TinkySD

http://www.angelfire.com/linux/tinky/rotational.xls NOTE: youll have to copy/paste the url into a new browser window to download it.

Probably only techies like me will like this but I figured I should post for anyone interested. I'm sure you've heard me ramble on and on about the best mod you can do being light weight wheels and tires. I found a spreadsheet a few months ago(I think from mazda6tech) that quantifies the effective static weight loss of a specified change in rotational mass through equations.

My findings:

When I got a flat tire on the freeway it turned out to be a blessing in disguise. I switched to some falken ziex 512 all season tires which weigh in at a slender 21lbs a piece.(a solid 5lbs per corner less than the stock tires) My car felt a little peppier, especially off the line. According to the equations the rule of thumb about 1lb of rotational mass being equal to 6-8lbs of static mass is very accurate for cars with our level of torque. Dropping 20lbs of that far edged rotational mass is the equivalent of about 160lbs of static weight... or dumping one of your friends out the back! (or about 6lb feet more torque)


A couple comments:
1) Losing rotational mass isn't EXACTLY like dropping static weight. The moment of intertia(or rotating mass) is more important the faster you are trying to accelerate it the rotating mass. Meaning the lower the gear the more you are going to feel the difference. Anectdotely this is why I felt the most difference off the line in first gear.

2) This kind of weight loss won't be measured to it's true degree on a dyno because you dyno in a high gear meaning the rate which you are trying to accelerate the mass is slower than in real world driving. However, as a rule of thumb drag racers figure 100lbs = .1 in the quarter mile for cars with more torque than ours. Meaning if you can drop the equivlanet of 200lbs static weight you are probably looking at a .2-.3 decrease in your 1/4 time which is HUGE.

3) Tire weight is the most important to shed. The farther a weight is from the axis of rotation the more torque it takes to turn it. With this said, the edge weighted model in the spreadsheet is the most accurate. I high reccomend toyo proxes t1-sfor summer tires or falken ziex 512 for a/s. They both weigh in at 21lbs. To be so much lower in weight they both have softer sidewalls than a heavier tire so you have to run a slightly higher inflation to get it to handle as precisely.

Dan Martin

All good info there Tinky.
I've been meaning to write a spreadsheet to do this but I'm glad someone beat me to it.

Edit: Linky no worky

CGTSX2004

iTrader: (1)

Nice

TinkySD

Make sure you copy/paste the url into a new browser! Angelfire doesn't like the direct linkage.....

Dan Martin

I rehosted it here: http://www.mtmsteel.com/rotational.xls

xizor

great post!

provench

Nice work Tinky

Stokeless_TSX

me like

sauceman

Cool stuff, Tinky. However, since your car is only an FWD, I don't believe you can include the back wheels in your calculations for acceleration. As for handling / unsprung weight, it's a different story.


Thus, if I'm not mistaken, you will need to divide your figures by 2, meaning it was as if you had stipped your car of 2 spares from the trunk, which will still allow you to feel a difference anyway.

TinkySD

Not true at all. All four wheels of the car have to turn to accelerate your vehicle....and at the same rate of accerleration.

I can't take credit for the spreadsheet though, I didn't write it.

sauceman

Yeah, the links for the spreadsheet didn't work, so I couldn't read it.

But isn't it only applicable for traction wheels? I would think the "towed" wheels are not subject to the same laws.

Let me illustrate this:

You are at the shopping mall walking around with a shopping cart. What shopping cart would require more energy from you to get moving, and keep rolling?

One with 3" lightweight wheels, or one with 6" wheels that weight twice as much?

You will notice the easier ones to get moving are the 6" wheels, because they require less rotation on the axis per feet travelled, than those of 3 inches.


Same laws apply to your rollerblades. You have ABEC5 70mm wheels, and another pair with ABEC 5, 78mm wheels. Which of the 2 pairs will let you roll with less effort? The ones with the higher diameter.

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Now, I know you will say that you are talking about the force needed to spin a wheel, so consider this: On an FWD car, the front wheels are spun from the center hub of the wheel, driven by the shaft. The hind wheels though, conversely are spun from the outer diameter of the wheel, and the hub will act as a resistance due to friction. What is more critical then, for non-tractionned wheels, is higher diameter, in order to reduce the friction of the wheel bearing. This is brought on by greater ft/rotation.

But the weight of the wheel will not be multiplied in like it would on the front wheels.


See what I mean?

TSX Hokie

A certain amount of energy is required to spin a wheel+tire up to speed. The energy is directly related to the moment of inertia of the rotating mass. Drive wheels or not, the wheels still accelerate at the same rate (assuming no slipping), so you have to take all four wheels into account.

Conservation of energy generally provides the answer to any physics question.

sauceman

Hmm. Let's pretend the TSX wheel has a demultiplication ratio of 10:1. With 150whlb-ft you will get the actual acceleration the TSX has. I

-If the wheel had a demultiplication of 5:1, the wheel would have the ability of spinning significantly faster to say 100RPM given the same amount of torque applied. (those same principles make a balerina spin faster as she closes her arms) In other words, it will require less torque to accelerate it at the same rate as a 10:1 wheel. This is because the exterior circumference of the 5:1 wheel will need to cover one half of the distance that a 10:1 wheel would.

-If the wheel had the same diameter (demultiplication ratio), but half of the weight of the original TSX wheel, it would have the ability to spin faster to 100RPM, because it is lighter and has less inertia (especially on the outer circumferences, where it is more critical, just like a balerina spins faster as she brings back the weight of her arms to a narrower cicrumference. Again, it would require less torque to accelerate it at the same rate as a TSX wheel. (Do you see the relation between weight and demultiplication ratio?)

So far, I am in agreement with what Tinky says, because the wheel is being driven FROM THE AXLE.

Now, on an FWD, the hind wheels are driven from it's exterior circumference, in direct contact with the ground. Thus, you have a 1:1 demultiplication ratio.

Does weight affect the speed at which that wheel will spin to 100RPM? YES! BUT at a 1:1 ratio. Not 6:1 or 8:1 (all weight distribution calculations taken into considerations). So 1lb lost on a rear wheel would be equivalent to 1lb in real life, not more.

If you want to understand what I am saying, put your car on a lift, try spinning your back wheel from the axle (put your fingers on the nuts) with your hands, then try spinning it from the tread of the tire, and repeat the same operation, but with a wheel 10lbs lighter. You'll see.

TSX Hokie

Sauce, you are thinking way too hard. Conservation of energy cannot be violated, thus it is the ONLY way to think about this problem or any physics problem.

In linear motion, Kinetic Energy=0.5*m*v^2. In rotational motion, Kinetic Energy=0.5*I*w^2, where I is the moment of inertia (m*r^2 for a simple ring) and w is the rotational speed (radians per second).

When you drive, your engine delivers power, which is energy delivered per unit time. This power goes to many different things, but during acceleration on a flat road it is primarily going to increase the kinetic energy of the vehicle. This is not only the linear kinetic energy of the entire car, but also the rotational kinetic energy of every rotating component. Since the radius term is squared in the kinetic energy equation, and the wheels+tires are the largest radius rotating component, they require the most power to accelerate. The flywheel is spinning faster than the wheels, so it stores a good deal of energy as well, but not as much as the wheels+tires.

When you lower the rotational inertia of your wheels+tires, more of the engine's power goes to increasing the linear kinetic energy of your car, and less goes towards increasing the rotational kinetic energy of the wheels+tires, so you accelerate faster.

Since it requires energy to increase the rotational speed of the rear wheels+tires, this energy can only be coming from one place, the gasoline that is powering your engine. Thus, the rotational inertia of the rear wheels matters just as much as the front wheels (actually more on our cars, since the rear brakes are so puny).

sauceman

Hmm I guess I don't get it. To me,

This concerns the front wheels only.

Right.

Indeed, but that would apply to braking and not accelerating!?


I really don't get it! Try spinning your rear wheel like I said, and you will see which method requires more strength to get the wheel turning: spinning it from the center, or from the tread. I don't have any equations, but my butt feel says I have a much harder time spinning the wheel from the hub! If I were to take 10lbs off one of these two wheels, I'd much rather take it off from the one I'm trying to spin from the hub, because I need a lot more torque to spin it

I

TSX Hokie

When you step on the gas, all four wheels accelerate (rotational and linear) at the same rate (assuming no slipping). Thus, more energy is required to accelerate ALL the wheels if they have a higher moment of inertia. Being drive wheels has no bearing on this.

It takes the same amount of energy to spin a rotating object whether you push from near the center or the outside. What changes is the force required (work=force*distance).

Sauce, you can almost think of it like the rear wheels are attached by belts or chains to the front wheels (since traction with the ground is forcing them to spin at the same speed). It will require twice as much torque to achieve a given rotational acceleration if a wheel is chained to a second, identical wheel.

sauceman

Ahh.. Now I see..

TinkySD

I come in on Tuesday and missed all the fun!

Note to sauce: make sure you copy and past the link in a seperate browser. Angelfire doesn't like direct linkage!

sauceman

Yeah, I got owned...

I was wondering why you kept silent, was afraid I had put you beside yourself.

Dan Martin

The link I posted in post #5 requires no copying and pasting.

TSX Hokie

Hey, I got 26 MPG on the last tank, so I am sure you 'owned' me there.

The best thing about these forums is that we can all learn form each others' knowledge. I might do OK with physics, but it took me half an hour to find the oil filter.

According to that spreadsheet, switching from the stock TSX wheels and tires to 17x7.5 SSR Comps and 225/45/17 Toyo Proxes T1-S would be akin to shedding 250-300 lbs!

sauceman

I tried it and all I got was something about Angelfire not allowing hotlinking.

Dan Martin

my file got

TinkySD

dan martin is teh n00b