Thanks I have, do I think.
D * D * pi / 4 * 20
0.6438mm * 0.6438mm * 3.14 / 4 * 20 = 6,507311508mm^2
6.63 = 9AWG
Is that right?
Yes that's true, but they give different results, I have tried several solutions, and manual calculation.
#22 is 0.0253" diameter.
American wire gauge - Wikipedia
Twenty of those mushed together round would be square-root(20) times larger diameter.
4.47*0.0253" diameter = 0.113" diameter
0.113" diameter is AWG #9.
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You can also go by area (see same table).
#22 0.642kcmil
20*0.642 = 12.84kcmil
#9 is 13.1kcmil, so 20#22 is say AWG#9.1.
You can also do it in metric, data in that table.
The table dimensions are for solid wire. Stranded wire of the same AWG# is approximately the same electrically, but will measure about 13% larger diameter due to air spaces.
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This has been about Resistance. If you are moving so much power the wire may get hot, you want Ampacity. Using a conservative temperature limit (and assuming Copper), #22 is 3A ampacity, so you would think twenty #22 would be 60A. That would only be true if each conductor was in FREE air. All together, heating each other, 20 conductors must be substantially derated, nominal 50%. The #9/10 conductor ampacity is 30A (not far from the derating for 20 #22).
Where the number of current-carrying conductors in a raceway or cable exceeds three, or where single conductors or multiconductor cables are stacked or bundled longer than 24 inches without maintaining spacing and are not installed in raceways, the allowable ampacity of each conductor shall be reduced as shown in the following table:
Number of Current Carrying Conductors ------ Percent of Values in Table
4 thru 6 --------------- 80%
7 thru 9 -------------- 70%
10thru20 -------------- 50%
The above derating factors do not apply to conductors in nipples having a length not exceeding 24 inches.
https://www.usawire-cable.com/pdfs/NEC AMPACITIES.pdf
I just realized you HAD the right answer, and only wanted confirmation.
Speedskater's finger-math says twenty #22 is fatter than #10 but much less than #7.
My figures lead to the phantom gauge #9.1. Even #9 is not a standard size and is omitted on many tables.
#9 seems right.
I'm kind of surprised that no one produced the 'simple answer'.
AWG gauge is a tenth-power-of-ten sequential numbering system. Whatever its hard numbers are, the 10N/10 scaling means that it is really easy to know how much larger, or smaller wires are of different AWG specificaitons, and vice versa.
20 cables, each of 22 GA, is '20 ×'. Figuring:
At least this formula works for any assortment of similar sized wires, as a bundle.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
PS… I meant to include some other examples.
AWG gauge is a tenth-power-of-ten sequential numbering system. Whatever its hard numbers are, the 10N/10 scaling means that it is really easy to know how much larger, or smaller wires are of different AWG specificaitons, and vice versa.
20 cables, each of 22 GA, is '20 ×'. Figuring:
ΔAWG = –10 × log10( 20 )
ΔAWG = –13.01 .. → –13
So, therefore, the 'net AWG' is just 22 - 13 = 9 gauge.ΔAWG = –13.01 .. → –13
At least this formula works for any assortment of similar sized wires, as a bundle.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
PS… I meant to include some other examples.
ΔAWG = 1 … 10(N = 1)/10 = 1.259 times the area.
ΔAWG = 3 … 10(N = 3)/10 = 1.999 → 2× the area. How convenient!
ΔAWG = 5 … 10(N = 5)/10 = 3.16 which also is √(10) = 3.16
ΔAWG = 10 … 10(N = 10)/10 = 10, which by now should be obvious.
And of course, it works in reverse exactly the same way.ΔAWG = 3 … 10(N = 3)/10 = 1.999 → 2× the area. How convenient!
ΔAWG = 5 … 10(N = 5)/10 = 3.16 which also is √(10) = 3.16
ΔAWG = 10 … 10(N = 10)/10 = 10, which by now should be obvious.
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