Drag tires for 20's?
11-30-2004, 05:42 PM
#31
Originally Posted by Todd@DiscountTireDirect
To answer with any certainty Id have to do soem Math which I really dont want to do, but theres a few variables here. The biggest issue with the bigger wheels is what was said above. Moving the mass farther from the center of the wheel. If that 55lb 16" wheel was a hideous design and most of the weight was in the center, then it had tiny lil spokes that went to the lip..maybe not. The 20"+ tires are also more reinforced as well so they are generally heavier regardless of overall diameter. This comes in to play with me. My wheels are light 22" 33lbs. But my tires are 45lbs. And all that weight is obviously as far from the center as possible
11-30-2004, 05:43 PM
#32
Originally Posted by Bryan TTM
its not blown...most of these guys are searchin for 100ths when they should be lookin for 10ths...just part of the learnin curve bein all caught up in the over-tech psych 
Editied my last post......this is some very interesting stuff we have stumbled onto.
11-30-2004, 05:53 PM
#33
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Id like some 20' drag radials. Some of us rarely see the track, but would like to have a little added traction when playing on the street. Im sure eventually some one will come out with some.
11-30-2004, 06:00 PM
#34
Originally Posted by greentahoe
Id like some 20' drag radials. Some of us rarely see the track, but would like to have a little added traction when playing on the street. Im sure eventually some one will come out with some.
and it came from above, so let it be written
11-30-2004, 06:01 PM
#35
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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
11-30-2004, 06:07 PM
#36
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
I'm of the mind that most people rollin 20's, are rollin a larger than stock tire diameter as well. That's where the most penalty will be found.
That said, buy light 20's to minimize the above effect, or is that affect?
11-30-2004, 06:20 PM
#38
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Originally Posted by BlownChevy
**EDIT** I just asked that question and you are 100% correct. Very interesting stuff if you ask me.
unklej stepped in and explained what I was trying to say.
11-30-2004, 06:27 PM
#40
Originally Posted by unklej
Not a problem.... It's nice to get a chance to dust off my engineering degree for a change... lol
Jeff
Jeff