Don't want to forget that thrust has been measured in the same axial direction relative to the thruster whether the thruster was pointed up, down, left or right. This speaks strongly against outside thermal drafts.

It does not speak to thermal expansion of course.

My theory on thermally generated thrust claims the cone section transmits heat energy to the surrounding air by conduction. The two ends do not have exposed Copper so the heat flow from each end would be much less. FR4 (high density fiberglass as used in PCBs) is a better insulator than Copper. The fact that March saw < 1 degree change in temperature would be expected because of the good heat transfer from the Copper cone section to the surrounding air. There is no mention of the surrounding air temperature so I am assuming this < 1 degree change refers to just the Copper section of the device.

...

You must have meant to write "transmits heat energy to the surrounding air by

~~conduction~~ convection" because heat transfer by convection in fluids like air occurs much faster than by conduction. Air has very low thermal conductivity and very low thermal diffusivity, so heat does not get transferred in air by conduction, but by convection.

Heat transfer by convection:

q/A = h dT

q/A = heat transfer per unit time per unit area

h = heat transfer coefficient

dT = temperature difference between the surface and the bulk fluid = Ts - Tf = 1 deg F = 0.56 deg C

Natural convection heat transfer coefficient = 2 to 20 Watt/((m^2)degC))

q/A = 0.56 deg C * 2 Watt/((m^2)degC)) = 1 Watt per square meter

q/A = 0.56 deg C * 20 Watt/((m^2)degC)) = 10 Watts per square meter

l = Sqrt [h^2 + (R - r)^2] = Sqrt [ 0.332^2 + (0.199 - 0.122)^2] = 0.341

Total surface area of truncated cone = Pi × ( r × (r+l) + R × (R+l)) = Pi × ( 0.122× (0.122+ 0.341 ) + 0.199 × (0.199+ 0.341)) = 0.515 m^2

Input Power was from 2.6 Watts to less than 16.9 Watts, so if 100% of the power would have gone into heat:

2.6/0.515 = 5 Watts per square meter

16.9/0.515 = 33 Watts per square meter

The lateral surface area is Pi×l×(R+r) = Pi×0.341×(0.199+0.122) =0.344 m^2

So that only increases the flux by 0.515/0.344 = 1.50, or 50% even if you consider both ends to be completely thermally insulated and that 100% of the heat escapes through the lateral surfaces (which would not be quite the case). It increases the flux so that it ranges from 7.5 W/m^2 to 49 W/m^2.

So, the delta T given by Paul March (1 deg F =0.56 deg C) makes sense, if a fraction of the input power went into heat that got transferred to the air.

The speed of an air convection current due to only 1 deg F temperature difference is so small, that any air convection naturally occurring within the chamber for other reasons could easily overwhelm it. (The vacuum chamber itself may have had a larger delta T within it for other reasons unrelated to the EM Drive).

Further bad news for explaining the measured forces as due to natural convection circulation is that this can only work with the warm part (the EMDrive) on the bottom of a natural convection circulation system such that the bottom part (the EM Drive) is warm and the top part of the chamber is cool and therefore the convection would be from the warmer EM Drive towards the cooler top of the chamber. (Natural convection taking place because of the extremely small amount difference in temperature: less than 1 deg F, producing extremely small differences in density of the air).

But at NASA Eagleworks the measured forces were in the horizontal direction. Furthermore, the direction of the force was always oriented towards the large diameter base, even when they flipped the EM Drive to point 180 degrees in the opposite direction. Furthermore, in the up and down test performed by Shawyer, the direction of the force should not have flipped (as reported by Shawyer) when Shawyer flipped the test article upside down, as natural convection always works such that the warmer part is on the bottom, and the air circulates from the warmer bottom part to the cooler top part of the chamber.

Further bad news for explaining the NASA Eagleworks response as natural convection from the warmer EM Drive is that the NASA Eagleworks test show a pulse response rapidly rising in 2 seconds which coincides with the inertial response of the inverted torsional pendulum, and is way too short a time compared with the Fourier dimensionless time based on the thermal diffusivity of the materials involved and the characteristic length. So the initial time-response cannot be explained in terms of thermal natural convection. The speed of heat transfer is restricted by the thermal diffusivity of the material.