I removed them today to put my RSB on and damn these suckers are heavy! So I put them on the scale and I was right, 50lb's each. It's like nissan is trying to perpicly slow down the 02 maxima so the altima will look better{speed wise of course}. It would be interresting to see how much the 02 SE Altima rims/tires weigh, anyone know? How about 2001 17' SE maxima wheels?

I thought my 17' Chrome bangers where real heavy at 43lb's..

 

It's a conspiracy I tell you!

 

Mine are 47...

 

Yep, thats what mine weighed a long time ago when I posted it! hehe

 

You do realise the 2002 SE rims are different from your 2K1 SE rims, correct? Now I have to ask, how did you post the weight of the new 6 spoke SE rims a year ago?

Greg, it look's like the weight is pretty close then. Thanks

 

Only difference is that on 2000 SE we have center piece that covers the bolts.
On 2002 SE they removed that center piece.

They should weigh the same.

 

I said a long time ago. It was like when I had 6k on the car and weighed it. They are identical rims in design. Yours have no caps, mine have caps, thats it.

 

You filled them with Helium! Next step, Hydrogen!!!

 

No, I have 225/45-17s.

 

and don't forget about the MOLE, if you have any gas at a certain PSI it will contain the same MASS of particals thus the same weight. helium or hydrogen or AIR at 31PSI all weighs the same.

 

Steve, you looked that up, didn't you...

 

no i have a VERY good memory. i get complimented on it all the time. i dont even write my orders down at friendly's, i'm a waiter to all those that didn't know. that's all i ever hear from people at work, OMG, your not writing any of this down, wow that's impressive. then i get locked in a conversation with them. i never even studied through high school, never.

 

They look pretty good, but they weigh a ton

I need some new rims... once the sucky stock tires wear out, the rims will probably be gone too....

 

So how do you afford a 2002 6 speed on as a waiter at friendly's?

Stereodude

 

He has two jobs, works in a bank too.

 

Ah... That explains it. I have two jobs also, but I don't get paid for the 2nd one. I need to stop doing things for free. (I make web pages for the 2nd one. PHP is so sweet.)

Stereodude

 

thanks greggy poo. i always forget to add that in and someone FREAKS out and wants to go work at friendly's even when i had my slow *** 95 it still had a lot into it and people wondered.

 

 

 

I am a street pharmacist

 

 

 

 

 

Do you wanna sellum? Where did you get them from?

 

New rims
just dont buy mine

 

thats why you go with hubcaps...they're stylish, they're light, weather-resistant, easy to replace

 

2000 stock rims w/ stock tires (the 17s) weigh in at a hefty 55lbs. so don't feel too bad about 50

 

i was gonna keep my stock 17's because well they look decent and are already 17's but i think i might be in for more than just tires in the spring. we'll see how the money is, but i know these re92's HAVE to go and prolly the rims too.

 

I got them from the car dealership when I bought my 2002 maxima . BTW I don't wanna sell them.

 

They are not exactly alike the spoke shape is sort of different as well between the 2k1 and 2k2se's. looking at an old photo of my 2k1.

 

Oh the humanity!

 

naah, the one with the hydrogen will be the lightest, followed by helium and then air...

 

ummm, no

 

My stock 17s weigh 50lb. Same as yours, but I have stock rubber.

 

yeah they would... well okay don't think I'm a pedantic d[ck, but this useless crap has been forced into my head so much that I have to let it out someplace before i burst.

so according to what you said, a balloon filled with helium/hydrogen would weigh the SAME as a balloon filled with air... therefore it wouldn't float/rise up (assuming both balloons are at the same pressure and volume)...
a mole is NOT a measure of mass.. it's an actual number (just like a dozen is equal to 12). A mole is equal to 6.022 x 10^23 (that's 10 to the 23rd power) particles, or Avogadro's number. Therefore one mole of ANY gas is equal to that number, regardless of how heavy/light the molecules in that gas may be. So given I have one mole of pure hydrogen gas (hydrogen being the lightest of all elements) in one balloon and one mole of a gas heavier/more dense than air (like some bromine gases) in a balloon of equal volume and pressure; the balloon filled with hydrogen would rise up (because it is lighter than air) and the balloon filled with the bromine gas would sink and stay on the ground (because it is heavier than air).
The volume that one mole of gas takes up is called the molar volume.
Now, in the ideal gas laws, variabls such as pressure, volume, and temperature and moles of a certain amount of gas are all related to an independent constant, i.e. since they are ideal gases, the properties of pressure of volume will be equal to a constant and do NOT depend on the nature of the gas.
This is proven in Boyle's law of ideal gases:
PV=k, where P is pressure in PSI, V is volume in liters, and k is a constant, in this case equal to about 1.206 PSIxL.

at 31 PSI:
PV=k
V=k/P
V= 1.206/31= 0.0389 L

so at 31 PSI, air, hydrogen, and helium will take up 0.0389 liters of volume at standard temperature (zero degrees Celsius)

now to take into account that we're not at standard temperature, say we're at like, ummm, 60 degrees Fahrenheit, or maybe about 16 degrees Celsius (I think), that the tires are inflated to 32 PSI, and that a tire has a volume of, say... umm ( I have NO clue so I'll guess) about 12 liters.

Using the ideal gas law:
PV = nRT (where n is the # of moles, R is a constant, and T is the temperature; I will convert the pressure from PSI to mm mercury and use the temperature in Kelvin)

PV = nRT
n = (PV)/(RT)
n = (1603 mmHg x 12 L)/(62.4 x 289 K)
n = 10.67 moles of gas

now say we have three of these tires (all of equal weight and volume), one filled with air (molecular weight of 29 grams/mol), one with hydrogen (MW of 1 g/mol) and one with helium (MW = 4 g/mol).

moles = grams/molecular weight, for any chemical compound:

n = g/mw
Air:
g = (n)(mw)
g = 10.67 (29)
total weight= 309.4 grams + 50 lbs (weight of tire and rim)
= 309.4 g + 22680 g = 22989.4g, or 23kg

Helium:
g = 10.67 (4)
g = 42.7 grams
total weight= 42.7g + 22680g = 22722.7g or 22.722kg

Hydrogen:
g = 10.67 (1)
g = 10.67
total weight= 10.67g + 22680g = 22690.67g or 22.691 kg

therefore, in order of weight:
hydrogen filled tires are lighter than helium filled tires, and both are lighter than tires filled with air...

*phew!*

 

I'll let you correct your rahter large mistake.

 

oops! sorry:
at 31 PSI:
PV=k
P/k = V
31/1.206 = V
V = 25.7 liters

is wrong...
should be:
PV=k
V=k/P
V= 1.206/31= 0.0389 L so it's a much smaller number and therefore makes sense... one mole of gas at that kind of pressure could not possibly take up 25.7 liters

but that still has no bearing on what I wrote. Sorry, I was in a hurry (late for class!)

 

You still haven't corrected it... HINT:How can a tire at 31 PSI weigh 15000+ pounds?

 

Using you're own formula I beleive it should read
Air+tire+rim=22989.4 or 22.989 Kg/50.58lbs

He+tire+rim=22722.7 or 22.723 Kg/49.99lbs

H+tire+rim=22690.67 or 22.691 Kg/49.92lbs

This seems to be a weight loss that would be not worth the effort.
1/2 lbs at the wheels at best, remove your spare tire at the track and you would see greater performance gains.

My .02