So it's Rt = 1 / ((1/R1) + (1/R2))
That's very easy.
So if Rt = 1 / ((1/2.8Ω) + (1/2.8Ω))
Solve R1 : 1/2.8Ω = 0.35714285714285714285714285714286
Solve R2 : 1/2.8Ω = 0.35714285714285714285714285714286
R1 + R2 = 0.71428571428571428571428571428571
1/ (R1 + R2) = 1 / 0.71428571428571428571428571428571 = 1.4Ω
Rt = 1.4Ω
Easy right?
Resistors in series : Rt = R1 + R2, Done. Even a child could do the math.
But here's my problem.
Each one of those resistors are two more resistors in parallel.
So R1 has two resistors that make it up, in parallel.
R1 : (R1a & R1b)
R2 : (R2a & R2b)
2.8 x 2 = 5.6Ω
2.8 x 2 = 5.6Ω
Rt = 1 / ((1/5.6Ω) + (1/5.6Ω))
Solve R1 : 1/5.6Ω = 0.17857142857142857142857142857143
Solve R2 : 1/5.6Ω = 0.17857142857142857142857142857143
Solve R1 + R2 : 0.35714285714285714285714285714286
Solve 1 / (R1 + R2) = 2.8Ω
Easy right?
*** Here's the problem ***
36 AWG, 316L Stainless Steel Wire : 17.80 Ω/Ft.
24 AWG, 316L Stainless Steel Wire : 1.101 Ω/Ft.
Target Resistance 0.7 Ω
Target Wattage (Power) : 100 Watts.
Supply Voltage (Vs) : 8.4Vdc
It's a dual coil build, with a supply voltage of 8.4V, total resistance is 0.7Ω
0.7Ω x 2 coils = 1.4Ω each.
Rt = 1 / ((1/1.4Ω) + (1/1.4Ω))
R1 : 1/1.4 = 0.71428571428571428571428571428571
R2 : 1/1.4 = 0.71428571428571428571428571428571
R1 + R2 = 1.4285714285714285714285714285714
1 / (R1 + R2) = 0.7Ω
Ohm's Law : Watts (Voltage & Resistance) P = E² / R
8.4 x 8.4 = 70.56 / 0.7Ω = 100 Watts.
But each coil (1.4Ω) each, has two wires that make up the coil.
2.8 Ω per wire.
Rt = 1 / ((1/2.8) + (1/2.8))
R1 : 1/2.8Ω = 0.35714285714285714285714285714286
R2 : 1/2.8Ω = 0.35714285714285714285714285714286
R1 + R2 = 0.71428571428571428571428571428571
1 / (R1 + R2) = 1.4Ω Single Coil (with two wires).
The 36ga. get's wrapped on a straight piece of 24ga. : Clapton Coil
36 AWG, 316L Stainless Steel Wire : 17.80 Ω/Ft.
24 AWG, 316L Stainless Steel Wire : 1.101 Ω/Ft.
The 24ga is very long, if I cut it at 2.8Ω
25.4mm * 12 In. = 304.8mm / Ft.
36ga : 17.80Ω / 304.8mm = 0.0583989501 Ω/mm
24ga : 1.101Ω / 304.8mm = 0.0036122047 Ω/mm
Solve the length of 24ga. needed to hit 2.8Ω
2.8Ω / 0.0036122047 = 775mm
That's a very long piece of 24ga.
The 36ga is fine I guess.
2.8Ω / 0.0583989501 Ω/mm = 47.9mm
It can't be 2.8Ω for the 24ga, and It has to be 24ga.
How do I shrink the length of 24ga. without missing 2.8Ω
I don't know the math, maybe I should just leave it alone. Because I'm also afraid that I wont be able to handle the math, as simple as it might be.
Doing the above math is easy when the numbers are identical. But if I need to shrink the length of wire, keeping the 24ga. shorter than the 36ga. I can't wrap my head around it.
People make these coils, and there's no math involved.
Resistors inside Resistors that are all parallel.
I guess it would look like something like this as a circuit drawing.
Dual Clapton Coils Build Video.
The people that make these, DO NOT KNOW THE MATH. I do.. but as you saw from the math, the 24ga. that the 36ga. get's wrapped on, is way too long for a single coil. The 36ga. would never wrap on the whole thing, and on top of that, again, the 24ga. is too long to begin with.
That's very easy.
So if Rt = 1 / ((1/2.8Ω) + (1/2.8Ω))
Solve R1 : 1/2.8Ω = 0.35714285714285714285714285714286
Solve R2 : 1/2.8Ω = 0.35714285714285714285714285714286
R1 + R2 = 0.71428571428571428571428571428571
1/ (R1 + R2) = 1 / 0.71428571428571428571428571428571 = 1.4Ω
Rt = 1.4Ω
Easy right?
Resistors in series : Rt = R1 + R2, Done. Even a child could do the math.
But here's my problem.
Each one of those resistors are two more resistors in parallel.
So R1 has two resistors that make it up, in parallel.
R1 : (R1a & R1b)
R2 : (R2a & R2b)
2.8 x 2 = 5.6Ω
2.8 x 2 = 5.6Ω
Rt = 1 / ((1/5.6Ω) + (1/5.6Ω))
Solve R1 : 1/5.6Ω = 0.17857142857142857142857142857143
Solve R2 : 1/5.6Ω = 0.17857142857142857142857142857143
Solve R1 + R2 : 0.35714285714285714285714285714286
Solve 1 / (R1 + R2) = 2.8Ω
Easy right?
*** Here's the problem ***
36 AWG, 316L Stainless Steel Wire : 17.80 Ω/Ft.
24 AWG, 316L Stainless Steel Wire : 1.101 Ω/Ft.
Target Resistance 0.7 Ω
Target Wattage (Power) : 100 Watts.
Supply Voltage (Vs) : 8.4Vdc
It's a dual coil build, with a supply voltage of 8.4V, total resistance is 0.7Ω
0.7Ω x 2 coils = 1.4Ω each.
Rt = 1 / ((1/1.4Ω) + (1/1.4Ω))
R1 : 1/1.4 = 0.71428571428571428571428571428571
R2 : 1/1.4 = 0.71428571428571428571428571428571
R1 + R2 = 1.4285714285714285714285714285714
1 / (R1 + R2) = 0.7Ω
Ohm's Law : Watts (Voltage & Resistance) P = E² / R
8.4 x 8.4 = 70.56 / 0.7Ω = 100 Watts.
But each coil (1.4Ω) each, has two wires that make up the coil.
2.8 Ω per wire.
Rt = 1 / ((1/2.8) + (1/2.8))
R1 : 1/2.8Ω = 0.35714285714285714285714285714286
R2 : 1/2.8Ω = 0.35714285714285714285714285714286
R1 + R2 = 0.71428571428571428571428571428571
1 / (R1 + R2) = 1.4Ω Single Coil (with two wires).
The 36ga. get's wrapped on a straight piece of 24ga. : Clapton Coil
36 AWG, 316L Stainless Steel Wire : 17.80 Ω/Ft.
24 AWG, 316L Stainless Steel Wire : 1.101 Ω/Ft.
The 24ga is very long, if I cut it at 2.8Ω
25.4mm * 12 In. = 304.8mm / Ft.
36ga : 17.80Ω / 304.8mm = 0.0583989501 Ω/mm
24ga : 1.101Ω / 304.8mm = 0.0036122047 Ω/mm
Solve the length of 24ga. needed to hit 2.8Ω
2.8Ω / 0.0036122047 = 775mm
That's a very long piece of 24ga.
The 36ga is fine I guess.
2.8Ω / 0.0583989501 Ω/mm = 47.9mm
It can't be 2.8Ω for the 24ga, and It has to be 24ga.
How do I shrink the length of 24ga. without missing 2.8Ω
I don't know the math, maybe I should just leave it alone. Because I'm also afraid that I wont be able to handle the math, as simple as it might be.
Doing the above math is easy when the numbers are identical. But if I need to shrink the length of wire, keeping the 24ga. shorter than the 36ga. I can't wrap my head around it.
People make these coils, and there's no math involved.
Resistors inside Resistors that are all parallel.
I guess it would look like something like this as a circuit drawing.
Dual Clapton Coils Build Video.
The people that make these, DO NOT KNOW THE MATH. I do.. but as you saw from the math, the 24ga. that the 36ga. get's wrapped on, is way too long for a single coil. The 36ga. would never wrap on the whole thing, and on top of that, again, the 24ga. is too long to begin with.
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