Wheel Weight - The Physics Way Options

[cobas]

Okay, so I've heard this before that heavier rims will hurt your car's acceleration due to added rotational inertia of larger diameter rims. So I decided to calculate (and in some steps approximate) the effect.

Conversions 2.2 lbs/kg, 1.6km/mile, 2.54cm/inch (Sorry, I do metric)
Using these, 3000lbs = 1360kg, 60mph = 26.7m/s, and 17"= 431mm so the total tire Radius is something like 431/2 + 205*.45 = 310mm for me (stock)
First the linear kinetic energy at 60mph KE = (1/2)mv^2 = .5(1360)(26.7^2) = 485kJ.

Second, the rotational kinetic energy KR = (1/2)Iw^2 where 'I' is the rotational inertia and w is the angular velocity. For the rotational inertia, I made the wheel a cylindrical shell and basically guessed the radius, so I=mr^2. I said m=20kg (44lbs) and r = 20cm (16" diameter). So I = 20*.2^2 = 0.8 (kgm^2).
Then the angular velocity is w = 2pi/T (radians/second), but T = circumference/velocity, which is 2pi*r / v, so w=2pi/(2pi*R/v)= v/R.
v is 26.7m/s and R is .31cm (for OEM tire diameter).
This gives .5(.8)(26.7/.31)^2= 3kJ per wheel.

Summary : The wheel/tire rotational energy accounts for 2.5% of the energy (and horsepower) required to accelerate. The linear kinetic energy is more important, since 45lb per wheel adds up to 6% of the car's weight. If somehow you cut that weight in half (by, for example... removing the tires), you should gain the equivalent of about 5.5hp. (Assuming a 3000lb / 135hp car) Anyway, this doesn't like a big difference to me given the price of light weight wheels, but I'm also not considering handling changes at all. One last caveat: I've seen in at least two places online significantly longer stopping distances with heavier wheels. But I don't understand why, unless the car can't lock the brakes at 60mph. Otherwise the heavier wheels would also increase traction and shorten braking distances.

[Biff]

Ahhh, you have to remember, heavier object roll faster. It is harder to stop a heavy rolling object rather than a light rolling object. That is just simple physics. Atleast that is what I think I remember from high school.

[elantragt]

Biff's got it, I think! It's about inertia...well I think


From Newton:

All objects resist changes in their state of motion. All objects have this tendency - they have inertia. But do some objects have more of a tendency to resist changes than others? Absolutely yes! The tendency of an object to resist changes in its state of motion is dependent upon mass. Inertia is that quantity which is solely  dependent upon mass. The more mass which an object has, the more inertia it has - the more tendency it has to resist changes in its state of motion.

[stauf]

And in English....

Heavier Wheel = More power to start, More brakes to stop..

Lighter Wheel = Less power to start, Less power to stop...

Heavy Wheels = Cheaper

Ligher Wheels = Pricier

just thought I'd be the smartass here..

[southpawboston]

i think cobas's estimate of about 5.5hp by cutting the wheel weight in half should not be taken lightly. how much money do people spend trying to get that extra 5.5hp using methods that not only make the car louder (exhaust, CAI), but also have the potential to destroy the engine (how many times have i heard of water getting sucked in through a CAI???).

plus there are added benefits to handling by reducing the weight of the wheels. the car should be much more responsive to changes in the road surface. wheels, outer CV's, strut assembly below the spring, and brakes are part of the "unsprung" weight of the car. this is the weight that is not supported by the suspension. the less weight of those components, the less inertia that has to be overcome when a wheel hits a bump. that means that the wheel can stay planted on the ground better.

for a number of years, audi tried to maximize this concept by having the front disc brakes placed inboard, basically at the inner CV joint. this moves the mass of the brakes to the sprung portion of the car. i don't know why they did away with this idea, except possibly for space and safety concerns (being that if an axle shaft broke you lost your brakes).

i may consider looking for really light rims instead of trying to extract that difficult 5 hp out of the engine, and enjoy added handling benefits to boot!

[oiml8]

Well this is the very concept that is why I am going to light wheels and back to 15". People have spent over $200 of a AEM or Injen CAI for 3-5hp. Light weight wheels that I am looking at (11lbs ea) will cost $640 shipped. That is less than twice what you might otherwise spend. You get 3 things for less than twice the money tho. You get new wheels AND you get hp too. The better suspension characteristics from less unsprung weight. So like SPB said the suspention reacts more quicly. You also reduce the effective vehicle weight. Every lb of rotational mass is very roughly equivelant to 8lbs of dead weight. So, in my case, I will be going from wheels and tires that are about 41lbs to a combo that is 31lbs. That is 10 lb reduction per wheel....times 4 wheels....times 8...equals and equivelant 320 weight reduction.

Now maybe the 320lb reduction in weight and the 2-3 ph you are talking about in parasitic loss re-aquired is the same thing. Too much for my simple mind to process this morning.

[Elantrick]

I still believe that the tires are going to give you the most feedback. Lowering your wheel weight by 10 lbs isn't going to give you enough feedback for you to feel. Is it better, yes, can you feel it?????

As I have said before. If you can feel the difference in 10lb of rotational mass, you should be paid quite nicely to drive other peoples race cars.

If you are drag racing or road racing and can time your runs or laps, then I think it would be good thing to get the lighter wheels.

Now, going to a 15" wheel with a low profile tire, you are lessening the contact patch vs a standard outside diameter wheel/tire setup. The smaller the circle, the smaller the contact patch.

[oiml8]

If this rough estimate of 1lb rotational = 8lb dead weight then a reduction of equivelant 320lbs is something most people ought to be able to feel. It is probably equivelant to 1-2 tenths at the drag strip. I would hope more (relatively) at the autox track since there is also the suspension benefit. I think an added benefit will be the better gear ratio with the smaller diamter wheels. Plus the tires are 50 series rather than 55. The overall height reduction from 1/2 for the wheels and then something like 1/8 for the series adding up to 5/8 drop in center of gravity. I don't know the square inch reduction in contact surface. Minimal I think. More than over come by the other benefits. I would go with a 215 instead but they don't make it in the Azenis or Kumho MX. Actually, Azenis will be coming out with more sizes this year. They should be available beginning in March.

Remember Jay stripping out the interior of his last elantra? That was all for a weight reduction of less than 300lbs if I remember right. I'm pretty sure that if it wasn't a big deal, he would not have gone through all that for nothing. It is a bit different because it was weight that was sprung and higher in relationship to the gravitational center.

[Elantrick]

The rotational mass being 8:1 is only for acceleration and deacceleration. The suspension benifits, you only have the 1:1 ratio.
So if you are at the drag strip, lighter wheels are better. If you are racing, yes they are better.

[JPJR]

So your saying the average person could not feel at least a 5 hp difference in performance? I agree, lite weight rims are not for everyone and not needed for the average daily driver. But for someone looking for more performance, they are a great idea. More power then a catback, almost as much as a CAI, plus better handling and looks. Perfect for someone who wants to customize their car and increase performance.

[cobas]

Well, as I said, I don't know how it affects the handling exaclty. I can see that the springs would react quicker when going over bumps, but on smooth surfaces? Say 60% of the 1360kg is up front, so each front spring carries 408kg. Normally this weight compresses the spring about, what, 6 inches? So that's a spring constant F/x = 26.8kN/m (assume not progressive springs). Assume the mass of everything (wheel, tire, lower strut, brakes, tie rod, CV, etc) is about twice the wheel/tire, or 40kg. The period of a oscillating spring is T=2pi * Sqrt(m/k), which gives 0.24 seconds for one cycle, so it goes top to bottom in half that, .12 seconds to recover from a speedbump. If the spring relaxation were the limiting factor, at .24 seconds per cycle you could slalom 4 cones per second. But ok, it's larger than I expected, so I could imagine that making a difference in a sharp turn.

I assumed the springs are "relaxed" when the car is lifted off the ground. I'm pretty sure that's not true, which means the spring rate is higher and that time is shorter. I also assumed the spring rate is constant, which I also don't think is true, and I'm not sure how that affects things.
One last point I haven't done: the change in angular momentum of the wheels when the whole car turns.

This post has been edited by cobas: Feb 22 2005, 01:07 AM

[stauf]

I believe springs are relaxed when the car is up in the air, because they're no longer compressed. This is why with RWD cars you can use a jack to remove the rear springs safely.. Spring rate should be a constant, but I don't believe it is due to how springs work. If you compressed a spring completely, the resulting bounceback would be much more forcefull then if the spring was only half compressed.. But I suppose figuring that out would reveal if the rate is constant or not. I am by no means a math guy, and I don't play one on TV either.. But these are just my observations on how things work.