PDA

View Full Version : OT: Airless Tires create Math Question



dalesvp
05-02-2013, 07:47 AM
Bridgestone Airless Tires
http://www.gizmag.com/bridgestone-airless-tires/20710/

Which got me to thinking what would be the effects if all existing tires were switched out to these airless tires?
According to <http://wiki.answers.com/Q/How_many_cars_are_in_the_world> "There are over 750 million motor vehicles in the world today." Which amounts to 3,000,000,000 tires on the road plus all the tires in process, storage, etc.
So my nagging question is how much captured air would be released back to the atmosphere if all these tires were replaced with airless tires?
The math question is how many cubic feet of air @ atm is contained in a single 15" automobile tire @ 50 psi? (The 15" @ 50 psi is a questimated average tire.)
I can't seem to wrap my mind around how to do this calculation. Do any of you math whizes have a solution? Thank you!

Black_Moons
05-02-2013, 08:07 AM
Volume of tire (Id consider a cylinder of the tire size to be close enough) * ((50PSI-1ATM) / ATM)
Assuming 15psi (rounded) per ATM, thats 35psi / 15psi or 2.3 extra 'volumes' of air inside a tire at 50PSI.

vpt
05-02-2013, 08:36 AM
I leave my compressor full all of the time.

fjk
05-02-2013, 08:47 AM
Bridgestone Airless Tires
http://www.gizmag.com/bridgestone-airless-tires/20710/

Which got me to thinking what would be the effects if all existing tires were switched out to these airless tires?
According to <http://wiki.answers.com/Q/How_many_cars_are_in_the_world> "There are over 750 million motor vehicles in the world today." Which amounts to 3,000,000,000 tires on the road plus all the tires in process, storage, etc.
So my nagging question is how much captured air would be released back to the atmosphere if all these tires were replaced with airless tires?
The math question is how many cubic feet of air @ atm is contained in a single 15" automobile tire @ 50 psi? (The 15" @ 50 psi is a questimated average tire.)
I can't seem to wrap my mind around how to do this calculation. Do any of you math whizes have a solution? Thank you!

For a rough approximation, think of the tire as a set of concentric cylinders with the axis of rotation parallel to the ground (that is, the axle).

Volume of a cylinder is pi x r^2 x h where pi is 3.1415... r is the radius, and h the height of the cylinder (or width of the tire in this case).

Calculate the volume of the whole tire -- the outer envelope of the thing, if you will -- V1. Calculate the volume of the wheel on which it's mounted, call that V2. The tire itself occupies a volume of V1-V2. You can calculate the volume of the treads and sidewalls in the obvious ways and subtract them out as well (should you be dedicated enough :-).

What's left is the air space in the tire.

This is a rough approximation as all these things are not perfect cylinders --- but it should give a good first approximation.

Then calculate the volume of the atmosphere... It's a set of concentric spheres. The volume is 5/3 x pi x r^3. Again, you calculate the volume of the outer envelope of the atmosphere and subtract out the volume of the earth. For the sake of argument, let's assume the atmosphere is 1mi thick.
The Earth's diameter is 7901 miles. Its radius is 3950.5 mi.
Therefore for our work, the atmosphere+earth is 7903 miles in dia, or 3951mi radius.
Or a volume of 2.583E11 cubic miles.
The volume of the earth (3950.5 mi radius) is 2.582E11 cubic

So the volume of the 1mi thick slice of the atmosphere is 98070547 cubic miles.
Or, in cubic feet, 1.444E19.
(And that's just taking into account the first mile of the atmosphere...)

You've Calculated that there are 3E9 tires in the world.

Those numbers are differ by a factor of about 5E10

To effect the atmosphere by 1% (in some way), you'd need something like 5E8 --- 500,000,000 cubic feet (a cube about 400' on a side) --- of air at 1atm in each and every tire in existence.


So it's safe to guess that if all tires in the world lost all their air the effect on the atmosphere would be unmeasurably small.

ed_h
05-02-2013, 09:41 AM
Approximate the volume inside the tire as a torus. Volume of a torus is cross sectional area x mean circumference.

Evan
05-02-2013, 09:57 AM
The one item missing from the calculation is the amount of gas released from solid compounds into the atmosphere by the manufacture of each tire. I suspect it greatly exceeds the amount of gas trapped inside the tire.

philbur
05-02-2013, 10:56 AM
It's not missing, it's a totally different question requiring a different equation. Stay with the program Evan.

Phil:)


The one item missing from the calculation is the amount of gas released from solid compounds into the atmosphere by the manufacture of each tire. I suspect it greatly exceeds the amount of gas trapped inside the tire.

philbur
05-02-2013, 11:09 AM
It's a volume of air at atmospheric pressure approximately equal to 1 square mile by 165 ft high.

assuming 16" mean diiameter and a 2" by 6" section.

Phil:)

bob_s
05-02-2013, 11:42 AM
Mean thickness of the atmosphere is about 20 miles (air pressure at the top of Everest is about 7psia, about 6miles above sea level, Chuck Yeager flew to over 123000 feet).

Canadian parliament is a good indication that there's more than enough to go around - most of that hot.

Forestgnome
05-02-2013, 11:58 AM
If you're wondering how much it would add to the volume of our atmosphere it would be nothing. The volume of those tires and the volume of the materials to make the new ones all currently take space in our environment. No volume change.

Guido
05-02-2013, 12:09 PM
Ah, as the tire itself wears out, where does that volume of tire material 'disappear to'?? Does it return to atmospheric vapor--- I've never seen it stacking up alongside the freeway?? Only approximately 25% of each wornout tires' original weight is considered as bulk landfill material.
--G

ikdor
05-02-2013, 12:58 PM
I'd be more concerned that the lower efficiency of that tyre will require more fuel to be converted to co2....
Igor

philbur
05-02-2013, 01:59 PM
The internet never ceases to amaze me, I just type"where does tyre wear go":

http://www.straightdope.com/columns/read/2661/when-the-rubber-meets-the-road-where-does-it-go

Phil:)


Ah, as the tire itself wears out, where does that volume of tire material 'disappear to'?? Does it return to atmospheric vapor--- I've never seen it stacking up alongside the freeway?? Only approximately 25% of each wornout tires' original weight is considered as bulk landfill material.
--G

Paul Alciatore
05-02-2013, 04:16 PM
I think you are missing a basic point here. The inflated tire still occupies space in the atmosphere. The air has just moved from outside the deflated tire to inside it after it has been inflated. The air will occupy a smaller space due to the pressure inside the tire. Here is a quick estimate of this effect:

Atmospheric pressure = 14.7 PSI (approximate, of course)
Estimate of average tire pressure = 40 PSI (most auto tires are less, many other types are more)(feel free to disagree)
Since tire pressure gauges read in PSI above the ambient atmosphere pressure, the real tire pressure is 14.7 + 40 = 54.7 PSI.
From Boyle's law, the product of volume and pressure is constant if the temperature is not changed. P1V1 = P2V2. Or, transposing terms, V2 = P1V1/P2
So for any unit of volume, we let V1 = 1 such unit and the equation becomes V2 = P1/P2.
Thus the volume of the compressed air is V2 = 14.7/54.7 = 0.269
Expressed another way, the volume of the compressed air is about 26.9% of the volume before it was compressed. Or about 1/4 the original volume.

For a completely accurate answer you would need to know the exact dimensions of all the inflated tires, INCLUDING the amount of flattening for the ones which are on vehicles. You would also need to know the exact pressure that they are inflated to and probably the exact temperature of each one. Then you could do a calculation for each and every one of them and add them together. It would take a lot of computing power to do this but there are a lot of PCs and other computing devices so it may be possible.

Anything less would be just an estimate and you need to determine the degree of accuracy you will accept. You could probably get a reasonable estimate with some reasonable estimates of the number of tires, the average volume of them, the average pressure, and the average temperature. Beyond that it is going to take a whole lot of work.

john11668
05-02-2013, 04:28 PM
Bear in mind that you blew the tyres up by compressing air taken from the atmosphere.
Let them down and it just goes back where it came from!
Matter is neither created nor destroyed ! Well this is true outside reactors!

philbur
05-02-2013, 06:30 PM
It should be (50+15)/15 or 4.33

Phil:)


Volume of tire (Id consider a cylinder of the tire size to be close enough) * ((50PSI-1ATM) / ATM)
Assuming 15psi (rounded) per ATM, thats 35psi / 15psi or 2.3 extra 'volumes' of air inside a tire at 50PSI.

Bob Fisher
05-02-2013, 06:39 PM
Does it make a difference? The air came from somewhere in the beginning And will return from whence it came. Remember, matter can neither be destroyed or created.Bob.

boslab
05-02-2013, 08:11 PM
Methinks the combined gas law more appropriate, ie Boyle's/Charles

P1.V1. = P2.V2
T1. T2
Does that make sense?
Mark

dalesvp
05-03-2013, 08:30 AM
I was simply curious about this and your many perspectives have answered my curiosity. Compressed air in tires is not that big a difference in the global scheme of things. Thank you all for your insightful conversation and inputs.

Paul Alciatore
05-03-2013, 01:50 PM
I used Boyle's law, which assumes constant temperature, because:

1. The temperature will eventually equalize and we are not talking about any short term effects here

2. It is a lot simpler and does properly account for what is happening.

Boyle's law is simply a special case of the general gas law.

Those who said the air comes from the atmosphere and goes back there are quite correct. The only change while it is in the tires is in the volume it occupies so I used Boyle's law to show what happens there.

"It should be (50+15)/15 or 4.33

Phil"

Both atmospheric and tire pressure varies. Many values can be assumed. I found a typical value for one atmosphere on Wikipedia and had to assume some average tire pressure. It would take a lot of work to find values that are better than either yours or mine.


Methinks the combined gas law more appropriate, ie Boyle's/Charles

P1.V1. = P2.V2
T1. T2
Does that make sense?
Mark