implosions

[JimS]

What no one has considered is the weight of the wheel assembly, the tire and rim combined.

We are assuming the unit would continually sink, but there is a possibility that at some point the air in the tire and density of the water achieve some sort of weird equilibrium and it no longer sinks.

I realize that this is outside the spirit of the question as we want to know if the tire will crush.

I think that could be answered by looking at ships that have airtight compartments. All ships have them to prevent or slow sinking. These compartments are designed to work at relatively shallow depths. What happens to them at extreme depths? I dont know the real answer, but I assume they crush.

Its kinda how depth charges work. They dont have to strike the hull. They just have to momentarily exceed the hulls pressure limit to begin the implosion.

My question is why in submarine movies do the occupants of the sub always look up at the ceiling of the sub? Do you look at your ceiling when you hear thunder? I dont look at the floor during an earthquake. Im certain real submariners dont look at the ceiling.

 

[john in la]

Now that is something to think about I did not consider.
On a tubeless tire where there is no cap on the valve stem the valve core now becomes the weakest link.
The water pressure would just push down on the valve core allowing water into the tire equalizing the pressure on the inside and the outside long before the bead would pop off the rim or the pressure pushed the entire valve stem into the tire.

With a cap in place I think it would be a toss up to the valve stem being pushed out or the bead coming off the rim.

 

[john in la]

Thunder and an earthquake are loud noises.
But if a tree branch was rubbing on the house I think you would look toward the faint sound as you try to hear it.
Like having a small air leak of compressed air.
You turn toward the sound trying to pick up the direction it is coming from because you can hear better looking toward the sound.
Much like a animal will pivot its ears toward a faint sound.

 

[redforlife]

So, your saying tube won't pop basically because of no ballooning effect. I'm following you.

But what if there was a ballooning effect somewhat reversed. When the valve stem is pushed into its hole, the tube is actually beginning to collapse in on itself. I'm not sure how things would go after that. If a run over by steam roller affect begins to happen, the rising air pressure would constantly be restricted to a smaller space, causing ballonage to that one area. As a result, the tube would blow in that one area. The air might not escape the main tire at that point (as in can't leave the valve stem hole because its above that), but nothing more would really happen because at that point nothing is really between the air and the water.

I guess it'd come down to weather the steam roller effect happens. I myself am unsure what would happen there. I could see the beginning phase of it happening. Not sure if it would continue until point of tube blowing. The rim holds the tube in a somewhat fixed position. Not sure if that is enough of a fixed position to allow completion of the steam roller effect. Would make a difference in that regard if the tire was horizontal or vertical. Being vertical would allow the tube to completely collapse in on itself at the bottom, beginning the chain affect. Then the question would be, what the collapsing would do once it reached the top of tire. Whether it would stop collapsing the tube in on itself, or if it would continue collapsing the tube against the air bubble inside. Saying it like that, almost makes me believe the collapsing would stop at some point, but I don't know that for sure.

 

[DoubleO7]

I am thinking the water psi would be from all directions at the same time.
Even when it is only a thin film between tire and tube. At some point the weakest spot would rupture and then psi would instantly equalize.

If the stem valve was gone the tube would collapse as if you hooked up a vacuum pump to the valve stem.

 

[redforlife]

I presume the OP (even though doesn't specify) means the wheel is takin down further than it would sink itself, atleast to the depth of the Titanic.

He does ask at what depth things would happen. To answer this, the wheel would have to be monitored on the way down. Or brought back up for inspection after being lowered to various depths. A couple hundred feet probably doesn't make much difference on implosions at the great depth the Titanic is at.

 

[redforlife]

A vacuum pump (if hooked up to the valve stem), gives the air a place to go. It's actually pulled out by vacuum. In this scenerio, the air remains inside. The question is, if it's reduced to a small enough space to allow ballonage. The more I think about this, the more I'm leaning to the air is reduced to such a small place, that there wouldn't be any ballonage. Just air compressing. As you say, like a vacuum pump hooked to the valve stem, but not completely. Leaving an area that is not fully collapsed, and contains the reduced air space. By pushing on the flexible walls of the tube, it would be pushing on the external hydraulic pressure. Much like air inside a hydraulic cylinder getting smashed into a smaller bubble.

 

[john in la]

Running over a tube with a steam roller will pop the tube.
The pressure from the roller will be lets say 6000 lbs (only as a good example for the water depth) and the pressure from the ground will be the same 6000 lbs.
But the sides will have no pressure (space between the roller and the ground).
So the tube will pop as it stretches out to the sides.
That is not true under water.

First if we send a metal cylinder down in the water (A sub)
The air pressure in the sub will be 14 psi same as we feel here on the surface.
As the sub goes down the cylinder stays the same size and the pressure inside stays the same while the water pressure outside increases the deeper we go.
Once we get deep enough the structure can no longer hold the pressure from outside the sub will imploded until a crack forms and water rushes in or the air pressure inside becomes the same as the outside water pressure.

But if we send a tire tube down with 25 psi of air in it.
The tube is not rigid like a metal sub.
The tube is flexible.
So as we go down the water pressure outside the tube will be pushing on every sq inch of the tube.
The tube will get smaller in size as we go down from the high water pressure pushing on the 25 psi air on the inside.
As the water pressure increases the air pressure inside the tube will increase to the same pressure.
Eventually we will hit bottom and the water pressure outside will be 6000 psi and the air pressure inside will be 6000 lbs with the shrunk tube just being a layer of rubber between the air and water.

Now lets take this same tube and rather than diving deep lets go up in the sky.
As we get higher and higher the air pressure on the outside of the tube will decrease.
This will cause the pressure inside to tube to expand.
Once the tube gets so big it can not expand any more the tube will rip.

So it is not the pressure in or out of the tube that allows the air to leak out.
It is the stretching of the tube beyond its limits that will pop it.
Stretching the tube is not a factor under the water.

 

[john in la]

With the water pressure on the outside being the same as the air pressure on the inside the valve core is not going anywhere.
And since the pressure is the same on the inside and outside the tube is not going to rupture.
The only way a tube can rupture is if you stretch it past its physical capacity.

To implode or rupture you first need a rigid outer shell.
A tire tube or balloon is not rigid so while it may shrink in size because compressed air takes up less space it is not going to rupture or implode.

 

[Texasmark]

Well we are talking about pounds per square inch of surface area. The surface area of a Shrader valve stem is quite small and if using a metal stem cap it negates the problem even more.... course the dimensional difference is quite large. Good question.

 

[GeneMO]

Pressure, volume, and temperature all play into it. Boyles law. At the surface of the water we are under one atmosphere of pressure, 14.7 PSI. in water, that pressure increases 14.7 psi for every 33 feet, (or 34 feet, I forget which is for salt water vs. fresh water) feet. So you dont have to go too deep to collapse that tire.

Gene

 

[Dave G9N]

(quoted from post at 08:20:07 07/01/23) just come up with a crazy cynario , if you somehow pulled your truck tire under water with 55 lbs of air in it, would the pressure push it
off the rim. at the titanic depth they say the pressure is 6000 psi . under about two and a half miles of water. and if it would at what
depth would this happen. i do not know how to even start figuring this out.

f you dunk it quickly, the bead should start to pop off around 120 feet.

I can go into agonizing detail, but in a nutshell, as you have stated, 6000 psi at 158,400 inches deep is 26 inches per psi. Not exactly correct, 2 feet per pound is close enough here.

As for damaging the tire, think about the load 55 psi puts on the tire. In the hoop direction, a 40" diameter tire sees 2200 pounds per inch of tread width. The sidewall might have an area of 600 to 800 square inches. 55x600=33,000. Any questions as to why split rims are dangerous?