Wheel weights: What is the real difference of 17 lbs (8 kg) weight

[goldNSX]

I now have a wheel weight of 96.7lbs with 8/17 and 9.5/18 and S-03. If I go with a lighter wheel/tire combo but the same tire sizes: Prodrive wheels and Dunlop SportMaxx 215/40/17 and 265/35/18 with a total of 79.4lbs. The difference between the two sets are 17lbs or 8 kg, 3/4 is due to wheel weight reduction, 25 % to tire.

What will be the difference in comfortable ride, performance, cornering and so on. Does the tire itself plays a larger role than the weight, so is it worth going to Dunlop instead of Bridgestones? I know that some even went with Kumho as they are very light. Lower weight is allways better but is it worth buying a new complete set.

Regards,
Thomas

 

[NSXDreamer2]

Thomas, are you asking the difference between the light weight wheels on track performance or just the overall experience?

I only had a short stint of driving my 17/17 stree set up once on track. It doesn't feel as agile as the OEM set up. But then my street setup is much wider both on wheel size and offset.

tires are tires, in my books if they are not unsafe to drive, it never worth to buy another set of them because of just a bit less desirable in performance or other reason. If you driving the car much, you will soon getting new tires, and then you could decide later... Unless selling and trading your existing tires is an option. But hearing that there's no many nsx in Europe, I don't know if that would be a sounded option.

S03 is relatively heavier than the others (They don't call themselves "Bridge Stone" for no reason :tongue: ) But it's a good tire which had been approved by most high performance car and track drivers. I could recommend the Kumho MX had as much, if not more dry grip as the S03, but the wet performance and road feel is not as good as the S03. They are lighter and only cost 1/2 of the S03. I had them on my street setting, as old track tires (retired), and on my daily driven Accord.

The new Yokohama Neova AD07 seems to be a very good tire and the new tire to get. Maybe it worths you doing some reseach on them.

The last note: On light weight wheels and tires combo, I found that it's all really comes down to personal preferences, skill level and driving habbits. Some traits that are absolutely unacceptable to you may not be detected by me, or vice versa......
Case in point, I do feel the ligher unsprung weight of my OEM 15/16 and the Volk Te37 lightness as soon as I release the clutch pedal, but then as soon as it rolls, more importantly when they turn, the tire contact patch, tread depth, offset and even the age of the tire compound complicated the whole feeling... IMO.

 

[goldNSX]

Thanks very muchf for your thoughts. Even if I don't track my car I asked my question in this forum as track people are more experienced.

Thanks again,
Thomas

 

[1BADNSX]

Back when I did the finite-difference acceleration calculations, I also calculated the effect of wheel and tire weight. The net result was losing 1 pound of wheel weight is similar to losing 1.5 lbs of sprung mass, while tire weight is about 2:1.

For pure acceleration, losing 100 lbs of sprung mass in the NSX is like gaining 7.5 HP. Based on the factors above you can figure the overall gain for your different wheel/tire permutations based on the loss in wheel or tire weight.

Bob

 

[goldNSX]

Thank you very much, Bob, I still have the gearing charts in mind which helped me very much in my decision.
I expected the performance gain in acceleration from lighter wheels marginal too.
So what about road contact due to lighter wheels. Heavy wheels and tires (Bridge-STONES ) make the car trampling more on even small irregularities. IMO the Bridgestone is a very good tire due to the soft compound and the stiff sidewall.

 

[1BADNSX]

Thank you very much, Bob, I still have the gearing charts in mind which helped me very much in my decision.
I expected the performance gain in acceleration from lighter wheels marginal too.
So what about road contact due to lighter wheels. Heavy wheels and tires (Bridge-STONES ) make the car trampling more on even small irregularities. IMO the Bridgestone is a very good tire due to the soft compound and the stiff sidewall.

I think ultimate grip and feel are the most important aspects for selecting a tire, but lighter weight is always good.

For accleration and the 17 lb savings you mentioned above, it would be like saving 28 lbs of sprung mass and that would be similar to gaining 2 HP. As you mentioned, lowering the unsprung mass probably more importantly allows the springs and dampers to control the suspension better.

Bob

 

[Tom239]

Back when I did the finite-difference acceleration calculations, I also calculated the effect of wheel and tire weight. The net result was losing 1 pound of wheel weight is similar to losing 1.5 lbs of sprung mass, while tire weight is about 2:1.

The 2:1 ratio applies (exactly) to mass that is at the periphery of the
wheel (the surface of the tire tread). The sidewall and beads in a tire are
closer to the axis of rotation, thus the effect for their mass is less than 2:1.
So the 2:1 ratio doesn't apply to the whole tire's mass.

1.5:1 also overstates the effect (slightly) for the OEM NSX wheel+tire sizes.

I explained how to do this calculation a while back in another thread here on prime.

 

[1BADNSX]

The 2:1 ratio applies (exactly) to mass that is at the periphery of the
wheel (the surface of the tire tread). The sidewall and beads in a tire are
closer to the axis of rotation, thus the effect for their mass is less than 2:1.
So the 2:1 ratio doesn't apply to the whole tire's mass.

1.5:1 also overstates the effect (slightly) for the OEM NSX wheel+tire sizes.

I explained how to do this calculation a while back in another thread here on prime.

Your explanation has the correct qualitative points, but your qualitative explanation is incorrect. Yes, the mass moment of inertia can be analyzed the way you stated, but you didn’t take into account the difference in angular velocity rate as the MOI reduces. My analysis was actually gear, speed and time dependent for various mass moment of inertias. Please explain what you did except for taking the radius ratio!

Bob

 

[Tom239]

Your explanation has the correct qualitative points, but your qualitative explanation is incorrect. Yes, the mass moment of inertia can be analyzed the way you stated, but you didn’t take into account the difference in angular velocity rate as the MOI reduces. My analysis was actually gear, speed and time dependent for various mass moment of inertias. Please explain what you did except for taking the radius ratio!

I asked myself the question, how much energy it take
to accelerate a wheel from zero to any given speed,
and how does that compare to the energy it takes to
accelerate non-rotating mass from zero to the same speed.
The result I'm looking for is a ratio of two energies,
be it 2:1 or 1.5:1 or whatever, and the same ratio
will apply no matter what the given target speed is,
or how much time it takes to accelerate to that speed.
It is also independent of gearing.

It takes energy to accelerate the wheel in a forward
direction (just like the non-rotating parts of the car)
and it also takes energy to get the wheel rotating.
The two can be calculated independently. If all
the mass is concentrated at the edge of the wheel,
the two components (energy of linear motion and
energy of rotational motion) are equal; thus the
wheel takes twice as much energy to accelerate
as non-rotating mass does.

If I understand your question, you're asking how
I concluded that the two energy components
(linear and rotational) are equal, for mass at
the edge of the wheel. In the other thread, I
briefly explained why that is so; a point on the
edge of the wheel has a tangential speed equal
to the forward speed of the vehicle.

But to show that the formulas of rotational
kinematics will give the same result:

given--
target vehicle speed v, meters/second;
wheel radius r, meters;
mass m, kilograms;

If the car's forward speed is v, the wheel's
rotational speed (omega, radians/second) is v/r

If the mass is at the edge (radius r), the
moment of inertia I = m * r^2

Rotational kinetic energy (in joules) E = 1/2 * I * omega^2
= 1/2 * m * r^2 * (v/r)^2
= 1/2 * m * v^2
which is the same as the linear (forward) kinetic energy
of a non-rotating mass m at a speed v.

(Or did I misundertand your question?)

 

[1BADNSX]

Rotational kinetic energy (in joules) E = 1/2 * I * omega^2
= 1/2 * m * r^2 * (v/r)^2
= 1/2 * m * v^2
which is the same as the linear (forward) kinetic energy
of a non-rotating mass m at a speed v.

Yes, your definitions are correct, but please show me what this proves other than r^2/r^2=1! How did you get 2:1 or 1.5:1?

The result I'm looking for is a ratio of two energies,
be it 2:1 or 1.5:1 or whatever, and the same ratio
will apply no matter what the given target speed is,
or how much time it takes to accelerate to that speed.
It is also independent of gearing.

I computed power, which as you know is energy per time, therefore by definition it is dependent on time. Because two actual vehicles with different acceleration rates are also required to shift at different points, this makes the computation dependent on gearing. I don’t have my original solution in front of me, but if I remember correctly I plotted equivalent horsepower gain versus speed and it wasn’t constant versus speed because of variables like the power curve and shift points. Then I believe I averaged the data over time to generate the generic values of 2:1 and 1.5:1.

Bob

 

[goldNSX]

Yes, your definitions are correct, but please show me what this proves other than r^2/r^2=1! How did you get 2:1 or 1.5:1?

Cool! I never expected a physical hour here on prime! Looks like the right people get to the basics.
The argument was that the rotational kinetic energy equals the linear (forward) kinetic energy under the assumption that the mass m of the wheel is completly at the edge. Therefore the hole energy is twice of the forward energy. This results in the same as if you put twice of the weight of the wheels in the trunk and see how much linear kinectic energy the weight has. But as the assumption is not fulfilled in the true world the true value is somewhere below 2:1, maybe 1.5:1 but not 1:1 as this would assume that the mass of the wheel would be centered in the axis which isn't fulfilled either.
The calculations by Bob convinced me that gaining more hp would be a better and maybe cheaper solution than reducing weight.

Besides that I'm still interested in the effect of a low mass wheel on the suspension response. There are some effects like wheel/tire-combo weight, compound softness (hard/soft), damping characteristics (hard/soft) and spring stiffness.
15 years ago I learned in school that the resonance is higher the stiffer the spring and the lower the mass is. I've the Eibach/Bilstein-suspension with a IMO heavy tire/wheel-combo. The Eibachs are soft, the Bilstein quite stiff and I suspect the Eibachs of giving the car too much initial roll in a turn-in with the result of loosing the inner rear wheel in even slow turns. It's a kind of setup that likes to be driven very softly but gives good control if something tends to go wrong. Therefore I'm definitly going to the complete and specially lowered Zanardi-setup with linear and stiffer springs. But I'd like to reduce the unsprung weight even further with a light wheel/combo which should help the suspensions response over irregularities even more.
I still have my Integra in mind with the very light 15'' compared to quite heavy 17''. The tire was a 40 compared 55 OEM. I know it's not fair to compare them but the 15'' felt very much lighter and the grip was nearly equal.
On my NSX I would keep the tire size to 215/40/17 and 265/35/18 but may change for weight reasons to the Dunlop SportMaxx and Prodrives.
Still my question: Is it worth it trying as suspension response is concerned?

Thomas

 

[Tom239]

Yes, your definitions are correct, but please show me what this proves other than r^2/r^2=1! How did you get 2:1 or 1.5:1?

The 2:1 is, I hope, obvious:

total energy to accelerate mass at edge of wheel =
energy to accelerate forward + energy to spin the wheel =
1/2 * m * v^2 + 1/2 * m * v^2

energy to accelerate non-rotating mass =
1/2 * m * v^2

For any given m and v, those energies are in a 2:1 ratio.

I can only estimate the ratio for a wheel; as the
wheel's mass is distributed at a range of distances
from the axis, it's a pain to calculate the wheel's
exact moment of inertia.

I modeled a 1995 OEM 16" front wheel as if all of its
mass were at a 7.5" distance from the axis. That's about
where the bottom of the wheel well is. There's some
metal farther than 7.5" out (bead seat and wheel lip) and
also a goodly amount of metal inboard (spokes and hub).
The radius of the full wheel (axis to tread surface) is 11.8".
Do the same calculations as I did last time, except
that the moment of inertia is lower by a factor of
(7.5 / 11.8) ^ 2, which is about 0.4. That gives an
energy ratio of (1 + 0.4) : 1.

I'd like to ask you a similar question. How did
you estimate the moment of inertia of the tire?
2:1 makes sense if you model a tire as having
all of its mass is at the surface of the tread,
but tires also have mass closer to the axis in
their sidewalls and beads.

I computed power, which as you know is energy per time, therefore by definition it is dependent on time. Because two actual vehicles with different acceleration rates are also required to shift at different points, this makes the computation dependent on gearing. I don’t have my original solution in front of me, but if I remember correctly I plotted equivalent horsepower gain versus speed and it wasn’t constant versus speed because of variables like the power curve and shift points. Then I believe I averaged the data over time to generate the generic values of 2:1 and 1.5:1.

That's certainly a more complete analysis, but when
all is said and done it should give an answer very
close to that of the energy ratio as I calculated.

I stuck with the energy ratio because it's easy to
understand and calculate, and it's a fair way to
contrast the effect on acceleration of rotating and
non-rotating mass. I think there's value in offering
a method that can be followed rather than just
giving the end result of a large complicated
calculation. Anyone who understands basic physics
can follow my reasoning and see for themself.

 

[1BADNSX]

How did
you estimate the moment of inertia of the tire?

That's certainly a more complete analysis, but when
all is said and done it should give an answer very
close to that of the energy ratio as I calculated.

I agree on this one. The “back of the envelope” approach is usually as good. I already had the acceleration algorithm; therefore it was very easy to determine the power consumption associated with different MOI wheel/tires.

I would have to give the power versus energy approach some thought, but it seems the delta-t would be way off with your approach compared to an analysis with the whole vehicle (rotating and non-rotating masses), because for a fixed power input, your approach would yield one-half the acceleration. But it seems you need to account for the different delta-t and the two types of masses together.

I thought you were dividing the energies and missed your casual observer 2:1. Do you know why some folks like bicyclist (large radius wheels) claim much greater numbers than 2:1?

I honestly don’t remember the exact values I assumed when estimating the MOI, but I probably made a rough integration like you did for the wheel.

Bob

 

[TTony]

I now have a wheel weight of 96.7lbs with 8/17 and 9.5/18 and S-03. If I go with a lighter wheel/tire combo but the same tire sizes: Prodrive wheels and Dunlop SportMaxx 215/40/17 and 265/35/18 with a total of 79.4lbs. The difference between the two sets are 17lbs or 8 kg, 3/4 is due to wheel weight reduction, 25% to tire.

Are you sure on the 96.7lbs. and 79.4lbs? I just installed a set of 17x8 (15.5lbs.) + 18x9.5 (16.35lbs.) TE-37's and 215/40/17 & 275/35/18. I weighed them total and the total weight was ~139lbs.

I just reduced my wheel/tire weight by 21 lbs. and don't really notice an acceleration difference (haven't been able to test it) but it is MUCH easier to turn the front wheels. Lack of power steering makes this very evident.