ARD T2

iTrader: (22)

Yes weight and diameter. You have to think of it in terms of "moment of intertia". So a 16" wheel and tire package that weighs the same as an 18" package will still be different. You will still lose power with the 18's. That's my lamen version. hehe.

Rishie

jgrewe

iTrader: (1)

Yes lower unsprung weight is better than more. I think the issue is rotational mass, think of rims/tires as 4 flywheels you have to spin up and then slow down. The only better place to lower weight is stuff that spins at engine speed. Also, smaller rims are better because part of the equation includes a 'distance from the center-squared' variable.

KevinK2

Regardless of using 16", or 18" rims, if you have the same OD tire, the inertial "equivelant dead weight" of the heavier rim and tire will be about 60% of difference in scale weights.

The inertal scaler for rims&tires does exist Jimlab, it's just a small effect.

A 10 lb heavier tire + wheel will also put 6 lbs in the trunk due to inertia. The theoreticl limit is if all the added weight is at the OD of the tire .... say embed extra 10 lbs of lead weights in the tread of your current tires. In this worst case, the inerta of the extra 10 lbs is like putting another 10 lbs in the trunk.

pm me an email address and I'll send my spreadsheet that looks at rim size, tire size, and gearing due to any change in the rolling diameter.

DamonB

Same applies to flywheels. Just because two flywheels weigh the same doesn't mean the engine sees the same load from them. The flywheel with more mass near the center will be accelerated faster than another even if they both weigh the same due to differences in their inertia moments.

For this same reason tire weight has a bigger effect than wheel weight on the moment of inertia of the wheel/tire combo. Let's take two wheel and tire combos of the same size that both total 40 pounds. One combo has a 15 pound wheel and a 25 pound tire. Another has a 20 pound wheel and a 20 pound tire. Even though the total weights are the same the combo with the heavier wheel will actually perform better. Its moment of inertia is less because more of the mass of the combo is concentrated towards the center.

You will absolutely never go wrong by getting the lightest wheels and the lightest tires you can though.

DamonB

If you don't have friends you can do the same thing with one person in an office chair. Have them spin themselves slowly with arms and legs extended. Pull arms and legs in and RPM increases. Extend them out and it slows down again. Same example, fewer friends required

Smilodon

Damon: Agreed. And. Agreed.

Now, as far as unsprung mass goes, the higher the ratio of sprung mass to unsprung mass, the faster the springs can slap the tire down onto the asphalt when it encounters a bump, or when the tire goes over a dip.

If a spring, for example has, at any given position, 650lbs of compression, and the wheel/tire/suspension thing has an effective mass of 32.5 lbs, the spring (ignore the shock damping, here) will be putting 20gs of acceleration on the tire to keep it or get it onto the asphalt.

This is very simplified, I know, but basic idea is the gist of what I am aiming at.

Now, if you take the same vehicle and reduce the unsprung mass to half, so that the wheel/tire/suspension has an effective mass of 16.25 lbs, the same spring will accelerate the tire downward with 40gs.

The rougher the road, the more important it becomes to have high sprung mass to unsprung mass ratio, and stiff springs.

The higher the lateral load, whether it is braking, acceleration, cornering, or any combination of the above, the more important it is to have low unsprung mass.

This is why solid axle rear ends are not as good as independent suspensions, because when a solid axle goes over a bump, the ring and pinion gear, pumpkin, and axle shafts all have to be moved by the bump upwards, and then by the spring downwards.

In an independent rear suspension, the ring and pinion and center pumpkin stay solidly mounted to the underside of the car, and do not travel with the wheel. Thus, they do not have to be pushed back down by the springs, but are instead part of the mass that the springs are backed up by to push the wheels down.

So, in short, for handling, braking and acceleration, try:
A) Reducing the polar moment (rotational inertia) of the wheels, tires, and brake rotors
B) Reducing the overall mass of the wheels, tires, and brake rotors.
C) Reducing the mass of the unsprung suspension pieces, such as half-shafts, rear axle, suspension A-arms, shocks, springs, anything that goes up and down on the suspension
D) Reducing overall vehicle mass

It is my guess that the Maxx-Blingg chromed cast/forged/whatever aluminum rims are one of the worst things that you can do for performance. Larger diameter, more weight, drastic increase in polar moment, wannabe gangsta dumbness factor squared, etc.

Better to leave all the soundproofing in your car and get some lightweight underpinnings than to strip all the soundproofing out of your car to shave that extra 45 lbs, thus making your interior sound like a cheap tin can taped to a bucket of iron tetrahedrons, which is rattling drastically while hanging from an open tractor exaust.

Then add all that weight back on and then some with 27 inch diameter wheels, 24 inch diameter triple-thick cast-iron rotors with so many pistons in the calipers that they completely hide the brake rotors from sight, and could stop a 747 upon landing without overheating.

I mean... really. How is it that the Shadow Can-Am racer in the 70s, with its, what, 1000 hp, 240 mph speeds, had ten-inch-tall front wheels and didn't explode in a thermonuclear rice cloud when they tried to stop it>?

Could it be that infinite wheel diameter and six-foot-tall brake rotors are not quite so important as your local parts salesman says they are, hmmmm?

Do more with less. Not less with more.

There ARE good brakes, wheels, and tires that ARE less than 24 inches in diameter. And less than 17 inches. AND less than 15 inches.

I know that a larger diameter tire puts a longer tread patch on the road without increasing width, rides over small bumps more smoothly, and attracts some of the most morally outstanding women in the history of the universe, but, performance-wise, they are not utterly necessary.

Formula One cars have tires that are about 25 inches tall. Amazingly, the wheels are not 23 inches tall. Now, why is it, that in a racing formula where the wheel height is only limited by the outer circumference of the tire, the wheels are not Maxx Blingged out to almost touching the outer circumference of the tires?

I'm not implying that people who spend hundreds of millions of dollars a year on tire and wheel technology know less than your average Maxx Blingg wheel and tire salesman. No, on the contrary, their wheels are what, fifteen inches in diameter? When they could have pretty well any diameter they want, why, oh, WHY!?!?!? Do they have wheels that are only one inch (or so) larger than a 1985 GSL-SEs????????

Somehow, they do not need 27 inch diameter, 55-pound brake rotors. And the matching 32-inch diamter cast lead chrome-dipped wheels with the 106-lb spinners. Perhaps you don't either.

KevinK2

DamonB -> "For this same reason tire weight has a bigger effect than wheel weight on the moment of inertia of the wheel/tire combo. Let's take two wheel and tire combos of the same size that both total 40 pounds. One combo has a 15 pound wheel and a 25 pound tire. Another has a 20 pound wheel and a 20 pound tire. Even though the total weights are the same the combo with the heavier wheel will actually perform better. Its moment of inertia is less because more of the mass of the combo is concentrated towards the center."

"Better" must be put in perspective. For this actual case, using 225-50-16" stock tire size, your worse case with the heavy tire will have a 2.5 lb dead weight penalty per wheel. Strait line car acceleration will suffer only .3%.

That is why the problem is more related to large changes in total mass, or rolling radius.

----------------------------------

Smilodon: Yes, response to bumps is degraded by mass, as I said before. But true analysis of the dynamic response must include dampers and spring rates, as the natural frequency of the unsprung system must be considered.

"Formula One cars have tires that are about 25 inches tall. Amazingly, the wheels are not 23 inches tall. Now, why is it, that in a racing formula where the wheel height is only limited by the outer circumference of the tire, the wheels are not Maxx Blingged out to almost touching the outer circumference of the tires?"

The FIA regulates the tire bead diameter to 13" !!!! That's why.

In F1, the tire is most of the suspension travel, especially up front.

DamonB

I don't follow your worst case scenario of a 2.5 pound difference? We see different tires of identical sizes with larger differences than that all the time?

Straight line acceleration may only see a small change, but handling will see a much larger one since the suspension is better able to manage the forces of the low inertia combo. Anyone ever held a spinning bicycle wheel in your hands and tried to move it around?

And thus most of the suspension travel on an F1 car is undamped. Those cars are strange beasts in many ways.

KevinK2

DamonB,

The 2.5 lbs is the effective extra dead weight due inertia only, per corner, for your example. You specified actual tire and wheel weights, with both cases adding up to 40 lbs. The advantage of your 20/20 combo is only like saving 2.5 lbs dead per corner.

Your post was only about acceleration. gyroscopic effects don't effect linear acceleration or response to bumps. I would not try to quantify the effect on tracsient turn in, although in theory it would just add trace amounts of oversteer to the mix.

I follow F1, and am always amazed at what little front end suspension motion there is, either in a 4 g stop, or launching the car when hitting the corner curbs. Also wild to see the tire displace laterally about 2" relative to the rim when curb wacking on a hard corner.

Smilodon

Accidental duplicate of following post erased! Sorry, guys!

Smilodon

From the official F1 website:

ARTICLE 12: WHEELS AND TYRES
12.1 Location:
Wheels must be external to the bodywork in plan view, with the rear aerodynamic device removed.
12.2 Number of wheels:
The number of wheels is fixed at four.
12.3 Wheel material:
All wheels must be made from an homogeneous metallic material.
12.4 Wheel dimensions:
12.4.1 Complete wheel width must lie between 305 and 355mm when fitted to the front of the car and between 365 and 380mm when fitted to the rear.
12.4.2 Complete wheel diameter must not exceed 660mm when fitted with dry-weather tyres or 670mm when fitted with wet-weather tyres.
12.4.3 Complete wheel width and diameter will be measured horizontally at axle height, with the wheel held in a vertical position and when fitted with new tyres inflated to 1.4 bar.
12.4.4 Wheel bead diameter must lie between 328 and 332mm.

ie: about 13 inches.

I shoulda dug deeper.

But, regardless, if an F1 car can survive, yea, even thrive, with 13 inch wheels, I would say that it would not be stretching the limits of technological imagination to think that I could too.

Smilodon

One more thing that is a factor in rotating mass: gyroscopic stability. The greater your angular momentum in any given object, the harder it is to initiate rotation about the other two axes.

In other words, the bigger, heavier, and faster-turning your wheels/tires/brake rotors, the more effort it takes to turn them with the steering wheel. Which can be a plus factor for high-speed stability, and a minus factor for making quick steering inputs at speed.

And, with an overall same weight, light wheels with heavy tires will yield worse inertia/momentum characteristics than heavy wheels with light tires, due to the fact that more of the mass is close to the outside diameter of the rotating body with heavy tires than with light ones.

KevinK2

Not always true. A dynoject drum has high angular momentum, but no gyroscopic effect. For a disc, if L is greater than .8D, then no significant gyro. effect.

If I remove a floor mat, the car will be lighter and quicker. True but an insignificant effect. It's easy to generalize, but specific quantitative answers, as I gave DamonB, are what's needed to show the true significance of the weight, as originally was asked.

sevensix

iTrader: (1)

so whats a good size for taking the fd to the track...

17 inch wheels?

18?

or stock 16s?

Smilodon

Unbelievable. No gyroscopic effect? There is no such thing as any rotating mass that has no gyroscopic effect.

My bicycle experienced it when driving fast down a steep hill. Hard to tilt the bike one one way or another due to the gyroscopic effect of the wheels.

Anyway, never mind. This thread is pretty dead anyhow.