Sceptimatic,

Buoyancy works in accordance with g (acceleration) and a solid body in liquid, basic formula:

Your 'Denspressure" has all of these except acceleration, what is your explanation for acceleration that's independent of mediums in which masses are displaced?

If you disagree with this, please provide a formula that describes it under your Denspressure idea.

So lets run through the predictions of Denspressure:

1. "Gravity" is just the push of the overlaying atmoplane above us, so air pressure gives acceleration. Prediction: Acceleration is proportional to air pressure on Earth, less pressure implies less push and so less force (basically acceleration).

2. The stack of air pressure is directional, and so objects fall opposite to the stacking direction of the air pressure gradient. This can only imply that pressure presses against objects meeting it. Prediction: Any falling object or airplane at high altitudes will meet the higher pressures on the ground, which will press up against it while the top overlaying pressure presses in the opposite direction. This means that the same force we observe that keeps us to ground at 1G is applied up to a descending object, which means acceleration will decrease and at some point where the stack above will be equal to below and so the object will levitate in equilibrium.

If this is not the case with Denspressure and something independent of the air pressure is pushing down for acceleration, then explain what that is.

The first one is debunked by the fact that acceleration is still consistent in a partial vacuum, as sokarul put out:

The second one is false since you'd shift weight significantly at higher altitudes, especially in a plane flight.

Barometer Pressure:

p = 101325 (1 - 2.25577 10-5 h)5.25588

Where h is the altitude above sea level, p is air pressure, and the '101325' is the Standard Temperature and Pressure and NTP, which will work for the basic purposes here.*

p = 101325 (1 - 2.25577 10-5 0)5.25588

p=101 kPa

At Sea level, the air pressure is 101 kPa.

p = 101325 (1 - 2.25577 10-5 [30,000])5.25588

p= 30.1 kPa

At 30,000 feet, it's 30.1 kPa

An air pressure drop from 101 to 30.1 kPa with passenger planes, the weight of them would drop significantly, flying much easier relative to the thrust they got.

So, why aren't planes dropping at least 50% of their weight at high altitudes or do they?

*http://www.engineeringtoolbox.com/stp-standard-ntp-normal-air-d_772.html