Ok, so I did some math given his numbers.
I will give numbers, but have neglected the margins for error deliberately.
The volume of the device is
[math]
V = 8.69 \cdot 10^{-3} m^3
[/math]
Buoyancy of an object is calculated as the weight of the gas it displaces. However, as the object fills with gas that is less dense, it still has a weight of its own.
Gas becomes less dense as it heats up and the same is true for air. I'm using the ideal gas law to describe air.
[math]
p \cdot V = \frac{m}{M} \cdot R \cdot T
[/math]
The quantities are
[math]p[/math] - air pressure
[math]V[/math] - chamber volume
[math]m[/math] - mass of air
[math]M[/math] - molar mass of air
[math]R[/math] - gas constant
[math]T[/math] - air temperature
I assumed an air pressure of [math]p = 10^5 Pa[/math] and took gas constant [math]R = 8314 \frac{J}{kmol}[/math]. Molar mas is [math]M = 29 \frac{kg}{kmol}[/math].
Rearranging the equation to find air density [math]\rho[/math]
[math]
\rho = \frac{p \cdot M}{R \cdot T} = \frac{K}{T}
[/math]
with [math]K = \frac{p \cdot M}{R} = 348.8 \frac{kg}{m^3}[/math].
Thus the change of force felt by the lever caused by buoyancy is given by
[math]
\Delta F = V \cdot g \cdot (\rho(T_1) - \rho(T_2))
[/math]
I made the first estimate using ambient air temperature of [math]T_1 = 20°C = 293 K[/math] and in-chamber temperature given in measurement 1 as [math]T_2 = 60°C = 333 K[/math]. The temperature [math]T_2[/math] is actually the temperature of magnetron, but it'll do as a fair estimate.
Under that assumption, the change of force felt by the lever is [math]\Delta F = 0.0122 N[/math].
The scale, however, feels a force about [math]2.89[/math] times greater, so [math]\Delta F' = 0.0352 N[/math].
The scale should then show a mass change of [math]\Delta m = 3.59 g[/math]. The scale shows a mass of change of about [math]1.2 g[/math] once it settles. So obviously the temperature inside the chamber doesn't quite reach [math]60°C[/math] or at least not in all places.
I don't know how he measured the temperature and what effect the microwaves might have on his measuring device.
How high is the temperature?
The power of the device is given as [math]P = 800 W[/math], but it takes more than [math]5[/math] seconds for the scale to go from [math]0[/math] to the initial jolt of [math]1.2[/math] and a total of [math]10[/math] seconds before the device settles.
We can find how by how much the temperature of air would change in that time.
The input of energy is
[math]
Q = P \cdot t
[/math]
with
[math]Q[/math] - heat
[math]t[/math] - time it takes to settle
The change of temperature is given as
[math]
Q = m \cdot C \cdot \Delta T
[/math]
with
[math]C[/math] - heat capacity of air
[math]\Delta T[/math] - change of temperature
The heat capacity of air isn't trivially easy to get. For dry air it's
[math]C = 1 \frac{kJ}{kg \cdot K}[/math]
However, for 50% humid air, it increase because the water vapor in the air needs to be heated as well. The fraction of mass of water in the air is small and the water is vapor, not liquid, so its heat capacity is lower. So overall, the heat capacity doubles.
[math]C = 2 \frac{kJ}{kg \cdot K}[/math]
The change of temperature is then
[math]
\Delta T = \frac{P \cdot t}{V \cdot \rho(T_1) \cdot C} = 387 K
[/math]
This would imply that the temperature of air inside can change VERY quickly and that most of the power doesn't go towards heating the air, but heating the metal.
So what about the shifting of the null point?
The mass change up is about [math]\Delta m = -1.2 g[/math] and down is [math]\Delta m = 0.25 g[/math]. If the buoyancy and the other unaccounted for force are the same in both cases, buoyancy accounts for [math]0.27 g[/math] and the other force about [math]0.72 g[/math].
Both number are given for the scale and are [math]2.89[/math] times smaller.
However, given how much energy is pumped into the machine and how much of a role air plays, I still think air has more contributions.
I've thought of one more contribution that can lift the machine and offset the null point:
If the metal heats up, it transfers heat to the surrounding air. Air on the top plate can rise, but air on the bottom plate can get trapped and pushes the machine up.
If the bottom plate is larger, more air will be trapped below it and the upward force will be larger.
In the other case, with the bottom plate smaller, less force will be contributed up.
However, the net effect in test No. 4 was still down. That means that whatever force is unaccounted for is still greater than buoyancy and lift.
I'm still skeptical, though. None of the explanations I've seen would even come close to the number we're seeing here. I think it's a measurement error from something that's unaccounted for.
EDIT:
Could you see any force on test 1 and 2, with the pendulum? Yes and no. If we assume mass of the device of about 1 kg and a length of the wire of 1 m, the movement would be in the 0.1 to 1 mm range. The device might begin to swing, but nothing conclusive...
