A heavy question about tires
10-12-2007, 05:03 PM
#1
A heavy question about tires
I've sold my set of wheels and tires, and in thinking about packing them up, I was wondering how much weight I'd save by deflating the tires instead of shipping them inflated to 35 psi.?
10-12-2007, 05:34 PM
#3
Seriously though.... can I put that quote in my sig?
Last edited by tvis; 10-12-2007 at 05:38 PM.
10-12-2007, 06:50 PM
#10
This one
Not trying to make fun..... but dude. How do u think that deflating your tires is going to save weight?
Not trying to make fun..... but dude. How do u think that deflating your tires is going to save weight?
10-12-2007, 08:12 PM
#16
Senior Member
iTrader: (11)
Join Date: Jun 2006
Location: Cherry Hill, NJ / Hoboken, NJ
Posts: 1,325
Ok, I can see some serious math is required, so move aside amateurs! 
215/55R16 tires
That means each tire is a cylinder 21.5cm high, and (16x2.54)+(0.55x21.5x2)=64.3cm across. Containing a volume of 2hr^2 = 2(21.5)(32.15^2) = 44446 cubic centimeters. Since we're only talking about the air in the rubber part, we'll subtract the volume of the wheel from that. which is 21.5cm high and (16x2.54)= 40.6cm across. With a volume of 2hr^2 = 2(21.5)(20.3^2) = 17720 cubic centimeters. Therefore the volume of the air just in the tire is 44446-17720 = 26726cc.
There are four tires, so that's a total displacement of 106904cc, or ~107 liters.
Now, if the tires are deflated, at atmospheric pressure and room temperature (68 degrees), by Van Der Waal's Law of Gases;
PV = nRT
(1atm)(107L) = n(0.08206)(293k)
n = 4.5mol
since air is primarily nitrogen, we can assume that 1 mol = 28g. 4.5x28 = 126g. That's 0.28 lbs for one atmosphere of pressure.
Now, 1atm is 14.7psi.If your tire pressure is 35psi, that's 2.38 atmospheres. Or 1.38 more. 1.38 atmospheres is just 0.28x1.38 = 0.39
So, if you deflate them fully, it will weigh 0.39 pounds less.
215/55R16 tires
That means each tire is a cylinder 21.5cm high, and (16x2.54)+(0.55x21.5x2)=64.3cm across. Containing a volume of 2hr^2 = 2(21.5)(32.15^2) = 44446 cubic centimeters. Since we're only talking about the air in the rubber part, we'll subtract the volume of the wheel from that. which is 21.5cm high and (16x2.54)= 40.6cm across. With a volume of 2hr^2 = 2(21.5)(20.3^2) = 17720 cubic centimeters. Therefore the volume of the air just in the tire is 44446-17720 = 26726cc.
There are four tires, so that's a total displacement of 106904cc, or ~107 liters.
Now, if the tires are deflated, at atmospheric pressure and room temperature (68 degrees), by Van Der Waal's Law of Gases;
PV = nRT
(1atm)(107L) = n(0.08206)(293k)
n = 4.5mol
since air is primarily nitrogen, we can assume that 1 mol = 28g. 4.5x28 = 126g. That's 0.28 lbs for one atmosphere of pressure.
Now, 1atm is 14.7psi.If your tire pressure is 35psi, that's 2.38 atmospheres. Or 1.38 more. 1.38 atmospheres is just 0.28x1.38 = 0.39
So, if you deflate them fully, it will weigh 0.39 pounds less.
10-12-2007, 08:29 PM
#18
Ok, I can see some serious math is required, so move aside amateurs!
215/55R16 tires
That means each tire is a cylinder 21.5cm high, and (16x2.54)+(0.55x21.5x2)=64.3cm across. Containing a volume of 2hr^2 = 2(21.5)(32.15^2) = 44446 cubic centimeters. Since we're only talking about the air in the rubber part, we'll subtract the volume of the wheel from that. which is 21.5cm high and (16x2.54)= 40.6cm across. With a volume of 2hr^2 = 2(21.5)(20.3^2) = 17720 cubic centimeters. Therefore the volume of the air just in the tire is 44446-17720 = 26726cc.
There are four tires, so that's a total displacement of 106904cc, or ~107 liters.
Now, if the tires are deflated, at atmospheric pressure and room temperature (68 degrees), by Van Der Waal's Law of Gases;
PV = nRT
(1atm)(107L) = n(0.08206)(293k)
n = 4.5mol
since air is primarily nitrogen, we can assume that 1 mol = 28g. 4.5x28 = 126g. That's 0.28 lbs for one atmosphere of pressure.
Now, 1atm is 14.7psi.If your tire pressure is 35psi, that's 2.38 atmospheres. Or 1.38 more. 1.38 atmospheres is just 0.28x1.38 = 0.39
So, if you deflate them fully, it will weigh 0.39 pounds less.
215/55R16 tires
That means each tire is a cylinder 21.5cm high, and (16x2.54)+(0.55x21.5x2)=64.3cm across. Containing a volume of 2hr^2 = 2(21.5)(32.15^2) = 44446 cubic centimeters. Since we're only talking about the air in the rubber part, we'll subtract the volume of the wheel from that. which is 21.5cm high and (16x2.54)= 40.6cm across. With a volume of 2hr^2 = 2(21.5)(20.3^2) = 17720 cubic centimeters. Therefore the volume of the air just in the tire is 44446-17720 = 26726cc.
There are four tires, so that's a total displacement of 106904cc, or ~107 liters.
Now, if the tires are deflated, at atmospheric pressure and room temperature (68 degrees), by Van Der Waal's Law of Gases;
PV = nRT
(1atm)(107L) = n(0.08206)(293k)
n = 4.5mol
since air is primarily nitrogen, we can assume that 1 mol = 28g. 4.5x28 = 126g. That's 0.28 lbs for one atmosphere of pressure.
Now, 1atm is 14.7psi.If your tire pressure is 35psi, that's 2.38 atmospheres. Or 1.38 more. 1.38 atmospheres is just 0.28x1.38 = 0.39
So, if you deflate them fully, it will weigh 0.39 pounds less.
10-12-2007, 08:38 PM
#19
Ok, I can see some serious math is required, so move aside amateurs!
215/55R16 tires
That means each tire is a cylinder 21.5cm high, and (16x2.54)+(0.55x21.5x2)=64.3cm across. Containing a volume of 2hr^2 = 2(21.5)(32.15^2) = 44446 cubic centimeters. Since we're only talking about the air in the rubber part, we'll subtract the volume of the wheel from that. which is 21.5cm high and (16x2.54)= 40.6cm across. With a volume of 2hr^2 = 2(21.5)(20.3^2) = 17720 cubic centimeters. Therefore the volume of the air just in the tire is 44446-17720 = 26726cc.
There are four tires, so that's a total displacement of 106904cc, or ~107 liters.
Now, if the tires are deflated, at atmospheric pressure and room temperature (68 degrees), by Van Der Waal's Law of Gases;
PV = nRT
(1atm)(107L) = n(0.08206)(293k)
n = 4.5mol
since air is primarily nitrogen, we can assume that 1 mol = 28g. 4.5x28 = 126g. That's 0.28 lbs for one atmosphere of pressure.
Now, 1atm is 14.7psi.If your tire pressure is 35psi, that's 2.38 atmospheres. Or 1.38 more. 1.38 atmospheres is just 0.28x1.38 = 0.39
So, if you deflate them fully, it will weigh 0.39 pounds less.
215/55R16 tires
That means each tire is a cylinder 21.5cm high, and (16x2.54)+(0.55x21.5x2)=64.3cm across. Containing a volume of 2hr^2 = 2(21.5)(32.15^2) = 44446 cubic centimeters. Since we're only talking about the air in the rubber part, we'll subtract the volume of the wheel from that. which is 21.5cm high and (16x2.54)= 40.6cm across. With a volume of 2hr^2 = 2(21.5)(20.3^2) = 17720 cubic centimeters. Therefore the volume of the air just in the tire is 44446-17720 = 26726cc.
There are four tires, so that's a total displacement of 106904cc, or ~107 liters.
Now, if the tires are deflated, at atmospheric pressure and room temperature (68 degrees), by Van Der Waal's Law of Gases;
PV = nRT
(1atm)(107L) = n(0.08206)(293k)
n = 4.5mol
since air is primarily nitrogen, we can assume that 1 mol = 28g. 4.5x28 = 126g. That's 0.28 lbs for one atmosphere of pressure.
Now, 1atm is 14.7psi.If your tire pressure is 35psi, that's 2.38 atmospheres. Or 1.38 more. 1.38 atmospheres is just 0.28x1.38 = 0.39
So, if you deflate them fully, it will weigh 0.39 pounds less.
The reason for my question was this:
I'm not trying to save weight as much as to fit the wheel/tire inside a rectangular box: a box for which one side is about an inch shorter than the diameter of the tire. Rather than distend the box to fit the tire, I think it would be easier to distend the tire to fit the box.
That's the motivation. The decrease in weight is just a thought question.
In the process of deflating the tire, I was curious as to whether there would be any noticeable change in weight, defined as a change of 1lb. or more.
Now, we see that there would be less than a 1lb change.
HOWEVER....
In any event, I am going to answer my question through experimentation. I will weigh my wheel/tire with pressure set at 35psi. Then, I will totally deflate the tire, and weigh it again. The resulting weight should be equal to the sum of the weight of the wheel (19.5lbs.) and the weight of the tire (26lbs.).
I will let you know what the real-world answer is. I have a digital scale that is accurate to +/- 1 ounce, or .0625lbs. -- which is less than the .0807lb. baseline.
If your calculations are right, Nalc, and I assume they are, then my test should confirm what you predict.
Or maybe not......
10-12-2007, 09:09 PM
#20
Sag.
You would be able to fit it in a smaller box (or at least one with different dimensions perhaps) than if they were all fully inflated.
10-13-2007, 12:44 AM
#21
Dude... are you f***ing kidding? They're not going to compress down after you deflate them like they were bicycle inner tube's. What are you gonna do... take out the valve stem core and compress the tire down with a strap and squeeze all the air out? Then keep them compressed (so they don't retain their shape) and squeeze them into a box of the same size all while saving weight and space?!? lol wow
... and why even ship them in a box anyways? It would be a pita getting an exact size box. Keep them inflated and put a piece or 2 of cardboard on either side and tape it all up. Use plastic wrap if you can. Tirerack ships them like that and I'm sure they have had no problems with it. I've gotten a set of wheels/tires from them like that. No they didn't save weight by deflating them, and no they didn't save room by deflating them either.
I've never seen questions and answers like this on any other forum before.
... and why even ship them in a box anyways? It would be a pita getting an exact size box. Keep them inflated and put a piece or 2 of cardboard on either side and tape it all up. Use plastic wrap if you can. Tirerack ships them like that and I'm sure they have had no problems with it. I've gotten a set of wheels/tires from them like that. No they didn't save weight by deflating them, and no they didn't save room by deflating them either.
I've never seen questions and answers like this on any other forum before.
Last edited by tvis; 10-13-2007 at 01:03 AM.
10-13-2007, 03:02 AM
#22
1997 Nissan Maxima SE
Join Date: Sep 2007
Location: Illinois
Posts: 110
Listen To tvis, Man!!! 
10-13-2007, 04:31 AM
#23
Thank you.
The reason for my question was this:
I'm not trying to save weight as much as to fit the wheel/tire inside a rectangular box: a box for which one side is about an inch shorter than the diameter of the tire. Rather than distend the box to fit the tire, I think it would be easier to distend the tire to fit the box.
That's the motivation. The decrease in weight is just a thought question.
The reason for my question was this:
I'm not trying to save weight as much as to fit the wheel/tire inside a rectangular box: a box for which one side is about an inch shorter than the diameter of the tire. Rather than distend the box to fit the tire, I think it would be easier to distend the tire to fit the box.
That's the motivation. The decrease in weight is just a thought question.
Clearly you changed your question. Your first post said "how much weight would I save".
I said none, because no shipping scale that your package would ever see will measure the difference. No need to scratch out mathematics to prove it.
Dave
Oh my,
10-13-2007, 08:32 AM
#25
Senior Member
iTrader: (11)
Join Date: Jun 2006
Location: Cherry Hill, NJ / Hoboken, NJ
Posts: 1,325
Thank you.
The reason for my question was this:
I'm not trying to save weight as much as to fit the wheel/tire inside a rectangular box: a box for which one side is about an inch shorter than the diameter of the tire. Rather than distend the box to fit the tire, I think it would be easier to distend the tire to fit the box.
That's the motivation. The decrease in weight is just a thought question.
In the process of deflating the tire, I was curious as to whether there would be any noticeable change in weight, defined as a change of 1lb. or more.
Now, we see that there would be less than a 1lb change.
HOWEVER....
In any event, I am going to answer my question through experimentation. I will weigh my wheel/tire with pressure set at 35psi. Then, I will totally deflate the tire, and weigh it again. The resulting weight should be equal to the sum of the weight of the wheel (19.5lbs.) and the weight of the tire (26lbs.).
I will let you know what the real-world answer is. I have a digital scale that is accurate to +/- 1 ounce, or .0625lbs. -- which is less than the .0807lb. baseline.
If your calculations are right, Nalc, and I assume they are, then my test should confirm what you predict.
Or maybe not......
The reason for my question was this:
I'm not trying to save weight as much as to fit the wheel/tire inside a rectangular box: a box for which one side is about an inch shorter than the diameter of the tire. Rather than distend the box to fit the tire, I think it would be easier to distend the tire to fit the box.
That's the motivation. The decrease in weight is just a thought question.
In the process of deflating the tire, I was curious as to whether there would be any noticeable change in weight, defined as a change of 1lb. or more.
Now, we see that there would be less than a 1lb change.
HOWEVER....
In any event, I am going to answer my question through experimentation. I will weigh my wheel/tire with pressure set at 35psi. Then, I will totally deflate the tire, and weigh it again. The resulting weight should be equal to the sum of the weight of the wheel (19.5lbs.) and the weight of the tire (26lbs.).
I will let you know what the real-world answer is. I have a digital scale that is accurate to +/- 1 ounce, or .0625lbs. -- which is less than the .0807lb. baseline.
If your calculations are right, Nalc, and I assume they are, then my test should confirm what you predict.
Or maybe not......
man thats some gangster a** math. 
10-13-2007, 02:40 PM
#28
Junior Member
Join Date: Apr 2007
Location: Toronto
Posts: 29
lmfao........ hahahah. seriously fill them up with helium and you might save some shipping weight
10-13-2007, 03:26 PM
#29
Actually, I was thinking of using hot air.
OK, I'm done here. I've got to go cut up these sturdy boxes I have so that I can put the cardboard pieces in between the tires and wrap the whole thing up with tape.
I'm also going to fill them all up to 44psi so that they'll be easier to roll off the truck.
Thanks for all the input, guys.
Last edited by dr-rjp; 10-13-2007 at 03:35 PM.
10-13-2007, 04:02 PM
#30
If you want to save money, ship them individually wrapped in plastic. Or in pairs. Put a couple of layers of cardboard on each side to prevent dings to the wheel.
Boxing wheels with tires on them automatically puts your package into the oversize category and adds weight (much more than a paltry .39lb) that adds to the bill.
Boxing wheels with tires on them automatically puts your package into the oversize category and adds weight (much more than a paltry .39lb) that adds to the bill.
11-08-2007, 12:35 PM
#33
Ok, I can see some serious math is required, so move aside amateurs!
215/55R16 tires
That means each tire is a cylinder 21.5cm high, and (16x2.54)+(0.55x21.5x2)=64.3cm across. Containing a volume of 2hr^2 = 2(21.5)(32.15^2) = 44446 cubic centimeters. Since we're only talking about the air in the rubber part, we'll subtract the volume of the wheel from that. which is 21.5cm high and (16x2.54)= 40.6cm across. With a volume of 2hr^2 = 2(21.5)(20.3^2) = 17720 cubic centimeters. Therefore the volume of the air just in the tire is 44446-17720 = 26726cc.
There are four tires, so that's a total displacement of 106904cc, or ~107 liters.
Now, if the tires are deflated, at atmospheric pressure and room temperature (68 degrees), by Van Der Waal's Law of Gases;
PV = nRT
(1atm)(107L) = n(0.08206)(293k)
n = 4.5mol
since air is primarily nitrogen, we can assume that 1 mol = 28g. 4.5x28 = 126g. That's 0.28 lbs for one atmosphere of pressure.
Now, 1atm is 14.7psi.If your tire pressure is 35psi, that's 2.38 atmospheres. Or 1.38 more. 1.38 atmospheres is just 0.28x1.38 = 0.39
So, if you deflate them fully, it will weigh 0.39 pounds less.
215/55R16 tires
That means each tire is a cylinder 21.5cm high, and (16x2.54)+(0.55x21.5x2)=64.3cm across. Containing a volume of 2hr^2 = 2(21.5)(32.15^2) = 44446 cubic centimeters. Since we're only talking about the air in the rubber part, we'll subtract the volume of the wheel from that. which is 21.5cm high and (16x2.54)= 40.6cm across. With a volume of 2hr^2 = 2(21.5)(20.3^2) = 17720 cubic centimeters. Therefore the volume of the air just in the tire is 44446-17720 = 26726cc.
There are four tires, so that's a total displacement of 106904cc, or ~107 liters.
Now, if the tires are deflated, at atmospheric pressure and room temperature (68 degrees), by Van Der Waal's Law of Gases;
PV = nRT
(1atm)(107L) = n(0.08206)(293k)
n = 4.5mol
since air is primarily nitrogen, we can assume that 1 mol = 28g. 4.5x28 = 126g. That's 0.28 lbs for one atmosphere of pressure.
Now, 1atm is 14.7psi.If your tire pressure is 35psi, that's 2.38 atmospheres. Or 1.38 more. 1.38 atmospheres is just 0.28x1.38 = 0.39
So, if you deflate them fully, it will weigh 0.39 pounds less.
11-08-2007, 03:11 PM
#35
Ok, I can see some serious math is required, so move aside amateurs!
215/55R16 tires
That means each tire is a cylinder 21.5cm high, and (16x2.54)+(0.55x21.5x2)=64.3cm across. Containing a volume of 2hr^2 = 2(21.5)(32.15^2) = 44446 cubic centimeters. Since we're only talking about the air in the rubber part, we'll subtract the volume of the wheel from that. which is 21.5cm high and (16x2.54)= 40.6cm across. With a volume of 2hr^2 = 2(21.5)(20.3^2) = 17720 cubic centimeters. Therefore the volume of the air just in the tire is 44446-17720 = 26726cc.
There are four tires, so that's a total displacement of 106904cc, or ~107 liters.
Now, if the tires are deflated, at atmospheric pressure and room temperature (68 degrees), by Van Der Waal's Law of Gases;
PV = nRT
(1atm)(107L) = n(0.08206)(293k)
n = 4.5mol
since air is primarily nitrogen, we can assume that 1 mol = 28g. 4.5x28 = 126g. That's 0.28 lbs for one atmosphere of pressure.
Now, 1atm is 14.7psi.If your tire pressure is 35psi, that's 2.38 atmospheres. Or 1.38 more. 1.38 atmospheres is just 0.28x1.38 = 0.39
So, if you deflate them fully, it will weigh 0.39 pounds less.
215/55R16 tires
That means each tire is a cylinder 21.5cm high, and (16x2.54)+(0.55x21.5x2)=64.3cm across. Containing a volume of 2hr^2 = 2(21.5)(32.15^2) = 44446 cubic centimeters. Since we're only talking about the air in the rubber part, we'll subtract the volume of the wheel from that. which is 21.5cm high and (16x2.54)= 40.6cm across. With a volume of 2hr^2 = 2(21.5)(20.3^2) = 17720 cubic centimeters. Therefore the volume of the air just in the tire is 44446-17720 = 26726cc.
There are four tires, so that's a total displacement of 106904cc, or ~107 liters.
Now, if the tires are deflated, at atmospheric pressure and room temperature (68 degrees), by Van Der Waal's Law of Gases;
PV = nRT
(1atm)(107L) = n(0.08206)(293k)
n = 4.5mol
since air is primarily nitrogen, we can assume that 1 mol = 28g. 4.5x28 = 126g. That's 0.28 lbs for one atmosphere of pressure.
Now, 1atm is 14.7psi.If your tire pressure is 35psi, that's 2.38 atmospheres. Or 1.38 more. 1.38 atmospheres is just 0.28x1.38 = 0.39
So, if you deflate them fully, it will weigh 0.39 pounds less.
11-08-2007, 03:57 PM
#38
I don't remember them asking that kinda question on the mcat. Maybe things have changed since I took it in '03. This's more AP physics or chem. Definitely not anything beyond what a high schooler should be expected to know.
However, in the current state of public education, maybe a high schooler won't be able to figure this out.
Jae
However, in the current state of public education, maybe a high schooler won't be able to figure this out.
Jae
