The comments below until this one are a part of a pm exchange.

It’s very difficult to be an effective inventor who is also a crackpot. I like this video on that: https://www.youtube.com/watch?v=11lPhMSulSU

You should go to a physics forum and ask where you are wrong in this quote for example.

My attempt:

You chose a special metal, tungsten for example, that is not melted at 2000 degrees Celcius. Tungsten has an emissivity of 0.46 at 2000 degrees Celcius.

To maintain a 1 m^3 cube of tungsten at that temperature in the vacuum of space, assumed to be at absolute zero, you would use this equation:

P = A \cdot M = A \ \varepsilon\ \sigma T^4

Example calculations, with now also the equation with the temperature of the room taken into account: https://testbook.com/physics/stefan-boltzmann-law

P = eσA (T⁴ – T₀⁴)

Let’s say the cube is in an oven that has been turned off and is now at a temperature of 1950 degrees Celsius, now the energy lost through radiation is:

P = (0.46) (5.67 × 10⁻⁸ W/m²K⁴) (6\ m²) × [(2273.15 K)⁴ – (2223.15 K)⁴] = 355674\ watts

This particular oven is likely very poorly insulated if it can maintain such a temperature difference with such a hot object in it.

The oven will likely be insulated. Let’s assume its outside temperature is at 300 degrees Celsius, it has an emissivity of 0.9, and a surface area of 12 m^2, in a room at 50 degrees Celsius.

P = (0.9) (5.67 × 10⁻⁸ W/m²K⁴) (12\ m²) × [(573.15 K)⁴ – (323.15 K)⁴] = 59404\ watts

You mention tungsten. Tungsten has an emissivity of 0.74 in its molten state. For a cubic meter block, it would radiate 46,000 kW, or 46,000,000 J/s at 3,400 C. The heat capacity of tungsten is 134 J/kg K, since we are radiating 46,000,000 joules per second, in the first 10 seconds the metal would drop in temperature by 240 K, such rapid cooling is impossible.

Why are you only referencing Gary Novak. Gary Novak is a complete and utter crackpot. Read what he writes. He thinks all of modern physics is wrong: About Gary Novak, Independent Scientist.

I did the math on one physics question you had. The physics question you have now you could ask on a physics forum. Here are search results: https://www.csc.com.tw/csc/ts/ena/pdf/no35/pages/2-Analysis%20of%20Temperature%20Dropping%20of%20Molten%20Steel%20in%20Ladle%20for%20Steelmaking.pdf

The exposed surface of the molten metal on top of the crucible would instantly solidify if the S-B law were the 4th power as claimed. Imagine a 50mm thick layer over a 1 square meter surface of molten metal, radiating 7600 kW/m2 or 7,600,000 J/s, the first 10 millimeters would cool at a rate of 405 K/s. The lid on the crucible might be magnesium oxide as a refractory metal, if radiated with 7.6 megawatts per square meter with an emissivity of 0.5 would instantly vaporize. In 60 seconds, a 50-millimeter thick magnesium oxide lid placed on top of the molten tungsten bath would reach 2000 C if its starting temperature was 20 C. If this refractory lid were already at 1000 C it would reach 4800 C in 120 seconds which is above its boiling point! This can easily be shown with the calculator below. Such heating is evidently not observed in real life suggesting the 4th power relationship between radiation intensity and temperature cannot hold. Science is an evolutionary process in which discoveries are made by observing discrepancies and paradoxes.

Using this radiative cooling calculator, the molten tungsten should cool down to 1000 C in only 140 seconds for the first 20 millimeters of thickness. Clearly, this would make melting it impossible unless heated at the rate of cooling which would require nearly 8 megawatts of thermal power for only 1 square meter of molten metal surface.

Do you mind me making this a public topic?

No not at all, as long as the debate is kept to the facts and does not become personal or antagonistic.

People are people are people, so I can’t guarantee that. You can expect the usual antagonism. If you think a comment crosses a line, you can flag it and I will remove it.

It doesn’t bother me. I think it would be of great public value to discuss, what I believe and others do as well, the probable deficiencies of the S-B law.

My physics isn’t good enough to be able to model all these different questions. Let’s just stick with tungsten at 2000 degree Celsius. Your source here is another crackpot, though not as atrociously delusional at the other one, so I can’t trust the model. He himself also says it is simplified.

https://www.researchgate.net/publication/267331267_Tungsten

Taking a shortcut, from the graph we see a thermal conductivity of about 100\ W m^{-1} K^{-1} at 2000 degrees Celsius. I didn’t find a source for its thermal conductivity at very low temperatures, so I am not going to try to calculate the heat loss to space, while in space, at 3 degrees Kelvin.

Let’s take a tungsten cube, and expose one side to either the vacuum of space or to some other environment, one side to a tungsten plate at a constant 2000 degree Celsius, and have the other 4 sides be perfectly thermally insulated.

At room temperature, I’ll say at 25 degrees Celsius, it has a thermal conductivity of about 175\ W m^{-1} K^{-1}, at 2000 degrees Celsius about 100\ W m^{-1} K^{-1}. It’s wrong to do so, but I’ll use the average of about 137.5\ W m^{-1} K^{-1}.

Thermal Conductivity, Stefan Boltzmann Law, Heat Transfer, Conduction, Convecton, Radiation, Physics

P = \dfrac{K \cdot A \cdot\triangle T}{l}

\dfrac{137.5 \cdot 1 \cdot 1975}{1}= 271562.5 \ watts

This is very much simplified, so let’s say this is in a vacuum at room temperature, and the room is infinitely large.

271562.5 watts needs to go into the plate to keep it at 2000 degrees Celsius, and the same 271562.5 watts is lost to the room.

I’m not sure I understand the result, or if it is right. Let’s try to see how hot the outside face of the cube needs to be to radiate at 271562.5 watts .

P = eσA (T⁴ – T₀⁴) => \dfrac{P}{eσA}= T⁴ – T₀⁴ => T = \sqrt[4]{T₀⁴ + \dfrac{P}{eσA}}

\sqrt[4]{298⁴ + \dfrac{271562.5}{0.46 \cdot 5.67×10⁻⁸ \cdot 6}}= 1149\ Kelvin \ / \ 876\ Celsius

867 degrees Celsius is not the same as 25 degrees Celsius, so maybe this equation only holds if the face I now exposed to a vacuum, I instead attach to a constant 25 degrees Celsius heat sink?

Also please show your work if you do calculations, so it can be checked. You’re probably not including the temperature of the environment in your calculations.

The radiative cooling and heating calculator provided seems perfectly accurate, it simply integrates the heat loss as a function of time.

P = ε σ A (T4 - Tambient4)

dT/dt = - (P / m c) = - (ε σ A (T4 - Tambient4)) / (m c)

dT/dt = - (ε σ A (T4 - Tambient4)) / (ρ A d c)

t = ∫ TinitialTfinal - (ρ A d c) / (ε σ A (T4 - Tambient4)) dT

Symbol Meaning
ε Emissivity of the tungsten slab (dimensionless)
σ Stefan-Boltzmann constant (5.67 × 10-8 W/m2K4)
A Surface area of the slab (m2)
T Temperature of the tungsten slab (K)
Tambient Ambient temperature (K)
m Mass of the tungsten slab (kg)
ρ Density of tungsten (kg/m3)
V Volume of the tungsten slab (m3)
d Thickness of the slab (m)
c Specific heat capacity of tungsten (J/kg·K)
dT/dt Rate of temperature change with respect to time (K/s)
t Time (s)

A few additional pieces of evidence that suggest the S-B law is greatly overstating the radiative flux produced by objects.
Gary Novak also points out that ice should radiate 300 W/m2, implying that an ice skating rink should heat the air above it or a table full of ice cubes should heat a room.
MIRVs should also be cooling down faster than they heat up since they reach around 2800 C as they enter the atmosphere. The MIRV is covered with an ablative surface made of phenolic resin. If the surface temperature of the MIRV reaches 2800 C, it should radiate around 4,800 kW, causing the material to rapidly cool beyond what the atmosphere can provide in heat.
The 4th power S-B also predicts unrealistically high radiation from extremely hot plasmas such as plasma torches. A plasma torch is thought to reach as hot as 26000 C, which would radiate 2,700 kW from a 1 cm2 surface. A typical plasma arc in a torch might be 1.5mm wide and 7mm long, giving a surface area of 0.8 cm2, which would mean the plasma torch should radiate 2.1 MW of power, which is 70 times greater than the power usage of a heavy-duty cutting torch of around 30 kW.

Where do you see this as being the function of the calculator? I see a description of what it claims to do, not how it does that. I do see this disclaimer in bold:

Which makes it pretty useless unless you want to cool your tungsten cube in outer space.

Let’s calculate the energy difference between a one meter cube of tungsten at 25 degrees Celsius and at 2000 degrees Celsius: Q=mc \triangle T

Heat capacity c is 132 J/(kg K) (value for solid phase), mass m is 19250 kg, so 19250\ \cdot 132\ \cdot \ 1975 = 5018475000\ J , or 5 GJ.

Here is a graph I made:

I evaluate P=eσA(T⁴–T₀⁴) every second and subtract that from the remainder of the initial 5 GJ, and then convert that back into a temperature value to use for the next second. This should show the cooling of the tungsten cube from its starting temperature at 2273.15 Kelvin in a constant temperature vacuum at 298.15 Kelvin. This looks at the first 100000 seconds / 28 hours. I ignore conduction inside and outside the cube, and also convection and insulation outside the cube. After 28 hours the cube is still at 377 Kelvin / 102 degrees Celsius.

The first 20 seconds:

Temperature Temperature^4 Seconds Radiative loss in watts Starting delta Q in Joules
2273.15 2.67001E+13 0 4177561.803 5018475000
2271.505938 2.66229E+13 1 4165487.559 5014297438
2269.866628 2.65461E+13 2 4153474.29 5010131951
2268.232045 2.64698E+13 3 4141521.556 5005978476
2266.602166 2.63938E+13 4 4129628.922 5001836955
2264.976968 2.63181E+13 5 4117795.958 4997707326
2263.356427 2.62429E+13 6 4106022.236 4993589530
2261.740519 2.6168E+13 7 4094307.334 4989483508
2260.129221 2.60935E+13 8 4082650.832 4985389200
2258.522511 2.60194E+13 9 4071052.315 4981306550
2256.920365 2.59457E+13 10 4059511.372 4977235497
2255.322761 2.58723E+13 11 4048027.595 4973175986
2253.729677 2.57993E+13 12 4036600.58 4969127958
2252.141089 2.57266E+13 13 4025229.928 4965091358
2250.556977 2.56543E+13 14 4013915.241 4961066128
2248.977317 2.55823E+13 15 4002656.128 4957052212
2247.402088 2.55107E+13 16 3991452.198 4953049556
2245.831269 2.54395E+13 17 3980303.066 4949058104
2244.264837 2.53686E+13 18 3969208.349 4945077801
2242.702772 2.5298E+13 19 3958167.67 4941108593
2241.145051 2.52278E+13 20 3947180.652 4937150425

Considering that the AWE field has now proven to have involved almost 100% crackpots - results, and especially lack thereof, are undeniable at this poiint. Not sure whether discussing the characteristics of tungsten at 2000 degrees C. could save it, but I’m not seeing how this could be relevant to the discussion. Brings us back to “really smart people” and, what was the term from Barnard, “amazing minds” or something to that effect? The problem with wind energy is, air is invisible so people (especially crackpots) can imagine the air doing whatever they want. Therefore, wind energy has always been, and continues to be, a magnet for crackpots. And airborne wind energy is a neodymium supermagnet!

You can ask this on a physics forum, I don’t understand this enough to be useful. This one for example: https://www.physicsforums.com/

From what I understand it, the reentry vehicle is kind of like a hammer and the atmosphere like an anvil at reentry speeds. The reentry vehicle is constantly hammering the air and heating up. Its resultant temperature comes from the balance of the things heating it up and cooling it down.

Here are videos:

Apollo - Atmospheric Entry Phase (1968)

Planetary Entry- Aerodynamic Heating & Deceleration (NASA)

If that doesn’t explain it enough or is too outdated, you could ask on the physics forum for example.

Gary Novak is an insane person. Listen to literally anyone else. Gary Novak is an insane person. Yes, ice radiates heat away. But it also receives radiation, and is insulated by the atmosphere.

Midweek Cuckoo:Gary Novak | moonflake