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Calculate bundle of cables AWG?
Calculate bundle of cables AWG?
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Old 14th February 2020, 08:02 PM   #11
FriedMule is offline FriedMule  Denmark
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Join Date: Apr 2018
I maybe had the answer, but not the additional knowledge you have given me, it has confirmed and expanded my knowledge of the problem, so it did help a lot!!
Please bear with me. English is self taught and it may look like I have ignored your reply. But please know, I am doing my best, but may overlook something.
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Old 14th February 2020, 11:26 PM   #12
GoatGuy is offline GoatGuy  United States
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Join Date: Dec 2012
Location: SF Bay Area
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:
ΔAWG = –10 ◊ log10( 20 )
ΔAWG = –13.01 .. → –13
So, therefore, the 'net AWG' is just 22 - 13 = 9 gauge.

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.
Burnished unicorn horn circuit-design only goes so far. When the magic falls flat, it is best to rely on solid maths. Just Saying... GG

Last edited by GoatGuy; 14th February 2020 at 11:32 PM.
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