A couple back of envelope calculations
Steel and copper glow red when hot, not aluminum, based on color charts for color vs temp, maybe the tip of the strip is 800 C. Maybe it’s 50 C next to the canister at the other end – I touched it and it was warm but not hot.
I burn 1/4 ounce of butane in 5 minutes. That’s 7 grams in 300 seconds. Heat of vaporization is 386 Ws/g. I need 9 Watts of power to vaporize my fuel.
I use canisters that are 8 ounces of butane and 5.4 ounces of steel. Maybe in the 5 minutes it heats this up 10 C – it starts slow and speeds up after a few minutes of heating. Specific heat of butane is 1675 J/kgK. Steel – 500. It’ll take 4589 J to heat up the canister 10 C. If it does it over 300 sec, that would be 15 W.
If the strip is 5 inches long and 1 inch wide, that would be 10 in2 area (both sides). Maybe the average temperature of the strip is 300 C. There’s an air layer of about 1/2 inch around the strip providing insulation. The thermal conduction of air is 0.024 W/mK. We are losing 3.6 W due to conduction to air.
So, there’s a total of 27.6 W flowing down the strip.
The cross sectional area of the strip is 0.02 in2. It’s 5 inches from tip to canister = 0.075 m. The conduction of copper is 401 W/mK. If 27.6 W is flowing, that would produce a 697 C temperature drop from the tip to the canister. That’s pretty consistent with my guess that the tip is 800 C at tip and 50 C at canister.
According to this calculation thermal conductivity from strip to air is a small effect compared to heating and vaporizing butane.
My aluminum strip was 1.75 x 7 inches – about 5 inches from tip to canister. 1.75 times bigger than Bob, but conduction is half, so it’s a little less. So I would expect a little more temperature drop. Bob’s seems to work a little better than mine so maybe this explains it.
Radiative heat loss is emissivity * sigma * area + (T strip ^ 4 – T air ^ 4). Maybe emissivity is 0.5. sigma is 5.67e-8 W/m2K4. Maybe approximate strip as a 800 C area of 1 inch^2 = 0.0006 m2. Air is 0 C. Then radiative heat loss = 22 W. If it was 600 C it would radiate 9 W. 400 C – 3 W. So, only the hottest part of the strip will have large radiative heat loss. So, most of the radiative heat loss is right at the tip where it’s hot. This is consistent with observation – it’s very hot at the tip but quickly cools down further away from tip.
Copper that’s been heated and heavily oxidized has an emissivity of 0.8. Heavily oxidized aluminum is 0.25. So maybe aluminum would be better? But maybe this isn’t important, because this is only important at the tip where it’s really hot.
One other question is how much heat is radiated from canister to the air. Let’s say canister is 0 C and air is -10 C (14 F). My 8 ounce canister is 4.25 inch diameter and 3 inches high so total area is 68 in2 = 0.04 m2. The boundary layer of air around it is 0.5 inch = 0.0125 m. Conduction of air is 0.024 W/mK. So it loses 0.8 W. This is insignificant compared to vaporizing butane and heating up canister. Don’t worry about insulating your canister.