Homemade Tire Rack
12-17-2010, 11:31 AM
#1
.
Thread Starter
Homemade Tire Rack
I just got myself a set of winter tires. However, I have no room to put my stock tires. So I decided to build myself a tire rack in my garage. It would be similar to this, but more supported:
But my question is can I deflate the tires to reduce weight, or deflating the tires doesn't make it lighter?
But my question is can I deflate the tires to reduce weight, or deflating the tires doesn't make it lighter?
12-17-2010, 11:37 AM
#2
Back From The dead
so, you're asking if taking air out of the tires will make them lighter?
12-17-2010, 11:44 AM
#4
Race Director
Considering buoyancy effect, looks like less than 1 oz of air per tire....
http://www.madsci.org/posts/archives...5256.Ch.r.html
http://www.madsci.org/posts/archives...5256.Ch.r.html
Last edited by nfnsquared; 12-17-2010 at 11:49 AM.
12-17-2010, 12:06 PM
#7
Back From The dead
if you're thinking that 32 lbs of air in each tire means 32 lbs of air?
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12-17-2010, 12:20 PM
#8
.
Thread Starter
12-17-2010, 12:52 PM
#11
Back From The dead
12-17-2010, 01:26 PM
#12
Three Wheelin'
I built a rack somewhat like that... you might consider running a cable to the roof from the end with the pole to help hold it up... I did and its crazy how tight it is now and empty its loose!
12-17-2010, 02:23 PM
#14
(pressure) / Pi (3.14)
It works out to a little over 10 pounds per tire. So if the rack you're building can support an additional 40 lbs (50 just to be on the safe side) you'll be fine.
Unless you inflated them during the warm weather using summer air....... If so, there's another element that needs to be applied to the calculation. (winter air is okay as it's density doesn't change much once the tire is removed from the car due to less humidity)
Last edited by Shalooby; 12-17-2010 at 02:30 PM.
12-17-2010, 03:37 PM
#16
Race Director
It's not a one-to-one relationship like that because tires are round. As such you have to apply the formula for calculating circles, radius, diameters, etc....
(pressure) / Pi (3.14)
It works out to a little over 10 pounds per tire. So if the rack you're building can support an additional 40 lbs (50 just to be on the safe side) you'll be fine.
Unless you inflated them during the warm weather using summer air, that is. If so, there's another element that needs to be applied to the calculation. (winter air is okay as it's density doesn't change much once the tire is removed from the car)
(pressure) / Pi (3.14)
It works out to a little over 10 pounds per tire. So if the rack you're building can support an additional 40 lbs (50 just to be on the safe side) you'll be fine.
Unless you inflated them during the warm weather using summer air, that is. If so, there's another element that needs to be applied to the calculation. (winter air is okay as it's density doesn't change much once the tire is removed from the car)
Here we go (someone can check my math on this....)
So it's tire volume x 2.18 x density of air = weight of air in each tire.
A tire is basically (but not exactly) a torus. Volume of a torus=
2*pi^2*R*r^2
where r is 1/4*(treadwidth + sidewall height) (yeah, I know, but since a tire isn't a true torus and doesn't have a constant radius or diameter, I choose an average measurement for r) and R = wheel radius + r.
So for OEM TL tire size and 17" rim, we get:
r=1/4*(9" + 2.1") = 2.775"
R=2.775" + 8.5" = 11.275"
Back to our formula:
2*[(3.14)^2]*11.275" *[(2.775")^2] = 1714 cubic inches
1714 x 2.18 compression factor = 3736 cubic inches = 61 liters of air in our tire.
61 liters x 1.29 gram/liter (density of air) = 79 grams
79 grams/454 * 16 = 2.8 oz of air per tire.
But factor in buoyancy, and the effective weight is even less than that.
Bottom line, air weight is negligible....
Last edited by nfnsquared; 12-17-2010 at 03:44 PM.
12-17-2010, 03:41 PM
#17
2011 ZCP E92 M3!!
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I just got myself a set of winter tires. However, I have no room to put my stock tires. So I decided to build myself a tire rack in my garage. It would be similar to this, but more supported:
But my question is can I deflate the tires to reduce weight, or deflating the tires doesn't make it lighter?
But my question is can I deflate the tires to reduce weight, or deflating the tires doesn't make it lighter?
12-17-2010, 05:29 PM
#20
.
Thread Starter
I will be making my own, and very similar to the one on the picture. I found the pic on the web when researching if anyone has attempt to do this. I will post my tire rack once its finished, probably tomorrow
12-17-2010, 06:16 PM
#21
Negative. Read the link I gave in post #4 or read on. Basically, standard sea level air pressure is 14.7 PSI (1 atmosphere). So 32 psi is 2.18 atmoshperes (or the volume of air compressed inside the tire is 2.18 x the inside volume of the tire.
Here we go (someone can check my math on this....)
So it's tire volume x 2.18 x density of air = weight of air in each tire.
A tire is basically (but not exactly) a torus. Volume of a torus=
2*pi^2*R*r^2
where r is 1/4*(treadwidth + sidewall height) (yeah, I know, but since a tire isn't a true torus and doesn't have a constant radius or diameter, I choose an average measurement for r) and R = wheel radius + r.
So for OEM TL tire size and 17" rim, we get:
r=1/4*(9" + 2.1") = 2.775"
R=2.775" + 8.5" = 11.275"
Back to our formula:
2*[(3.14)^2]*11.275" *[(2.775")^2] = 1714 cubic inches
1714 x 2.18 compression factor = 3736 cubic inches = 61 liters of air in our tire.
61 liters x 1.29 gram/liter (density of air) = 79 grams
79 grams/454 * 16 = 2.8 oz of air per tire.
But factor in buoyancy, and the effective weight is even less than that.
Bottom line, air weight is negligible....
Here we go (someone can check my math on this....)
So it's tire volume x 2.18 x density of air = weight of air in each tire.
A tire is basically (but not exactly) a torus. Volume of a torus=
2*pi^2*R*r^2
where r is 1/4*(treadwidth + sidewall height) (yeah, I know, but since a tire isn't a true torus and doesn't have a constant radius or diameter, I choose an average measurement for r) and R = wheel radius + r.
So for OEM TL tire size and 17" rim, we get:
r=1/4*(9" + 2.1") = 2.775"
R=2.775" + 8.5" = 11.275"
Back to our formula:
2*[(3.14)^2]*11.275" *[(2.775")^2] = 1714 cubic inches
1714 x 2.18 compression factor = 3736 cubic inches = 61 liters of air in our tire.
61 liters x 1.29 gram/liter (density of air) = 79 grams
79 grams/454 * 16 = 2.8 oz of air per tire.
But factor in buoyancy, and the effective weight is even less than that.
Bottom line, air weight is negligible....
12-17-2010, 07:48 PM
#22
personally to the OP, i would run the recommended cable to the ceiling, so you can leave the tires inflated, and try and prevent any type of possible flat spotting that they could potentially develop over the season
Last edited by friesm2000; 12-17-2010 at 07:52 PM.
12-17-2010, 07:51 PM
#23
http://www.tirerack.com/accessories/detail.jsp?ID=164
Last edited by csmeance; 12-20-2010 at 08:49 PM.
12-17-2010, 10:30 PM
#24
Suzuka Master
all you need to do is cut a couple pieces of plywood in a triangular shape that aligns with the outside of your wood framing and screw it on 6 - 8 inches on center and it will be bulletproof. lag to wall studs
12-18-2010, 01:35 AM
#26
2011 ZCP E92 M3!!
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like this..., but out of wood instead though...
http://www.tirerack.com/accessories/detail.jsp?ID=164
http://www.tirerack.com/accessories/detail.jsp?ID=164
Last edited by csmeance; 12-20-2010 at 08:50 PM.
12-20-2010, 04:20 PM
#27
Instructor
I thought about buying that but my cousin came over for a couple hours with his tools and we knocked this out.
The tires are not resting on the wall, there is another 2x4 close to the wall that forms the cradle.
It was maybe 20 bucks in parts? (wood, screws, cable, anchors)
The tires are not resting on the wall, there is another 2x4 close to the wall that forms the cradle.
It was maybe 20 bucks in parts? (wood, screws, cable, anchors)
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