Rotational Mass - Performance

[OldSchoolFormula]

I was having a rather heated discussion with a few friends of mine on the benefits of reducing rotational mass and the old 1 pound of rotational mass is worth 10(?) pounds elsewhere on the car. What it came down to was the difference in the front wheels versus the rears in a RWD application. I argue that reducing the weight of the front wheels is just as important as reducing the weight of the rears as rotational mass is rotational mass. They argue that the front wheels are not nearly as important because they are not connected to the drivetrain and thus are effectively negligible in decreasing performance. Basically the extra weight of the wheels would be the same as the weight anywhere else on the car. As far as I'm concerned the drivetrain has to move the front wheels at the same rate as the rears and thus a heavier front wheel is just as detrimental as a heavy rear.

Input?

[Blown353]

You are correct, saving weight up front is just as important as saving it out back. Barring any wheel spin and assuming the OD is the same between the sets, and for now ignoring the speed differences experienced in a cornering situation both the front & rear wheels/tires accelerate, decelerate, and rotate identically. There may be less energy tied up in the front wheels & tires if they are narrower (which means lighter) than the rears, but saving weight in the front wheels and tires is just as important as saving it out back. If the car has 4 identical wheel/tire combos at each corner the same amount of energy is tied up in each one while rotating.

The "1 in 10" number you cite is a somewhat meaningless ratio. Reducing rotational mass is definately more beneficial than reducing "dead" mass because with rotational components you are penalized in two departments for the mass. The exact ratio depends on where exactly the weight is saved. Just as an example, if you save 1lb at the very outermost point in the tire/wheel combo that will do more good than saving 1lb right on the hub centerline because of the difference in moment of inertia.

Your friends are incorrect... unless the car is pulling a wheelie the entire time and then the front wheels can be considered and treated as "dead" (non-rotational) mass. Please throw their foolishness back at them in the most satisfying way possible.

[baz67]

Another way to put it to them is ask them to think about the brakes. Thinking about it that way brings the front wheel package into better focus. With everything else being the same you will lose perfomance in the brakes with more mass at the wheel.

[BA.]

This is an interesting question, and one that a friend (with a PhD) and I have had before.

Blown553, if you don't mind my asking, what is the data for your answer?
EDIT: Never mind, I see that you HAVE gone ahead and edited it with more detail.

(back to main topic now)
I hate to 'dissect' the topic, but OBVIOUSLY having extra rotational mass is different physics than plain old dead weight.
I would never call front wheel weight "effectively negligable".

That being said, I *am* under the recollection that having 100lb wheels/tires on a the back of a RWD car IS more detrimental than having 100lb wheels/tires on the front. (exaggeration to show effect)

Sounds like you and your buddies at least agree that having extra weight on the drive wheels is worse than static weight in the trunk. So, the only remaining question is on the non-drive front wheels, is it the same effect?
I'll ask the Doctor tomorrow.(ok, it's 2am, it'll be later today!)

[chicane67]

All of this is actually dependant on the application and where the mass actually is. In drag racing, sure.... it would be more detrimental for the weight to be in the rears.

But its not necessarily good/meaningful for any other application.

In reality, the front suspension takes more and does more than the rear suspension. Turning, redirection, load and braking.... the majority is all done on the front end. Accleration is merely one event.... and it would benifit from a lower MOI, but, does it make a bigger difference in the front or rear ?? For accleration specific applications.... yeah, I would think so... but only if they were of same/same size/weight.

But, mass is mass. You have to rotationally turn the front wheels and move the sprung mass, driven or not. Just like you have to turn the rears..... and it takes power to turn either of them.

Now what the real question is..... what impact does it have on the independant systems front and rear ?? Considering that the unsprung mass is greater on the rear (in most cases) than the front. Or is it ?? Obviously, each system would have to be screwtinized independantly to get an emperical answer. Of course, most of us would initially say that the rear would benifit the most by where the weight was in the diameter... but put some thought into it before you answer.

Plain theory has it merit here, but more information is necessary to give an intelligent answer.

[scoobienoobienoo]

I was having a rather heated discussion with a few friends of mine on the benefits of reducing rotational mass and the old 1 pound of rotational mass is worth 10(?) pounds elsewhere on the car. What it came down to was the difference in the front wheels versus the rears in a RWD application. I argue that reducing the weight of the front wheels is just as important as reducing the weight of the rears as rotational mass is rotational mass. They argue that the front wheels are not nearly as important because they are not connected to the drivetrain and thus are effectively negligible in decreasing performance. Basically the extra weight of the wheels would be the same as the weight anywhere else on the car. As far as I'm concerned the drivetrain has to move the front wheels at the same rate as the rears and thus a heavier front wheel is just as detrimental as a heavy rear.

Input?

At BA's request, I have come to shed some light on the topic....

Blown353 has the right idea. I will explain:

Think about this in terms of energy. Power is energy dissipated per unit time (how fast you can do work). Therefor, your horsepower applied integrated over time is your total energy applied to the system. In simple terms, power x time is approximately energy.

You have a finite amount of energy.

Total energy in (TE) - Total energy from losses (TL) = Energy available.

Energy available = rotational energy (RE) + kinetic energy (KE).....
or what's left over for motion after subtracting for losses, wind resistance, etc.

To go fast, you want kinetic energy (KE) big.

TE - TL = KE + RE

But if you have rotational energy eating up energy, you have..

TE - TL - RE = KE

KE = 0.5*m*v*v

RE = 0.5*I*w*w

where m is the TOTAL mass, v is the velocity of the car, I is the moment of inertia for the rotating objects and w is the angular velocity (how fast it rotates...not independent of v. In fact, w = v/r where r is the radius of the wheel).

I is dependent on several factors, but the idea is that the farther the mass is located from the center of rotation, the bigger I gets. Hence worse for acceleration.

Anyway, you can solve for v in terms of m and energy and it looks like:

sqrt[(2/m)*[TE-TL-RE]] = v

Or if you like, you can replace I = me*r*r where me is the effective mass of the rotating objects (in this case wheels).

sqrt[2(TE-TL)/(me+m)] = v

Basically, what this says, is more weight in general = slower velocity, and the higher RE is (or larger rotating effective mass), the slower your velocity is.

You get a double whammy from any rotating mass period. Whether it's front or rear wheel, doesn't matter. If it's rotating, you lose.

PS. If you REALLY want me to, I can quantify how big of an effect this would be for wheels and tires, but it'll have to be next week when I get back from TDY.

Cheers.

[BA.]

Aahhh. I see that Blown353 has massively edited his post. Glad to see that, it reads much better now.

Thanks to you and Scoobie too.

[Blown353]

Aahhh. I see that Blown353 has massively edited his post. Glad to see that, it reads much better now.

Yeah, added some more verbage before I went to bed last night. Decided I was too terse and lacking some description.

I didn't feel like throwing any equations around, even basic ones, because the question is just so "common sense" I didn't think it was needed... and they'd probably be over the heads of his friends anyways. (no offense intended towards your friends Ian, please talk some sense into them-- I'll keep my fingers crossed there's hope for them yet.)

[baz67]

At BA's request, I have come to shed some light on the topic....

Blown353 has the right idea. I will explain:

Think about this in terms of energy. Power is energy dissipated per unit time (how fast you can do work). Therefor, your horsepower applied integrated over time is your total energy applied to the system. In simple terms, power x time is approximately energy.

You have a finite amount of energy.

Total energy in (TE) - Total energy from losses (TL) = Energy available.

Energy available = rotational energy (RE) + kinetic energy (KE).....
or what's left over for motion after subtracting for losses, wind resistance, etc.

To go fast, you want kinetic energy (KE) big.

TE - TL = KE + RE

But if you have rotational energy eating up energy, you have..

TE - TL - RE = KE

KE = 0.5*m*v*v

RE = 0.5*I*w*w

where m is the TOTAL mass, v is the velocity of the car, I is the moment of inertia for the rotating objects and w is the angular velocity (how fast it rotates...not independent of v. In fact, w = v/r where r is the radius of the wheel).

I is dependent on several factors, but the idea is that the farther the mass is located from the center of rotation, the bigger I gets. Hence worse for acceleration.

Anyway, you can solve for v in terms of m and energy and it looks like:

sqrt[(2/m)*[TE-TL-RE]] = v

Or if you like, you can replace I = me*r*r where me is the effective mass of the rotating objects (in this case wheels).

sqrt[2(TE-TL)/(me+m)] = v

Basically, what this says, is more weight in general = slower velocity, and the higher RE is (or larger rotating effective mass), the slower your velocity is.

You get a double whammy from any rotating mass period. Whether it's front or rear wheel, doesn't matter. If it's rotating, you lose.

PS. If you REALLY want me to, I can quantify how big of an effect this would be for wheels and tires, but it'll have to be next week when I get back from TDY.

Cheers.

Um, that's a good first post. Nice and welcome to the site.