# Calculate bundle of cables AWG?

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#### FriedMule

I have tried several sites, via google and have got different formala's and results, all from 1 AWG to 9 AWG. So I please need your help 20 x 22 AWG = X AWG when collected in one bundle?

#### Lojzek

Seek official data on awg standard to be able to feed the formula with wire diameter. Cross section of a round wire is a circle. Find the formula to calculate area of a single circle and multiply it by 20.

#### FriedMule

Seek official data on awg standard to be able to feed the formula with wire diameter. Cross section of a round wire is a circle. Find the formula to calculate area of a single circle and multiply it by 20.
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?

#### Destroyer OS

Paid Member
There's calculators on the internet.

#### FriedMule

There's calculators on the internet.
Yes that's true, but they give different results, I have tried several solutions, and manual calculation. #### Speedskater

two 22AWG = one 19AWG
four 22AWG = one 16AWG
eight 22AWG = one 13AWG
sixteen 22AWG = one 10AWG
thirty two 22AWG = one 7AWG

most 22AWG wires have a rather large, diameter tolerance.

#### FriedMule

Thank you, it, helped a lot! #### PRR

Paid Member
...20 x 22 AWG = X AWG when collected in one bundle?

#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.
______________

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.
__________________

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

#### FriedMule

Thanks so much, I learn a lot from all of you! I'll save your explanation! #### PRR

Paid Member
... = 9AWG
Is that right?

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.

#### FriedMule

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!!

#### GoatGuy

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.

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