derelict

I had a conversation a few nights ago in EQ with a guildmate and we got to talking about my car and wheel sizes. Here's an excerpt and I'm curious how true it is:

him:"If you can stay low with them 18's, or better yet 19's, you'd get a slightly faster take off than the 17's'"
me:"How's that work?"
him:"Torque basically. The moment of torque is created by the perpendicular distance from the point of application of the force, to the outside edge of yer tire, multiplied by the force applied. You can't do a lot to incrase the torque very easily, but you can increase the distance with bigger wheels.'
me: "Now my head hurts"
him: "Hehe. It's just leverage"

Calling all engineers....

mclarenf3387

Theoreticly yes, but rotational inertia is hard to over come, and since most 19's weigh a lot more than 17's, you'll usually be slower.

edgalang

True...in a perfect world where you do NOT consider mass. In this case...that statement is false.

darrinb

i was thinking with 19's u will spin less cause they are heavier

mclarenf3387

The spinning can be decreased with some nice sticky rubber. Being heavy means they won't spin as much, but due to the increased weight, its harder to geet them moving.

Mechauto

Sorry for the length, but it should clear this problem up.

First, don't most people usually compensate for the larger rims by buying narrower sidewalled tires? If the overall diameter of the TIRES are the same you will have NO change in acceleration, regardless of rim size.

Secondly, The mass of the rim will likely have a very minimal effect in this case. However, if you want to get really picky about this, you would also have to consider where the mass is distributed on the rim, not just its overall mass. For example, if two rims have the same overall mass, but one has more mass near the edge and one has more mass at the center, the one with more mass at the edge will take more torque to get it started.

Finally, your friend's explanation of torque is correct, but it works opposite to what he/she thinks. His equation says the following:
Torque=(radius)*(Force), which is correct.
But, by increasing the radius, you increase the torque REQUIRED to turn the wheel, not the torque you get. His esplanation works if you are talking about the length of wrench necessary to loosen a bolt (increase the length and you get more torque). But think of at as if you are turning a bolt instead. Would you rather turn a 5/8" bolt, or a 5/8" bolt with a 10ft. pole attached to it. I'll take the bolt without the pole attached.

GEB

All this makes my head hurt!

If you like the way the 19's look, go for it. Plus, I always thought that going with larger rims decreased the size of the wheel wall and therefore decreased the amount of sideways roll thus increasing the grip in the corners ...

sgmotoring

I thought you need more side wall to have a better launch? Look at those Top Fuel Dragster.

Titand19

My head hurts too, i guess i'll have to treat it with some herbal remedies

pimpscls

the bigger the better for handling but for launching and cornering 18's are the way to go even though 19's look good

NiteQwill

A good launch requires 17's or even 16's. The less rotational mass and unsprung weight the wheels have, the less energy required to spin.

A tire's sidewall must 'give' when torque is applied to it, as well as running a higher pressure. The reason you see draggers run high profile tires, not low...

EricL

Yes, they do...

If you look at the front-to-rear footprint of a 19" vs. 18" vs. 17" vs. 16" (etc), the tires with the smaller wheels generally have larger front-to-rear footprints. As NiteQwill mentioned, the tire's sidewall on dragster's (etc) are pretty compliant and you can see them "wrinkle"/"crinkle" on take-off. They have very low pressure and have a very long front-to-rear foot print (they simulate a car with under inflated tires [low loaded radius]).



The other stuff related to the very first posting in this thread:

RE: Wrench analogies -- a long wench gives more leverage that a short one, but you move the end of a longer wrench more distance than a short one. (The wrench is traversing the same arc length, but the lever arm is increased.) Note: it would probably be better to just note that the poor student was thinking about “levers” and confused this with car gearing; it’s related, but…

However, tall tires actually == taller gearing. If you had a much taller tire, it would be like starting out in second gear. That is not a good thing most of the time (might be useful on an icy road). If you want lower gearing, and MORE torque to the road, you go with a smaller tire *and* this has *nothing* to do with wheel size!!!

A smaller tire/wheel (17" wheel + 235/45-17) combo will generally have lower rotating mass (rotational inertia) that a 19" wheel + 235/35-19 combo. Less mass means you can accelerate more quickly. Heavy wheels are not a good thing, and heavy wheels with large diameters are even worse. The inertia is directly related to the radius of the wheel. Since rubber is generally lighter than steel and/or aluminum, you’re putting heavy stuff towards the outside of the wheel where it will hurt performance most.

vinarnold

the 17 would be better.

Linner

is this narrowing down your options on a future purchase of a rim size? if it is who cares, go for what u like/looks the best, if not you made my head hurt too with all this physics!!

Linner

is this narrowing down your options on a future purchase of a rim size? if it is who cares, go for what u like/looks the best, if not you made my head hurt too with all this physics!!

acuraman32cl

that true, get a ruller and put a weight 6' away from one end and hold the other, then do that same at 12', the rotaional force doubles. it is harder however to move a larger object, as the larger the diamter gets, the larger the rotational inertia is, so the engine does more work to turn the wheels. Also. with a large circumfrence, you travel more per wheel turn, making you go fast at comparative engine rpms in relation to you old 17s

OUTRAGEOUS CL TYPE S

my 19inch tires and rim weight 50lbs on the dot, i want to measure my stock wheels and rims and c. i have a feeling its 50lbs as well,

EricL

The tire/wheel combo might weigh the same, but if the wheel is bigger (19" vs. 17"), and doesn't have "special" construction, you will lose acceleration.

The weight on the outside of the wheel is really bad for acceleration and it's effect is like adding 3x static weight to your car (like adding 3 times as much weight in the passenger seat, trunk, etc). As you move to the inside of the wheel, you reach a point at the very center of the wheel where the "weight" (actually mass) is equivalent to just 1x static weight.


Some formulas and pictures to make your brain hurt:

I cut, edited, and pasted the following from a very old post that I put in the FAQ topics section a couple of years ago (it only talks about 18" vs. 17" wheels, but the general idea can be applied to 19" wheels):


However, assuming that an 18" wheel has the same construction as a 17" wheel, it needs to be about 2-5lbs lighter (depending on weight) to have the same rotational inertia. The rotational inertia is only indirectly related to weight.

For example, I start with two 17" wheels that both weigh 20 lbs.

Wheel #1 has lead spokes, but a very, very light outer flange/rim made of carbon fiber or titanium (so the weight/mass that makes up the 20 lbs is concentrated at the inside of the wheel).

Wheel #2 has carbon Kevlar spokes, but a very, very heavy outer flange rim made of a lead like compound (so this wheel is very heavy at the outside of the wheel). (The 20lbs is concentrated at the outside of the wheel.)

Wheel 1 would be much easier to spin up than wheel 2, but both wheels weigh the same.

If two wheels are identical in construction, it is possible to use a simplified equation to find the difference that a change in wheel size will make, here are some pictures, equations:

Simple example:



"Hoop About Central Axis"

I = MR^2

I = rotational inertia (more is bad)
M = mass of rim
R = radius

More complicated version:



"Hoop/Ring with Width (you know ID and OD)"

I = 1/2 M (R1^2 + R2^2)

I = Rotational Inertia
M = Mass of the wheels outer rim
R1 = ID of rim/flange
R2 = OD of rim/flange

blader

FYI, when I swapped my rims on I measured my stock rubber .. roughly 50lbs. but my new 18's with new rubber weighed 42.5 ..

jucee187

what if the larger rim/tire weighs LESS than the smaller rim/tire?

will it still affect the launch?

EricL

If you have really heavy 17" rims (cast aluminum with chrome for example) and replace them with some really light forged alloy 19's, you could end up with better acceleration from an "inertial" standpoint.

And, if you had factory stock 17" tires (not that sticky), and replaced them with some *very* sticky, lightweight 19" tires (max-performance and/or even stickier R-rated rubber), you could end up getting more traction *and* reduce weight at the very outside where it is very important.

For the heck of it, I'll oversimplify like crazy, and plug in the following numbers into the MR^2 equation (mass * radius squared) for 17" (8.5" radius) and 19" (9.5" radius) wheels.


Solving for mass_of_19in_wheel (relative to a 17" wheel):

mass_of_17in_wheel * 8.5^2 = mass_of_19in_wheel * 9.5^2

mass_of_19in_wheel ~= 0.8 * mass_of_17in_wheel

And, the conversion for an 18” vs. 17” wheels would be:

mass_of_18in_wheel ~= 0.89 * mass_of_17in_wheel


So, if you had a 17" wheel that weighed 25lbs, you would want a 19" wheel that weight 25 * 0.8 = 20lbs.

This makes a lot of assumptions like:

1. Wheels are of identical construction, design, and mass is distributed in identical proportions from center to outside (this is going to be the exception).
2. The tires' construction and weight are not considered.

jproy

Ahhh physics! This thread is very accurate and correct. It is not that difficult to understand. I have a minor in this area.