Lightning

Technical drag racing question

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Old Jan 21, 2002 | 07:31 PM
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soap's Avatar
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From: Motor City
Technical drag racing question

OK so I know all of you have seen the new SRT-10 and I was actually "lucky" enough to see it at the auto show. I also remember alot of people ******* the 24"ers saying they were nasty.......and I do agree that they are a bit much, but......

Now for comparison sake lets use a 16" wheel add a tire and get 28" outside diameter. Take the Dodge's 24" wheel add a tire and get a 33" outside diameter. I am not sure of outside diameter of that combo maybe it is more maybe less, but this is just for comparison sake. Now since our trucks make gobs of torque I know we don't need a smaller tire than stock to increase our gears.

OK here is the technical part (kinda).......after one of "his" rotations he goes (2*pi*r=) 103.67". Now one of our rotations only goes (2*pi*r=) 87.96". So for equal rotations he is already pulling on us.........

Am I at all on track here or am I delirious.......

--Joe
 
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Old Jan 21, 2002 | 07:42 PM
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I think you are always delirious!! J/K
 
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Old Jan 21, 2002 | 07:48 PM
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Are you trying to say that the diameter of the wheel determines the final gear ratio? Well it doesent. The actual circumference of the tire determines gear ratio. You could have a 10 inch rim or a 30 inch rim, but if the tire crucumference is the same, you have the same gearing. If you are talking about rotational weight, it all depends on how heavy the wheel is......and stuff.

And what is "2*pi*r=) 87.96"
 
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Old Jan 21, 2002 | 07:52 PM
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I know.......that's why I mentioned the 28" o/d and 33" o/d.

But I am saying that since Dodge has a 24" wheel chances are pretty damn good that his o/d is larger than our too.

2*pi*r = one rotation of a circle!!!!!!!!!!



--Joe
 
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Old Jan 22, 2002 | 07:36 PM
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From: Motor City
ttt
 
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