Voron TempFluct: Unterschied zwischen den Versionen
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This results in P~4W (= 250W/30/2) (half of the power of one of the resistors flows through that area to the heat bed surface). | This results in P~4W (= 250W/30/2) (half of the power of one of the resistors flows through that area to the heat bed surface). | ||
4*35*1000/220/700 K ~ 1K | delta T = 4*35*1000/220/700 K ~ 1K | ||
I would assume, that my thermal cameras resolution is not sufficient to detect such small variations in the temperature over the bed. | I would assume, that my thermal cameras resolution is not sufficient to detect such small variations in the temperature over the bed. | ||
Will give it a try in the next few weeks. (Beginning of 2026) | Will give it a try in the next few weeks. (Beginning of 2026) | ||
Aktuelle Version vom 24. Dezember 2025, 20:30 Uhr
The thermal conductivity of aluminium is lambda=220W/m/K The max distance of the resistors is 70 mm
I estimate the area A[mm^2]=70*10mm^2, through which the power of the load-resitor passes
I estimate the distance from the middle of the resistor to the resistor itself l=35mm
At 110° ~ 30% of the 750W Heatbed, which was installed before was sufficient.
This results in P~4W (= 250W/30/2) (half of the power of one of the resistors flows through that area to the heat bed surface).
delta T = 4*35*1000/220/700 K ~ 1K
I would assume, that my thermal cameras resolution is not sufficient to detect such small variations in the temperature over the bed.
Will give it a try in the next few weeks. (Beginning of 2026)