trod2902

Awesome job man, keep up the hard work!

-Audio Engineer's Reference Book If I'm reading that correctly, the absorbent material would help attentuate the panel resonances and standing waves, and it allows the inside temperature to remain nearly constant. So the reduction of standing waves causes an increase in the propagation of energy, which results in a dQ without the concomitant dT? while the box without poly fill would experience changes in T without changes in Q, caused by dP Also, the sound is slowed because the medium is more compressible than air alone and also more dense? I'm not sure what any of your terms such as dQ and dT are referring to. For this scenario: The polyfill changes the interior "air operation" in the box. Allows for the temp. to stay consistent, instead of rising. Lower temp = lower speed of sound (velocity). I don't know how the temp. change effects how the enclosure acts, sorry. Polyfill also diffuses standing waves (resonance), which could reduce sound coloration, but I don't believe that's the reason why it makes the enclosure's volume more effective. I could be wrong here, but I think it's just another issue. I know in HT, multiple subwoofers can be used in a way to create interference that cancels out standing waves, thus improving the FR. dQ= change in heat dT= change in temp Im just trying to understand this from the things ive learned in my physical biochem class but idk enough about waves to really grasp whats goin on Is it the temp change that has the most effect to cause the slowed wave or the change in medium through which the wave propagates? I see the ds as deltas, meaning change, but didn't know Q stood for heat, and so on. Not sure which. I do know sound travels faster in liquids and solids then air. We'll have to do some more reading. Haha yep, i have some studying to do now though so it will have to wait I have a test tm and really dont give a shit rn, not good...

-Audio Engineer's Reference Book If I'm reading that correctly, the absorbent material would help attentuate the panel resonances and standing waves, and it allows the inside temperature to remain nearly constant. So the reduction of standing waves causes an increase in the propagation of energy, which results in a dQ without the concomitant dT? while the box without poly fill would experience changes in T without changes in Q, caused by dP Also, the sound is slowed because the medium is more compressible than air alone and also more dense? I'm not sure what any of your terms such as dQ and dT are referring to. For this scenario: The polyfill changes the interior "air operation" in the box. Allows for the temp. to stay consistent, instead of rising. Lower temp = lower speed of sound (velocity). I don't know how the temp. change effects how the enclosure acts, sorry. Polyfill also diffuses standing waves (resonance), which could reduce sound coloration, but I don't believe that's the reason why it makes the enclosure's volume more effective. I could be wrong here, but I think it's just another issue. I know in HT, multiple subwoofers can be used in a way to create interference that cancels out standing waves, thus improving the FR. dQ= change in heat dT= change in temp Im just trying to understand this from the things ive learned in my physical biochem class but idk enough about waves to really grasp whats goin on Is it the temp change that has the most effect to cause the slowed wave or the change in medium through which the wave propagates?

I dont think spelling needs to be a Important part of these threads. U onet peepl to spale not more better and hav a ferom fool of trash

-Audio Engineer's Reference Book If I'm reading that correctly, the absorbent material would help attentuate the panel resonances and standing waves, and it allows the inside temperature to remain nearly constant. So the reduction of standing waves causes an increase in the propagation of energy, which results in a dQ without the concomitant dT? while the box without poly fill would experience changes in T without changes in Q, caused by dP Also, the sound is slowed because the medium is more compressible than air alone and also more dense?

29 My dad told me if i made a 30 he would buy me a new guitar amp, well i didnt get the amp lol Also i have the MCAT coming up on April 13th, been preparing for it with practice tests and this thing is a pucking beast.

i started duplicating my tab to see if it posted before i remove the text from the reply box. Its pretty annoying when you think you posted something and it didnt, or when you think you didnt post something and it did.

infuse?

Is it an old optimas optimus? i have an old receiver thats powering my home setup rn

I like the sony amps

quadruple post FTW! Is that a record?

You cant go wrong with either, they are both excellent woofers made in the US and designed for similar applications.

If that is what you listen to then id definitely choose the BL.

This is what I meant about misapplying Ohms Law. In this scenario P=V^2/R and it's variants apply to the output of the amplifier. It has nothing to do with the power input, the power drawn by the amplifier from electrical system. Yes, if we assume two amplifiers have the same efficiency and output the same power, they will draw exactly the same power (both voltage and current) from the electrical system. For a given supply voltage, they have no choice but to draw the same current from the electrical system. P=VI. If they are both drawing the same power from the electrical system (which they will if their power output and efficiency is the same) at the same supply voltage (the voltage of your electrical system), then it is required by physics that their current draw be the same also. If amplifier A is 1kw @ 4ohm @ 80% efficiency and amplifier B is 1kw @ 1ohm @ 80% efficiency, their current draw will be identical. The two amplifiers will output different voltage and current based on Ohms Law (P=V^2/R, I^2R), but their draw from the electrical system will be identical. Yes, this is fully consistent with Ohms Law. You are misapplying Ohms Law and relating the load impedance to the current drawn by the amplifier with the formula P=I^2R, when the two are not related in this way. That's not how that formula is applied. Thats what i was missing

The difference in efficiency is the only thing that impacts current draw (well, that and the supply voltage to the amplifier along with power output). You are misunderstanding the application of Ohms Law. The impedance of the load has nothing at all to do with it other than, for the same amplifier, as impedance of the load decreases so does efficiency (generally speaking). This does not mean that amplifier A operating at 2kw into a 1ohm will be more or less efficient than amplifier B operating at 2kw into a 2ohm or 4ohm load. That completely depends on the respective designs of the amplifier. It has nothing to do with Ohms Law other than Volts * Amps = Power. If amplifier A has a lower efficiency, then for a given supply voltage it will need to draw more current than amp B.....if it has a higher efficiency, then it will draw less current. To make this a little more clear; If amplifier A is outputting 2kw (impedance doesn't matter) and is 60% efficient, then it needs to draw 3333 watts from the electrical system in order to output 2000 watts. If the supply voltage is 12V, then it needs to draw 277.8 amps from the electrical system. If amplifier B is outputting 2kw (impedance doesn't matter) and is 80% efficient, then it needs to draw 2500 watts from the electrical system in order to output 2000 watts. If the supply voltage is 12V, then it needs to draw 208.3 amps from the electrical system. Impedance has nothing to do with it. It's efficiency, power output, and supply voltage. Thanks, i understand why amplifier efficiency plays a huge factor i just dont understand why P=I^2R doesnt really apply Heres where im confused... if two amps are theoretically 100% efficient and supply the same power (say 1kw) but do so at different loads (amp A 1ohm, amp B 4ohm) can they have the same current draw? This would violate ohms law correct? maybe i need to read about amplifiers lol

2kw at 4 ohm would be nice on your electrical Though there is a general correlation between efficiency and load impedance for the same amp, there's no reason to assume those amps running at 2ohm each (strapped into a 4ohm load each amplifier would be operating at 2ohm) would be any more or less efficient than another amplifier operating at a lower impedance. It depends on the design and efficiency of the amplifiers, not just the impedance of the load. i was just saying that regarding ohms law and current draw But difference in efficiency between two different amps can have just as much impact on current draw?

2kw at 4 ohm would be nice on your electrical

My roomate has a pair of beats and my $40 yamaha headphones sound better lmao.

I said it's an average. Its always different based on enclosure, vehicle, woofer(s), and frequency. ya but enclosure, vehicle, and woofers are constants in this example. I see that the 3x rule is an average, but it does deviate somewhat, and i was just wondering if it deviates the same amount for each option when frequency changes.

that is for any sine wave regardless of frequency? Changing frequency wouldnt change the relationship between the two options?

Single [email protected] would make more power then 2 3500@2 strapped. Can you explain why?

I consider people who I have consistent conversations outside of the forum friends. Notice I used the word label People you know where there heart is and your relationship don't require a label. And the existence of those people is the only reason I am at all active in the OT here after the first Kevin debacle about his amps. Well thanks for staying man even though that douche is still here, because your knowledge and directness make you an indispensable part of this forum

Looks like a nice build ahead, also nice work on the past builds