Reducing Rotating Mass: Ditching Dead Weight [Archive] - Bimmerforums

M3formeM3foryou

03-01-2003, 03:43 AM

a lot of mechanical engineers who study here are car buffs, so this potential argument always comes up. one particular dynamics professor lays the issue to rest during lecture. the mechanic is right in that the polar moment of inertia will be decreased significantly by the change. so, say if your rear tires were jacked up off the ground, and you were seeing how quickly you could accelerate each set of wheels, the lighter ones would have a marked advantage. but, since you are still accelerating about 3000 pounds of car that is attached to the wheels, the wheels could even weigh nothing and the difference in acceleration given a constant force (ie-same engine) would be tiny.

this is very similar to the big debate over light flywheels. the big difference is that flywheels get accelerated to much higher speeds in far shorter periods of time. in this case, the weight loss is far, far more perceptible, and even then, when you get into 4th or 5th, the engine speed isn't changing fast enough for one to notice a big difference over stock.

M3formeM3foryou

03-01-2003, 03:55 AM

there are plenty more things that i left out, but one thing is if you decreased the overall diameter of your wheels, given the same torque to your wheels (same engine, once again), you will have more force transmitted to the ground, which equals increased acceleration.

this stuff is confusing, but the best way i can think to sum it up is that by minimizing rotational inertia, you are reducing the power you put into giving your wheels angular momentum (not turning slower--just reduced weight) and putting it toward the linear momentum of your car. but cars are so heavy and wheels so light in comparison that this won't do much.

EvanL

03-01-2003, 04:07 AM

You're also forgetting the advantages that lighter wheels have for your suspension. The lower unsprung rate will translate into less linear momentum in the y-plane, and thus allow your suspension to respond much quicker.

M3formeM3foryou

03-01-2003, 04:15 AM

of course. but the 7 pounds thing in terms of acceleration or handling doesn't make sense.

EvanL

03-01-2003, 04:26 AM

Well, not really, no...but I just thought it was necessary to point out that suspension is pretty much the only reason to buy lighter wheels.

I love my SSR Comps on the track...I believe they're part of the reason that my car seems to be set up so well.

frayed

03-01-2003, 10:26 AM

Originally posted by EvanL
Well, not really, no...but I just thought it was necessary to point out that suspension is pretty much the only reason to buy lighter wheels.

So, two things at play: reduction in polar moment of inertia helps acceleration, braking, turning. I think the delta in acceleration is not really that noticable, but braking and turning can be.

The second is reduction in unsprung weight and its benefits on suspension action. I haven't been able to discern any differences in how my car hooks up due to better suspension action.

KILLR B

03-01-2003, 11:01 AM

Do not understand "y planes", "polar moment of inertia" and "the delta in acceleration". All I know is the '99 M Coupe last night had no chance from 55 mph roll even with my SSR Integrals and SO3's. I have taken about 400 pounds out of the car which allows the car to accelerate, brake and dance infinitely better than it did!!:redspot :awink:

frayed

03-01-2003, 11:19 AM

Originally posted by KILLR B
Do not understand "y planes", "polar moment of inertia" and "the delta in acceleration". All I know is the '99 M Coupe last night had no chance from 55 mph roll even with my SSR Integrals and SO3's. I have taken about 400 pounds out of the car which allows the car to accelerate, brake and dance infinitely better than it did!!:redspot :awink:

LOL. Well, shit, that's what counts in the end! Good for you. :)

Def

03-01-2003, 11:47 AM

Originally posted by M3formeM3foryou
a lot of mechanical engineers who study here are car buffs, so this potential argument always comes up. one particular dynamics professor lays the issue to rest during lecture. the mechanic is right in that the polar moment of inertia will be decreased significantly by the change. so, say if your rear tires were jacked up off the ground, and you were seeing how quickly you could accelerate each set of wheels, the lighter ones would have a marked advantage. but, since you are still accelerating about 3000 pounds of car that is attached to the wheels, the wheels could even weigh nothing and the difference in acceleration given a constant force (ie-same engine) would be tiny.

this is very similar to the big debate over light flywheels. the big difference is that flywheels get accelerated to much higher speeds in far shorter periods of time. in this case, the weight loss is far, far more perceptible, and even then, when you get into 4th or 5th, the engine speed isn't changing fast enough for one to notice a big difference over stock.

I agree to a certain extent. Don't buy lightweight wheels expecting to now run 1 second faster in the quarter mile, but if you add up enough mole hills - you will eventually get a mountain. :)

An even BIGGER difference in getting lighter wheels is getting lighter tires. Really stiff tires like S03's are proverbial bricks compared to most other street tires. Losing 3-5lbs on the VERY outside edge of your tire will make more of a difference than probably losing 6-8lbs on your rim lip. Even then, most lightweight rims save it around the hub area, so the inertia of the body is usually very close to a seemingly much heavier wheel. So don't overlook those tire weights guys! Manufacturers like to hide them, but you can dig around and usually find pretty reasonable values.

I also doubt lighter weight wheels will cause any discernable benefit in suspension reaction. I mean - you're dealing with LARGE forces with performance oriented springs and dampers, is a 40lb wheel assembly moving at a few inches per second(MAX) really going to be anything more than basically negligible when tossing around a 3000lb car?

Matthew C Smith

03-01-2003, 12:37 PM

not sure if this will really help anybody who's a little confused about this but you want the moment of inertia (I) to be as low as possible (it'll take less energy to spin it). Moment of inertia (I) is proportional to the mass of the object spinning times the distance from the point it is spinning squared. So removing weight from the farthest point away from the center (ie. the tire) will show the best gains.

Hellrot Sachin

03-01-2003, 02:40 PM

this post really couldn't have come at a better time. I'm pretty convinced that my having gone to lighter tires made a dicernable different in acceleration and such, but now I'm consider bailing out on 18" RKs (don't remember their exact weight) and possibly going to 17" contours. The new rims will be considerably heavier but my main reason for holding back is my perception of acceleration. I really don't want the car to feel much slower, but if the increased weight is that focused around the center of rotation and not all that significant considering the weight of the car, no biggie.. Maybe I won't feel a difference in performance with this.

lkstaack

03-01-2003, 05:49 PM

The details of this thread is way above my current level of understanding. My perception is that (all things being equal) a 17" wheel will accelerate and handle better than a heavier 18" wheel. If this is true, will a 16" wheel perform better than a heavier 17" wheel? Why or why not?

nsk223

03-01-2003, 09:47 PM

The '7 pound' rule sounds about right to me. Some people have argued that reducing the weight of wheels and tires by say 40 lbs is the equivalent of reducing the overall weight of the car by 40 lbs. Not true: While the the car would clearly be 40 lbs lighter, the fact that this weight came from the rotational mass of the the wheels means it has effectivley reduced the force necesary to accelerate the given car at a given rate. In other words, given two identical cars with one with 40 lbs removed from the actual car of one and 40 lbs removed from the wheels of the other, the car with lighter wheels will be faster. This car could, concievabley, be the equivant of 280 lbs lighter (7x40 lbs wheel weight). Another component of this thread which I would like to clarify is the term 'polar moment of intertia.' Automotively speaking, this term describes the relative willingess of a certain car to change directions. A car with all the mass centered around the center of the car will be much more responsive that a car with the same amount of mass spread out around the car. This is why many/most purpose built race cars are mid-engined. The aforemtionted '7 pound,' regardless of whether or not 7 is the actual number, relates to much more than a cars ability to accelerate. I'm sure that most of you are familiar with the concept of 'unsprung weight' and it should be stressed that increases in handling performance with lighter wheels and tires will be much more signigicant than increases in acceleration.

SI///M3

03-01-2003, 11:20 PM

Originally posted by Hellrot Sachin
this post really couldn't have come at a better time. I'm pretty convinced that my having gone to lighter tires made a dicernable different in acceleration and such, but now I'm consider bailing out on 18" RKs (don't remember their exact weight) and possibly going to 17" contours. The new rims will be considerably heavier but my main reason for holding back is my perception of acceleration. I really don't want the car to feel much slower, but if the increased weight is that focused around the center of rotation and not all that significant considering the weight of the car, no biggie.. Maybe I won't feel a difference in performance with this.

I don't remember the exact weight of RKs but if they're anywhere close to the weight of my RCs it shouldn't be TOO much of a difference. When I got my RCs I did some basic calculations (not the most accurate since I made quite a few assumptions, I know), but the moment of the contours was almost exactly the same as my RCs (23lbs for the contours, 18.7lbs for the RC). Of course I just did the wheels since I had no ideas how much my tires weighed, but it wasn't a huge difference really. With tires I bet the contours are actually ahead (lower moment) of the RCs. Weight is very important in looking at wheels, but diameter is alot more since it the moment goes up by distance^2.

SI///M3

03-01-2003, 11:24 PM

Originally posted by lkstaack
The details of this thread is way above my current level of understanding. My perception is that (all things being equal) a 17" wheel will accelerate and handle better than a heavier 18" wheel. If this is true, will a 16" wheel perform better than a heavier 17" wheel? Why or why not?

Yep a 16" wheel is better in terms of how big the moment is. Smaller is better when looking at that. :) Thing is with 16s that most rims won't clear M3 brakes, and even though your moment is smaller sidewall flex and such starts to work against you. I remember a magazine article a long time ago comparing the best size for performance, 17 was it. Any bigger and the moment starts to kill your power, any smaller and tirewall flex starts to kill your handling.

scottycs

03-01-2003, 11:28 PM

I went from 26 pound wheels, to 16.5 wheels. There was NO difference to be felt in acceleration. Don't hope to feel anything different. The 7 pound rule is WAY off, it is no where near the equivalent of 280 pounds lighter. Don't follow the hype.

Bernman

03-01-2003, 11:40 PM

I went from a 40 pound wheel and tire combo to a 47 pound one. The acceleration difference is *definately* noticeable.

It is not hype, though the 1:7 pound rule of thumb may not quite apply.

Acceleration is not everything (but it sure is nice) and I am very pleased with the handling and looks of this combination.

scottycs

03-01-2003, 11:45 PM

Mine was noticable, but it was all psychological.

nsk223

03-01-2003, 11:47 PM

You can't really argue with physics here, if your wheels are lighter they will have less inertia and the car will accelerate better as a result. It should be stressed, though, that improvements in handling resulting from lower unsprung weight will be much more significant than improvements in acceleration. With respect the wheel diameter and its respective level of inertia, diamater is very important, but equally important is the height of the tire sidewall and the weight of the tire. For example, assume an overall wheel/tire diameter of 20 inches, not all cars with a 20 inch overall diameter will have the same level of rotational intertia: what matters is where the mass is concentrated on the this diameter-if the wheel weighs very little and the tire weighs a lot or the other way around.

stevegee

03-02-2003, 02:06 AM

Ok, lets do some math here. Hopefully one of the Mech. Eng's will check my math since it isn't so good.

If you just want to trust my calculations, jump to the end...

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.

M3formeM3foryou

03-02-2003, 05:38 PM

*dusts off the ol' dynamics book*

one little problem Steve... the moment of inertia you used is for a uniform hollow cylindrical shell. for a solid cylinder, it's mr^2/2.

i didn't verify your calculations at all or recalculate the results, but let's all think about this. rather than losing 13 pounds off of the outer most radius, the weight lost is distributed evenly across the wheel. in terms of the calculations, the difference between the before and after moment of inertias will be smaller, and thus, so will the change in acceleration.

i'm not saying that using a solid cylinder is a more accurate approximation of the actual moment of inertia; the hollow cylindrical shell approximation might actually be better. it may even be better to try using mr^2 for the tire and adding mr^2/2 for the wheel... you can simply add the two since they're spinning about the same axis.

number junk aside, this is an awesome thread. thanks for everyone's input.

----------------------

At a divorce hearing for Mickey and Minnie Mouse, the judge asks Mickey, "So you’re saying your wife is crazy?" Mickey responded "No, I said she is fucking Goofy."

Nimble

03-02-2003, 06:57 PM

Originally posted by scottycs
I went from 26 pound wheels, to 16.5 wheels. There was NO difference to be felt in acceleration. Don't hope to feel anything different. The 7 pound rule is WAY off, it is no where near the equivalent of 280 pounds lighter. Don't follow the hype.

I do agree the 7lb rule is way off...closer to 4 in my opinion.....but you have to realize one other thing.....our perseption of the accelerative difference of adding/removing weight from wheels/tires is easier to tell when going to a heavier wheel.

So in your case, Scotty, if you were to put on those heavy 26lb wheels after being used to the lightweight RC's you WOULD feel a difference in acceleration. I mean a 10 lb PER corner weight difference is huge. It is a possibility, however, that the RC's you put on your car had heavier tires than did the 26lb wheels, thus negating the positive benefits of the lightweight RC's....just a thought.

KILLR B

03-02-2003, 09:27 PM

What about the gear ratio changing with the diameter of the tire? Is it not true that changing from an 18 inch wheel to a 17 inch wheel will reduce torque? And then does this not also effect the perception(or reality) of faster acceleration? Another way to look at it is to have two sets of wheels weighing the same and with same weight tires, but the difference being 17's v. 18's, do not the 18's allow the car to accelerate faster because of the effect on gearing?:95 :az: :dunno

stevegee

03-02-2003, 10:09 PM

Originally posted by M3formeM3foryou

one little problem Steve... the moment of inertia you used is for a uniform hollow cylindrical shell. for a solid cylinder, it's mr^2/2.

I knew I'd be off somewhere. Though even though I had intended to use a solid cylinder, a hollow shell might be closer to reality since the tire and outer rim are essentially at the edge of the cylinder. But anyway...

If we use the solid cylinder equation, it makes the acceleration difference less, down to 2.5%. Which then makes every pound of rotational mass equal to only 1.5 lbs of non-rotational weight reduction.

I ran similar numbers for the M3 flywheel just to see how they'd come out. I'm calculating a 1% acceleration difference in 1st gear going from a 26lb fly to a 12 lb fly, ignoring all other rotational mass (gearbox, wheels, driveshaft, accessories, etc). In this case that's like dropping 31.5lbs off the car's weight, or 2.24lbs for each pound of rotational mass drop. This doesn't seem like the dramatic difference that people seem to feel.

Def

03-02-2003, 11:00 PM

Originally posted by stevegee
I knew I'd be off somewhere. Though even though I had intended to use a solid cylinder, a hollow shell might be closer to reality since the tire and outer rim are essentially at the edge of the cylinder. But anyway...

If we use the solid cylinder equation, it makes the acceleration difference less, down to 2.5%. Which then makes every pound of rotational mass equal to only 1.5 lbs of non-rotational weight reduction.

I ran similar numbers for the M3 flywheel just to see how they'd come out. I'm calculating a 1% acceleration difference in 1st gear going from a 26lb fly to a 12 lb fly, ignoring all other rotational mass (gearbox, wheels, driveshaft, accessories, etc). In this case that's like dropping 31.5lbs off the car's weight, or 2.24lbs for each pound of rotational mass drop. This doesn't seem like the dramatic difference that people seem to feel.

There are some pretty big assumptions you used to arrive at your figures. I'm going to contend you *can't* ever put a fixed ratio number as to what the reduction in rotational inertia would 'feel' like in terms of static weightloss.

First off, most wheels save weight around the hub area. BMW wheels might seem really heavy(even the Forged LTWs are ~20lbs a wheel) - but look at the area around the hub(like my sig for example), then compare that to an SSR Comp(very lightweight). The rim of the wheel will be basically the same weight since they're both forged, which is the most important part with respect to inertia. So all the weight is saved at the hub, which has to be accelerated MUCH less. There really isn't an easy way to approximate this in calculations, and the only way is to do a pretty nasty integral which I couldn't even begin to formulate.

In plain English - with the same diameter wheels, you will not appreciably change the rotational inertia by just going to a lighter weight wheel. Forging vs. Casting will produce a lighter weight rim section, which is probably the biggest benefit there is to be had.

BTW - you also forgot centipetal acceleration - w(omega)^2*R. But it's all a wash IMO. ;)

As for things 'feeling' slower. If you make your overall gearing taller by going to an 18" wheel - you drivewheel torque will be less IN EVERY GEAR. Therefore you car will accelerate slower, but longer - not because of the increase in weight or inertia(well, this might be a slight factor - but not noticable IMO), but because of the proportional decrease in force pushing your car along. I really don't know WHY people go to 18" wheels for their E36, especially when they don't have GOBS of power to benefit from their taller gearing. If they made a 235/35-18 tire then it might not change it much - but to my knowledge no tire is made in this size, so your gearing has to get taller.

On the flywheel issue - keep in mind a flywheel has a HUGE amount of angular acceleration, especially in the first 3 gears. It is not just spinning up to the ~700-800RPM of your wheels at 60-70mph, but all the way to 7000RPM in a matter of a few seconds. Therefore a reduction in rotational inertia is quite profound here.

Now I'm not saying it doesn't matter that you get some 49lb bling-bling 20" Chrome rims, but obsessing over a few lbs of weight on your wheel is pointless. Especially when everybody LOVES to spring for the HEAVIEST street tire out there - the Bridgestone S03. BTW - the tires rock in the traction department, but I think there are some that grip just as well, yet don't weigh nearly as much(or lighten up your wallet quite as much :biglaughb ).

stevegee

03-02-2003, 11:44 PM

Originally posted by Def
There are some pretty big assumptions you used to arrive at your figures. I'm going to contend you *can't* ever put a fixed ratio number as to what the reduction in rotational inertia would 'feel' like in terms of static weightloss.

I beg to disagree. This behavior is relatively simple physics. Whether inertia is rotating or static, its all inertia.

The real analysis is significantly more complicated than what I've done, but I think its pretty close.

Originally posted by Def
First off, most wheels save weight around the hub area. BMW wheels might seem really heavy(even the Forged LTWs are ~20lbs a wheel) - but look at the area around the hub(like my sig for example), then compare that to an SSR Comp(very lightweight). The rim of the wheel will be basically the same weight since they're both forged, which is the most important part with respect to inertia. So all the weight is saved at the hub, which has to be accelerated MUCH less. There really isn't an easy way to approximate this in calculations, and the only way is to do a pretty nasty integral which I couldn't even begin to formulate.

Also true. Although this can be approximated rather easily. My math showed that lightweight wheels don't do much for acceleration. Your point shows that they do even less.

Originally posted by Def
BTW - you also forgot centipetal acceleration - w(omega)^2*R. But it's all a wash IMO. ;)

I think you're joking, but its been a while since I took physics. Why would this affect anything, other than ripping the the wheel apart if spun too fast?

Originally posted by Def
On the flywheel issue - keep in mind a flywheel has a HUGE amount of angular acceleration, especially in the first 3 gears. It is not just spinning up to the ~700-800RPM of your wheels at 60-70mph, but all the way to 7000RPM in a matter of a few seconds. Therefore a reduction in rotational inertia is quite profound here.

That's right. What I did was multiply the angular acceleration that the wheels experience by the gear ratio and final drive. That new number is the angular acceleration of the flywheel. The thing is that the flywheel is also signifcantly smaller than a wheel. I assumed it was about 11" in diameter, since the clutch is about 9.5". In contrast, I assumed the wheel+tire was 25". The flywheel is also a lot lighter than a wheel+tire, ~26lbs.

Since moment of inertia is proportional to radius^2, the flywheel is much easier to accelerate than both rear wheels. I'm getting the equivalent to be 70lbs of wheel+tire in the 25" diameter size..

joshE46

03-03-2003, 12:44 AM

I'll leave the physics to the smart guys here, I'm just an accountant!:redspot But hey, KB, I want my SSR's back when you take them off...the Integrals...so let's talk before you do anything with 'em. Also, in case you were wondering, they weigh about 18.8lbs each, IIRC when I asked Tirerack when I bought 'em. Of course, the S03's don't help. But, I don't think you'll see a 50lb weight improvement. Maybe around 30lbs, if you can save 2-3lb per wheel, and 3-5lb per tire.

See you Thursday for lunch!

Josh

Def

03-03-2003, 01:39 AM

I beg to disagree. This behavior is relatively simple physics. Whether inertia is rotating or static, its all inertia.

The real analysis is significantly more complicated than what I've done, but I think its pretty close.

I understand inertia, and your analysis was some simple algebra. Yet your model is fundamentally flawed by grossly oversimplifying the distribution of weight in the wheel. If one did measure the inertia of the wheel through emperical observation, I think you'd find the results for two far different weight wheels of the same diameter to be much closer than you are expecting.

Also true. Although this can be approximated rather easily. My math showed that lightweight wheels don't do much for acceleration. Your point shows that they do even less.

I'd say a few lbs less on each wheel giving a few percent more acceleration significant. Hell, I'd spend the money on it. Although I think in the real world you'd be lucky to get even a few TENTHS of a percent difference in acceleration in even 1st gear. We both agree, we're just arguing degrees of difference here.

I think you're joking, but its been a while since I took physics. Why would this affect anything, other than ripping the the wheel apart if spun too fast?

Us engineers have no sense of humor - I do not know the meaning of this... 'joke'? Maybe BMWITSRacer will see that and get lil' chuckle out of it. :biglaughb

Seriously though - no I guess it doesn't matter in your analysis - but I was saying it as more of a reminder that this force is what helps determine a body's inertia. Therefore saving weight on the hub of a wheel does you almost no good except by reducing the load of the suspension's work(which isn't much).

In case you want to get a refresher in dynamics, I'm bored so I'll ramble on for a few minutes and bore any hapless victim that manages to read this far in the thread to death. You have been warned!!! :laugh

You have two vector components of centripetal acceleration. Looking at the profile of a rotating body you have - Alpha*R in the horizontal and Omega^2*R in the vertical. Think about it, a point out on the wheel will experience an acceleration equal to the angular acceleration(Alpha) times the radius - the further out you go(bigger R) the more the point will acceleration horizontally. Now you also have all points being CONSTANTLY accelerated inwards due to circular motion. This is affected by the angular velocity of the wheel - since even when the wheel is experience no angular acceleration(Omega = constant, d(Omega)/dt = 0 etc. etc.), you will still have any point on the wheel being forced to accelerate inwards at any time.

Now wasn't that interesting.

Basically - all that huff and puff only means that since lightweight wheels rarely take much material off the rim of the wheel, diameter of street wheels is the biggest factor in how much 'general' rotational inertia the wheel will have.

That said - the lightweight SSR Comp is still THE BEST WHEEL EVAR!

About the flywheel - most lightweight flywheel designers are smart enough to save mass where it matters most. At the very edge of the flywheel. So the change in inertia is much more IN FAVOR of lowering the inertia than between a 'heavy' and 'light' wheel of the same size. I still contend that the savings is more than 'meets the eye' by simple analysis in the flywheel situation. I have felt the difference of going to a 18lb FW in a Civic to a 9lb FW and it was quite drastic. Also felt the difference between big 17's and much lighter weight 15's and it was barely noticeable compared to the flywheel.

M3formeM3foryou

03-03-2003, 03:57 AM

*dynamics text in hand*

well the uniform cylinder approximation was optimistic for a wheel, and the hollow cylindrical shell is the worst case scenario. so, the actual polar moment of inertia should lie somewhere in between. either way, as i said before, the resulting acceleration isn't changed much since you don't get around moving the 3000 pound car.

anyway, go check your dynamics text-- by definition the centripetal acceleration has only one component, and it points to the axis of rotation. the "horizontal" acceleration, as you called it, is the coriolis acceleration. and for rotation about the center of mass (this case) the sum of the moments = I(polar moment of inertia) * the angular acceleration. the centripetal acceleration is implicit in I.

i feel like i should be taking a midterm now or something.

positivepower

03-03-2003, 08:47 AM

Ok now I'm wondering. What is a better tire for the money that is lighter then the S03. There is a group buy going on for the S03. If there is a better lighter tire let's get that one. Any ideas.?

nicot

03-03-2003, 10:01 AM

Kumho MX are suppose to be lighter than S03's.. well I went from 17's BMW radials to 18's BBS RSII and now 19's all in the space of a month.. for handling I personally prefer 18's but for acceleration yes the 17's were quickest I could feel it..19's around hill climbs werent as responsive..

but all this maths makes me feel I should have payed more attention in class..hehe and in physics if they use this as an example I definitely would have done better in it.

For me I think losing 25-30 pounds of body weight is the cheapest and best way to get rid of weight..

stevegee

03-03-2003, 01:47 PM

Originally posted by Def
I understand inertia, and your analysis was some simple algebra. Yet your model is fundamentally flawed by grossly oversimplifying the distribution of weight in the wheel. If one did measure the inertia of the wheel through emperical observation, I think you'd find the results for two far different weight wheels of the same diameter to be much closer than you are expecting.

Granted that the analysis was simplified, but I think only enough to illustrate the point. I believe that the general analysis is good enough to show that the rotational mass drop to be not as important as everyone thinks (I realize that you agree with me on this).

And as M3formeM3foryou wrote, the two cases done with varying moment assumptions were on both sides of the weight distribution skew, and yet the results didn't change much. The wheels are just too small in relation to the overall car's mass to have much of an effect.

Originally posted by Def
Us engineers have no sense of humor - I do not know the meaning of this... 'joke'? Maybe BMWITSRacer will see that and get lil' chuckle out of it. :biglaughb

I honestly haven't done any sort of physics analysis in about 10 years, so I tend to forget things. Couldn't tell if you were just throwing something in to bust my chops or not. :)

Originally posted by Def
You have two vector components of centripetal acceleration. Looking at the profile of a rotating body ... Now wasn't that interesting.

Thanks for the physics lesson. I do appreciate it. I'm here to learn all that I can.

Originally posted by Def
About the flywheel - most lightweight flywheel designers are smart enough to save mass where it matters most. At the very edge of the flywheel. So the change in inertia is much more IN FAVOR of lowering the inertia than between a 'heavy' and 'light' wheel of the same size. I still contend that the savings is more than 'meets the eye' by simple analysis in the flywheel situation. I have felt the difference of going to a 18lb FW in a Civic to a 9lb FW and it was quite drastic. Also felt the difference between big 17's and much lighter weight 15's and it was barely noticeable compared to the flywheel.

I'm sure the designers of the flywheel are trying to optimize, but at the end its still a relatively uniform disc. The one's I've seen look just to use a materials change to lighten the weight.

But the point is that even a heavy flywheel is a small mass compared to the weight of the car, or even the moment of the wheels. Even big changes to small factors will still lead to small results.

[i]Originally posted by M3formeM3foryou
i feel like i should be taking a midterm now or something.

Wait 'til you start working. :) When you do engineering analysis on a daily basis, then it becomes natural to tackle all problems that way. :D

KILLR B

03-03-2003, 07:17 PM

Actually, Matt at Orr weighed each set of tires and came up with about 48 pounds difference:dunno. You let me know on the Integrals/SO3's(still have 50% tread). See you 3/7

.

UD///M

03-03-2003, 09:23 PM

Awesome thread indeed. One small input from me. All the talk about uniform cylinders and approximate equivalents to a wheel/tire combo on a car left out one thing. In order to use a uniform cylinder you need to take into consideration the location of the mass of the wheel/tire. This will effectively create a smaller radius when an equivalent uniform cylinder is computed. For all intents and purposes you could probably just use the diameter of the rim itself. Again that is just a guestimate. If you have really tall tires that gestimate goes out the window.

Nonetheless, this is good stuff.

Since the 17" wheels have a better "gear" ratio, they will allow for better acceleration over an 18" counterpart of equal mass. The same effect of having a numericly higher gear ratio. Such as going from a 3.15 to a 3.45 or something. Higher number= lower gear. Shorter tires will affect this. Simple levers.

Now that I think about it, mass is what is important i.e. kg. Not weight. Weight has an implied force behind it. Such as you weigh more on earth than you do on the moon but your mass remains the same. Of course that has already been touched on in the posts mentioning material distribution.

Bernman

03-03-2003, 09:58 PM

This brings up a point that I neglected in my earlier statement about my acceleration change with the new wheels and tires. The brand new 18" S-03's are significantly taller than the worn out 16" RE-730's. On top of that, the effective rolling radius for the 16" is likely to be even smaller since the 18" tire appears to compress less.

The "gearing" difference is probably considerable. Certainly worth calculating. I will see what I can do... :)

UD///M

03-05-2003, 06:34 PM

Something I have been mulling over in my head is since both masses (body and wheels) are being accelerated at the same rate in the same direction eg. forward, their total mass should be additive. I don't know if this is completely correct though. The inertia of the car is linear but the inertia of the wheels would appear to have an angular component as well. Therefore the so called sprun/unsprung mass equivalent ratio seems moot. In otherwords 1:1.

I am not a mech eng and don't know the math behind this complex system but from my perspective this makes sense.

Another musing about this rotating mass stuff. The reason an engine benefits so much from reduced rotating mass is because the engine itself is a rotational machine. Reduce its inherent mass and you will increase its ability to rotate quicker dr/dt. (not faster r/t)