soundigital SD16K

[nastyaudioman 10+ year member]

Just goes to show you dont know wtf u r talking about... SD12K and SD16K both have dual 1/0 Positives and Grounds:laugh:

How many inputs dos their 6K-12k plus amps have? .....

 

[plugitin 5,000+ posts]

I guess they expect someone to weld/solder inputs to the battery terminal. No problem, you guys have fun with that.

gtfo //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif

 

[nastyaudioman 10+ year member]

lol

 

[TurdFergueson2 5,000+ posts]

http://bcae1.com/images/swfs/wirepowercalculator.swf

nice calculator. I run 2/0 though.

 

[plugitin 5,000+ posts]

nice calculator. I run 2/0 though.

hm, never thought of that...

im not saying the calculator is exactly right or anything...but it should help some

 

[TurdFergueson2 5,000+ posts]

hm, never thought of that...
im not saying the calculator is exactly right or anything...but it should help some

2/0 welding is close in size to 1/0 kicker. So being that we don't know what the calc is based off of it is a little inconclusive.

But the point was that a shorter run would handle more and the calc proves that.

 

[murilo 10+ year member]

EVERYONE always says this...but that is for a piece of 0 gauge that is 15 feet long, or something like that. Now what if we use a piece 1 foot long, or shorter...what now? Hmmm...

Correct

example, look at this 600A fuse

http://www.littelfuse.com/part/CNN600..html

check how thin is the fusible, it is like a 12GA cable, but is very short.

 

[sub-FATHER 10+ year member]

i guess.
see what you do on music on the tl.

View attachment 26514810 WOULD A 149.9 BASS RACE ON THE TERM LAB COUNT?

THAT WAS DONE WITH ONLY ONE AMP

AND MY 155.2 WAS ALSO ON THE TERMLAB

AND MY 156.3 WAS ON THE TERMLAB

 

[sub-FATHER 10+ year member]

http://bcae1.com/images/swfs/wirepowercalculator.swf

THAT SPECIFICALLY STATES

330amps OF CURRENT

SO....STUPID....

THIS IS HOW YOU CALCULATE IT.....

FOR EVERY 330 AMPS OF CURRENT YOU WILL LOOSE .05volts

SO IF I AM PULLING SAY 990AMPS. WHICH IS 3 * 330 = .15VOLTS OF LOSS

 

[plugitin 5,000+ posts]

THAT SPECIFICALLY STATES330amps OF CURRENT

SO....STUPID....

THIS IS HOW YOU CALCLATE IT.....

FOR EVERY 330 AMPS OF CURRENT YOU WILL LOOSE .05volts

SO IF I AM PULLING SAY 990AMPS. WHICH IS 3 * 330 = .15VOLTS OF LOSS

you fail once again...learn to read

shoulda tought you that in jail //content.invisioncic.com/y282845/emoticons/hilarious.gif.02a037aad04aa96f19982b298a3d70a8.gif:hilariou://content.invisioncic.com/y282845/emoticons/hilarious.gif.02a037aad04aa96f19982b298a3d70a8.gif:hilariou:

and to spell and use proper grammer

 

[basscort2009 5,000+ posts]

THAT SPECIFICALLY STATES330amps OF CURRENT

SO....STUPID....

THIS IS HOW YOU CALCLATE IT.....

FOR EVERY 330 AMPS OF CURRENT YOU WILL LOOSE .05volts

SO IF I AM PULLING SAY 990AMPS. WHICH IS 3 * 330 = .15VOLTS OF LOSS

lolololol yea u got it bud //content.invisioncic.com/y282845/emoticons/rolleyes.gif.c1fef805e9d1464d377451cd5bc18bfb.gif

 

[sub-FATHER 10+ year member]

If we had a 1/2 horsepower motor designed to run on 120 volts, nominal, with a F.L.C. of 9.8 amp as per NEC Table 430-148 , and we lost the 3.6 volts due to a long distance run causing the above calculated voltage drop this motor would now be pulling at the full load amps as follows;

Special Notes: You must change the Amps into Volt-Amps, in order to accomplish the conversion, in the change of voltage, due to the voltage drop calculated above.

9.8 AMPS x 120 VOLTS = 1,176 VOLT AMPS

1,176 VA DIVIDED BY 116.4 VOLTS AT END OF LINE = 10.103 AMPS

Formula to change amps into approximate volt - amps = amps x volts = volt amps

Example of Voltage Drop, Approximate "K" And Exact "K"

Calculating the Resistance of a Conductor = (K) (Approximate K & Exact K)

Approximate K = 12.9 for copper.

Approximate K = 21.2 for aluminum

VD=2xKxDxI/Cm

I= Current

2= round trip loss (you'd use 1x if it was a balanced wye circuit)

KxD/Cm represents the resistance of this wire.

Voltage Drop = 2x Current x Resistance

D=length of wire

K/Cm=resistance per unit length.

K is the unit resistance of copper per circular mil

Cm is circular mills of the conductor. (roughly 2.8 Ohms/kcmil/100 feet DC) This equation is only accurate for DC. You can approximate AC (K is roughly 1.3 for AC), but it varies by cable size due to skin resistance and isn't terribly accurate.

Really, you should also be adjusting it forpower factor and the type of conduit it's in and wire temperature, too, but usually if you have to go into that much detail, you should be upsizing anyhow.

http://www.paigeelectric.com/tools/metric_awg_conversion.aspx

 

[sub-FATHER 10+ year member]

you fail once again...learn to read
shoulda tought you that in jail //content.invisioncic.com/y282845/emoticons/hilarious.gif.02a037aad04aa96f19982b298a3d70a8.gif:hilariou://content.invisioncic.com/y282845/emoticons/hilarious.gif.02a037aad04aa96f19982b298a3d70a8.gif:hilariou:

and to spell and use proper grammer

SORRY.. i also learned to type 110wpm as well.

oops... i forgot one letter.

 

[plugitin 5,000+ posts]

you are an idiot guy

 

[sub-FATHER 10+ year member]

you are an idiot guy

do me a favor and stop being a ********. im all about helping people but your just stuck on being stupid.