motor

jz04r1 · 01-30-2005, 12:31 PM

I'm trying to convince my friend that it's true but I haven't been able to find any articles through google. Can anyone help me or prove me wrong??? I think Car and Driver had a article about it but I don't have that mag anymore. Thanks in advance

Vansquish · 01-30-2005, 12:36 PM

I can prove it from a theoretical standpoint, but not with actual test results. Basically, as you increase the size of the rim, the moment of inertia of the wheel grows. This moment of inertia is the wheel's resistance to movement. In other words, if you have a very large wheel, the moment of inertia will be large and it will be hard to get it rolling in the first place, then, when you want to stop, it will have more angular momentum and will be more resistant to stopping than a smaller wheel.

What all this means is that the larger the wheels on your car, the larger their moment of inertia, and the harder it will be to get them rolling, or to stop them. So...if you fit large wheels to your car, you're going to increase your stopping distances, sometimes by a significant margin (esp. in the case of fitting 26" wheels to something that might've only had 17"'s to begin with).

gigdy · 01-30-2005, 01:27 PM

Go freshman physucks.

Vansquish · 01-30-2005, 01:35 PM

lol...definetely stuff from freshman year and HS....at this point we're talking quantum mechanics, tensor calculus and string theory.

graywolf624 · 01-30-2005, 01:40 PM

Inertia depends on object but always involves: mass * squared(radius) In case of a wheel shaped object: 1/2 m squared(r)
Torque= inertia * angular accelleration
torque=1/2 mass * squared(radius) * angular acceleration

Thus larger radius and maybe even addition larger mass of those larger wheels = increased inertia which means for the same torque less angular acceleration occurs.

For a similar reason smaller and larger tire/wheel ratios have a large effect on gearing and hense also top speed, acceleration, ect.

jz04r1 · 01-30-2005, 01:47 PM

cool thanks for the help

Vansquish · 01-30-2005, 02:08 PM

OK...we have a little bit of confusion still going here. Graywolf, the moment of inertia is more or less as you have described it except for your specific case. In fact, a wheel is a combination of the characteristics of a hollow cylinder, and a series of rods rotated about their midpoints, or if you prefer, something partway between a hollow cylinder and a solid disk. In any case, I=(1/2)mr^2 is not an accurate depiction of what happens with a car's wheel, as that is the solid cylinder case.

If one really wanted to calculate the moment of inertia of a car's wheel, it would be a sum of the different parts, i.e. for a simple 6 spoke wheel with solid spokes and a solid rim, the moment of inertia would be something like:

I = I(spokes) + I(rim) = 6*(1/3)ML^2 + mr^2 where L is the length of a single spoke (from the hub out to the rim of the wheel), M is the mass of a single spoke, r is the radius of the cylindrical part of the wheel, and m is the mass of the cylindrical bit. Then of course you have to add the tire onto that, and that comes to another term of the form, mr^2 but with different masses and different radii. Of course this is still quite inaccurate, as any wheel you might find on a car is a much more complicated shape than this simple depiction. If you really wanted to calculate the moment of inertia from a theoretical standpoint, you'd have to consider the hub, the bearings, the driveshaft, the reciprocating masses of the engine, the brake disk and so on and so forth.

As for torque being simply inertia times the angular accel, that's also not entirely accurate. The sum of all torques present is the product of the angular acceleration and the moment of inertia. Thus, for the case of a wheel on the road, you have to account for the torque applied by the coefficient of static friction, as well as that of the brakes, driveshafts, reciprocating masses of the engine etc...etc...etc...

graywolf624 · 01-30-2005, 02:14 PM

Im well aware. but I think your going over the audiences head if you get more technical then I did. When presenting this particular concept to a class in freshmen physics or likewise.. you need to factor out the extra shit to simplify. Esepeciallly in the case that not all wheels have spokes.. some literally are a solid cylinder with holes cut in the middle of moderate size. Obviously making the equation even more complicated..
Same with the friction and other forces involved.

In other words.. keep it simple when your explaining it. Otherwise I could write you a 50 page paper.. including the suspension influence from the new ic height. Theres a reason engineers use modeling software.

Assume a basic cylinder wheel of solid mass distribution and all other things being equal. Thats the way it was actually worded on my physics 1 final in college. Otherwise I have to write out math after finding out the specific makeup of his wheels, tires, suspension design, ect.

Vansquish · 01-30-2005, 04:50 PM

Originally Posted by graywolf624

Im well aware. but I think your going over the audiences head if you get more technical then I did. When presenting this particular concept to a class in freshmen physics or likewise.. you need to factor out the extra shit to simplify. Esepeciallly in the case that not all wheels have spokes.. some literally are a solid cylinder with holes cut in the middle of moderate size. Obviously making the equation even more complicated..
Same with the friction and other forces involved.

In other words.. keep it simple when your explaining it. Otherwise I could write you a 50 page paper.. including the suspension influence from the new ic height. Theres a reason engineers use modeling software.

Assume a basic cylinder wheel of solid mass distribution and all other things being equal. Thats the way it was actually worded on my physics 1 final in college. Otherwise I have to write out math after finding out the specific makeup of his wheels, tires, suspension design, ect.

If you look at my original post on this page I think you'll find I did explain it in the simplest terms possible, using no equations and no attempt at a mathematical explanation. The slightly more in-depth discussion was directed more at you and the other people here on JW with a higher understanding of the basic physics involved with answering the question posed. Since the explanation you gave was somewhat flawed even at the level at which you wrote it seemed logical to make things right.

graywolf624 · 01-30-2005, 05:14 PM

didnt read it .. sorry..

But yeah.

5vz-fe · 01-30-2005, 10:47 PM

Braking distance increase (also leads to higher brake wear)
Acceleration decrease (also leads to less gas mileage)
ride comfort decreases
cornering ability decreases

That's all I can think of.

DeMoN · 01-30-2005, 11:08 PM

As far as I know, the bigger the rim, the bigger the wheel = more grip... but thats the end of story. Worst accelleration, decelleration etc.

antonioledesma · 01-30-2005, 11:12 PM

at last... you were taking this to a "technical" level for an untrained mind

now, back in the real world, just checking out some raw numbers without a calculator and no more than your explanations (had those classes years ago, so I don't remember nothing

)... the braking distance that will increase is almost meaningless; doesn't it?
those are the kind of calculations that you consider zero or simply don't care.

this seems like my car's gas mileage of I consider I use a goatee

"oh shit, I lost $1 in this year"

SFDMALEX · 01-30-2005, 11:14 PM

Originally Posted by DeMoN

As far as I know, the bigger the rim, the bigger the wheel = more grip... but thats the end of story. Worst accelleration, decelleration etc.

The wider the rim the more grip, not bigger as in radius/diameter.

T-Bird · 01-30-2005, 11:17 PM

OK we actually tested this in my brakes class over the summer and there is a significant difference in braking distance between a 16" rim and a 19"/20" combo without upgraded brakes afterward he did add larger brakes though. And yes width makes grip but height can also but not nearly aswell as width.