How much heavier will my new 18 inch rims be compared to my stockers?
05-30-2003, 01:56 PM
#1
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Join Date: Nov 2000
Location: Calgary
Posts: 676
How much heavier will my new 18 inch rims be compared to my stockers?
Today I'm getting my 18 inch Tubolares with 275/35/18 Kumhos mounted on my TA, and of course garaging the stock 16 inch 5 spoke rim.
How much heavier are my new rims going to be compared to my old? I was only jsut thinking of this when I was lugging around my new rims.
Thanks,
Leo
How much heavier are my new rims going to be compared to my old? I was only jsut thinking of this when I was lugging around my new rims.
Thanks,
Leo
05-30-2003, 04:45 PM
#5
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Join Date: Jun 2002
Location: Hagerstown, MD
Posts: 736
The actual weight is not much different,what will slow your car down is ineartia which it the ditribution of mass around the axis of rotation ineartia is mass times radius squared, the radius is the major factor so your going from 8" to 9" so 64 to 81 times the mass difference. but all these numbers are small compared to your cars acceleration. torque equals rotational acceleration x ineatia. Take your tires cicumference and use a 14 sec 1/4 mile as example and figure out the angular acceleration of your tire in radians/sec. if your tire is 24" this is 30 rads/sec2. Lets assume your stocker is 45# and the 18 is 55#. inertia for stock is 0.62 vs 0.96. so your torque loss will be the differnce of the torque equations. so 28.8 - 18.6 = 10.2 Ft# will be your torque loss per tire, approximitly which is really addtional load creted by the larger mass when you run a 1/4 mile. This probably won'tbe the case though because your new wheels are probably dsigned well enough that they weigh less than the stock ones.
Last edited by WhoBetter?; 05-30-2003 at 04:51 PM.
05-30-2003, 05:48 PM
#7
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Join Date: Jun 2002
Location: Hagerstown, MD
Posts: 736
Ahh, I'm just goofing around, seems resonable though, doesn't it?
A 3600#car doing a 14 sec 1/4 mile will require a constant appllied force at the wheels of 10.31 ft/sec2 X 3600/32.2 = 1,153# (F=MA), you got like 3.5 axle gears so you would need 330 ft# engine. not including wind drag. I'm just a college student, i can't explain why a dyno chart shows 300/ whatever ft# torque when like I said you need a force of 1153#, it would take a few hours to completly figure it out and match the numbers in both directions and figure out the reltionships and what's really happening.
A 3600#car doing a 14 sec 1/4 mile will require a constant appllied force at the wheels of 10.31 ft/sec2 X 3600/32.2 = 1,153# (F=MA), you got like 3.5 axle gears so you would need 330 ft# engine. not including wind drag. I'm just a college student, i can't explain why a dyno chart shows 300/ whatever ft# torque when like I said you need a force of 1153#, it would take a few hours to completly figure it out and match the numbers in both directions and figure out the reltionships and what's really happening.
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