X is a better approximation. A 10lb/wheel mass increase would not hurt acceleration worse than carrying RinTinTin.



Article 2

http://www.bimmerforums.com/forum/sh...525#post766525


Assumptions:
1. M3 Curb Weight (except wheels): 2950 lbs

2. M3 Peak Wheel Force in 1st Gear: 2830 lbs (actually doesn't really matter, but just to make realistic numbers. I'm assuming stock wheel sizes and gearing, and 200 lb/ft pk torque after drivetrain losses)

3. All wheels same size, running 225/45-17" tires. Wheel+tire radius R=1.04ft.

4. No friction forces (wheel bearings, etc)

5. No wheel slip

6. Wheels+tires have evenly distributed mass (uniform disc).


Math:
Angular acceleration of wheel = aa.
Car acceleration = A = aa*R

F (tire-to-ground) = mass(M3) * A + I(wheels)*aa
= mass(M3) * A + I(wheels)*(A/R)
= (mass(M3) + I(wheels)/R) * A

Or otherwise

A = F / (mass(M3) + I(wheels)/R)


Moment of Inertia of 16lb wheel + 20 lb tire:
I = mr^2 = (36/3.215)*(1.04ft^2) = 12.111

Moment of Inertia of 24lb wheel + 25 lb tire:
I = mr^2 = (49/3.215)*(1.04ft^2) = 16.484

Mass of M3 = (curb weight) / G
With 16lb wheel combo: = (2950+ (36*4)) / 3.215 = 962.36
With 24lb wheel combo: = (2950+ (49*4)) / 3.215 = 978.54


Results:

Acceleration with 16lb wheel+20 lb tire combo
= 2830/(962.36 + 4*(12.111)/1.04)
= 2.804 ft/sec^2 = 0.872g

Acceleration with 24lb wheel+25 lb tire combo
= 2830/(978.54 + 4*(16.484)/1.04)
= 2.716 ft/sec^2 = 0.845g


Conclusions:

With the assumptions used, lighter wheel+tire combo gives 3.1% acceleration improvement. Equivalent to dropping a 6.2s 0-60 to 6.0s.

If stock M3 is then 3146 lbs (24lb wheel+25 lb tire combo) then 3.1% weight reduction is 97.53 lbs. Actual weight reduction was 52 lbs. So in this case, 1lb of removed rotational weight is equivalent to removing 1.87 lbs from the body. That's not even close to the 7lb number tossed around. Though the assumption of even distribution of wheel mass does skew this number down some.



Any math/physics nerds want to check these guys work?

Mike

[B8_Jim]

Here are some old posts I used to reference for possibly getting close to calculating unsprung weight loss from wheels....

I got a factor of about 2x making a bunch of conservative assumptions on where the weight increase was located (i.e. all out very near the tread of the tire)...

http://www.audizine.com/forum/showth...=1#post7888404

[eurotic]

Here are some old posts I used to reference for possibly getting close to calculating unsprung weight loss from wheels....


Article 1

http://www.audiworld.com/tech/wheel13.shtml

Effect of Wheel Mass on Acceleration Eduardo Martinez 2001
What is the effect of a change in wheel/tire mass? Several people have quoted factors of x6 or x8 to determine the equivalent non-rotating mass variation. That is, an increase of 10lbs/wheel could be as detrimental to performance as carrying Arnold Schwarzenegger and Scooby Doo as passengers.

I decided to refresh my Physics and came up with the following conclusion: from the point of view of acceleration, an increase of X in wheel or tire mass is no worse than an increase of 2X in passenger mass. Not 6x, not 8x, just 2x worst case. This is why.

At any given speed/gear combination there is maximum torque T available at the wheels. The torque does two things: (1) opposes the rolling resistance and the aerodynamic drag and (2) accelerates the car. The equation that describes the equilibrium is:

T = I*u + M*r*a + D*r

*: denotes multiplication
T: torque at the wheels
I: total moment of inertia of the rotating parts
u: angular acceleration of the rotating parts
M: total mass of the vehicle
r: external radius of the tires
a: linear acceleration
D: total drag and rolling resistance

The first term, I*u, is the torque used in making the wheels rotate faster. The second, M*r*a, is the torque used to accelerate everything (wheels included) in the direction of movement. The third is the torque used to cancel the drag and rolling resistance.

Another way of writing the equation above is:

T/r = I*u/r + M*a + D

Now the terms are forces. The left side is the force available at the contact patch.

If the tires are not slipping, the angular acceleration and the linear acceleration are related in the following way:

a = u*r

Replacing in the equation and moving things around, we get

T/r - D = (I/r^2 + M)*a

^: denotes exponentiation (r^2 means "r squared")

We can say that the left side is the force available for acceleration. Such force accelerates an "non-rotating equivalent mass" E,

E = I/r^2 + M

Now suppose that we increase the mass of the wheels and tires by an amount X. Both the total mass and the moment of inertia will increase; let's call the new mass M' and the new moment of inertia I'. Obviously,

M' = M + X

What about I'? Well, the moment of inertia of a "punctual mass" [m] (a mass concentrated in a point) at a distance [r] from the axis of rotation is [m*r^2]. That is,
the moment of inertia depends critically on the distance between the mass and the axis of rotation.

In a real wheel+tire combination the mass is distributed in different amounts at different
distances from the center. In order to compute the total moment of inertia we would need to know the mass distribution and use integral calculus. We can do a simpler thing though. We can make the pessimistic assumption that all of the mass increment is
located on the periphery of the tire, that is, at a distance [r] from the center. This assumption is pessimistic because in a real wheel some of the mass will be located closer that [r] and will contribute less to the total momentum (it is not too pessimistic though: most of the mass is located pretty far from the center, if not at the periphery). So now we can compute I',

I' = I + X*r^2

The new "equivalent mass" is,

E' = I'/r^2 + M' = I/r^2 + M + 2*X = E + 2*X

In other words, from the acceleration point of view, the equivalent non-rotating mass increment corresponding to an increment X in rotating mass is - at worst - 2X.

NOTE: After doing some approximations and assumptions about mass distribution in a typical wheel+tire combo, I believe that 1.7X is a better approximation. A 10lb/wheel mass increase would not hurt acceleration worse than carrying RinTinTin.



Article 2

http://www.bimmerforums.com/forum/sh...525#post766525


Assumptions:
1. M3 Curb Weight (except wheels): 2950 lbs

2. M3 Peak Wheel Force in 1st Gear: 2830 lbs (actually doesn't really matter, but just to make realistic numbers. I'm assuming stock wheel sizes and gearing, and 200 lb/ft pk torque after drivetrain losses)

3. All wheels same size, running 225/45-17" tires. Wheel+tire radius R=1.04ft.

4. No friction forces (wheel bearings, etc)

5. No wheel slip

6. Wheels+tires have evenly distributed mass (uniform disc).


Math:
Angular acceleration of wheel = aa.
Car acceleration = A = aa*R

F (tire-to-ground) = mass(M3) * A + I(wheels)*aa
= mass(M3) * A + I(wheels)*(A/R)
= (mass(M3) + I(wheels)/R) * A

Or otherwise

A = F / (mass(M3) + I(wheels)/R)


Moment of Inertia of 16lb wheel + 20 lb tire:
I = mr^2 = (36/3.215)*(1.04ft^2) = 12.111

Moment of Inertia of 24lb wheel + 25 lb tire:
I = mr^2 = (49/3.215)*(1.04ft^2) = 16.484

Mass of M3 = (curb weight) / G
With 16lb wheel combo: = (2950+ (36*4)) / 3.215 = 962.36
With 24lb wheel combo: = (2950+ (49*4)) / 3.215 = 978.54


Results:

Acceleration with 16lb wheel+20 lb tire combo
= 2830/(962.36 + 4*(12.111)/1.04)
= 2.804 ft/sec^2 = 0.872g

Acceleration with 24lb wheel+25 lb tire combo
= 2830/(978.54 + 4*(16.484)/1.04)
= 2.716 ft/sec^2 = 0.845g


Conclusions:

With the assumptions used, lighter wheel+tire combo gives 3.1% acceleration improvement. Equivalent to dropping a 6.2s 0-60 to 6.0s.

If stock M3 is then 3146 lbs (24lb wheel+25 lb tire combo) then 3.1% weight reduction is 97.53 lbs. Actual weight reduction was 52 lbs. So in this case, 1lb of removed rotational weight is equivalent to removing 1.87 lbs from the body. That's not even close to the 7lb number tossed around. Though the assumption of even distribution of wheel mass does skew this number down some.



Any math/physics nerds want to check these guys work?

Mike

Mike, as always, you have something awesome to contribute.

I actually threw the calculations in a spreadsheet so I could substitute numbers and the results were pretty cool.
I plugged in 3900 for the weight of the car without wheels and tires, the actual weights of my different wheels and tire, and the time my car took on the summer tires as the acceleration time (I took the best time of 6.01 seconds). The result said that the calculated time for my winter setup would be 5.92 seconds...that's damn close to my measured values. I wonder if it is just coincidence? I think it is good enough for me to do some more guesstimating on what wheel/tire package I should go for.

[Victor Newman]

yes and no.

to one side - you could run the same test with 200lbs in the back seat, and 200lbs not in the back seat and compare your results against OP's first test.

the difference would be 200lbs of static weight against 32lbs of rotational mass.

now I get that when you dive into specifics 32lbs does NOT translate 100% as you have the weight placement, etc. but, you certainly have the comparison of how much of a difference static vs rotational.

No, my point was that db12 is talking about a ratio of rotational weight translated to weight. Of course removing 1 lb. from the wheel removes 1 lb. of weight. The difference that he's referring to is ALL rotational, which is solely determined by inertia. Because every single tire and wheel combination would have different inertia values, there is no "common number" that could be generally applied that would be reasonably accurate.

What you say is true, it just doesn't address my point. The OP's experiment could be run with added weight in the car, but that would ONLY apply to the wheel and tire change that he made specifically, and it would not apply to ANY other combination, even if the first and second wheel/tire weights were different by the same weight.

[Victor Newman]

Couldn't the analysis be simplified by just doing a ratio of torque = inertia x angular acceleration? If applied shaft torque stays the same, you could set inertia 1 x angular acceleration 1 = inertia 2 x angular acceleration 2 to find out the ratio. Convert it to linear acceleration using the wheel radius, and then applying that to the wheel and tire weight difference.

For example, when I did that using specs of a stock peeler and PSS tire, and then going to a Vorsteiner VFF106 (5.5 lb. difference) and PSS tire, I got a 1.16 ratio for linear acceleration, which only made a 22 lb. wheel weight loss (sum of all four wheels) perform like a 25.6 lb. standard weight loss. Makes sense that the 60% estimation from the articles above would be decreased if only the wheel was changed, assuming its weight loss is closer to the center of the wheel, the "additional" benefit from lower rotational inertia would only feel like another 16% of an improvement. The tire kept the total inertia high - improvement in inertia was only 7%).

I could be wrong though. Just thinking out loud

[mr shickadance]

No, my point was that db12 is talking about a ratio of rotational weight translated to weight. Of course removing 1 lb. from the wheel removes 1 lb. of weight. The difference that he's referring to is ALL rotational, which is solely determined by inertia. Because every single tire and wheel combination would have different inertia values, there is no "common number" that could be generally applied that would be reasonably accurate.

What you say is true, it just doesn't address my point. The OP's experiment could be run with added weight in the car, but that would ONLY apply to the wheel and tire change that he made specifically, and it would not apply to ANY other combination, even if the first and second wheel/tire weights were different by the same weight.

but he wasn't.

he was talking about static weight versus rotational weight, and getting a comparison of the two, which is possible.

if you shave off 40lbs of rotational weight (saved from lighter wheels/tires) and compare it against dropping 200lbs of static weight, and the results are the same, you can loosely argue (going back to your point) or draw a conclusion that there is a 1:5 ratio, and then you can go from there.

is it easier to shave off 200lbs of static weight, or run lighter wheels? or do both?

Either way, I think you are reading into db's comment too far and missing the point he was trying to make.

[db12]

but he wasn't.

he was talking about static weight versus rotational weight, and getting a comparison of the two, which is possible.

if you shave off 40lbs of rotational weight (saved from lighter wheels/tires) and compare it against dropping 200lbs of static weight, and the results are the same, you can loosely argue (going back to your point) or draw a conclusion that there is a 1:5 ratio, and then you can go from there.

is it easier to shave off 200lbs of static weight, or run lighter wheels? or do both?

Either way, I think you are reading into db's comment too far and missing the point he was trying to make.

Correct as most of the threads discussing rotational weight drops are trying translate that to a static weight drop. Maybe my terminology was off but it seems in all the threads I have seen that translated number is all over the board from 1lb rotational weight drop = 5Lbs static or 10 lbs static to 20 lbs static weight etc

For people that track their vehicles this is crucial as having a 200 lb passenger in your car vs solo translates to a significant time decrease. I dropped almost 7 lbs per tire with my track set up using 18" light weight rims and lighter rotors. I have been trying to get a rough estimate of the static weight loss equivalent.

[SportVier]

As far as I can tell, all all of this discussion is only about forward linear acceleration after initial weight transfer (rolling start). So maybe somewhat relevant if you are really focused on measuring "strip" times down to tenths of a second. But once you complete the picture and bring in weight transfer, braking, lateral acceleration, and ultimate grip (in other words, "actual driving"), the relationships between rotational, static, sprung and unsprung masses get much more complicated and murky. And when you start with a total sprung mass somewhere far north of 1.5 tons... Let's just say that will tend to dominate the conversation.

[Victor Newman]

but he wasn't.

he was talking about static weight versus rotational weight, and getting a comparison of the two, which is possible.

if you shave off 40lbs of rotational weight (saved from lighter wheels/tires) and compare it against dropping 200lbs of static weight, and the results are the same, you can loosely argue (going back to your point) or draw a conclusion that there is a 1:5 ratio, and then you can go from there.

is it easier to shave off 200lbs of static weight, or run lighter wheels? or do both?

Either way, I think you are reading into db's comment too far and missing the point he was trying to make.

Yes he was. He used the word "debate" which means relating to more than one person or situation, and you're doing the same thing above. Of course you can run the experiment for one specific setup, but it doesn't contribute any accurate information for anyone else unless they're making the exact same tire and wheel change (if inertias change by the same exact amount).

If the OP ran a test on removing X lbs. of rotating weight from the wheels, and added Y weight to the car and the acceleration results were even, you can get a ratio from that, and it's accurate for that one setup change. But it doesn't mean that same weight ratio (rotational to static) can be applied to any other setup accurately "and go from there," because the inertia is the real differentiating factor, not weight. That's why I said in a previous post that of course removing 1 lb. from a wheel is equivalent to removing 1 lb. from the car, but beyond that the additional amount between rotational and static (what everyone is debating about) is related to inertia only.

If you could use published inertia specs, you COULD do what you and db12 are suggesting. But since all we're given is weight, you can't. So you can't "go from there." If you could, that one number or small range of values would be posted all over the internet and would be common knowledge. But it isn't.

For example you've got two lightweight flywheels. One is 18 lb. and another is 20 lb. Which one will spin up faster? The answer is "not enough information." To say that the 18 lb. flywheel spins up 11% faster (20 lb. / 18 lb.) is not correct. It is quite possible that the 20 lb. flywheel spins up faster than the lighter one. That's why you can't use a weight ratio accurately in the wheel example above - relating rotational to static is only due to inertia, and since weight does not give that piece of information, you can't use a "remove 1 lb. of wheel weight equals 1.6 lb. of weight out of the car" weight ratio to predict how another wheel/tire setup would perform.

[Victor Newman]

As far as I can tell, all all of this discussion is only about forward linear acceleration after initial weight transfer (rolling start). So maybe somewhat relevant if you are really focused on measuring "strip" times down to tenths of a second. But once you complete the picture and bring in weight transfer, braking, lateral acceleration, and ultimate grip (in other words, "actual driving"), the relationships between rotational, static, sprung and unsprung masses get much more complicated and murky. And when you start with a total sprung mass somewhere far north of 1.5 tons... Let's just say that will tend to dominate the conversation.

I think the answer is pick the best tire for grip, and then use the lightest wheel that you like. In my example, the tire influences the inertia value far more than the wheel, so it's a smaller return by just changing wheels. My conclusion is that yes, lighter wheels are better than heavier wheels for performance, but not by the amount I would have thought before.

[SportVier]

My conclusion is that yes, lighter wheels are better than heavier wheels for performance, but not by the amount I would have thought

... or by an amount that you are likely to feel from the driver's seat.

[mr shickadance]

Yes he was. He used the word "debate" which means relating to more than one person or situation, and you're doing the same thing above. Of course you can run the experiment for one specific setup, but it doesn't contribute any accurate information for anyone else unless they're making the exact same tire and wheel change (if inertias change by the same exact amount).

If the OP ran a test on removing X lbs. of rotating weight from the wheels, and added Y weight to the car and the acceleration results were even, you can get a ratio from that, and it's accurate for that one setup change. But it doesn't mean that same weight ratio (rotational to static) can be applied to any other setup accurately "and go from there," because the inertia is the real differentiating factor, not weight. That's why I said in a previous post that of course removing 1 lb. from a wheel is equivalent to removing 1 lb. from the car, but beyond that the additional amount between rotational and static (what everyone is debating about) is related to inertia only.

If you could use published inertia specs, you COULD do what you and db12 are suggesting. But since all we're given is weight, you can't. So you can't "go from there." If you could, that one number or small range of values would be posted all over the internet and would be common knowledge. But it isn't.

For example you've got two lightweight flywheels. One is 18 lb. and another is 20 lb. Which one will spin up faster? The answer is "not enough information." To say that the 18 lb. flywheel spins up 11% faster (20 lb. / 18 lb.) is not correct. It is quite possible that the 20 lb. flywheel spins up faster than the lighter one. That's why you can't use a weight ratio accurately in the wheel example above - relating rotational to static is only due to inertia, and since weight does not give that piece of information, you can't use a "remove 1 lb. of wheel weight equals 1.6 lb. of weight out of the car" weight ratio to predict how another wheel/tire setup would perform.

lol wut?

so just by saying debate you can talk about whatever you want?

Debate!

Sandwiches - personally, i am a reuben guy


on a more serious side though, I still feel like as long as you put a disclaimer in that 1) weight is coming from the wheels, and 2) the tires are the same you can still run that loose ratio and get similar results from different wheel setups.

[Victor Newman]

lol wut?

so just by saying debate you can talk about whatever you want?

Debate!

Sandwiches - personally, i am a reuben guy


on a more serious side though, I still feel like as long as you put a disclaimer in that 1) weight is coming from the wheels, and 2) the tires are the same you can still run that loose ratio and get similar results from different wheel setups.

i'm directly addressing what db12 brought up, and i understand why you don't think so, no worries 👍

[Jay@JXB]

https://youtu.be/20AV1ugqdIE

I did a similar thing last winter. Here are the results in a format that's easy to see.


Sent from my iPhone using Tapatalk

[eurotic]

https://youtu.be/20AV1ugqdIE

I did a similar thing last winter. Here are the results in a format that's easy to see.


Sent from my iPhone using Tapatalk

Neat video and great results. I found it a little hard to see at times when the camera moved. You should do a side-by-side or one on top of another.

[Jay@JXB]

Neat video and great results. I found it a little hard to see at times when the camera moved. You should do a side-by-side or one on top of another.

Yeah I was having difficulty mounting the camera. Ended up going to the steering wheel, which go figure moves a lot even when you're trying to go straight...the time difference at the end is the real takeaway from this. Granted you'd never bang through gears from 2200 rpm to almost redline like this, but it just shows a measurable effect. I forget what the exact delta was; I should go through and pick ranges next time that relate to a quarter mile run. LWFW and clutch going in right now so I'll have another data point to add to this soon.


Sent from my iPhone using Tapatalk

[eurotic]

Yeah I was having difficulty mounting the camera. Ended up going to the steering wheel, which go figure moves a lot even when you're trying to go straight...the time difference at the end is the real takeaway from this. Granted you'd never bang through gears from 2200 rpm to almost redline like this, but it just shows a measurable effect. I forget what the exact delta was; I should go through and pick ranges next time that relate to a quarter mile run. LWFW and clutch going in right now so I'll have another data point to add to this soon.


Sent from my iPhone using Tapatalk

That would be great if you can do a before and after as I wold really love to see if there is a measurable difference assuming you can do the runs in similar conditions as I suspect it will take some time to get through the break-in period.

[Victor Newman]

I got a chance to chicken scratch some more, and got a bit more info that might be useful.

In the previous articles that were posted, it seems like similar conclusions are being reached of a "ratio" of wheel weight to car weight of 1:1.6 or so realistically and 1:2 theoretically. IF you're using the following assumptions of 1) the tire diameter does not change, and 2) the wheel diameter does not change (so you're still using a 26" tire and a 19" wheel for the S4, just changing those to different tires or wheels of the same size), you can get the same rough estimate of a wheel/tire weight comparison to static weight in the car by applying the same driveshaft torque to the wheel/tire and setting the torque equations equal to each other.

With static weight, you're dealing with one component of linear motion: force = mass X acceleration. With the wheel/tire, you have this same component, but you have an additional component of angular acceleration: torque = inertia X angular acceleration. So if you take 1 pound out of a wheel/tire or the car itself, you don't have to accelerate that 1 pound linearly (think about the wheel not spinning - just moving along with the car as a dead weight). But there is an additional torque needed to spin the wheel/tire as well.

Since the wheel/tire is spinning on a moving vehicle, it's inertia is going to resist acceleration or deceleration, and there is additional torque that is needed for this - torque that is going to be stored in spinning the wheel/tire instead of accelerating the car. It's also working in reverse - making the brakes work harder to stop the car.

Torque = inertia X angular acceleration, so, applying the same torque, and saying that wheel/tire 1 is heavier, and wheel/tire 2 is lighter, you get:

Inertia 1 X angular acceleration 1 = inertia 2 X angular acceleration 2

Which basically just says that inertia and angular acceleration counteract each other. Rearranging, you get:

Inertia 1 / inertia 2 = angular acceleration 2 / angular acceleration 1

So the ratio of the inertias is opposite to the ratio of angular acceleration. Now linear acceleration = angular acceleration X radius, so the right side of the equation above can be substituted with acceleration 2 / acceleration 1 with no other adjustment needed in this case. we're just looking for a ratio here

For a tire, inertia can be represented by the equation for a ring or hollow cylinder, which is just mass X radius ^2 (squared), and for a wheel, it's probably better to use the equation for a disk, which is 1/2 mass X radius ^2. But since the assumed radius of the tire or wheel does not change in this example as part of our assumptions, the ratio of the inertias is just 1 (the weights will change, but the ratio for our purposes is 1).

So what you end up with is a theoretical factor of 1:1 for linear acceleration, and a factor of 0:1 for angular acceleration, to end up with an overall theoretical factor of 1:2 for wheel/tire weight to car weight. It basically says that because you're spinning a tire, but not changing the radius, you've got a lighter tire that is easier to spin by a factor of 1:1. Because you're spinning a wheel, but not changing the radius, you've got a lighter wheel that is easier to spin by a factor of 1:1. Add that you don't have to accelerate that tire or wheel weight linearly either, and you end up with 1:2.

Actual numbers will change because the assumption that the weight distribution of the different wheels and different tires is not realistic, and the effecive radius will not be constant, will not cancel out, and will affect the end result, lowering the ratio below 1:2 in some cases, and raising it in others. That would also account for going from a 19" wheel to an 18" wheel, where the radius has a "squared" effect on the inertia, and would make a larger difference. Adversely for going from 19's to 20's too. It is possible to have a higher ratio than 1:2 when the radius changes.

That's what I've got so far, but I could be wrong. I'm an engineer, but this isn't my area of expertise. It just logically makes sense that you've got to move the wheel/tire from one place to another as a static weight, and you also have to use torque to spin it. Torque applied is the same, and because the only squared term in the equation is radius, which stays constant, inertias cancel out to just a 1:1 weight factor, and 0:1 for spinning "weight" to static weight for a total ratio of 1:2 of static weight to spinning weight.

For example, you switch from a Continental Conti Sport Contact 3 to a Michelin Pilot Super Sport, and the tire weight goes down 1 pound. Because they are the same size tire, the theoretical effect would be 1 pound removed from the tire would be like removing 2 pounds from the car. Same with the wheels. If you went from a stock peeler to a Vorsteiner VFF106, weight goes down 5.5 pounds and it should be equivalent to 11 pounds removed from the car. This is really rough, and does not apply when the radius of the tire or wheel changes (which would be the case even if you kept a 19" setup, but went with a wider wheel, for example). So it's a very narrow, rough estimate, but it gives a ballpark. Could be wrong though - anybody that can add to or correct any of the above is welcome to it.