Every once and a while I get a dig about my 15/16 wheels. I usually get the dig from the NSX owner BEHIND me
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LarryB
. LarryB
Something wrong with my NSX?
- Thread starter sgvnsxrated
- Start date
Not by choice, but I am now running the stocker pizza cutter 15/16 combo with new Hankooks. My previous setup was 18/19. The car is so much quicker now. I will never run 18/19 boat anchors again. 17/18..yes...anything bigger is just robbing these cars of performance.
I agree with almost everything stated above, including:
However...
Yup. Putting on tires (it's actually the tires rather than the wheels) with a larger diameter has the same effect as swapping your gears for taller ones. For example, swapping the stock tire sizes on a '91-93 NSX for 275/30-19 is like swapping the stock R&P gear with a 4.062 ratio for one with a 3.938 ratio.
However...
There are many reasons why people run the OEM wheels at the track, and I think that one is way down on the list. The three most common reasons are (a) lots of OEM wheels available, thanks to aftermarket ones, so you can get used sets for cheap; (b) lots of R compound track tires and supersticky street tires available in the OEM wheel sizes, not in the bling bling big wheel sizes; and (c) you can pack the set of track tires inside the NSX and drive to the track on your street tires.
OK. I understand the theory of a larger outer diameter tire and the effect on gearing. However, I just went out to the garage and measured my 19" rear wheel on my '91 and my 17" OEM rear wheel on my '97 and they are almost exact in height (outer diameter). Any difference in wheel size is compensated for by a shorter side wall height on the tire.
I think weight has more to with it than you think. At least from a engineering standpoint....unsprung rotational weight is more detrimental than static weight by almost 3:1 (increases as the weight moves away from the center axis).
It would be interesting to see a controlled study on the effects of both weight and outer diameter.
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+1 And we will learn something
Yes, but the further out the weight of the wheel is from the center the worse it is. I can't imagine that the sidewalls of your tires weighs the same amount as all that metal around the outside edge of the wheel.
I'm not sure what your saying.....I think you're agreeing with me? I already stated that the weight is worse the further away from the center axis (see quote above). My point is that MY different size wheels have the same outer diameter......and that I think the less weight on the wheels the better as far as accelaration goes.
You may have picked rim+tire sizes where this is the case, but often for looks or due to tire sizes available people buy combos where this isn't the case. The profile may be shorter to compensate, but not by enough, so the end result is taller...not by inches, but by enough to make a difference.
Ken's example of going from 225/50-16 to 275/30-19)...
25.496" - 24.858" = 0.638 (2.6% increase in diameter)
80.098" - 78.094" = 2.004 (2.6% increase in circumference)
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Weight is the enemy. Also the width.
Those who are running 275 or 285 R 19 will definitely slow the car down, change the handling characteristic, and
wearing out your clutch faster too!!!
Those who are running 275 or 285 R 19 will definitely slow the car down, change the handling characteristic, and
wearing out your clutch faster too!!!
I doubt that. But a lot of things can affect the outer diameter (and, correspondingly, the sidewall height). Heck, just the difference in tread depth between a new tire and one that's ready to be replaced can be half an inch in diameter. The inflation in the tires can differ, resulting in a different amount of deflection by the pavement. And tires labeled with the same size can actually differ significantly in size. Bottom line, though, is that, if you keep all the variables the same - which it sounds like you're not doing - the actual difference between two tires of different sizes usually is roughly equal to the calculated differences in their outer diameters.
That's funny, because I've heard other people claim that it's 1.5:1 and still others claim that it's 2:1. And a few others claim that it's even more than 3:1.
I'm not saying that weight doesn't matter; if you can take a couple hundred pounds off your car, you can feel the difference pretty easily. But the difference in wheel weights generally tends to be nowhere near as important as many other factors. If you want to know for sure, though, just do what others above have suggested - swap the wheels and drive the cars again.
You "doubting" is irrelavent. It's true.
Blah...Blah....Blah...could you please explain that NA1 and NA2 thing one more time? I just love that....
That's because it's a dynamic figure based on how far away from the axis the weight is. They're all right.
The car not only needs to propel the "dead weight" of the wheels, but also must generate an angular velocity and acceleration of the wheels, which is determined by calculating the mass moment of inertia. so a heavier "dead weight" wheel would increase the moment of inertia, decreasing acceleration and lowering gas mileage.
It's easy to say it doesn't matter where the mass is, a # is a #, sprung or un-sprung. Weigh the entire car and apply the equation a=F/m to the whole mess. Weight reduction on the overall vehicle is critical obviously but thanks to the inertia it takes to propel not only the entire mass but actually rotate the wheels.......wheel weight becomes more critical.
The equivalent weight of a wheel and tire assembly would be equal to the rotating inertia of that assembly divided by the square of the effective radius of the tire. (The effective radius is the distance up from the track surface to the axle centerline as the car launches.) Which brings into the equation what you had mentioned in a previous post about tire diameter. The formula says it all.
So thanks for your opinion. Of course with over 22,000 posts it doesn't look like you withhold your opinion much. You're not gonna tell me to use the search function now are you?
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