Drag tires for 20's?
11-30-2004, 06:46 PM
#51
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Originally Posted by moregrip
I suppose with your above formula, someone; anyone could make a generic QRG to see what effect that larger/different tire/wheel combo might have on performance vs. stock (base #'s)
Here is a web site that had a whole bunch of equations...
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html
I hope it helps
11-30-2004, 06:46 PM
#52
Originally Posted by unklej
Funny thing was I asked Blownchevy for a job yesterday. I guess I will just have to settle working for my current company (its not all bad, my business is booming 
)
Jeff
)Jeff
11-30-2004, 06:47 PM
#53
Originally Posted by unklej
Here is a web site that had a whole bunch of equations...
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html
I hope it helps
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html
I hope it helps
but thanks.
11-30-2004, 06:54 PM
#54
Originally Posted by unklej
Funny thing was I asked Blownchevy for a job yesterday. I guess I will just have to settle working for my current company (its not all bad, my business is booming )
Jeff
)Jeff
Had nothing to do with work!
All personal
damn women
11-30-2004, 06:55 PM
#55
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Originally Posted by unklej
Well, its all about moments of inertia. The futher the weight is from the axis of rotation, the more inertia you have and the more force it takes to get wheel spinning. Now if you want to bring in overall tire diameter, that throws everything off. You would really need to do a hand calculation to figure out which is the lesser of two evils.
But, to break it down, a 30" od tire on a 16" rim with a weight of 55# will have less inertia than a 30" od tire on a 20" rim with a weight of 55#. Thus, the 16" combo would require less force to break the inertia forces....
The Moment of inertia equation for a hoop is I=mr^2 (m = mass, r = is the radial dimension where the mass is located) So lets just guess that the mass is located on a 16" rim combo at r = 10" and r = 12" on a 20" rim combo.
I(16") = 55*10^2 = 5500
I(20") = 55*12^2 = 7900
For linear motion F=ma, for rotational motion t=I*angular acceleration.
So with the same anguler acceleration you would need more rotational force.
Just like Newton said, bodies in motion want to say in motion and bodies at rest want to stay at rest. Inertia is the key.
Jeff
But, to break it down, a 30" od tire on a 16" rim with a weight of 55# will have less inertia than a 30" od tire on a 20" rim with a weight of 55#. Thus, the 16" combo would require less force to break the inertia forces....
The Moment of inertia equation for a hoop is I=mr^2 (m = mass, r = is the radial dimension where the mass is located) So lets just guess that the mass is located on a 16" rim combo at r = 10" and r = 12" on a 20" rim combo.
I(16") = 55*10^2 = 5500
I(20") = 55*12^2 = 7900
For linear motion F=ma, for rotational motion t=I*angular acceleration.
So with the same anguler acceleration you would need more rotational force.
Just like Newton said, bodies in motion want to say in motion and bodies at rest want to stay at rest. Inertia is the key.
Jeff
r= radius correct ?
Well last time I checked. 16/2=8" ?
And wouldnt the 20" diameter be a 10" radius ?
So the equation would go like this:
I(16") = 55*8^2 = 3520
I(20") = 55*10^2 = 5500 ???
I mean maybe I am wrong......
11-30-2004, 07:01 PM
#58
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Well, yes and no. I was making a guess where the heart of the mass would be on the wheel and tire combo. So the overall diameter would be 30". Anyway, that equation is really not valid if you want to get the actual moment of inertia for a rim and tire, it was just used to simplify the explanation.
If you want true moments of inertia, I would recommend modeling the rim in tire in a 3-D CAD software and use the mass calculator to fine the true moment of inertia.
Jeff
If you want true moments of inertia, I would recommend modeling the rim in tire in a 3-D CAD software and use the mass calculator to fine the true moment of inertia.
Jeff
11-30-2004, 07:03 PM
#59
Originally Posted by unklej
Well, yes and no. I was making a guess where the heart of the mass would be on the wheel and tire combo. So the overall diameter would be 30". Anyway, that equation is really not valid if you want to get the actual moment of inertia for a rim and tire, it was just used to simplify the explanation.
If you want true moments of inertia, I would recommend modeling the rim in tire in a 3-D CAD software and use the mass calculator to fine the true moment of inertia.
Jeff
If you want true moments of inertia, I would recommend modeling the rim in tire in a 3-D CAD software and use the mass calculator to fine the true moment of inertia.
Jeff