Then at sea-level the density of air is 1.225 kg/m^{3}, but at 10,000 ft is only 0.4135 kg/m^{3}.

I have been at that altitude and untold others live much higher than that,

*but at the top of **Tioga Pass, CA*, I most certainly did not weigh only 34% of my normal weight.

In other words, the whole idea that air pressure causes weigh is total bunkum!

Scales also suffer the same change. Think about it.

Sorry for butting in, just saw this bit and started thinking about it. I imagine Rabinoz's immediate response is going to be that when you're at those altititudes you can't jump 34% higher, and with that I'd just like to double check something. I remember a discussion we had a while ago about equal and opposite forces in vacuum, so for the mountain pass question would it be right to say that someone who tries to jump wouldn't be able to jump as high as one might think in your model precisely because there's less downwards force, and thus less to brace against the ground with?

So while in theory one could jump 34% higher, doing so would require being able to jump with the same amount of force that you can jump with at sea level, which wouldn't be possible. While you might be able to jump higher at greater altitudes, the ease won't increase nearly as quickly as it seems at first look.

Just checking in to see if I understand your model, it's fun to figure out.

If anyone does understand it, please explain how to *calculate the weight* from the effect of "Atmospheric pressure upon dense mass".

Well it'd be based on some specific constant of denpressure, an acceleration based on how much you're making air molecules stop their expansion (ultimately that's a velocity you're decelerating to zero, hence acceleration). Let's call the constant d.

Then you know that the denser an object is, the more force denpressure exerts. And on the flipside, the less dense an object is the more air can fit inside it so the less air overall is displaced. Let's call the density p.

And equally the larger an object is, the more air is displaced, so the more force denpressure exerts. Let's call the volume v.

So, increase either p or v and you increase denpressure, and they're the main factors, along with the constant d that depends on air itself.

Logically, the force due to denpressure, F, may be found by

F=dpv

Where d is some constant to be calculated. I think d=9.8m/s/s is a good value.

Sound good?