Circle.

I just put 4g wheels and they fit nice.

er... 3.7 * 10 * 4 * 6 <> 24... it = 888. >.> lol I am confused.

EDIT: Ah ok, rereading the info there is a typo on the example.

it should be (3.7 * 10)/6 *4 = 24.7

(Note: You cna do the caclculations in any order, I just wrote it that way because rule of thumb si to follow the order of operations)

I didn't write that equation. But one thing will have to be changed for it to be applied to our cars.
It's like one of those critical thinking questions from highschool math class. You have to know how to look at it. But the rotational inertia formula only applies to the tires receiving drive. Newtons 1st law of motion says, in a nutshell, that whatever moving state an object (tire) is in, it will tend to remain in that state untill a force is acted upon it. In reguards to automobiles,the external force will inherently be the engine, acting only on the tires it drives, along with other rotating components such as the axles, flywheel, ect. (But for now were just focusing on the tires, and rims). Therefore the lighter the object, the less force is needed to cause it to move. In other words your engine dosen't have to work as hard to make the tires spin, in which case the leftover horsepower can then be utilized to overcome other forms of resistance a vehicle creates.(Wind drag, gravity, ect.)
My argument is thas as the formula for rotational inertia (case in point) and "dead" weight are very different, it is plain to see that the rear (trailing) tires'/rims' weight reduction will yield a much less significant advantage with respect to minimizing resistance. So unless the vehicle is awd (which our cars are not) the formula should read as follows
( 3.7 lbs savings
x 10 static rule of thumb )
x 2 wheels driven
x 6 horsepower rule of thumb
------------------------------------------
12.3 horspower gain on 2wd models

I should have caught it before I posted it before. SRY (IMG:style_emoticons/default/auto.gif)

LMAO good call, I didnt even catch that, myself.

Also, just a little bit of knowledge being passed down, but if you are wondering how the speed sensor on a car works:

The programming of a speed sensor is actually very simple and strait forward. a constant is declared in the script which is the wheel diameter. For this example we will use 17" (Remember, you have to include the tire too, but for simplicities sake, we will ignore the tire). the sensor then divides the diameter by 2 resulting in the radius. (8.5"). Since the circumference of a circle = the distance traveled in 1 rotation, we can find use the following formula to find this:

C= 2 * (pi) * r

Where C = the circumfrence and r = the radius

So we have:

C = 2 * (pi) * 8.5

Thus,

C = 53.38"

After that calculation is performed, the sensor counts the number of rotations per second. This various with horespower, wind conditions, elevation, etc.. etc.. For our example we will use 3 rotations per second (Yea, I pulled that number out of my ass... lol) After the rotations are counted the sensor multiplies that number by the circumfrence and stores the result in inches per second, like so:

speed = 53.38 * 3

Thus,

speed = 160.14 in/s

After all this is done, the sensor simply converts this data to the appropriate units. (MPH for the US, and KmPH for everywhere else.)

Speed = 9.09 mph

So in conclusion, a 17" wheel (wihtout the tire lol) rotating 3 times per second is, in fact, traveling 9.09 mph. Seem slow right? Well think of high small that wheel is if you include the tire. :x

You can substitute the radius variable circumfrence formula, with the diameter variable and skip two steps.

Ex. 53.38=2*3.14*8.5 = 53.38=17*3.14

If you really want to get technical, a loaded tire is an ellipse, not a circle. So you will have to find the elliptic integral and write out about a page and a half of equations converting expansions. And after that decide which of the two approximations you are most confortable with. (depending on how good at COS you are). And after all that it's still not accurate because turning, hills, wind (creating downforce), will change the elliptic integral of specific tires hundreds of times a second. (IMG:style_emoticons/default/wacko.gif)

It is cool to see the way putting a larger or smaller tire on a car can confuse the hell out of a speed sensor. And that only couple of inches on a tire can make a huge difference in percieved MPH. (Im leaving out the tires too) (IMG:style_emoticons/default/tongue.gif)

Ex. 16in. rim (on my car now) * 3.14 (pie) has a circumfrence of approximately 50.24in.
So if the sensor reads my tire is spinning 15 times a second it would calculate 15*50.24=753.6in./sec.

And convert it (5,280ft in a mile = 63,360 in. in a mile) (60 secs. in a minuite)
So it would take me 84.07 seconds to travel a mile based on 63,360 (inches in a mile) / (753.6 / 12)
Totaling 62.8 ft/sec
And since 1ft/sec=.68 mph the formula 62.8*.68=42.7mph at 15 RPS(Revolutions per second)

But lets say I put a 12in. rim on the car
Following the same thing I did above the MPH will be reduced to 34.69mph at 15 RPS(Revolutions per second)

This difference increases with speed
Lets say my sensor is reading 25RPS with a 16 in. rim
It will register 71.17 mph
On a 15 in. 66.72MPH
On a 14 in. 62.27MPH
On a 13 in. 57.82MPH
And on a 12in.
It will register 53.38mph
NOW THATS A DIFFERENCE