Been thinking about getting new wheels and I'm leaning towards chrome. How much does the extra weight actually affect performance compared to stock 10 spokes??? I mean about how many horses do you actually lose??? I would think that it would be only a couple right???Thanx

 

not enough to make a thread on it

 

There's plenty of debate on this, but here's the facts.

You don't lose any horsepower with heavier wheels, even if they weighed 100lbs more each. Dyno's may show that you've lost hp, due to the way dynos calculate hp, but this is just a phantom measurement and also a reason that dyno numbers are not hard numbers, just a baseline measuring tool. It is impossible to lose power due to weight......that's not up for debate, it's fact.

However, with any increase in unsprung weight, you're going to lose acceleration. But that's only part of the story. If you get a heavier wheel with a lower polar moment of inertia, that's going to actually be better than a lighter wheel with a higher polar moment.

All else equal, the racers math I've seen have been .1sec in the 1/4mile for each 10lbs extra wheel weight. I think the actual effects are less than that.

 

Oh ok I get it. So its not really in power but acceleration loss? I'm just curious why people spend sooo much money on light wheels and why I've heard "Man I would have made so much more power if I didn't have these damn heavy wheels!"

 

I mean but it goes hand and hand cause power loss from the extra weight means less power for acceleration. I guess heavier wheels aren't good

 

LMAO....AT LEAST 200 horses

 

F=ma

Or as it applies to a car, horsepower = weight x acceleration. (paraphrased)

Less weight means faster acceleration with the same horsepower.
Heavier wheels may not cost you any flywheel horsepower, but you will put less power to the ground because more power is used turning the extra mass of the wheels.

It's also why your car will run faster times if you're not towing a boat.

 

I have a 98 Camaro v6, so it has the factory painted 16s(five spokes). Tonight, I swapped them off for ss chrome 10 spokes. Putting them on I found out that they are heavier, even by a good little bit. But, after driving it, I found that there really was not that big of a difference in acceleration. I actually didn't even notice a difference(besides better response). So...go for them.

 

I know it's come up several times, but take me back to school: if I go from 16" rims to 17" rims, and the rims stay the same weight and the tire stays the same overall diameter, will that effect performance, and if so in what way and why?

Thanks.

 

Lol seriously one of the morst mistakes you could make we might as wheel call rice on this one.

 

You can easily off-set the heavier weight of a set of chrome wheels by taking a lug nut or 2 off per wheel. It all depends on how much quicker you wanna go or how much power you wanna gain back compared to when you had the stock 10 spokes.


..................................

 

ok ok but what about me? i have 18"s on the back and 17"s on the front...what does that mean? huh does that mean im screwed cuz of all that rotational mass on the rear wheel?? does it slow the front one down..oh wait its factory.. omg hahahahaha

 

Well it is a dynomometer phenomena, you will "make" more power on a dyno with lighter wheels as nj02vette said.

I had this happen when I swapped from stock 16s to much heavier 17s, showed less power (butt dyno was negligible.)

However insignificant, lighter wheels= better acceleration (in terms of both increasing and decreasing).

Is it worth spending 3k for race wheels for a street car? No. If you are trying to get competitive though I would seriously look at getting some stock or lighter wheels. (Like GTO spare tires for skinnies at the strip.)

 

You obviously won't lose BHP, that is ludicrous, but you will lose RWHP, it is physics. However, if you have 16's and go to 17's where the 17's weigh the same, you could actually see an increase in power because the tires will have a smaller mass making the over all combo less.