Effect of unsprung/rotating mass

[B8_Jim]

So - looking at tires and noticed 2 of my top choices (Michelin PSS and Bridgestone S04) differ by 5 lbs. PSS = 24 lbs and the S04 is 29 lbs.

Wondering how much that extra mass might effect things - I calculated the effect on acceleration.

I assumed the weight difference to be out at 25" diameter (thinking most of the difference is in the tread and/or the structureright under the tread, not in the sidewall) and I get the result below for a 2nd gear pull. Does this look about right? Equivilant to losing 3 ft-lbs of torque at peak acceleration.

The rotational "loss" is all but the same as the "loss" due to just putting on 20 lbs of non-rotating weight, so it's equivalent to adding 40 lbs of non-rotating weight. That is about 1% of the car weight, and 3 ft-lbs is close to 1% of torque - so it seems reasonable.



For wheels ... assuming weight change would be near the barrel diameter - 5 lb heavier/lighter would result in about 1/2 of the rotational loss as the tires - a result of the weight difference moving from 25" in to 18" (barrel of my 19" wheels). Half the wheel rotational loss and same linear loss is around 2.25 ft-lbs. That's for 5 lb lighter wheels on all 4 corners.

For brake rotors, assuming 13" diameter - it's only about 0.2 ft-lb due to the rotational effect - so it's nearly all just the linear portion of the weight - totals about 1.7 ft-lbs for a set of 4 rotors that are 5 lbs lighter each.

If I haven't made any mistakes, the extra weight isn't likely to be a large enough difference in acceleration to influence my purchase, but what does 5 lbs on the tires/wheels do to handling? - no simple F=ma calculation that I know of for that...

Comments/thoughts?

[esz61]

So - looking at tires and noticed 2 of my top choices (Michelin PSS and Bridgestone S04) differ by 5 lbs. PSS = 24 lbs and the S04 is 29 lbs.

Wondering how much that extra mass might effect things - I calculated the effect on acceleration.

I assumed the weight difference to be out at 25" diameter (thinking most of the difference is in the tread and/or the structureright under the tread, not in the sidewall) and I get the result below for a 2nd gear pull. Does this look about right? Equivilant to losing 3 ft-lbs of torque at peak acceleration.

The rotational "loss" is all but the same as the "loss" due to just putting on 20 lbs of non-rotating weight, so it's equivalent to adding 40 lbs of non-rotating weight. That is about 1% of the car weight, and 3 ft-lbs is close to 1% of torque - so it seems reasonable.

For wheels ... assuming weight change would be near the barrel diameter - 5 lb heavier/lighter would result in about 1/2 of the rotational loss as the tires - a result of the weight difference moving from 25" in to 18" (barrel of my 19" wheels). Half the wheel rotational loss and same linear loss is around 2.25 ft-lbs. That's for 5 lb lighter wheels on all 4 corners.

For brake rotors, assuming 13" diameter - it's only about 0.2 ft-lb due to the rotational effect - so it's nearly all just the linear portion of the weight - totals about 1.7 ft-lbs for a set of 4 rotors that are 5 lbs lighter each.

If I haven't made any mistakes, the extra weight isn't likely to be a large enough difference in acceleration to influence my purchase, but what does 5 lbs on the tires/wheels do to handling? - no simple F=ma calculation that I know of for that...

Comments/thoughts?

Interesting ... but you lost me a little.

1) Are you just changing tires? You mention that at first but then you mention wheels and brakes.
2) The effect of changing the rotational mass is related to the change in mass as well as to the location of the additional mass relative to the axis of rotation, as I think you may have considered (you mention diameter but radius is the distance you're concerned with).
3) You state that 'the rotational "loss" is all but the same as the "loss" due to just putting on 20 lbs of non-rotating weight, so it's equivalent to adding 40 lbs of non-rotating weight". How did you calculate that? Generally speaking, adding rotational mass will have a greater negative effect than just adding that same mass elsewhere on the car.
4) Increasing the unsprung weight is always detrimental to the cars handling, particularly it's ability to respond to road imperfections.

Bottom line, the effect on performance of adding 20 lbs of tire weight to a 3800 lb car with 300+ hp will be negligible and not likely something that you would be able to notice.

[Info@EuroCode]

The change in rotational mass has both detriments and benefits. Here is a good read from tire rack http://www.tirerack.com/wheels/tech/...jsp?techid=108

[Jimminez]

On my old B6A4, when I had the 19" CH-R, I had an >10% difference in gas milage between the 28lb Bridgestone RE760 and the lighter Michelin Pilot Super Sports which were 23lb. Not taking in to consideration what kind of rolling resistance they had, the extra 5lb on the outside had a huge impact.

That's one of the reasons I won't be buying Bridgestones anymore.

[MrFunk]

I see you comparing two tires - one being 20lbs more than the other...

BUT - did you compare each tire to the one already on the car? That would be more of a beneficial comparrison. If each tire is less than your original than don't worry about which one is lightest - you are going from a heavier to lighter tire.

Also - with all your very precise and calculated approach - you did not factor in which tire may have more grip. Perhaps one will be essentically giving you 3lbs tq more than the other, but if it has less grip then who cares and you decision based on calculations may be all for nothing? Or maybe you get lucky and the lighter tired in fact has the better grip, you will not know until you buy them both and try them both and now we're just speculating making all that fancy calculations above worthless... :)

But I admire your approach

AKA - 5lbs difference on a 3k+ heavy car with over 350hp is not going to affect anything.

[esz61]

On my old B6A4, when I had the 19" CH-R, I had an >10% difference in gas milage between the 28lb Bridgestone RE760 and the lighter Michelin Pilot Super Sports which were 23lb. Not taking in to consideration what kind of rolling resistance they had, the extra 5lb on the outside had a huge impact.

That's one of the reasons I won't be buying Bridgestones anymore.

In all likelihood, the change in gas mileage that you experienced was related entirely to the difference in rolling friction between the two tires. Adding 20 lbs to a car is not going to impact the gas mileage.

[esz61]

I see you comparing two tires - one being 20lbs more than the other...

BUT - did you compare each tire to the one already on the car? That would be more of a beneficial comparrison. If each tire is less than your original than don't worry about which one is lightest - you are going from a heavier to lighter tire.

Also - with all your very precise and calculated approach - you did not factor in which tire may have more grip. Perhaps one will be essentically giving you 3lbs tq more than the other, but if it has less grip then who cares and you decision based on calculations may be all for nothing? Or maybe you get lucky and the lighter tired in fact has the better grip, you will not know until you buy them both and try them both and now we're just speculating making all that fancy calculations above worthless... :)

But I admire your approach

AKA - 5lbs difference on a 3k+ heavy car with over 350hp is not going to affect anything.

Calculations aren't speculating, they are calculations ... real numbers with real meaning.

[B8_Jim]

1) Are you just changing tires? You mention that at first but then you mention wheels and brakes.
2) The effect of changing the rotational mass is related to the change in mass as well as to the location of the additional mass relative to the axis of rotation, as I think you may have considered (you mention diameter but radius is the distance you're concerned with).
3) You state that 'the rotational "loss" is all but the same as the "loss" due to just putting on 20 lbs of non-rotating weight, so it's equivalent to adding 40 lbs of non-rotating weight". How did you calculate that? Generally speaking, adding rotational mass will have a greater negative effect than just adding that same mass elsewhere on the car.
4) Increasing the unsprung weight is always detrimental to the cars handling, particularly it's ability to respond to road imperfections.

1) & 2) I am changing tires which led to this - I calculated the torque it would take to accelerate that extra mass both in rotational motion, and in linear motion. While I had the spreadsheet, I looked wheels/brakes too, since I know those are other things that people change to save weight - I was just curious how much it affects things - you can see from the rotor #'s that since they are so close to the axle there is little required to spin them up - it's almost like putting weight in the trunk for acceleration. And yes, I used radius for the calculations.

3) The torque required to spin up the tires was about 1.5 ft-lbs at peak, and the torque required to accelerate that 20 lbs forward (linear) was also very close to 1.5 ft-lbs - the of 3.0 ft-lbs is the same as having 2x as much non-rotating added weight.

The change in rotational mass has both detriments and benefits. Here is a good read from tire rack http://www.tirerack.com/wheels/tech/...jsp?techid=108

Thanks for the link.

I see you comparing two tires - one being 20lbs more than the other...

BUT - did you compare each tire to the one already on the car?

I have Continental DW's on the car - same weight as the PSS. The Pilot super sport's test and are rated VERY highly, so there is no doubt they are better than the DW - they are also about $60 per tire more though. /shrug/ not a huge deal, but they are max perf summer tires (like the DW) and will likely wear out in 20-25k miles with my driving.

The RE970 are UHP all season, but also highly ranked and top of their test at tire-rack. $30 cheaper per tire and have a 40,000 mile treadlife warranty - including that, it might be effectively about $100 per tire difference. I wouldn't call what I did as "fancy calculations" - I guessed at a few numbers and multiplied/divided them appropriately in a spreadsheet to get a rough feel for the effect. I honestly didn't know if it would be .01% or 2+% when I started.

I'm leaning toward the PSS, but just wanted to look at all aspects. This simple calculation is just one piece - I learned something about the relative amount of torque required for rotational and linear acceleration and that for these cars it's fairly small. Unfortunately I can't test drive both tires before buying.

Again - thanks to everyone for the info/comments.

[Akatsuki...]

On my old B6A4, when I had the 19" CH-R, I had an >10% difference in gas milage between the 28lb Bridgestone RE760 and the lighter Michelin Pilot Super Sports which were 23lb. Not taking in to consideration what kind of rolling resistance they had, the extra 5lb on the outside had a huge impact.

That's one of the reasons I won't be buying Bridgestones anymore.

Thanks for pointing that out. I never realized how heavy those tires are. Super Sports must be a real bang for buck, light weight and good grip

[jfabes]

i have read on numerous other forums throughout my car modding life that the effect of unsprung weight translates into 6-8 lbs of static weight. i read a great thread about a year ago where some actual testing was done and they supported the results with calculations. there is no way that a 5lb lighter wheel results in the insignificant gains poseted above. if that were the case, wheel and tire weight wouldn't be as important as so many trackers and racers make it out to be. if you just use 6 lbs of static weight per 1 lb of unsprung weight, and looking at a wheel/tire combo that is 5 lbs lighter, that's 30 lbs/corner, or 120 lbs static. That is something that will reflect 1/4 mile times.

[A4 Centaur]

Can't wait to get BST Carbon Fiber Wheels for my Ducati!

[B8_Jim]

i have read on numerous other forums throughout my car modding life that the effect of unsprung weight translates into 6-8 lbs of static weight. i read a great thread about a year ago where some actual testing was done and they supported the results with calculations. there is no way that a 5lb lighter wheel results in the insignificant gains poseted above. if that were the case, wheel and tire weight wouldn't be as important as so many trackers and racers make it out to be. if you just use 6 lbs of static weight per 1 lb of unsprung weight, and looking at a wheel/tire combo that is 5 lbs lighter, that's 30 lbs/corner, or 120 lbs static. That is something that will reflect 1/4 mile times.

I've seen some of those, and I've also seen some posts saying that those numbers are bunk. I'm trying to get to the bottom of it. My math doesn't support 6-8 (I've seen as high as 10-20 claimed) - it shows 2. Why the difference?

For 0.5 G acceleration, you need 2.5 lbs to for 5 lbs (5lbs * 0.5). The B8 has a 26" diameter tire, or 13" radius = 1.08 ft - so 2.5 lbs at 1.08' is 2.7 ft-lbs - that's the "sprung" mass torque required.

At 0.5 G, the rotational acceleration is 2.36 revolutions/s^2, or 14.85 radians/s^2. The inertia of that 5 lbs at 25" diameter I=mr^2 - so (5/32.174)*(25/2/12)^2 = .1686. Torque required, T, is inertian * rotational acc. So .1686*14.85 = 2.5 ft-lbs.

Maybe I'm doing something wrong, but I can't see how you would get 6-8 lbs per pound. ? I show something slightly less than 2.

[Info@EuroCode]

Jim, your calculations appear to not have a coefficient for friction added, they are simply based upon weight. I think the friction coefficient as well as other rolling resistance nuances ( temperature of tire, temperature of wheel bearings and temperature of the substrate the tire is in contact with.

There are multiple slopes (nearly infinite) that would need to be taken into account.

[B8_Jim]

Jim, your calculations appear to not have a coefficient for friction added, they are simply based upon weight. I think the friction coefficient as well as other rolling resistance nuances ( temperature of tire, temperature of wheel bearings and temperature of the substrate the tire is in contact with.

There are multiple slopes (nearly infinite) that would need to be taken into account.

Not for a difference analysis, I don't think. Everything you mentioned would be very close to the same value and have nearly zero effect on torque required to maintain 0.5G of acceleration. Given that acceleration and no wheel spin, the substrate temp, etc are basically irrelevant.

I agree there are hundreds of things that will affect the actual acceleration value and you could throw them in here and obscure things - but the basic physics are pretty simple and there seems to be a lot of bad "facts" out there.

[Akrion]

AKA - 5lbs difference on a 3k+ heavy car with over 350hp is not going to affect anything.

Its a start :).

Lightweight tires + lightweight wheels + lightweight rotors + lightweight battery can give you weight savings of more than 80lbs. I do think that would be noticeable ... no?