Impious

Then just wire each subwoofer to a 2ohm load, then wire each sub individually to the amplifier. The amplifier will internally parallel the connections and output 1ohm power, each driver receiving 800w.

Does your monoblock amplifier have one or two pair of speaker connections? What amplifier do you have?

If you have a signal that's 50V peak, run a sine wave and then a square wave at that same peak voltage into a 1ohm load you'll have average power of (50*.707)^2/1 = 1250w with the sine wave but 50^2/1 = 2500w with the square wave. Both signals will be within the rail voltage of the amplifier. The peak voltage of the signal hasn't changed, but the average power of the signal has increased significantly with the square wave. I guess my point is.....even if you take the same amplifier and run the same peak signal voltage for a sine wave and square wave through it.....the square wave will have significantly more average power than the sine wave even though the peak voltage of the signal hasn't changed (and hence both signals are within the rails of the amp). That 1000w amplifier is a 1000w amplifier with a sine wave, not a square wave. Average power will increase with the square wave, which leads to the build up of heat. Granted you may hit current or power supply limitations which will not allow the amplifier to output a full 2500w continuously.....but you'll be at the limits of the amplifier, which will result in an increase in average power with a square wave compared to a sine wave with the same amplifier, same peak voltage, within the rail voltages of the amplifier. If you are going to do a normalized average power comparison, which is what we are trying to compare, you have to reduce the peak voltage of the square wave so that both the sine wave and square wave have the same RMS voltage, which would give both signals equivalent average power. So a 50V peak sine wave would have an RMS voltage of 35.35V. The square wave would need to be reduced to a level of 35.35V for an equal average power comparison. So a square wave wouldn't hit the peak 50V and stay there, because it would peak at 35.35V. Only the sine wave would peak at 50V. This is what Sean, 95Honda and myself are arguing. With the same average power (i.e. the same RMS voltage, not peak voltage), there won't be a significant difference in heat and/or failure time within the driver. We can't just talk in terms of voltage alone, we have to differentiate between peak and RMS signal voltage, because that dramatically affects the average power of the signal. And you aren't going to make this comparison by only flipping the signal generator between sine and square wave, as the peak and RMS voltages of the signal would need to be adjusted to properly compare average power. I was thinking about something else on my morning drive......the square wave assumes essentially an instantaneous change in voltage level. If the signal is 5V peak to peak, the signal has a 10V swing in level with virtually zero change along the y-axis of time. Loudspeakers have inductance and energy storage......I would imagine there has to be some sort of "lag" present in the driver's ability to respond to an instantaneous change in voltage like that. The driver isn't going to be able to respond instantaneously and go from a position of +5V to -5V in almost literally no time. That's almost exactly what inductance describes...the ability of the driver to respond to a change in voltage. So I wonder how much "rounding" of the square wave there is in the actual response of the loudspeaker to the signal, as this would affect (reduce) the amount of time the driver was actually at the plateau of square wave, which would reduce the amount of heat build up you are suggesting occurs. By the time the speaker finally gets up there to the plateau of the square wave, the signal is turning around heading back down in the other direction, so the speaker turns around and follows it back down (obviously overly simplified). I wonder if it would almost "track" the same path as a sine wave due to the inductance induced lag. You're the speaker engineer Nick What do you think?

Right. The value of the graph would be in volts. The coil would track the motion of the graph. If you were wanting to compare average power levels, as we are, you would have to factor the RMS voltage of the sinewave and make that value the voltage of the graph for the squarewave, as I did in my graph. Which would equate to a 10% difference.

I'm no electrical engineer or mathematician, so I very well could be wrong, but I do not believe you are factoring it correctly. Peak and RMS voltages are not calculated the same for a sinewave and squarewave. What you have represented there would be a sinewave and a squarewave of equivalent peak voltage, not RMS voltage (or average power). If you have a 5V peak sinewave (the area under the curve there is peak, not RMS), then you would need to compare it to a 3.535V RMS squarewave to arrive at equivalent average power. (For a sinewave, RMS voltage is Peak voltage * .707, for a squarewave RMS voltage = Peak Voltage. Since average power is Vrms * Irms, we would need to compare RMS voltages not peak voltages for an accurate comparison of normalized average power) http://www.wolframal...rom+x%3D0+to+pi http://www.wolframal...rom+x%3D0+to+pi The difference there is 10%, not 50%, for a sinewave and squarewave at normalized average power. With a 50hz signal, the subwoofer is completing one cycle every 20ms. How much more quickly will a 10% increase in heat ("area under the curve") build in a coil that's being actively cooled via coil movement every 20ms, or 50 times per second? The difference in area under the curve would be the same as comparing a 5V peak sinewave and a 5.5V peak sinewave. If I'm wrong here, please show me where, as it's possible I'm misunderstanding. But I do not believe I am.

You're test is invalid, equipment(not actually subwoofers) and tones(freqeuncy) were wrong. I like that you atleast try it, but you shouldn't us this as a good scientific test, because it wasn't. And subsam(and nick with the pictures) have already explained why, sqaure wave vs sine wave are just physics, the wave is more "powerfull" with a square wave.(in terms that everyone understands) EDIT: what do you mean with the next comment? (That's awesome.. Lol.. ^^^ It obviously wasn't scientific....... ) Yes you tried to be scientific, but it wasn't scientific enough, you need to have the correct enviromnemt.(and a controlled one.) The pictures shows normalized voltage, not normalized power. No one would disagree that a squarewave will deliver significantly more power over time than a sinewave of a normalized voltage. That's not what is at issue. What is at issue is a difference with normalized power. Nick explained his experiment in short detail, and 95Honda is spot on that flipping a switch, without normalizing power, doesn't answer the question. If he's operating the amplifier at 200V peak voltage, then the RMS voltage with a sinewave is ~141V. If you flip a switch to generate a squarewave, that same 200V peak voltage equates to 200V RMS voltage, significantly increasing average power (doubling it, in fact). We all agree, that's bad. But that's not what is the disagreement is over. The disagreement is over normalized power, which I don't believe has actually been settled yet. Nick's stated his opinion, I'm not sure if he's actually tested it as what he described wasn't comparing normalized power (I could be wrong, maybe he has tested it but just didn't describe in detail how).

I live my life a quarter-mile at a time.

Well, I'm bored, so we're going to go through the math anyways. Let's say you have a 1000w amp and two dual 2ohm subwoofers wired to a final 2ohm load. First, let's find out the total voltage and current output of the amplifier. Using basic ohm's law, we find: Voltage = sqrt(voltage * resistance) = sqrt(1000*2) = 44.72V Current = Watts/Volts = 1000/44.72 = 22.36A And, to verify we are right, 44.72 * 22.36 = 1000w. Cool, step 1 complete. Now let's find out out much voltage and current each speaker is receiving. Let's just look at the two drivers on the "whole", forget about individual coils right now. Each driver is 4ohm (since the individual coils are wired in series) and the two drivers are wired in parallel. One thing we know about parallel circuits is that the voltage is the same across the resistors, so we know each driver is going to be seeing 44.72V. To figure out how much current each driver is going to see, we divide the voltage by their resistance; Current = Volt/Resistance = 44.72/4 = 11.18A So how much power is each driver receiving? Simply multiple voltage and current; 44.72V * 11.18A = 500w. Cool, so we know each driver is going to see 500w. Now, how much power is each coil going to receive? Well, in a series circuit, the voltage is "split" between the resistors and current stays the same. So current to each coil is 11.18A. Voltage = Amps * Resistance = 11.18 * 2 = 22.36V And power to each coil is voltage multiplied by current; 22.36 * 11.18 = 250w to each individual coil. So, there it is. Basic Ohms Law. Each coil will receive 250w. Or, since all of the "resistors" (coils) are the same resistance......we can skip all of that headache and just divide the power by the number of coils; 1000/4 = 250w. You would really only need to bust out Ohms Law to figure this out if you had drivers that were different impedances. And then it gets a little more hairy.

The simple answer is that if each coil is the same resistance, then each coil will receive 1/4 of the power output from the amplifier. Or did you want the math behind it?

No, it's not. Nick: Have you managed to determine a higher (quicker?) rate of failure for a clipped signal operating at equivalent average power levels as a non-clipped signal? Obviously a heavily clipped signal of the same RMS voltage as a non-clipped signal is going to cause a driver to fail sooner (if you breach the thermal or mechanical thresholds) as the clipped signal will deliver significantly more power over time. But I've not yet seen a test that determined a quicker failure rate for a clipped signal vs a non-clipped signal at the same average power level.....and I've never personally had enough drivers on hand to intentionally blow stuff up just for shits and giggles

If you are getting significant roof flex your best option is bracing rather than damping. As for rattles......it depends on what's rattling and why. If you are getting rattling from, for example, plastic panels vibrating against each other you would want to isolate them with foam. So, really, we need more info about the problems you are trying to fix.

The funniest part of all of this is that you think you've proven anybody wrong. To each their own. But don't interpret anything you're doing as "proving" anything other than you don't understand how to consider good advice and constructive criticism when it's given.

Does it matters why he wants it to peak at a low freq? I don't know why you are ranting about it. He just sets a goal and tries to get to it. Some people think it's a failiure, i think it's a (expensive) lesson to learn Sometimes some people don't know any better or don't think for themselves, and making them think can inspire their own thoughts, potentially making them realize how dumb of an idea it was. It's been talked about at length with him. He wants it because he wants it. You can't argue reason with an unreasonable person, so we've all given up on that front.

im not buying them to use them im going to buy them to sell them for more money .if i was buying them to use them i would have told you guys more info on install Now look at how much back-and-forth nonsense you would have saved everyone if you would have just followed everyone's advise and stated that from the beginning? How difficult was that to simply say? How hard would that have been to include in your original post? You made this way more difficult than it needed to be for no reason other than you didn't want to take the initial advice and give us more details about your intentions.

First thing you need to realize is that everything is application specific. They might be a decent sub for $140 for 2 drivers. But if the driver doesn't fit your application or goals, then it's a waste of money and the wrong purchase for you to make. So no, it's not a simple question. No, it's not a simple answer. Yes, the details are important. No, we can't tell you whether or not they are worth your purchase as we don't know your application or goals. If you want to just randomly purchase equipment and hope they are a good fit for you, then stop asking us and just start purchasing equipment. If you want us to actually help you select the proper equipment for your application and goals, then please provide more information. Despite the fact it might look like we're being jackasses.....the simple fact is that we are actually trying to help you, but you aren't giving us the necessary information to allow us to do that and when we ask for it, you come back with some nonsense.

If you have the proper crossover ranges, I'd just run them active.

Were these installed in a boat? Why is everything so rusty and corroded? Were any of the components actually bad, or are things just so corroded that you aren't getting good connections? Those things probably aren't worth saving unless you put a lot of work into cleaning out all of the rust. EDIT: And yeah, like Sean said you're missing an inductor not a capacitor.

I was thinking that as well... I think I need to wait for a reply from edesign before I make a decision on that. I have one good crossover so I could purchase a similar capacitor and solder it in place... Any ideas on that part? Should I take some photos of it for you guys to look at? Sorry, my reply was talking about the passive crossovers, not the crossovers on the amplifier. As I said in my 1st reply......Pull the values off the components in the passive crossovers and post a couple pictures. They more than likely are pretty basic/standard crossovers. Shouldn't be too hard to figure them out.

I would guess the crossovers are probably pretty standard. If you can pull the values off of the components within the crossovers and get a picture of the crossovers, probably wouldn't be too difficult to figure out the xover point of the passives. And yes, you could run them active from those two amplifiers as long as the amplifiers offered the necessary crossover ranges.

This question was recently asked. Please see thread HERE (click me)

Link doesn't work. What sub are you using? What are your goals for the system? Hard to know if it fits your application, if we don't know what your application is

Skin effect is also related to frequency. It's generally not a problem until well outside of the audio bandwidth.

Sell it all and start over. If you give us a budget and goal, we can get you headed in the right direction. I don't know what your goals are, but I don't think you're really on your way to achieving that goal with your current production selections. As M5 said, none of that equipment will compliment each other in the manor that they should.

Agreed. When I and many others unload equipment, we have a variety of items for sale at once. It'd drive me insane having to post, update and track 4 different threads in 4 different subforms because I have a variety of things for sale at once. I haven't read past the 1st page, so I'm not sure if this has already been settled or not.

Some things have no rationalization