Does speaker cable length matter?

[ThisBass]

Originally Posted by gumtownbassman

I agree.

I don't agree

Was there a query?

[dmusic148]

Originally Posted by Got2SadowskyNYC

I've got the dough and wouldn't spend it on hype. There's other things I rather have than an over priced power cable that does nothing more than get power from the wall to my amp. Kind of like what came with it, for free.

You misunderstand. I meant the amp. The cable is pure bs.

[Ragoo]

I skimmed over the manual a bit and saw the "Italian Headphone Chorus" part. It reminded me than my active Ibanez P has a chorus effect on all the strings. The controls are just Volume and that Phat II EQ boost Ibanez puts in most of their basses. I would play an open E or some notes higher up on the string and I would hear a chorus-like shimmer. It's not the amp; my B600H and the 4x10 and 1x15 don't do that with my G&L or my Squier. It's not a bad thing by any means (I certainly dig the piano-like low E), but is there a reason for it?

[dspellman]

FWIW, Speaker cables are all about wire gauge.
All (as in ALL) lengths of speaker cable degrade performance and efficiency. The only real question is to what degree.

The power loss for a 50' cable run also depends on the impedance; the lower the impedance, the greater the power loss a 50' run of cable represents as a percentage of total power.

So for an 18 AWG cable to an 8 ohm load, using a 50' cable, the power loss would be 0.71%.
To a 4 ohm load, the power loss would be 5.71%, and to a 2 ohm load, it would be 24.21%.

For a 14 AWG cable, the loss to an 8 ohm load is 0.34%, to a 4 ohm load 2.72%, and to a 2 ohm load 11.19%.

A 2 ohm load requires four times the copper of an 8 ohm load.

A 50% drop in power would drop volume by roughly 3 dB.

In short, a 6' length of 18AWG would have a fairly negligible effect on total power output at any load from 8 ohms to 2 ohms.

[dspellman]

Originally Posted by thejumpcat

Does the length of the speaker cable matter?

I've got a Monster cable -- a Prolink Performer 500 Ultra Wide Bandwith High Power Speaker Cable with Magnetic with Magnetic Flux Tube. Yeah, that's a mouthful.

It doesn't indicate anywhere on it what the guage is. I've had the thing forever, and it sounds fine. It's probably 20 feet long, but I've always got it coiled up with a velcro strap.

So yes, the length of the speaker cable matters -- the shorter the better.

There's no data on the Monster site about the gauge of the cable you're using, but there's certainly a lot of gobbledegook about Carbon infused polymer shielding and MicroFiber® dielectric and Two Time Correct® multiple gauge wire networks for an even frequency response and accurate phase reproduction.

[dspellman]

Originally Posted by ThisBass

That's wrong because you did not considered the Peak To RMS Ratio of the Audio Signal.

I did, however, consider the binary framus of the overhand camshaft relay valve.

I'd appreciate you redoing those calculations, then, considering the Peakaboo to RootMeanSquare Ratio. I think all you really need to know is the DC Resistance of copper wire:

In ohms per 100' length:

18AWG: 6.39
16AWG: 4.02
14AWG: 2.52
10AWG: 0.9988

[megafiddle]

I believe that should be per 1000 feet

[ThisBass]

Originally Posted by dspellman

FWIW,
A 2 ohm load requires four times the copper of an 8 ohm load.

P = U*I
P = U^2 / R
P = R * I^2

I = SQRT(P/R)

P = 100 Watt

R = 8 Ohm
I = SQRT(100Watt/8Ohm) = 3.54 Ampere

R = 2 Ohm
I = SQRT(100Watt/2Ohm) = 7.07 Ampere

A 2 ohm load requires two times the current of an 8 ohm load.
A 2 ohm load requires two times the copper of an 8 ohm load.

edit:
for PA sub 1000 Watt multiply by 10

[megafiddle]

Originally Posted by ThisBass

P = U*I
P = U^2 / R
P = R * I^2

I = SQRT(P/R)

P = 100 Watt

R = 8 Ohm
I = SQRT(100Watt/8Ohm) = 3.54 Ampere

R = 2 Ohm
I = SQRT(100Watt/2Ohm) = 7.07 Ampere

A 2 ohm load requires two times the current of an 8 ohm load.
A 2 ohm load requires two times the copper of an 8 ohm load.

edit:
for PA sub 1000 Watt multiply by 10

No

The power loss in the wire is related to I squared R (I x I x R).
Double the current and the power loss goes up 4 times (for a given resistance).
Therefore you have to reduce the resistance by 4 times, not 2.

[ThisBass]

Originally Posted by megafiddle

No

The power loss in the wire is related to I squared R (I x I x R).
Double the current and the power loss goes up 4 times (for a given resistance).
Therefore you have to reduce the resistance by 4 times, not 2.

Ähm yes, now I understand but, I did not intend to grill a chicken with this wire heating ...
Your increase of losses in the wire are obsolete like junk food

[megafiddle]

You're right, in most cases it simply is too small to matter.

About the junk food, it's often unadvisable to get between anything and it's food.

[Arjank]

Originally Posted by gumtownbassman

I agree.

There is indeed a lot of misinformation on this forum (like 80%), but not the point that I made about the impedance of the cable having influence on the Qes if the driver/woofer. It;s the same as adding a resistor in series with a woofer, this will have impact on the drivers TSP's. Why do you think that they use inductors with large gauge wire in high-end designs? same reason...
A lot of people here on this forum need to do their homework 1st before making comments on someones post!

[ThisBass]

Originally Posted by megafiddle

About the junk food, it's often unadvisable to get between anything and it's food.

Yes you are aright.
I should have known better.

[Codger]

Originally Posted by Arjank

There is indeed a lot of misinformation on this forum (like 80%), but not the point that I made about the impedance of the cable having influence on the Qes if the driver/woofer. It;s the same as adding a resistor in series with a woofer, this will have impact on the drivers TSP's. Why do you think that they use inductors with large gauge wire in high-end designs? same reason...
A lot of people here on this forum need to do their homework 1st before making comments on someones post!

The driver's TS parameters are what they are and they cannot be changed by adding external resistance. What these external resistances can affect is the damping of the SYSTEM, Qtc, which includes all system resistances, mechanical and electrical. The external resistances include:

• speaker cable
• crossover inductor
• connectors
• the amplifier output resistance (related to the essentially meaningless spec called damping factor)

The effect of these external resistances on Qtc is minimal, being almost entirely dominated by the driver's voice coil resistance.

An analysis for a sealed box 2nd order Butterworth alignment showed that an increase in external resistance of 0.75 ohm generated a peak of only 0.1dB in the output response.

The effect of speaker wire resistance, amplifier damping factor, crossover inductor resistance is minimal on the system Q.

There is a typo in the analysis. It should read:

                 Re + Rg 
     Qec' = Qec --------- 
                    Re

-- bradley

[Arjank]

Originally Posted by Codger

The effect of these external resistances on Qtc is minimal, being almost entirely dominated by the driver's voice coil resistance.

An analysis for a sealed box 2nd order Butterworth alignment showed that an increase in external resistance of 0.75 ohm generated a peak of only 0.1dB in the output response.

The effect of speaker wire resistance, amplifier damping factor, crossover inductor resistance is minimal on the system Q.

Like I allready said, it only has minimal effect with sealed cabinets. But ported designs are more affected by this.
It indeed affects the Qtc.
A general rule is to not exceed 10% of the speakers voicecoil impedance with external (inductors, wiring) parts.

There's a german speaker magazin that, when reviewing drivers/woofers, shows the optimal volume and tuning frequency for that particular driver with 0.2ohms and with 1ohm of additional resistance. You can clearly see the difference between 0.2 and 1ohm of extra resistance in the simulated plots. Any good simulation program can show you this (e.g. Ajhorn)

[thejumpcat]

This is what I ended up getting:

http://www.ebay.com/itm/NEW-ITEM-NEU...item415e763178

[delta7fred]

Originally Posted by thejumpcat

This is what I ended up getting:

http://www.ebay.com/itm/NEW-ITEM-NEU...item415e763178

Perfect!

[ThisBass]

May be somebody ever kept an eye inside a pro 5..10k Watt power amp.
The contact track to the speakon output plug looks like rubbish.
I never calculated the resistance but, regarding to somebody's opinion here at the forum the track has to burn of at full rated output power.

Ähm no, I don't mean low budget gear but pro gear you have to charge a couple of k$ for it.

[gumtownbassman]

Originally Posted by ThisBass

May be somebody ever kept an eye inside a pro 5..10k Watt power amp.
The contact track to the speakon output plug looks like rubbish.
I never calculated the resistance but, regarding to somebody's opinion here at the forum the track has to burn of at full rated output power.

Ähm no, I don't mean low budget gear but pro gear you have to charge a couple of k$ for it.

Speakon is rated at 30amps RMS and 250 volts AC
http://cpc.farnell.com/1/1/4574-spea...c-neutrik.html
so a short run track with a cross-sectional area of 1.5mm squared will do.
The most it will handle (by design) is 7.5kW (RMS).
I would hope that "pro" equipent designers put their gear through very rigorous tests.

[ThisBass]

Originally Posted by gumtownbassman

so a short run track with a cross-sectional area of 1.5mm squared will do.

yes but the short track is probably less then 1/2 of 1.5mm^2 (as far as I remember).

IMO that's not a problem at all cause audio program material contents a very a large peak to rms ratio which pushes down the RMS current and therefore the losses as well.