2 + 2 doesn't = 4? An original proof created by myself.

[CamaroSkooter]

Originally Posted by ajp

Disagree.

Lets say I have a container that holds 2 quarks. I put 1 quark in it.

Its now 1/2 full/empty.

Also - the absence of a unit is equivalent to the presence of a unit, so 2+2 is always 4.

That's kind of a bad analogy.

In reality, every container is always full, regardless of what's in it.

In the typical, "is the glass half full or half empty," argument, the answer is always neither.

A 200mL beaker filled with 100mL of water is full because the other 100mL is made up of air

[ajp]

No, sorry that is too simplistic. Try your beaker analogy in a true vacuum.

Also the 1/2 full argument is more about the container than what is in it.

I am not a physicist, but I do watch "Through the wormhole".

Originally Posted by CamaroSkooter

That's kind of a bad analogy.

In reality, every container is always full, regardless of what's in it.

In the typical, "is the glass half full or half empty," argument, the answer is always neither.

A 200mL beaker filled with 100mL of water is full because the other 100mL is made up of air

[DGthe3]

Originally Posted by ajp

No, sorry that is too simplistic. Try your beaker analogy in a true vacuum.

Also the 1/2 full argument is more about the container than what is in it.

I am not a physicist, but I do watch "Through the wormhole".

Ah, but such a vessel cannot exist. A perfect vacuum would have 0 particles in it. The vessel also contains 100ml of water, in addition to the 100 ml of void space you want. Therefore, it cannot be a perfect vacuum.

[ajp]

Originally Posted by DGthe3

Ah, but such a vessel cannot exist. A perfect vacuum would have 0 particles in it. The vessel also contains 100ml of water, in addition to the 100 ml of void space you want. Therefore, it cannot be a perfect vacuum.

I will use your argument, but add a linear component.

You start with a vessel that is a perfect vacuum. This can be done. You then fill with desired amount of liquid, and only the desired amount of liquid.

Done!

[SSE 4 2SS]

Originally Posted by ajp

I will use your argument, but add a linear component.

You start with a vessel that is a perfect vacuum. This can be done. You then fill with desired amount of liquid, and only the desired amount of liquid.

Done!

In a perfect vacuum, water can't exist in a liquid state... It would boil and turn to steam... Expanding at roughly 1600/1 thus removing the vacuum and overfilling the vessel....

[ajp]

Originally Posted by SSE 4 2SS

In a perfect vacuum, water can't exist in a liquid state... It would boil and turn to steam... Expanding at roughly 1600/1 thus removing the vacuum and overfilling the vessel....

The vessel is not in a vacuum. The container content is empty, i.e. create a true vacuum within the container. Then you fill with the desired amount liquid. Even if it expands, it is still 1/2 full/empty.

[DGthe3]

Originally Posted by ajp

I will use your argument, but add a linear component.

You start with a vessel that is a perfect vacuum. This can be done. You then fill with desired amount of liquid, and only the desired amount of liquid.

Done!

The flaw is that it is not possible to create a perfect vacuum. In a perfect vacuum, there is not even a single stray electron floating around inside. On top of that, there is the fact that the structure of the vessel cannot be emitting any radiation of any sort. The only conditions where that is possible is at absolute 0, which is impossible. So since absolute 0 temperature is impossible, it is also impossible to have absolute 0 pressure.

[SSE 4 2SS]

Originally Posted by ajp

The vessel is not in a vacuum. The vacuum is within the container.

Regardless, the water still cannot stay in a liquid state in a vacuum, therefore the water put into the vessel will boil and expand reducing the vacuum.... until a point of stasis is reached where the pressure reaches a point that it no longer boils.... and you will no longer have a half full container... as was mentioned, you will have a full container, part water and part water vapor... pull more vacuum and you increase the boiling.... until stasis is again reached...and where is the additional vacuum pulled, at or below the water line... it doesn't matter other than in the water level as opposed to the vapor volume....

[ajp]

Originally Posted by DGthe3

The flaw is that it is not possible to create a perfect vacuum. In a perfect vacuum, there is not even a single stray electron floating around inside. On top of that, there is the fact that the structure of the vessel cannot be emitting any radiation of any sort. The only conditions where that is possible is at absolute 0, which is impossible. So since absolute 0 temperature is impossible, it is also impossible to have absolute 0 pressure.

You have failed to provide proof that is impossible, you are merely suggesting in a subtle way, that you dont know how. On that we agree.

However I think I understand the limits you are placing on your own idea process. When you position something as not possible, you limit the possibilities.

[its a syn]

I don't have time for this because now the clock doesn't exist, right?

[CamaroSkooter]

The problem I have with vacuums is, if you could create one, and there are zero particles inside it...

Then what's inside it?

The vacuum itself has a volume, right? So what's filling that volume if not particles of something?

[Mean Green Z28]

shit, had I known this, I would've graduated with my BSME much faster! I'd have answered 90% of the engineering questions like this!

[hrpiii]

Why am I thirsty and feel the urge to vacuum the living room floor?

I hate engineers!

[trademaster]

A number without units is considered a dimensionless quantity and is assumed to be of dimension 1. Essentially, the implied unit is "all" unit measurements as the equation applies to all possible units when the dimensional quantity is constant across the function. Any constant unit can be applied to all of the numbers in the equation and the equation will return true.