View Single Post
  #8   Report Post  
Posted to rec.audio.car
Mariachi Mariachi is offline
external usenet poster
 
Posts: 174
Default Replace positive wire on HU?

On Feb 25, 4:01 pm, Matt Ion wrote:
Mathematically, there may be a theoretical difference... REALISTICALLY you can
be 99.9% sure the difference would barely be measurable in your deck's output,
and certainly not audible or noticeable, especially when you factor in the
ambient noise and the fact that your HU's built-in amp isn't actually giving you
anywhere near 25W/ch.

Mariachi wrote:
At any rate, you'd have to know the length of the wire--there's
nothing magical about the gauge of the wire, it's all a matter of the
total resistance that wire has. So, would be fine with 1" of 18 AWG on
a 750 Watt amplifier, but you'd be in really bad shape if you had to
go 15 feet with the same wire on that same amp. :-)


yes I agree...


Resistance of the wire = Resistivity * Length / Area
R(wire) = p * (L/A)


If you lower the wire by 3 gauges then you have a cross sectional area
increase by a factor of 2.
Therefore, the total resistance decreases by a factor of 2 given that
the resistivity and length are still the same.


So if you want to make a power wire twice as long, you need to
increase A by 2 times also. Therefore it would be the same ratio and
therefore it would be the same total resistance.


So if I have positive power wire with a length 2 meters with a 16
guage cross sectional area...
Resistivity of copper = 1.7 microohm/cm = 1.7 microohm/cm * (1 *
10^(-6) ohms / 1 microohm) * (1 m / 100 cm) = 0.000000017 ohm/m
L = 2 meters
A (16 gauge)= 0.8107 m^2


R(16 GWG wire) = (1.7e-8)*(2/.8107) ohms = 4.194e-8 ohms
R(14 GWG wire) = (1.7e-8)*(2/1.291) ohms = 2.634e-8 ohms


4.194 / 2.634 = 1.59


Resistance decreased by 1.59 times


R(16 GWG) / R(13 GWG) = 2
R(16 GWG) / R(14 GWG) = 1.6


my head unit is rated to get a continuous power at 24 Watts per
speaker, but with a 5% THD. It gets 17 Watts per each speaker with a .
01 % THD.