So, for

Papa Bear, 240 = 67 x R and R = 3.58 Ohms

Mama Bear, 240 = 55 x R and R = 4.36 Ohms

Baby Bear, 240 = 37.5 x R and R = 6.4 Ohms

OK let's just look at Papa Bear here for a minute. The resistance of 12 gauge Kanthal A1 wire is 0.134 Ohms/Foot so to get 3.58 Ohms we only need 3.58/.134 or 26.72 feet of wire. Hey that's a very elegant solution and it won't cost much either. Just 27 feet of wire and we can fire the kiln. But there is a major problem and we need to understand what it is and what we can do about it. Just because it works mathematically does not mean it will work in real life.

Let's consider that our 27 feet of wire is a cylinder, where the diameter of cylinder is .081 inches -- that's the diameter of 12 gauge Kanthal A1 -- and the height is 27 feet. So we have a cylinder of diameter 81 thousandths of an inch and a height of 27 feet. The surface area of the cylinder is the circumference of the circle times the height of the cylinder. So we have .081 x 3.14 x 27 x 12 = 82.4 square inches of surface area. Note I had to multiply 27 times 12 so that everything is in inches.

So we have 82.4 square inches of surface area and we need to generate 16,000 watts of power so that each square inch of element must generate 194.17 watts of power. (16000/82.4)

The online Kanthal manual

(http://www.kanthal.com/C12570A7004E2D46/062CC3B124D69A8EC1256988002A3D76/D1D355F37C940491C12572B9003FD970/$file/1-A-5B-3%20UK%20resistance%20alloys.pdf?OpenElement)

would like for the watt loading figure of the elements to be around 15 instead of 194. That is, a good long lasting element will only have to generate around 15 watts per square inch of surface area.

My guess is that our 27 foot element would last only milliseconds with a full 240V passing through it.

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