Nuke

Damn! Beat me to it...

MoneyShot

thats what she said... oh wait...

forensicsteve

You'll also get less side wall flex when trying to hook up. Less side wall to bend means eaier to spin the tires out and loose your traction.

+1. 20" may have low profile tires...not good for dragracing. More sidewall helps with absorbing the shock at launch, reducing spin and wheel hop.


1st pic shows closeup of our rear tires taking the hit of a transbrake launch (1.4 short time)

2nd pic is complete image, showing both fronts off the ground

Norm Peterson

Ummm, rotational mass (rotational moment of inertia, actually) is entirely separate from gear ratio. What rotational inertia effectively does is "bleed off" a bit of the torque that would otherwise get turned into forward traction force down at the contact patches of the drive tires. Call it a parasitic loss.

For anything other than timed competition (whether that be drag racing or autocross), a couple of lbs per wheel won't matter at all. Even 5 lbs/wheel only ends up being worth a tenth of a second or two on a 40-ish second autocross (by actual test).


Rotational moment of inertai is partly about weight and partly about how that weight is distributed. Mass that's further away from the axle represents relatively more rotational inertia than mass that's close to the center hole. That's really all that can be said about any comparison between/among wheels without knowing more.

If you like to do math (or feel like punishing yourself), a rough approximation that seems to be good enough most of the time divides the wheel weight 2/3 to the barrel (using the MOI equation for a cylindrical ring) and 1/3 to the wheel center section (MOI equation for a circular disc).

As hinted at by 157db, and if you really like to do MOI math, you can also split the tire up into separate tread and sidewall masses for better overall comparison purposes.


Norm

rpietro

To answer 157db, here are the Webster's dictionnary definitions of the words "wheel" and "rim"

Wheel:

With its cycling/compaction of momentum, reduction of friction and resulting lower amount of work necessary to move items, the wheel is essential in transport. On many wheeled vehicles (for instance, automobiles and motorcycles), a rigid wheel does not come directly in contact with the surface, but instead serves as a mounting point for a flexible tire, usually filled with compressed air.

Rim:

The outer part of a wheel to which the tire is attached.

AmericanSpeed

Can always rely on Norm to give a concise engineering perspective to answer to the question.

Riptide

+1 Norm is a great resource.

He's the one I always pester with questions when something like this comes up.

157dB

jUST FUNNIN. Geeeze. The wheel is the tire/rim assembly according to Ford.
Ford dont work by Websters definitions. They have their own.

Yep Norm has the suspension answers
right up front wit out thinking bout it.

danzcool

Okay, here is how I think the issue comes into play.
Assume you have a 17" rim and a 20" rim that both weigh the same, the outside of the rim has to travel 283" per revolution on a 17", where the 20" has to travel 314", the low down is that the 20" has to be moved 10% farther , therefore requireing 10% more energy with the weight being equal... problem is that the weight is rarely equal.
For another example think back to break dancing or ice skating, when the guy is spinning on his back or the ice skater just spinning, they pull their weight in closer to the body which results in the ability for the energy already expended to increase the rotational speed.

I think that's the theory anyway.

Cusp

Nice. Now what about the tires? Say the factory ones. The 17s come with 235X55 and the 18s come with 235X50s... so less tire. If the 50 series tire is lighter than the 55, by how much. 3.5 pounds worth?