[QUOTE]Originally posted by NFRs2000NYC
Not *too* likely.

 

Didn't mean to hijack the thread either... sorry I just figured it was kinda related.

 

good thread...

looks like i missed all the fun...

as a side note, i'll throw in... if you do go above the alternator's surplus capacity, as long as you're peaking over the alternator's output, the battery will fill in the gaps, and recharge w/ whatever is left from the alternator the rest of the time. it's when you're never dropping below the alternators capacity that you'll get into battery problems.

and Evan, i'm not sure if that shortens the life of your alternator or not... (nor do i really feel like thinking about it) but historically, i have gone through a few alternators, and generally never paid for it (always covered under the 3 yr warranty)... so, i can't help but think it takes it's toll on the alternator, as your average alternator probably should last longer than 3 yrs.

personal keyword: evenflo

 

[QUOTE]Originally posted by PJK3
good thread...

 

[engage smart-ass mode]

I am really confused by what I'm reading here, so I'm going to stir things up and ask questions. So far, about the only thing I agree with in this whole thread is NFRs2000NYC's statement that you'll almost never be using your amplifier's "rated watts" in the real world.

I understand the "40% available current" concept, which sounds reasonable, even if it's a bit conservitive. But WTF is the "3" used for? What does it represent? Why do you multiply the alternator's available current (44A) by 3?

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[QUOTE]Originally posted by MacGyver
From the looks of it, I really screwed the pooch on my initial in-my-crazed-head calculations... <sigh>

The two equations any engineer should memorize...P=I*V and V=I*R.

 

Well, Im not an electrical engineer by any means, so I def cant prove you wrong mathematically.....

first of all, to answer your question....what is that 3?

"Since the average music signal requires about 1/3rd of the average power in a test tone, divide by 3 "

---From crutchfields amp install guide.

I however, do not understand where you get your numbers. How did you get 1500watts? 110A x something???

And I didnt say amplifiers dont produce rated power...I said amplifiers arent drawing maximum power 100% of the time....they only draw big power on biiiiiig long hits.

 

Modifry, I think this quote from the BCAE website answers this.

"A power amplifier takes an input signal, usually a preamp level signal, which has both low current and low voltage characteristics, and produces an output which will have higher current and voltage levels. Most amplifiers have a special circuit (switching power supply) to boost the available battery/charging system voltage to a higher voltage. The higher voltage developed in the amplifier's internal switching power supply will allow the audio output voltage swing to be greater. This allows the amplifier to produce more power into the speakers connected to the amplifier's output terminals."

Mathematically I cannot illustrate this without knowing for a specific amp what the increases in voltage and current are but this would explain how you get out more than what you put in. Also, I believe the 50% efficiency is in regards to the fact that in a perfect world the amp would take a 31A input and increase it to a certain output Amp and volt to make a given power, however due to the inefficiencies of the amplifier (loss of power through the creation of heat and the such) the amp would require a larger amount of current from the alternator in order to create that same amount of output power. So in this example it might need closer to 42a input in order to have enough to ramp up after heat loss and the such to make the rated 950w of output power.

This is my understanding of it, and I might have missed some stuff but I think it is the general idea.

Hope this helps a little and doesn't lead to more confusion.

 

I'm not an electrical engineer either, or any other kind of engineer for that matter.

I was agreeing with you that amplifiers "only draw big power on biiiiiig long hits". I guess I didn't make that clear, sorry. I was confused by the "3" because I didn't see an explanation for it.

I agree that music normally only uses about 30% the power capacity of an amplifier, because the amplifier's power capacity is measured using a steady sine wave. Since music is not a continuous tone it "averages" out to much less than steady tone tests.

But that doesn't increase the amount of power available from the alternator, though I can see how you did the math. Since the "1/3 rule" decreases the load on the amplifier and alternator, you chose to triple the alternator capacity as a means of factoring in the "1/3 rule". I would have done the math differently and reduced the amplifier load by dividing it by 3. The result is the same (determining if the alternator can run the amplifier at 1/3 rated capacity), but I think tripling the alternator output capacity is misleading. Somebody's going to see "132 amps" (from your previous calculation) and think they can get that much current out of the alternator.

In our example, a 950 Watt amplifier at full power output would consume 1900 watts (50% efficiency rule), which is more than the alternator's output capacity. But since it will only average 1/3 that, or 633 watts most of the time, you can get away with it.

I got the "1500 watts" right out of the book. Somewhere in the Haynes manual it gives the alternator output in watts, and I think it was 1500. You can reverse-calculate it by taking the 110 amps (your number) and multiply it by 13.5 volts to get 1485 watts, pretty close to my number.

I think we are on the same page (now that I understand the "3"), just describing it differently. If I were to make a "rule of thumb" I would factor in all the "rules" and say "Take your amplifier's rated watts and multiply it by 1.33. If that number is less than the total output capacity of your alternator (in watts) then you're OK."

I base that on the "50% amplifier efficiency" rule, the "you'll only use 1/3 the amplifier's capacity" rule, and the "50% of the alternator's output is available" rule. Note that my rule #3 is where I think Crutchfields's "40% of the alternator's output is available" rule is somewhat conservitive.

.

 

Ok.....now I see where we're at.

So basically, our numbers are pretty much in sync.....

so like we said....900 continuous watts would translate into about 650 or so REAL used watts, so thats where we stand.

So I guess its safe to say, that you can have a 900 watt system in an s2k, since it will create a 650watt load.....

does this make sense?

 

[QUOTE]Originally posted by daedalus