typeR

gaining horse power to the wheels from going to a lighter wheel? i realize what we're really talking about here, is loosing less horspower with lighter wheel than with heavier, due to rotational mass...also is there a formula for: each pound of rotational mass removed is like removing x number of pounds in unsprung weight?

Starter

Not sure what you're asking here. Are you talking about a lighter flywheel?

If so, I don't know that there is a formula for it, but you're right - you just lose less HP than with a stock flywheel. It also revs faster.

Unsprung weight has nothing to do with HP. Your total weight affects how much HP is needed to overcome the inertia of same.

civic4982

losing weight in the flywheel is similar to losing weight in the wheel weight. Same idea of being able to spin more quickly and more easily. It does add hp, how mcuh? I have no idea

SiGGy

Well rule of thumb with unsprung weight is for every 170-200lbs removed it will take off .1 from your 1/4 mile time.

I don't know the rotational mass equation off the top of my head. But I bet a search on google will help you find it.

Now go run the 1/4!!!

CO-CL-S

I think someone once said that for every pound of "wheel weight" removed, is like removing 4 pounds of car weight?

typeR

see iheard for every pound rotational is like 20lbs unsprung...

CO-CL-S

Could be, I'll see if I can find something on the net..

CO-CL-S

From.. http://www.magnesiumwheels.com/

I got:
Magnesium engineered alloys offer a variety of strengths coupled with ductility. However, the most appealing characteristic is magnesium's inherent low density, which makes magnesium the lightest structural metal available, being two-thirds the weight of aluminum alloys, with equaling strength-to-weight ratios. In automotive wheel applications, magnesium offers enhanced performance by reducing un-sprung weight, specifically the rotational mass of the wheel, where it is noted that for every pound of un-sprung weight saved is equal to, or greater than five pounds of chassis weight. This type of weight reduction with rotating wheels (mass) greatly enhances the performance of automobiles by requiring less energy to accelerate, thus, actually adding horse power with no engine modifications. The elite automotive racing teams (F-1, CART, and IRL) utilize forged one-piece magnesium wheels for these reasons.

EricL

It depends on the construction of the wheels!

You could end up with an equivalent reduction of static weight close to reduction in wheel mass (if weight loss was concentrated near the center of the wheel/rim).

So, example: If the tires are the same, rim diameters the same, but the verymost inside of the wheels weight 2 lbs less per wheel, you would have 8lbs less weight and you could just subtract this from the car's weight and reconvert your power-to-weight. (3500 - 8 = new weight). You then figure your new power-to-weight.

There is no such thing as HP gain from loss in rotational inertia or loss in static weight.

The outside of a wheel is moving with a rotational component and translational component. The inside of the wheel is moving at the same speed as the rest of the car (it's call translational motion).

IF the "bulk" of the tire/wheel weight were concentrated at the outside of the wheel, the "effect" of this weight loss would be close to 3x.

The max equivalent gain is close to 3X (So, if you lost 60lbs from tires and wheels you could consider that to be 180lbs of lost weight from the car at the most). If that same 60lbs was only "lost" at the inside of the wheel, the gain would only get equal to 60 lbs. Take the weight and subtract it from the car’s weight and plug in the NHRA acceleration calculator and that’s about as close as your going to get without getting into some real work.

Every wheel is made differently, so there is no such thing as getting xx HP from xx lbs of lost wheel weight.

See the following picture:




LINK: http://dept.physics.upenn.edu/course...on4_1_4_3.html

The wheels and tires on a vehicle are translating at 1x car's velocity and the outside is rotating at 2x car's velocity); this makes for a possible upper bound in equivalent static weight going of 3X (1x + 2x).

EricL

No, 1x to 3x...

typeR

thanks...now does this upto 3x count for all four wheels or only the one that drives the car? and lets use the 17/8 +48 SSR comp as the example...
and to be clear eric...are you saying that the same car the same day wouldnt dyno higher going from stock wheels to the SSR? again i realize you wouldnt be gaining any horse power but instead loosing less through the drive system

mattg

it should dyno higher.

EricL

The whole issue of what you get on a dyno vs. real world is "messy". The dyno rollers are not going to have the same weight as a given car. It also can't tell about weight loss at the center of the wheel.


Example:

My Toyos are about 2.2 lbs lighter than each stock tire. My 17x8" SSR wheel is about 12 lbs lighter than the stock rim. I could just add 12 + 2.2 = 14.2 as the weight loss PER tire/rim. I just rounded down to an even 50 lbs for the set of 4. (12.5 lbs less per wheel * 4 = 50 lbs).

The 50 lbs gets multiplied by 1 (worst case = 50 lbs static equivalent)

The 50 lbs get multiplied by 3 (best case = 150 lbs static equivalent.

Think of 150 lbs "equivalent" as removing the back seat, or changing the front seats (whatever). You wont see the effect of any static weight loss on a dyno! The dyno will not show you the decreased weight of a wheel that is sitting near the center of the wheel's axis of rotation. But, if you hit the track, and you have 100lbs of Uranium at the very center of the wheel, it will slow YOU WAY DOWN!!!!! Remember, the car still needs to drag that extra weight at the center of the wheel down the track. The wheels are both "translating" and "rotating" as the car moves (you are “effectively dragging the wheels down the track and having to spin them up too)

The whole weight loss affects the whole car when all 4 wheels are in motion. If you got on a DynoJet and it you noticed that a set of really light tires and wheels gave you 10HP more on every run, you would have to multiply that by 2 to get the "effective" gain of all 4 wheels. (And that figure is "deceptive" and doesn't fully account for weight on the inside on the wheel).

That 1x / 3x figure has nothing to due with dynos and HP. It's all about power-to-weight.

I didn't say that. With the SSRs and Toyos I'd expect the dyno to show a mild bump (as to how much -- I haven't a clue).

If the bulk of the weight loss were at the OUTSIDE of the wheel, you would expect the dyno to show "some" gain in HP (this being do to a reduction is the amount of mass that you need to spin-up). Just to be "difficult", I could make a set of wheels that showed little or no gain on a dyno, but showed a lot of improvement on a track. If the weight were located within 1/4" of the center of the wheel, you would expect only a minor improvement in dyno figures. However, if that weight gain made all four wheels 100lbs heavier, you are adding 100lbs (4x25lbs). If you run down a quarter mile, do you want a 100-lb passenger in the car? (That would be the 1x example/extreme for a wheel)