Wondering if anyone has weight of the stock wheel by itself, the stock tire by itself, and the weight the tire mounted on the stock wheel. Just looking around for lighter wheels. I hear the Dropstars are only 30lbs for 22" but not sure. Thanks

Derek

I'm pretty sure the rears are much lighter than 30 lbs. I could be completely wrong as I have no data or measurements to back that up with, but I thought I read somewhere that our stock wheels were forged.

Wheels = Front , 36.2 lbs. Back, 37.6 lbs.
Tires = Front, 36 lbs. Back, 40 lbs.

Well, there you have it, looks like I was completely wrong! :o

Thanks alot!!!!

Now begins the quest for a 22" tire and wheel combo that is either lighter or the same weight as stock.

Derek

Dropstars (or ANY cast 1 piece wheel, for that matter) will weigh considerably more than the stock one-piece forged wheel.

Thanks alot!!!!

Now begins the quest for a 22" tire and wheel combo that is either lighter or the same weight as stock.

Derek

I am doing so eventually as well. Dropping weight on wheels tires will make more of a difference than intake/exhaust. You not only get better acceleration, but better braking & handling. Not to mention looks.

My new rims shipped from Davin yesterday. Should be in on Friday..... my new wheel/tire combo (22 x 9.5 front and 22 x 10.5 rear) will weigh 67 pounds each in the front and 73 each in the rear. So not much lighter, but lighter none the less.

Thanks alot!!!!

Now begins the quest for a 22" tire and wheel combo that is either lighter or the same weight as stock.

Derek
Just remember that a larger diameter wheel, even if it is slightly lighter, will hurt performance. It is all about mass moment of inertia. Here is a technical explination:
http://en.wikipedia.org/wiki/Moment_of_inertia

And if the OD is nearly identical?

Then the mass moment of inertia should be the same assuming same OD and same weight.

And if the OD is nearly identical?

And if the OD is nearly identical?

Great question. The link uses old school wheels as an example - old school wheels do not have tires.

If your OD is the same and the overall weight is less, who cares if you have a 22" rim. You could roll 15's or 24's, as long as your OD is identical to stock.

And if the OD is nearly identical?
Same thing. The larger the wheel, the farther out the heavier metal mass (the actual rim, not the spokes) is placed, even if the TIRE OD is the same. Here is a post from another forum I frequent:
Consider only the wheel/tire combo (I will refer to the wheel/tire combo as simply the “wheel”). To get the wheel rolling, you must do two things: accelerate the wheel’s mass in the direction of straight-line travel, and spin the wheel. I’m sure all of you know the equation F=ma, or (Force) = (mass)*(acceleration). This applies to the straight-line acceleration part. To spin the wheel, you use the equation (moment) = (mass moment of inertia)*(angular acceleration). The mass moment of inertia depends on the mass and diameter of the wheel and where that mass is distributed in the wheel (i.e. how much of the mass is close to the center and how much is closer to the perimeter). For simplicity (and to assume the worst case scenario), assume that ALL of the wheel’s weight is concentrated along the outer perimeter of the wheel. Then (mass moment of inertia) = (mass)*(radius)^2. Putting these two equations together gives you (moment) = (mass)*(radius)^2*(angular acceleration). Because (moment) = (force)*(radius), you can substitute this into the equation to get (force)*(radius) = (mass)*(radius)^2*(angular acceleration). Divide by (radius) on both sides to get (force) = (mass)*(radius)*(angular acceleration). Since (angular acceleration) = (acceleration)/(radius), you can substitute in and get (force) = (mass)*(radius)*(acceleration)/(radius). Radius cancels, and you get (force) = (mass)*(acceleration). (Note that this is the same equation as the one we used for straight-line acceleration). So… the TOTAL force needed to get the wheel rolling equals the force required to accelerate the wheel’s mass in the direction of straight-line travel, plus the force required spin the wheel. (Total force) = (straight-line-acceleration force) + (spin-the-wheel force) = (mass)*(acceleration) + (mass)*(acceleration) = 2*(mass)*(acceleration). In other words, a pound of weight in the outer perimeter of the wheel (in the tire’s tread) is worth 2 pounds attached to the frame.

Where tire OD comes into effect is your effective gear ratios. By putting on a larger tire (increasing the radius of the "wheel") you are in effect making your gearing higher.

But the axles cannot decipher between tire and wheel, to the vehicle, it is just combined weight. True, the center of gravity moves with a 22" upgrade, but GE2 is buying lighter wheels.

If you guys really want the answer, plug your numbers into one of the aforementioned equations.

Really interesting reading indeed. I know the serious power-heads here will be concerned about the possibility of a fractional power loss with larger diameter wheels, even if the wheel tire combo is lighter. To me it's just interesting mathematics. Not sure I totally understand or agree until I do the math myself.

Good thing I care more about looks than power :)
edit: another thought...... even though the new barrel (or rim, as you call it) is an inch further out, I'm not certain I buy the "heavier metal mass" argument. Yes, the metal mass is an inch further out on 22s, but the run flat tires (on the stock 20s) weight considerably affects the equation, shifting a much greater percentage of overall wheel and tire weight to the out edge of the OD. So moving the metal an inch outward, but lightening the outer load drastically doesn't fit well in your equations.

But the axles cannot decipher between tire and wheel, to the vehicle, it is just combined weight. True, the center of gravity moves with a 22" upgrade, but GE2 is buying lighter wheels.
?? What do you mean ? I am not sure you understand the concept being described here. The farther out that the mass of the wheel is placed the harder it is to accelerate.

Good thing I care more about looks than power :)
You may feel some power loss, but not a huge amount. If you are not a "power head" you should be fine. :)
One other thing, while your acceleration may be hurt somewhat, your unsprung weight will be less, so that should help your handling and ride a tiny bit (as long as the tire sidewalls are not stiffer). :)

Just add more horsepower.

Addmittedly, I do not mathematically understand all the properties you're speaking of, but I do know that a good portion of mass is determined by weight. The outer 5-7 inches (I'm guessing here) of the stock setup (including barrels and run-flats) --if the same OD-- would weigh considerably more than the outer most 5-7 inches of the 22s and rubber. Granted, the actual metal barrel is 1 inch further from the center point, but with so much less weigh on the outer edge of the OD, I really can't see the above equation working the same. If tire weights were equal between stock and aftermarket, and barrel weights were also equal, then increasing rim size would DEFINITELY fall into your equation.... but that simply isn't the case here.

Addmittedly, I do not mathematically understand all the properties you're speaking of, but I do know that a good portion of mass is determined by weight. The outer 5-7 inches (I'm guessing here) of the stock setup (including barrels and run-flats) --if the same OD-- would weigh considerably more than the outer most 5-7 inches of the 22s and rubber. Granted, the actual metal barrel is 1 inch further from the center point, but with so much less weigh on the outer edge of the OD, I really can't see the above equation working the same. If tire weights were equal between stock and aftermarket, and barrel weights were also equal, then increasing rim size would DEFINITELY fall into your equation.... but that simply isn't the case here.

You statement hinges around what I bolded. I think you would find that most of the weight of a wheel is in the barrel. Also, what tires are you using? I would be interested in the weight. Generally lower profile tires weigh more becasue of the extra reinforcements in the sidewall.

No, no.....I absolutely agree with you that the majority of the wheel's weight is in the barrel. But compared to the barrel of the stock rim, it would be lighter. My new tires will be 34 pounds in the front and 36 pounds in the rear. a few pounds lighter than the run flats, each.

Maybe you are right. Without testing it would be hard to say. Let us know what you feel after you change them.

I feel stupid after reading that. :eek: Any pics ge2?

No-- no pics yet. But I know they shipped from back east on Monday, so they are on their way. Tires are already here, and the new rotors and pads will also go on at the same time. Maybe Friday or early next week. The wheels should be a pretty unique look.

Anyone know why these are so much heavier than the LX SRT8 20's?

Anyone know why these are so much heavier than the LX SRT8 20's?


Another good question. I have been wondering the same since I sold my Charger SRT. That wheel set up felt like a difference of 10lbs+ per for sure. I know the tire weight between the F1 and the run flat RSA is 4-5 lbs difference on its own. I think the rims are another 4-5lbs for sure as well. Possibly to do with stabilizing a heavier 4700lb SUV vs. the 4200lb Charger?

What is the width of the LX rims? Ours are 20x8's and 20x10's. Also, we run 255/285's vs. their 245's.

What is the width of the LX rims? Ours are 20x8's and 20x10's. Also, we run 255/285's vs. their 245's.

I believe they are 20x9 all 4. They run 245 front and 255 rear stock with the F1 tire trim.