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Topic # 108586 2-Sep-2012 09:09
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Hi All,

QUESTION 1: If I have a 4 core 14 AWG cable and a 2 core 14 AWG cable do they contain the same cross sectional area of copper or does the 4 core 14 AWG cable have twice that of the 2 core 14AWG cable ?

I am leaning to "Yes the 4 core has twice the area of copper" as I have seen a few posts on here that state that twisting a 4 core cable in to 2 cores will drop a 14AWG to 11AWG (double the cross sectional area).

QUESTION 2: If a cable is quoted as 14AWG is that 14AWG per core or per cable?

QUESTION 3: If a cable is quoted as having an impedence of 0.01ohm/m (100ohm/km) is that per core or for the entire cable?

I'd be able to answer the next questoin with a definitive answer to QUESTION 3.

QUESTION 4: If I have a 2 core speaker cable rated as above and a 20m cable run to a 8ohm speaker does the following hold true?

Cable impedence/Load = (0.01*20)/8 = 0.2/8 = 2.5%

(or do I need to double the cable length as it has 2 cores?)

QUESTION 5: The above calculation is meant to hold the lowest speaker load.  I have read somewhere that a typical 8ohm speaker has a lowest load of 3ohm.  Is this correct?  Or should 8ohms remain the value of the variable in the above equation?

I ask this as I have also read that for "audiophiles" this value should be below 5%.

Thanks for any help to any of the answers!

EDIT: I forgot this question:

QUESTION 6: What is this all about "(7/15/0.16BC)" - I'm guess, strands per core/twists per ? and something to do with area?

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Reply # 680335 2-Sep-2012 11:05
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  Reply # 680577 2-Sep-2012 21:49
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Q1: Per core so twice the copper.

Q2: per core.

Q3: typically per core, so the round trip is double.

Q4: you would typically double it as it is 2 ways, yes that would hold true

Q5: impedance curves vary massivly between speakers, typically if they are described as "easy to drive" then they will be flatter than a challenging speaker. The variance in impedance coupled with different amplifiers is what causes different amps to have different sonic characteristics. You should be able to get a graph for the speaker, so look for the lowest point in the freq range you care about.

Q6: that is the correct way to express cable construction, whereas AWG is pretty meaningless as to the constriction since it is just the outside diameter of the bundle and doesnt allow for all the gaps between them, that would be 7 bundles of 15 wires, each at 0.16mm diameter. which should be quite visible when you strip the jacket back. You need to convert the diameter to mm^2 and multiply it out by all the bundles to get the total area, but be aware if you then convert that back to AWG you will be getting a smaller size so terminals that will not fit possibly, since AWG is the hole size that it will fit into.

And if you see cables for sale locally that are only listed as gauge or similar they are probably lying.



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  Reply # 680676 3-Sep-2012 09:41
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Thanks Rich,

Some of those answers were merely for interest sakes.

I'm planning the wiring for my new build.  I'll be wiring for multi-room audio.  The distances for each zone from the central cupboard are estimated as: 10m, 20m, 9m, 23m, 20m, 14m, 17m and 22m.

I've read numerous DIY cable sites, forums, comparisons etc, flirted with the idea of doing something crazy and then decided to purchase this cable: Dynamix 4 core 14AWG  As I only need two cores I will twist the red and white/blue and black cores to create a cable which is equivilant to a 11AWG cable that has the characteristics of the TNT Star sells this cable for $342.20 per 100m - which is the current best price on the net.  They are out of stock and non is due until mid-November.  This is about the perfect time I will need it.

According to Wikipedia I will have enough copper to cover these distances. 14AWG covers 24m.

Thanks for helping me close out my understanding of this topic.

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  Reply # 681202 4-Sep-2012 10:07
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That's what I used. It's decent enough. I didn't use it for my fronts, but it was OK for the rears and second/outdoor zones.

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Reply # 681208 4-Sep-2012 10:29
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I can't seem to do the maths on this cable.  It is 14awg (15/7/0.16) 4 core.

As stated above, each core is 14AWG.  Each core has 7 bundles of 15 conductors.   Are the conductors or the bundles 0.16mm dia?  If bundle I get a cross-sectional area for each bundle = 0.16*pi() = 0.503mm2

Multiplying this by the bundles I get a cross-sectional area for the core = 0.503*7 = 3.52mm2.  Accordin to a table from tnt-audio this core is actually about 12AWG.

Twisting the cores together will result in a 2 core cable and each core will have an area = 7.04mm2.  That puts it in as roughly 8.5AWG terittory.

Can anyone explain?

If this is correct this cable is EASILY large enough to work for me.

On that, I was wonder Bazaar why you didn't use this for your fronts?

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  Reply # 681239 4-Sep-2012 11:43
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Your maths is off, I think. Area = pi*r^2 = 3.14 * 0.08mm^2 = 0.02mm^2. Presumably this is per conductor because 0.02 * 7 * 15 = 2.11mm^2 ~= 14AWG.

I didn't use it for my fronts (and center) because I had better quality "audiophile" cable that was long enough. Incidentally, that also has four cores.

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Reply # 681282 4-Sep-2012 12:26
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Hilarious. I've been learning VBA and coding in a formula for calculating vertical acceleration at a remote lever arm (for work) with formula like (induced heave):

heave_ind = -1 * (-dY * Sin(Radians * P) + dX * Sin(Radians * R) * Cos(Radians * P) + dZ * (1 - Cos(Radians * R) * Cos(Radians * P)))

and I forget the area of a circle!

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