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| Wheels | Tires | Spacers | Hub -- Sponsored by The Tire Rack Specific topics relating to wheels and tires. |
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#29 |
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Lighter wheels do not store as much inertia so you can not coast as easily. At cruising speeds lighter wheels hurts your mpg. But it is a small amount....
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#30 |
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No no no no no. At constant speed cruise it does not matter at all how much the wheels weigh. For in-town stop/go driving, more rotational mass does hurt fuel economy a bit (extra energy spent to spin up heavy wheels/tires is totally lost to braking heat).
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#31 | |
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If you happen to figure out 86 wheel rates sometime feel free to share those as well. The thread I started on it fell completely silent which was surprising for how many track this car. Granted the car is new, but the Integra and Civic boys have all the data on wheel rates they need pretty much. If the wheel is wider w/ a larger section width it increases drag and rolling resistance. Most fuel is consumed in start stop traffic so the benefit is much larger there. |
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#32 | |
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If anything, the wider aspect ratio will increase rolling resistance, which can lower MPG. Right? |
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#33 |
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#34 | |
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Lighter weight equals less stored energy so the motor works more to keep a constant speed Sent from my Nexus 4 using Tapatalk 4 |
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#35 |
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Correct. But you actually mean section width, not aspect ratio unless you mean it relative to section width and then you also have aerodynamic drag with a larger frontal area pushing through air.
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#36 |
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Get the wheels you want and can afford. Then buy the best tyre you can afford as that is where you will feel the biggest difference
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#37 | |
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Wheel weight.. does it Matter?
Quote:
y = Diameter of 18" wheels w/ 225/40/18 = 637.2mm % Error = (|x - y|/x)*100 = 1.9% So assuming I'm getting 21.0mpg with my new aftermarket 18" wheels/tires, I'm actually getting 1.9% better, or 21.4mpg. Does that math sound correct or close enough? Or I should I circumference instead of diameter? Update: Just had to look up some old high school formulas to remind me C = 2R(Pi) = D(Pi). So the error would still be 1.9%. |
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#38 |
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i think they actually went faster with the wheel that was 3 lbs heavier on a miata, a car that is much more responsive to wheel weight.
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#39 | ||||
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#40 |
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Actually I was referring to increased rate of effective weight/force, not speed at the polar moment.
Last edited by Anaxilus; 08-21-2013 at 12:21 AM. |
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#41 | |
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Quote:
Polar inertia is something independent of traction, and far more relevant to the street than to track. Track cars maintain more high speed even in deceleration than street cars, so they have less work to do to spin the wheels back up. Plus they tend to have much more power/torque than street cars and move less mass overall. It's a question of balance versus your driving environment. In terms of absolute timed track performance, you always go with the most rubber you can put on the road (unless the weather is bad). Street is different and will depend on user preferences and driving style. The ideal track car above will absolutely suck to drive in street conditions. |
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#42 |
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where did you get 1.7 from? Because all the research I come up with say that there is no number and that the gain is proportional to the part, its location and how soon it breaks inertia. For example the tire and wheel have a bigger impact then the brakes and pads would. Apparently the real life gains after removing 1lbs from a wheel ``are`` greater then removing 1lbs of non rotating sprung mass ( something that doesn't spin and is after the suspension) but nowhere can I find any kind of value for sprung vs un-sprung. Also I read that removing 20lbs from your wheels would be more of a direct improvement on acceleration and breaking compared to shaving 100lbs from the body even tough loosing 100 lbs will help more ``over all``!?!?
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