Lets say you inflate your tire to 32 PSI. You climb on. Even with your added weight, the pressure in the tire is still 32 PSI.
HOW CAN THAT BE???
Where does your weight go? Aren’t you (through gravity) ‘pushing down’ on that tire? Shouldn’t the pressure in that tire immediately jump up to, say, 190 pounds?
The same is true in a car. Three thousand pounds of car is being held up by thirty pounds of air!
It’s like, MAGIC!
I’m so confused…
The measurement is the volume of air in the tire. You aren’t changing that by sitting on it.
…was this a serious question?
O.o
o.O
come on dude…psi is volume, not pressure
you pushing down on the tire just changes the shape of it…it doesn’t add or take away air…the pressure inside the tire will still be 32 psi…
Was this a real question?
If a tree falls in the woods and no one is there to here it, does it still make a sound?
I believe your weight adds pressure, however, it adds pressure to the entire wheel, not each square inch. the PSI is Pounds per Square Inch, so if 150lbs is added to the tire, you need to divide that by the number of square inches in the tire.
Maybe this is why I study business and law instead of natural sciences, but I am confused.
Sitting on a uni, adds pressure to the tire right??? Just like how the added weight of the air in the atmosphere adds pressure as we get lower in the atmosphere??
If that wasn’t the case, how is it that a tire pops eventually if you put too much weight on it?
Maybe I’m just not getting this. I always thought it added pressure… but perhaps PSI isn’t measuring that kind of pressure?
32psi+[150/(i)(i)] > 32psi
no… wait… that inequality is completely flawed.
And I’m an economics major… oye.
Given a constant temperature, the pressure of the air in the tyre will only change if the volume changes.
As you put additional weight on the tyre, the tyre deforms, and inevitably the volume changes - but only very slightly. The shape of the tyre is only changing at the bottom, where the tyre is in contact with the road.
Say that the wheel is 26 inches diameter, then it has a circumference of 26x Pi inches = 81 inches approximately. The deformed part of the tyre is probably 3 or 4 inches of that at most, and maybe only an inch or two. Assume 4 inches.
4/81 = around 5%. That means that only around 5% of the tyre is deformed.
That part of the tyre that is deformed does not lose all of it’s volume. It might only lose, say 1/4 of its volume due to the change in shape.
So the total loss of volume is (on these very rough figures) 1/4 of 5% which is around 1.25%. That means that there will be a very slight change of pressure - something like 1.25%. On a pressure of 32 psi, that’s barely detectable.
So, how does it support your weight? because the pressure is acting equally on every square inch of the inside of the tube/tyre.
Making some more very rough assumptions:
The circumference of the tyre is around 81 inches.
The circumference of the tube at any given point is probably 4 inches.
Therefore you have a total of around 4 x 81 square inches.
4 x 81 = 324 square inches of inner tube, each resisting 32 psi of pressure.
324 x 32 is 10,368 = a lot of pounds. Similar principle to a hovercraft - a large amount of low pressure = a lot of lift. However, if the weight isn’t distributed evenly (put all the payload in one corner of the hovercraft; focus all the weight of the unicyclist on one small part of the tyre, where it touches the road) and it may “bottom out”.
Compressing air is fairly easy so when you sit on the uni, rather than it rigidly supporting you, the air moves about and finds a new equilibrium - a slight deformation of the tyre, and a very slight increase in pressure.
air pressure is pounds per square inch (psi)
air pressure = (weight in pounds) / (contact patch in square inches)
increase weight on the tire, and the size of the contact patch will increase in equal proportion so that air pressure remains unchanged (ignoring other factors like tire characteristics)
That is demonstrably untrue.
interesting. Makes sense though now that I think about it.
ha, yeah the equation was a bit of a stretch but it should help simplify the concept for him.
Simple demonstration: I get a different pressure reading if I test the air pressure in my scooter’s back tyre with the scooter up on its centre stand, and then again with it on its side stand.
Simple demonstration 2: Open the valve on your unicycle tube, listen to the hissing, then sit on the uni and see if the hissing gets louder/faster. If the air pressure increases, then the air will come out of the valve faster.
Yes but nobody would here it.
Hmmm. You haven’t really got the hang of this Zen thing, have you?
There is also the supportive strength of the tyre carcass, which is by no means insignifcant in a car. I can quite happily sit all my moderately substantial weight on a car tyre without it collpasing even when it’s not fitted to a rim. Fitting it to a rim adds lateral stiffness and thus increase the load it can carry. Run-flat tyres are designed to take the entire car’s weight in the side walls, and while ye, they are especially thick for this purpose, they are not massively different from standard car tyres.
Because by pushing down on the tire, you change the shape of it. By doing so, the air that is inside moves within the tire, wherever there is still room for it to go. If you push down on all parts of the tire, to such a degree that there is not enough room to withhold the air pressure, then it has no choice but to find a way out (popping the tire).
However, sitting on your uni does not change the air pressure in the tire. Only adding more, taking out, or temp changes will do that.
Sitting on the uni will change the pressure, be it ever so slightly. If the tyre deforms, then the available volume for the air is reduced. (There might be an exception if the tyre is so hard that it doesn’t deform at all, but the initial question postulated only 32 PSI.
When full of air, but not bearing any weight, the inner tube is almost perfectly circular in cross section.
A circle contains the maximum space for a given circumference.
Squash the bottom of the tube into an ovoid, and you have the same amount of rubber, but a slightly reduced volume available.
The best way to see this is to take it to extremes. Put so much weight on the hypothetical uni that the rim is bottoming out. It is then easy to see that the very bottom part of the tyre and tube (around the contact patch) has virtually no volume. All that air has to go somewhere.
The rest of the tube/tyre doesn’t expand to accept that displaced air, therefore the air has to be slightly compressed. The pressure increases by a small amount.
For many years I was a hardcore scuba diver, diving in closed cell neoprene one day, and in an air-filled membrane drysuit the next, using either a horse-collar buoyancy compensator (ABLJ) or later a waistcoat style buoyancy compensator (stab jacket). I used lifting bags from time to time to retrieve stuff from the sea bed. I spent many many hours in lectures about gas laws, and a hell of a lot of my life relying on my understanding of those laws to keep me and other people alive. (Even the boats we used were inflatable.) The tyre example is a very simple application of the very simple and logical gas laws.
Given a constant mass of gas, and a constant temperature, pressure will increase if volume decreases.
Given a circular cross section, you have maximal volume.
Given a flattened cross section, volume decreases.
Therefore pressure increases.
In the example given - the unicycle tyre - the effect is small, but it is there.