Rims? Does the weight of rim have an effect on the performance??
11-19-2004, 03:21 PM
#1
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Thread Starter
Join Date: Jul 2004
Posts: 32
Rims? Does the weight of rim have an effect on the performance??
hi, I want to buy a pair of rims for my car really bad, but im not sure if the weight really plays a key factor. I want 19's really bad but im a big performance freak. Right now I have the stock 17's, and have herd they are really heavy. So I started to look up different types of rims, and would like to know what is average weight for a 18, and what is a average weight for a 19. So I can try and get the light rims im looking for, in hopes that the rims I get either a 18 or a 19, that it will be lighter then a the stock 17. Thanks for the time!
11-19-2004, 03:42 PM
#3
stock wheels weigh around 25 lbs, so try shooting for lower than that. for every pound of unsprung weight such as wheel weight, it is believed to eliminate 8lbs of sprung weight. the less the energy the engine has to use to turn the wheels, the quicker the car can accelerate. there is a well put together article discussing this topic on the main page under shop talk then reading room under technical articles.
11-19-2004, 03:53 PM
#4
Originally Posted by sloppymax
stock wheels weigh around 25 lbs, so try shooting for lower than that. for every pound of unsprung weight such as wheel weight, it is believed to eliminate 8lbs of sprung weight. the less the energy the engine has to use to turn the wheels, the quicker the car can accelerate. there is a well put together article discussing this topic on the main page under shop talk then reading room under technical articles.
Aren't wheels considered sprung weight?
11-19-2004, 04:02 PM
#5
...every single pound you add to your wheel and tire combination is equivalent to adding eight pounds to the curb weight of your car. However, you don't just have one wheel and tire. You have four of them, and all four wheels require energy to accelerate.
11-19-2004, 04:11 PM
#6
Originally Posted by Mizeree_X
Aren't wheels considered sprung weight?
11-19-2004, 04:12 PM
#7
Originally Posted by Arnold K.
Hmm, I thought our rear wheels just get dragged along? Can anyone specify if that's correct? I also always thought it was 4lbs for every 1lb your rim weighs. 8lbs is downright incredible. My OZ Racing Superleggeras should give me insane gains...
11-19-2004, 06:10 PM
#10
i'm running 18x8.5 with 245/40/18....the total combo weighs around 41lbs.....before i was running on stock 2k2 wheels.....but when i added this combo on there.....i couldn't really notice a difference.....there was a slight difference in acceleration.....but i'd say most of it is in my head.......i guess its cuz of the fact that i went +1 in diameter..and i know if the diameter is further away from the HUB....it makes it harder for the engine to turn it regardless if the rim is lighter than stock....
if you want strictly performance....stay with the stock rim diameter....for example if you have a 2k2....get a rim thats 17" tire combo that weighs like 30-35lbs....i know for sure u will feel a difference....but as with my setup....since i went +1 i can't feel the difference that much even tho my rims weigh almost 10lbs off the stock one..
if you want strictly performance....stay with the stock rim diameter....for example if you have a 2k2....get a rim thats 17" tire combo that weighs like 30-35lbs....i know for sure u will feel a difference....but as with my setup....since i went +1 i can't feel the difference that much even tho my rims weigh almost 10lbs off the stock one..
11-19-2004, 07:11 PM
#11
also remember that the radius (diameter) affects performance the bigger the slower. E=M*R^2*W^2 r being radius and M being weight. the bigger this number the slower the car and the less efficient the power transfer.
11-19-2004, 07:20 PM
#13
Originally Posted by Matt_01SE_20th
also remember that the radius (diameter) affects performance the bigger the slower. E=M*R^2*W^2 r being radius and M being weight. the bigger this number the slower the car and the less efficient the power transfer.
which is why i keep my stock 16..can you put that equation into numbers for me so i can see the actual benefit of my 16s.
how much less energy am i using?
11-19-2004, 08:33 PM
#14
Originally Posted by nickk817
YEA, because I just recently found 18" rims that only weighs 15.5 pounds each, so im guessing if I put these bad boys on, I should deff see a difference right?
11-19-2004, 08:49 PM
#15
Originally Posted by sloppymax
nope, thats article should clear it up. brakes would also be considered a contributor to unsprung weight. the front wheels are the only ones being driven so that calcuation would work for the two up front but im not sure if it improves anything for the rear.
As for unsprung weight, it includes anything not held up by the cars suspension. You could say that if you jumped up and down on your car, only the sprung weight would move straight up and down.
Your suspension pieces are actually calculated as being part sprung and part unsprung weight. While they do move, it isn't the type of linear up and down movement as say your bumper.
11-19-2004, 08:54 PM
#16
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Join Date: Sep 2004
Posts: 426
Originally Posted by nickk817
So I can try and get the light rims im looking for, in hopes that the rims I get either a 18 or a 19, that it will be lighter then a the stock 17.
11-19-2004, 10:45 PM
#17
Originally Posted by soonerfan
which is why i keep my stock 16..can you put that equation into numbers for me so i can see the actual benefit of my 16s.
how much less energy am i using?
E=I*w^2
I=M*R^2
to correctly do this i would need the mass of the rim but lets call it 12kg. (it is so much easier to do it in metric)
so if the rim is a 16 in. and the tire is about 4 in. that is about 24 in. diameter, we will say that the center of mass is probably at about 8 in. from center. (about .204m) and a 17 in. rim has center of mass about 8.5in.(.216m) and masses about 14kg.
I=12*8^2= 768 for the 16 in. rim
I=14*8.5^2=1012 for the 17 in. rim
now say you are getting the car up to 60mph. (1056 in/sec)
75.4in./rev (16)
78.5in./rev (17)
your wheels are turning at a rate of:
14rev/sec (16)
13.45rev/sec (17)
2*pi rad/rev so:
88rad/sec (16)
84.52rad/sec (17)
so finally we have:
E=768*88^2=5947392 J (16)
E=1012*84.52^2=7339354 J (17)
1391962 J of difference.
one gallon of gas contains 132,000,000 J (engines are only about 25%efficient so really only 33,000,000 J get to the wheels.)
1391962/33000000 = 4.22%
a 17 in wheel requires about 4.22% more energy to spin the wheels than a 16 in rim.
NOTE: all numbers used in this example are estimations and should be taken as approximations only. this is a general equation and there is alot more to the whole picture than just the wheel size. this is meant to give a general idea of the energy required to turn the wheels.
P.S. this equation would be multiplied by 4 in each scenario but the percent difference will remain the same.
11-19-2004, 11:09 PM
#18
Senior Member
Join Date: Jul 2002
Posts: 251
Originally Posted by Matt_01SE_20th
E=I*w^2
I=M*R^2
to correctly do this i would need the mass of the rim but lets call it 12kg. (it is so much easier to do it in metric)
so if the rim is a 16 in. and the tire is about 4 in. that is about 24 in. diameter, we will say that the center of mass is probably at about 8 in. from center. (about .204m) and a 17 in. rim has center of mass about 8.5in.(.216m) and masses about 14kg.
I=12*8^2= 768 for the 16 in. rim
I=14*8.5^2=1012 for the 17 in. rim
now say you are getting the car up to 60mph. (1056 in/sec)
75.4in./rev (16)
78.5in./rev (17)
your wheels are turning at a rate of:
14rev/sec (16)
13.45rev/sec (17)
2*pi rad/rev so:
88rad/sec (16)
84.52rad/sec (17)
so finally we have:
E=768*88^2=5947392 J (16)
E=1012*84.52^2=7339354 J (17)
1391962 J of difference.
one gallon of gas contains 132,000,000 J (engines are only about 25%efficient so really only 33,000,000 J get to the wheels.)
1391962/33000000 = 4.22%
a 17 in wheel requires about 4.22% more energy to spin the wheels than a 16 in rim.
NOTE: all numbers used in this example are estimations and should be taken as approximations only. this is a general equation and there is alot more to the whole picture than just the wheel size. this is meant to give a general idea of the energy required to turn the wheels.
P.S. this equation would be multiplied by 4 in each scenario but the percent difference will remain the same.
I=M*R^2
to correctly do this i would need the mass of the rim but lets call it 12kg. (it is so much easier to do it in metric)
so if the rim is a 16 in. and the tire is about 4 in. that is about 24 in. diameter, we will say that the center of mass is probably at about 8 in. from center. (about .204m) and a 17 in. rim has center of mass about 8.5in.(.216m) and masses about 14kg.
I=12*8^2= 768 for the 16 in. rim
I=14*8.5^2=1012 for the 17 in. rim
now say you are getting the car up to 60mph. (1056 in/sec)
75.4in./rev (16)
78.5in./rev (17)
your wheels are turning at a rate of:
14rev/sec (16)
13.45rev/sec (17)
2*pi rad/rev so:
88rad/sec (16)
84.52rad/sec (17)
so finally we have:
E=768*88^2=5947392 J (16)
E=1012*84.52^2=7339354 J (17)
1391962 J of difference.
one gallon of gas contains 132,000,000 J (engines are only about 25%efficient so really only 33,000,000 J get to the wheels.)
1391962/33000000 = 4.22%
a 17 in wheel requires about 4.22% more energy to spin the wheels than a 16 in rim.
NOTE: all numbers used in this example are estimations and should be taken as approximations only. this is a general equation and there is alot more to the whole picture than just the wheel size. this is meant to give a general idea of the energy required to turn the wheels.
P.S. this equation would be multiplied by 4 in each scenario but the percent difference will remain the same.
11-20-2004, 01:21 AM
#19
Originally Posted by Matt_01SE_20th
E=I*w^2
I=M*R^2
to correctly do this i would need the mass of the rim but lets call it 12kg. (it is so much easier to do it in metric)
so if the rim is a 16 in. and the tire is about 4 in. that is about 24 in. diameter, we will say that the center of mass is probably at about 8 in. from center. (about .204m) and a 17 in. rim has center of mass about 8.5in.(.216m) and masses about 14kg.
I=12*8^2= 768 for the 16 in. rim
I=14*8.5^2=1012 for the 17 in. rim
now say you are getting the car up to 60mph. (1056 in/sec)
75.4in./rev (16)
78.5in./rev (17)
your wheels are turning at a rate of:
14rev/sec (16)
13.45rev/sec (17)
2*pi rad/rev so:
88rad/sec (16)
84.52rad/sec (17)
so finally we have:
E=768*88^2=5947392 J (16)
E=1012*84.52^2=7339354 J (17)
1391962 J of difference.
one gallon of gas contains 132,000,000 J (engines are only about 25%efficient so really only 33,000,000 J get to the wheels.)
1391962/33000000 = 4.22%
a 17 in wheel requires about 4.22% more energy to spin the wheels than a 16 in rim.
NOTE: all numbers used in this example are estimations and should be taken as approximations only. this is a general equation and there is alot more to the whole picture than just the wheel size. this is meant to give a general idea of the energy required to turn the wheels.
P.S. this equation would be multiplied by 4 in each scenario but the percent difference will remain the same.
I=M*R^2
to correctly do this i would need the mass of the rim but lets call it 12kg. (it is so much easier to do it in metric)
so if the rim is a 16 in. and the tire is about 4 in. that is about 24 in. diameter, we will say that the center of mass is probably at about 8 in. from center. (about .204m) and a 17 in. rim has center of mass about 8.5in.(.216m) and masses about 14kg.
I=12*8^2= 768 for the 16 in. rim
I=14*8.5^2=1012 for the 17 in. rim
now say you are getting the car up to 60mph. (1056 in/sec)
75.4in./rev (16)
78.5in./rev (17)
your wheels are turning at a rate of:
14rev/sec (16)
13.45rev/sec (17)
2*pi rad/rev so:
88rad/sec (16)
84.52rad/sec (17)
so finally we have:
E=768*88^2=5947392 J (16)
E=1012*84.52^2=7339354 J (17)
1391962 J of difference.
one gallon of gas contains 132,000,000 J (engines are only about 25%efficient so really only 33,000,000 J get to the wheels.)
1391962/33000000 = 4.22%
a 17 in wheel requires about 4.22% more energy to spin the wheels than a 16 in rim.
NOTE: all numbers used in this example are estimations and should be taken as approximations only. this is a general equation and there is alot more to the whole picture than just the wheel size. this is meant to give a general idea of the energy required to turn the wheels.
P.S. this equation would be multiplied by 4 in each scenario but the percent difference will remain the same.
What the hell did I just read. Now my head hurts.
11-20-2004, 05:00 AM
#20
Originally Posted by Matt_01SE_20th
E=I*w^2
I=M*R^2
to correctly do this i would need the mass of the rim but lets call it 12kg. (it is so much easier to do it in metric)
so if the rim is a 16 in. and the tire is about 4 in. that is about 24 in. diameter, we will say that the center of mass is probably at about 8 in. from center. (about .204m) and a 17 in. rim has center of mass about 8.5in.(.216m) and masses about 14kg.
I=12*8^2= 768 for the 16 in. rim
I=14*8.5^2=1012 for the 17 in. rim
now say you are getting the car up to 60mph. (1056 in/sec)
75.4in./rev (16)
78.5in./rev (17)
your wheels are turning at a rate of:
14rev/sec (16)
13.45rev/sec (17)
2*pi rad/rev so:
88rad/sec (16)
84.52rad/sec (17)
so finally we have:
E=768*88^2=5947392 J (16)
E=1012*84.52^2=7339354 J (17)
1391962 J of difference.
one gallon of gas contains 132,000,000 J (engines are only about 25%efficient so really only 33,000,000 J get to the wheels.)
1391962/33000000 = 4.22%
a 17 in wheel requires about 4.22% more energy to spin the wheels than a 16 in rim.
NOTE: all numbers used in this example are estimations and should be taken as approximations only. this is a general equation and there is alot more to the whole picture than just the wheel size. this is meant to give a general idea of the energy required to turn the wheels.
P.S. this equation would be multiplied by 4 in each scenario but the percent difference will remain the same.
I=M*R^2
to correctly do this i would need the mass of the rim but lets call it 12kg. (it is so much easier to do it in metric)
so if the rim is a 16 in. and the tire is about 4 in. that is about 24 in. diameter, we will say that the center of mass is probably at about 8 in. from center. (about .204m) and a 17 in. rim has center of mass about 8.5in.(.216m) and masses about 14kg.
I=12*8^2= 768 for the 16 in. rim
I=14*8.5^2=1012 for the 17 in. rim
now say you are getting the car up to 60mph. (1056 in/sec)
75.4in./rev (16)
78.5in./rev (17)
your wheels are turning at a rate of:
14rev/sec (16)
13.45rev/sec (17)
2*pi rad/rev so:
88rad/sec (16)
84.52rad/sec (17)
so finally we have:
E=768*88^2=5947392 J (16)
E=1012*84.52^2=7339354 J (17)
1391962 J of difference.
one gallon of gas contains 132,000,000 J (engines are only about 25%efficient so really only 33,000,000 J get to the wheels.)
1391962/33000000 = 4.22%
a 17 in wheel requires about 4.22% more energy to spin the wheels than a 16 in rim.
NOTE: all numbers used in this example are estimations and should be taken as approximations only. this is a general equation and there is alot more to the whole picture than just the wheel size. this is meant to give a general idea of the energy required to turn the wheels.
P.S. this equation would be multiplied by 4 in each scenario but the percent difference will remain the same.
The rotating mass would also be very close (given similar rim design). The tires tread would still be the same distance from the center of the wheel. With more rubber being spun around, the rotating mass would be similar.
I could be off but that is how it looks to me.
11-20-2004, 07:32 AM
#22
Member
Join Date: Nov 2003
Posts: 146
yeah, the overally circumfence of the tire and wheel combo is what should seal it, im no math wiz honestly, but if you go from a stock 16 to an afterarket 18, keep the same aspect ratio, and the 18 wieghs less, in theory you would gain a little bit
when i put 18s on my thunderbird, i didnt noticed the diff between the stock 16s, i believe the stock 16s weigh in close to 25lbs maybe more as my 18s are 21lbs a piece, and with the lower pro tires, with tires weigh a whole lot less b/c theres less rubber on there
when i put 18s on my thunderbird, i didnt noticed the diff between the stock 16s, i believe the stock 16s weigh in close to 25lbs maybe more as my 18s are 21lbs a piece, and with the lower pro tires, with tires weigh a whole lot less b/c theres less rubber on there
11-20-2004, 08:01 AM
#23
Originally Posted by joeyicu
yeah, the overally circumfence of the tire and wheel combo is what should seal it, im no math wiz honestly, but if you go from a stock 16 to an afterarket 18, keep the same aspect ratio, and the 18 wieghs less, in theory you would gain a little bit
when i put 18s on my thunderbird, i didnt noticed the diff between the stock 16s, i believe the stock 16s weigh in close to 25lbs maybe more as my 18s are 21lbs a piece, and with the lower pro tires, with tires weigh a whole lot less b/c theres less rubber on there
when i put 18s on my thunderbird, i didnt noticed the diff between the stock 16s, i believe the stock 16s weigh in close to 25lbs maybe more as my 18s are 21lbs a piece, and with the lower pro tires, with tires weigh a whole lot less b/c theres less rubber on there
11-20-2004, 08:12 AM
#24
Senior Member
Join Date: Oct 2002
Posts: 1,554
What about the frictional forces the torques are faced by pavement. Energy is often release in the form of sound, or heat.
Then consider the mass distribution of a fancy racimg rim. A very good rim will have it as close to the center as possible. Remember that torque is the cross product of force and radius (or radius cross angular velocity is it i forget).
This angular velocity should be exactly perpindicular to the exact axis of rotation, in this case if this is not so, your situation is all to ideal and doesn't make sense in real life. Since property i'm talking about now is momment of ineritia. An angular mass is not the same since velocity is always changing about an axis, and of course the translation to actual car velocity is effected by the wheels radius, but don't forget the whee's mass, and it's mass distribution, as well as that of the tire, and the coefficent of friction the tires impose, as well as that of the drivetrain.
Also you talk about 1 gallon of gas like it's always going to produce that much work. You are awere cars vary in performance levels, efficency, and of course drive train loss.
Then consider the mass distribution of a fancy racimg rim. A very good rim will have it as close to the center as possible. Remember that torque is the cross product of force and radius (or radius cross angular velocity is it i forget).
This angular velocity should be exactly perpindicular to the exact axis of rotation, in this case if this is not so, your situation is all to ideal and doesn't make sense in real life. Since property i'm talking about now is momment of ineritia. An angular mass is not the same since velocity is always changing about an axis, and of course the translation to actual car velocity is effected by the wheels radius, but don't forget the whee's mass, and it's mass distribution, as well as that of the tire, and the coefficent of friction the tires impose, as well as that of the drivetrain.
Also you talk about 1 gallon of gas like it's always going to produce that much work. You are awere cars vary in performance levels, efficency, and of course drive train loss.
11-20-2004, 03:12 PM
#25
when doing this calculation i just kept the tire size at a 225/50 and changed the rim size. this is like the difference between the 4th gen and 5th gen or the 16" 5th gen and the 17" 5th gen. a lower profile tire still sets teh CENTER of massfurther out because it is more concentrated in one area, further away from center.
Rim weight: i was simply comparing a estimated mass of the stock 16's vs. the stock 17'. i am trying to say that teh larger radius requires more energy to spin.
Joeyicu: the reason you prbably notice no difference is because the mass difference offsets the radius difference.
mjg: this is why i put teh NOTE at teh bottom, like i said there is much more to it but this is GENERALLY how it works.
the mass distribution is very ballenced around the center but i am talking about the "ring of mass". basically the center pf mass is calculated for an infintesimaly small angle and then applied to the rim as a whole. (force and radius are correct)
you are correct in saying this situation is too ideal but that is how you do an example withou teaching the entire class. all i am talking about is the wheels radius, the wheel's mass, and it's mass distribution, as well as that of the tire. NOT the coefficent of friction the tires impose, as well as that of the drivetrain.
you are correct in the fuel energy conversion being too general but this is the scientifically accepted estimation of bothe energy quantity of gasoline and tyhe efficiency of the IC engine
EVERYONE PLEASE READ THE NOTES AT THE END OF POSTS. the wole reason i put that note in was so that people wouldnt take this into too much detail. if you want to really know all the factors you need to know alot of phisics.
Rim weight: i was simply comparing a estimated mass of the stock 16's vs. the stock 17'. i am trying to say that teh larger radius requires more energy to spin.
Joeyicu: the reason you prbably notice no difference is because the mass difference offsets the radius difference.
mjg: this is why i put teh NOTE at teh bottom, like i said there is much more to it but this is GENERALLY how it works.
the mass distribution is very ballenced around the center but i am talking about the "ring of mass". basically the center pf mass is calculated for an infintesimaly small angle and then applied to the rim as a whole. (force and radius are correct)
you are correct in saying this situation is too ideal but that is how you do an example withou teaching the entire class. all i am talking about is the wheels radius, the wheel's mass, and it's mass distribution, as well as that of the tire. NOT the coefficent of friction the tires impose, as well as that of the drivetrain.
you are correct in the fuel energy conversion being too general but this is the scientifically accepted estimation of bothe energy quantity of gasoline and tyhe efficiency of the IC engine
EVERYONE PLEASE READ THE NOTES AT THE END OF POSTS. the wole reason i put that note in was so that people wouldnt take this into too much detail. if you want to really know all the factors you need to know alot of phisics.
11-20-2004, 11:26 PM
#26
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Thread Starter
Join Date: Jul 2004
Posts: 32
Ok, im not trying to be an ***, I appreciate everybody's input with their math skills and all, but all I want to know is will I feel a difference between the stock 25lb 17 and the 15.5lb 18, and is it worth it going for the lighter 18 or should I just get a 19 that weighs like 28. Keeping in mind I'm a performance freak lol.
11-21-2004, 07:32 AM
#27
Originally Posted by nickk817
Ok, im not trying to be an ***, I appreciate everybody's input with their math skills and all, but all I want to know is will I feel a difference between the stock 25lb 17 and the 15.5lb 18, and is it worth it going for the lighter 18 or should I just get a 19 that weighs like 28. Keeping in mind I'm a performance freak lol.
11-21-2004, 11:49 AM
#28
Originally Posted by nickk817
Ok, im not trying to be an ***, I appreciate everybody's input with their math skills and all, but all I want to know is will I feel a difference between the stock 25lb 17 and the 15.5lb 18, and is it worth it going for the lighter 18 or should I just get a 19 that weighs like 28. Keeping in mind I'm a performance freak lol.
Being a performance freak you should value analytical data over "will my butt dyno feel different" but I guess that's just me. Anyways it depends on how accurate your butt dyno is, I know I can feel a difference switching back and forth between my wheels. I know you have no use for "numbers" but if you went to the track and compared the two under otherwise similar conditions, you would definately see a difference assuming your wheel/tire combos have a similar difference as just the wheels themselves do (on the order of 9.5lbs or so). My timeslips indicate a clear difference when I switch from one set of wheels from another at the track. Whether the difference is "worth it" depends entirely on how hardcore of a performance freak you are. Since you have no use for peoples hard data and calculations, I'm going to go out on a limb and say for you, no, it's not worth it. But I could be wrong.
11-22-2004, 08:25 AM
#31
Senior Member
Join Date: Jul 2001
Location: state of confusion
Posts: 1,341
Originally Posted by nickk817
Ok, im not trying to be an ***, I appreciate everybody's input with their math skills and all, but all I want to know is will I feel a difference between the stock 25lb 17 and the 15.5lb 18, and is it worth it going for the lighter 18 or should I just get a 19 that weighs like 28. Keeping in mind I'm a performance freak lol.
If performance is truly of absolute priority, whether or not you can feel the difference does not matter - it is there and it will show up in the clocks at timed events (dragstrip, auto-x, or timed road course lapping).
As to the separate issue of whether you should choose style or performance - how serious are you about the performance (read: how much of it are you willing to trade away for the sizzle)?
Norm
11-22-2004, 11:02 AM
#32
In auto-x both unsprung weight and gearing matters greatly - much more so than dragstrip. Most everybody of national stature runs 14"-17" rims depending on vehicle type. I've found 15"-16" to be ideal for G3-G5 Maxima's.
11-24-2004, 08:37 AM
#33
Member
Join Date: Nov 2003
Posts: 146
Originally Posted by nickk817
Ok, im not trying to be an ***, I appreciate everybody's input with their math skills and all, but all I want to know is will I feel a difference between the stock 25lb 17 and the 15.5lb 18, and is it worth it going for the lighter 18 or should I just get a 19 that weighs like 28. Keeping in mind I'm a performance freak lol.
11-24-2004, 11:08 PM
#34
I am about to put some 2K2 17's on my 96 Maxima SE, man do they weigh a lot!!!! something like 50+LBS for each wheel/tire combo. I already have the 97-99 SE rims on it right now.
I am going to also check gas mileage with these 2k2's to see the difference. A lowered 4th gen with 2K2 wheels look really nice though, I ll probably take the 2k2's off if the car suffers a lot in performance and gas mileage and get something light, maybe even a 16' rota slipstream.
I am going to also check gas mileage with these 2k2's to see the difference. A lowered 4th gen with 2K2 wheels look really nice though, I ll probably take the 2k2's off if the car suffers a lot in performance and gas mileage and get something light, maybe even a 16' rota slipstream.
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