Does rotational mass affect non-driven wheels?
 

That's it. So on my FF 1.5l Mazda if I wanted to go with really light weight rims such as the TE37 and didn't care about looks would I be better off just buying 2 rims for the front wheels because of the reduced power loss?

Rotational weight is rotational weight.
Remove as much as you can, where you can. Even though the rear wheels are not driven they have to be turned, so less weight to turn means more efficancy, or better preformance.

I understand that but that does not help my question. To rephrase it this way, if I got lighter rims on the back, would that still be like removing double the weight (that's what everyone says about rotational mass), or would it be the same at removing the back seat?
I was just confused because of the double weight deal about RMass and I always thought that the reason why it helps to get lighter rims is so that you lose less power to the ground. Hope my rambling helps.

Yes removing rotational weight is more effective than removing stationary weight,but in my experiance it's more expensive. It is even more effective in the drivetrain, I have been told 1lb off the flywheel is like removing 7 from the car.

Thanks! I realize it's expensive but I guess if anyone wants to chime in, this is my question broken down. Is it like
1) RMass on drivetrain>RMass unpowered>Stationary weight or
2) RMass on drivetrain=RMass unpowered>Stationary weight or
3) RMass on drivetrain>RMass unpowered=Stationary weight or
something else that I'm not thinking of. Sorry. Don't mean to be a PITA, just curious.

What you are trying to do is figure out how much energy is being absorbed by trying to spin up the wheels - and it doesn't matter if the wheels are driven or not. The energy required is the same.

The term you are looking for is "Inertia" - the resistance an object has to movement. Normally inertia is proportional to mass - except when we talking about rotational inertia - where it is proportional to the radius of rotation squared times the mass. In wheels, much of the mass is concentrated in the rim area (as opposed to the spyder) - BUT - differences in mass (weight) might not be. Put another way, a lighter wheel might not have a lower rotational inertia since it is possible for the differences in weight to be in the spyder area of the wheel.

And lastly, consider that if a wheel is 4 pounds lighter and the vehicle weighs 4,000 pounds, that's only 0.1% - and considering 4 wheels and double the rotational inertia loss, we're talking about about a 1% change. You may want to consider that when costing these things out.

Nice looking wheels
 

Forged aluminum wheels can have the same strength as steel with less weight, which is very interesting albeit expensive. How much weight will you save?

The total weight of the vehicle is the primary factor for energy and power calculations, especially when looking at the acceleration phase. But 10 lbs off of 3000 lbs makes little difference in acceleration power and even less at constant speed.

The rotational weight (or mass) of the wheels, also known as the rotational inertia, is the primary factor for calculating the energy and power required to accelerate the wheels. So dropping 10 lbs off of 80 lbs in the drive wheels would make a significant difference in acceleration power of the wheels. For equal acceleration of the vehicle the lighter wheels would be slightly more fuel efficient. Not sure if you could measure it in normal driving, but it might show up in drag racing.

So it will take less power to accelerate a lighter wheel up to a given speed, but a heavier wheel will have a higher angular momentum and kinetic energy and will coast longer.

Thank you. I believe that answers my question!

Counterpoint.

If it's a lightweight car with heavy wheels, it can work out to 2%. (my car - 2500 lb, lose 6 lb per wheel) That's in the same ballpark as power increases claimed from air intakes or exhaust mods, BUT it also benefits ride quality, handling and braking.

You can also move to a smaller diameter wheel. That will improve rotational inertia even more, along with usually being lighter due to the smaller size.

This is an upgrade I'm considering, downsize from my heavy 16" alloys to a 15" sport wheel. Stock rims are 17 lb each, Konig Helium 15" are 11 lb and cost under $100 each. I'd love a set of TE37's but the price... :eek:

BUT not reducing the overall diameter of the tire.
infact please consider increasing the overall diameter of the tire. (many threads on this topic.