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Old Apr 18, 2001 | 08:19 PM
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Does anyone know how much the factory SE 17" wheels weigh?
Old Apr 18, 2001 | 09:42 PM
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i belive around 30lbs each... not sure.. try a search.
Old Apr 19, 2001 | 04:44 AM
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I don't know if anyone came up with a solid answer to that...30 lbs seems a bit heavy though
Old Apr 19, 2001 | 07:26 AM
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I weighed them with tires and I don't remember exactly what it was but I want to say it was 46lbs.
Old Apr 19, 2001 | 06:18 PM
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My 18" wheels and 245/40-18 tires only weigh a total of 44lbs each. I think thats pretty good.
Old Apr 19, 2001 | 06:23 PM
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Well, the stock nissan 17" and RE92 combo are extremely heavy! If you want to reduce weight, get the Pilot XGT Z4 and it will save some weight.


I don't know how much the rim weights though....
Old Apr 19, 2001 | 06:53 PM
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MYcar..

Do you have any fender rubbing with the 245/40/18's? I'm looking at some 18's, but i'm not sure if 245's are the way to go. Can you please advise. Thanks!

GC
Old Apr 19, 2001 | 07:43 PM
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Heh.. hate to bring this up, but u guys do know that there is more than just weight when talking about a wheel... the distance of that weight from the center is important... BIG TIME... (i gotta get goin so i'm not gonna explain.., but if you're interested, and wanna sound smart in front of all your friends, look up "moment of inertia" on the web... i'm sure you could find a really cool explanation somewhere)

-Tom
Old Apr 19, 2001 | 08:08 PM
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I've tried searching for this and have gotten nowhere. Same thing for the 16" wheels. Doesn't anybody know this?
Old Apr 20, 2001 | 04:38 AM
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Originally posted by beaglemax
I've tried searching for this and have gotten nowhere. Same thing for the 16" wheels. Doesn't anybody know this?

well you can do the experiment yourself... take a weight... lets say 5lbs and attach it to a string of lets say 2ft... now spin it with an equal steady amount of force... now take that string and turn it into a 6ft string... now spin it with the same amount of force... it is going much slower now... you dont even know if it will raise up all the way and tighten the string... that is what he is trying to explain... maybe when i get home from school if im not lazy i will search for it or i will just type up the definition from the physics book...
Old Apr 20, 2001 | 06:30 AM
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I was referring to the wheel weight question - not what a polar moment of inertia is. I've had enough years of college to answer that.

Originally posted by psing23



well you can do the experiment yourself... take a weight... lets say 5lbs and attach it to a string of lets say 2ft... now spin it with an equal steady amount of force... now take that string and turn it into a 6ft string... now spin it with the same amount of force... it is going much slower now... you dont even know if it will raise up all the way and tighten the string... that is what he is trying to explain... maybe when i get home from school if im not lazy i will search for it or i will just type up the definition from the physics book...
Old Apr 20, 2001 | 11:06 AM
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Ya... I sorta picked up on that as I was reading the last post...

My ques, is why you need to know the weight? (unless you plan on shipping it somewhere), I would think the moment of inertia about the center would be a much better indicator of the performance increase/decrease you could expect by switching rims/tires..

Interesting ques. tho either way... i've never really thought about how much my wheels weighed...


-Tom






Originally posted by beaglemax
I was referring to the wheel weight question - not what a polar moment of inertia is. I've had enough years of college to answer that.

Old Apr 20, 2001 | 11:59 AM
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Originally posted by beaglemax
I've tried searching for this and have gotten nowhere. Same thing for the 16" wheels. Doesn't anybody know this?
I don't know but I bet the dealer does. Call them and find out what the shiping weight is. Figure on a few lbs for packing and such. :P:

Paul
Old Apr 20, 2001 | 06:33 PM
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The heavier the wheel/tire package the more unsprung weight the vehicle has to move there fore a decrease in performance.

On the subject of inertia, if the weights are equal (17" wheel with a tire diameter of 25.4" versus an 18" wheel with a tire diameter of 25.4") and the overall diameter of the wheel/tire package are the same the level of inertia should be the same.
Old Apr 20, 2001 | 08:57 PM
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Originally posted by mykmar10
The heavier the wheel/tire package the more unsprung weight the vehicle has to move there fore a decrease in performance.

On the subject of inertia, if the weights are equal (17" wheel with a tire diameter of 25.4" versus an 18" wheel with a tire diameter of 25.4") and the overall diameter of the wheel/tire package are the same the level of inertia should be the same.
No. The moments of inertia are not the same.
The larger diameter wheel has its weight further from the axis of rotation. Moment of inertia, for a 3 dimensional object (like a wheel and tire), is defined the integral of rho*radius^2*dV over its volume. (rho is the volume density of the object, radius is the distance the mass is from the axis of rotation) The important thing to take away from that is that the moment of inertia is proportional to the radius squared. So by moving the mass further away from the axis or rotation the moment of inertia increases. Since it is a square law relationship, an increase in the radius can have a BIG effect.

How does this relate to performance?
torque=I*alpha, where I is the moment of inertia and alpha is the angular acceleration. The larger moment of inertia requires more torque to reach a given angular acceleration. Basically, your engine has to work harder to get the car moving. As alpha gets bigger, you go faster (unless you are spinning your wheels and have no traction). I hope that this clears it up some.
Old Apr 21, 2001 | 10:30 AM
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BeagleMax, I stand corrected.

The heavier in object the more torque it takes to move that object. That also slows performance.
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