John95cab

I have the stock 17" and am considering going to the 18" MY97 5 spokes. Anyone know whether they are going to be lighter or heavier?

Edward

Do you mean "MY02" 5-spoke wheels? Or do you mean "MY997" 5-spoke wheels? The latter certainly looks like a very close cousin of the MY02's, and the 02's are pretty light ...within a couple of pounds of the Cups, IIRC, which is excellent considering it's larger width and diameter. Perhaps someone knows the actual weights of the 997 5-spokers, but they don't "appear" any more massive than the 05s and I would be surprised if their weights were appreciably different.

Edward

campbesl

Gents: Let me take a stab at this.

Newton's First Law says, "An object at rest tends to stay at rest and an object in motion tends to stay in motion... unless acted upon by a force." Hmmm, this could apply to Porsches.

Galileo took it another step and described "inertia" - that is, an object's need for energy to be influenced to change it's 'state' or velocity. He reasoned that the more mass an object has, the more energy required to influence it. And he concluded that the relationship is directly proportional. This could apply to Rennlist Member owned Porsches.

Newton's second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. Now if one assumed that the net force (horsepower and gearing) remained constant, mass would be the only variable (hint - think wheel/tire mass on a Rennlist Member owned Porsche 993).

Now, taking a leap and combining the work of Newton, Galileo and some geometry freak named Euclid, and applying it to a donut (i.e. tire/wheel) the equation for inertia of a rotating object can be derived as pi/32 * (outer diameter raised to the fourth minus inner diameter raised to the fourth). Now we realized that changing the diameter of that spinning wheel/tire mass has a definite effect (because it's raised to the fourth power). But what is the change in effect for a wheel/tire combo change on a Rennlist Member owned Porsche 993 C2?

For the sake of argument, let's assume that David replaced 265/35-18 wheels/tires with 255/40-17 wheels/tires. Using the formula presented by Tire Rack's Tech section, the total diameters are 25.3 and 25.0 inches respectively. That is, the total reduction in overall diameter was only 0.3 inches. BUT, the centroid (officially known as the radius of gyration or k0) of the mass moved inward.

k0 is calculated as 0.7071* sqrt (R2+r2). For David's '97 C2, this means that the move from 18" to 17" setups moved his k0 from 11.0 inches to 10.7 inches from the axle. Now, using Bill Verburg's 10 lbs per corner weight change number, this causes a change in mass moment of inertia (mass times effective radius) from 439.1 inch-pounds (18" setup) to 320.9 inch-pounds (17" setup). Hey, that looks like a big change!

That is a 27% decrease in torque (force) required to accelerate (act upon) his 17" setup versus the 18".

Finally, a few beers got consumed during this project. If it makes no sense what so ever, thank Anheiser Busch.

Scott

dhicks

Wow Scott......I am impressed with what you can do on a few Anheiser Busch...jus think waht you could have done with some "good" beer inside you.....(a joke dont take offense).....appreciate all the work and research that went into this....so what you are saying is ... I am feeling an improved acc'lr'n due to a 27% reduction in required torque to get the car moving...very good, thanks...

campbesl

David,

You restated correctly.

Now if there are any engineers out there, feel free to correct any and all errors.

And for the record, it was Budweiser (Anheiser Busch product) a.k.a. The King of Beers. The No. 1 Selling Beer in the World. Why did they name them tastebuds? Make mine a Bud. This Bud's for you!

Ok. Back to coffee.

Scott

clib

Hey, really nice discussion Scott. I would not have had the background to put it into mathematical terms, but it was the same point I was trying to make earlier. I've seen reproducible dyno changes on motorcycles ranging up to 5-7% from changes in wheels based purely on weight and moment of inertia. This is all part of the need to power the drivetrain, that reults in power numbers measured at the crank being significantly higher than those measured at the rear wheels.

now can you mathematically summarize the effect on accelaration due to changes in final gearing due to perhaps a 2% change in outside tire circumferance which was the other factor discussed? we could reemburse you for the beer

bill

Bill Verburg

Question: What is the effective additional mass of wheels/tires due to their rotation?

For the vehicle in motion, the kinetic energy is given by:

Ekinetic = ½ * (Mtotal*V2 + 4*Iwheel*w2)

Where
Mtotal = Total vehicle mass including wheels
Iwheel = moment of inertia of a single wheel/tire combination
V = Velocity of vehicle
w = rotational velocity of the wheel (in radians/sec)

Now, V = Rtread * w where Rtread = effective radius of the tire's tread

Combining, we find:

Ekinetic = ½ * (Mtotal + 4*Iwheel / Rtread2) * V2

So the effective additional mass added to the vehicle due to the wheels rotation = 4*Iwheel / Rtread2

Now, Iwheel = Mwheel * Reffective2

Where
Mwheel = mass of the wheel/tire combination
Reffective = radius of gyration (which is always less than Rtread )

Now, finally, the effective additional mass added to the vehicle due to the wheels rotation is:

Mdue to rotation = 4 * Mwheel * (Reffective / Rtread)2

On a per wheel basis, the EFFECTIVE TOTAL wheel weight is given by:

Mwheel, effective = Mwheel * (1 + (Reffective / Rtread)2 )

Reffective / Rtread is always less than 1 and probably somwhere around 80% by my guess. This ratio is a function of wheel and tire weight disribution.

So in this case, Mwheel, effective = Mwheel * 1.64

The absolute maximum (impossible) case would be Mwheel, effective = Mwheel * 2

Example: New wheels and tires are fitted, lowering the weight of each wheel / tire combo by 10 lbs. The apparent effect is to lower the weight of the car by 65.6 lbs (16.4 lbs per wheel).
________________________________________________________________________

Well, let's make it easy. THERE IS NO (significant) CHANGE IN THE POWER OUTPUT OF THE ENGINE, STEADY STATE.

What you are doing is reducing the rotational inertia, and due to a rather neat effect to do with geared systems, it can have a big effect on your acceleration.

The rule is that rotational inertias in a geared system have an effect proportional to the SQUARE of the gear ratio, so if you have 3:1 first gear and 4:1 in your diff you will see a reduction in rotational inertia 144 times greater than if you had reduced the inertia of the wheel by the same amount.

For various reasons it is easier to do a power sum, so we get

power at wheels=rate of change of kinetic energy+rate of change of rotational energy-rolling resistance*v-airdrag*v

rateofchangeofKE=d/dt (1/2*m*v^2)=1/2*m*2*v*a=m*v*a

rateofchangeofRE=d/dt(1/2*I1*w^2+1/2*I2*w^2*AR^2+1/2*I3*w^2*AR^2*GR^2)

w*rollingradius=v

hence

rateofchangeofRE=1/radius^2*d/dt(1/2*I1*v^2+1/2*I2*v^2*AR^2+1/2*I3*v^2*AR^2*GR^2)

rateofchangeofRE=v*a/radius^2*(I1+AR^2*I2+AR^2*GR^2*I3)

airdrag=1/2*airdensity*Cd*A*v^2

rollingresistance=Cr*m*g

Cr is about 0.015

I1 is the rotational inertia of the wheels and halfshafts, I2 is the propshaft and output shaft of the gearbox and I3 is everything that rotates at engine speed. The I of a uniform cylinder is mass*radius^2/2

AR is the axle ratio and GR is the gear ratio.

OK, I think those are all the equations you need to get a solution for the change in acceleration.

so at a given speed

change in acceleration=-1/(m*v+(v/radius^2)*(I1+AR^2*I2+AR^2*GR^2*I3))+1/(m*v+(v/radius^2)*(I1+AR^2*I2+AR^2*GR^2*I4))


where I4 is the new rotational inertia of the engine, and I3 is the old one.

Somehow I doubt that helps!

________________________________________________________________________

was about half way through writing that lot when I realised it was going to get very ugly, very quickly.

OK, here is the useful little nugget:

for a change in the rotational inertia of the flywheel "I" the reduction in mass you have to accelerate at the contact patch is (take a deep breath)


(1/rollingradius^2)*(AR^2*GR^2*I))

So, if your flywheel inertia drops from say 0.2 kg m^2 to 0.1 and radius =.4 gives

1/.16*144*.1

or the equivalent of almost 90 kg less linear inertia. So in a 1 tonne car you might see an improvement of 5% or so in acceleration in first gear. The acceleration in first gear is already heavily constrained by the rotational inertia of the powertrain, so its effective inertia is much higher than 1 tonne. You can use the same equations to find out how much, 25% is the number I remember.

There is no appreciable difference in dyno power, just transient responses.

None of the above takes into account the deformation of the tire under a load and due to centripital forces, both cause significant cahnges to the tires rotational properties.

As far as the change in acceleration due to the smaller rolling radius of the tires, the rev/mi is the most accurate predictor.

Not knowing the specifisc tires in question a generic guess of 265/40x18 has 822 rpm and 255/40x18 has a rpm of 931 this is a 13% difference in every gear that is multuplied by the trans and final drive rations to give the effective drivetrain multiplication factor.

note that inerial effects are magnified more in lower gear sthat upper and are not multiples whereas rolling radius is a constant multiple in every gear at all speeds.

I guarantee you that the change in rolling radius of the tire us the difference that is felt.

epj993

Bill, that's an incredible answer - I'm still digesting & trying to fully understand. This would be a great model in a spreadsheet with the ability to plug in wheel weights & tire sizes and see the resulting changes - even if theoretical.

Bill Verburg

Years ago when I was fooling around w/ the setup on my older Carrera and tring to decide what wheels, tries and brake to use I did set up a very detailed ss model of the effects that each change would produce, It was all integrated w/ the dyno data. and gave me some very edifying info.

here's one showing the net thrust change in gear from my old C3 to the new 993RS

campbesl

Bill,

How many beers did it take you to do all that analysis? Judging by my experience, you did more than a case of analysis ...possibly a case and a half.

Nice work. Are you a rocket scientist?

Scott

BS911

A case and a half to write it??? My head was swimming about 1/10th of the way into JUST READING it and I had to start drinking. It's ok though, I'm working off of a keg so I only had ONE.