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impedence rise
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<blockquote data-quote="Immacomputer" data-source="post: 2773441" data-attributes="member: 570419"><p>I'm going to break this done for you one last time. First, I will state that you do not really understand as much as you would like to think you do. Second, we can all see this and you're not fooling anybody.</p><p></p><p>Type what you mean next time. If you know what inductance is, don't type this: "i think it was pretty safe to say (without knowing wat inductance is) that..." That is what you typed but you edited to make yourself seem more educated than you really are.</p><p></p><p></p><p>Let us read something I have already typed:</p><p></p><p><em>"In a DC environment, your current is constant (hence the Direct Current) and thus the part of the voice coil that is acting like an inductor is now acting like a short (think of the resistance and inductance of the sub as if they're in series with each other). If the inductance is a short, you can say that its value of resistance is 0."</em></p><p></p><p>Yep, inductance is equal to 0 in DC, just like helotaxi also said.</p><p></p><p></p><p>I never stated this. I DID state multiple times that your amp will never see a load lower than your DCR. Read and understand; don't put words in my mouth.</p><p></p><p>In a DC environment, it will be. This is the context I was talking about. When you're working with a DC source, inductance will be 0 so DCR + 0 = impedance = DCR. Just like 1 + 0 = 1. Read in context and understand.</p><p></p><p>And that is similar to what I said here: <em>"With this flux, we're going to have inductance added to the DCR . . . So we now have the inductance (which will vary over frequency) added with the DC resistance of the coil. This value, over the usable range of the sub while free-air, will be roughly 2 ohms per voice coil."</em></p><p></p><p>The problem is that impedance is going to change because inductance changes over frequency and it will change based on the environment the sub is in. This value of inductance will only increase though and will never become negative in a sub. Also, it's not equal to the nominal impedance as you will read here later.</p><p></p><p>Nominal does not equal lowest... sorry. That value is also taken while free-air. When you load down the sub in an enclosure, your nominal impedance will be different from it's rated value; this new value will almost always be higher.</p><p></p><p></p><p>Actually, that "i" value is used to describe the phase angle of the impedance when in polar form. "i" is used as the 90* phase angle portion of the impedance. Like I have said before, impedance is the summation of the purely resistive load and purely inductive load. This is not a normal sum like you would think it to be though. Because we have an imaginary value, we switch to a polar coordinate system instead of our standard geometric system. This form of addition is done by taking the square root of the sum of each value squared. This will give you the magnitude of the impedance. The phase angle of the impedance is going to be the tan^1 of the imaginary value over the resistive (real) value.</p><p></p><p>Example:</p><p></p><p>Something has a DCR of 2ohms and an inductance of 2i ohms at some random frequency. To find the actual impedance, we solve this in our calculator: sqrt[2^2 (real value) + 2^2 (imaginary value)]. This is equal to the sqrt(8) which equals ~2.83ohms. This is the magnitude or total value of resistance. Now, the phase angle is found by taking the tan^-1(2/2) which is equal to 45*.</p><p></p><p></p><p>If your amp was worth a ****, it will handle a load slightly lower than it is rated for. I have a shitty profile amp from the early 90s that can handle a 2 ohm load bridged for hours and it is fine. It sounds like you have wired your amp to .5 ohms. //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif</p><p></p><p>I'm not taking it as arguing but rather you trying to defend your ignorance. Trust me, you're failing hard at this. You're young so I wouldn't expect you to know this stuff. It's honorable that you're trying but there comes a point when you are not fooling anybody and you just need to stop. You are way past this point and you're too far over your head.</p><p></p><p>Oh, and when you sum DCR and inductance, you don't get the nominal impedance but the magnitude and phase of the impedance at that frequency. There IS a difference but again, you don't understand it.</p></blockquote><p></p>
[QUOTE="Immacomputer, post: 2773441, member: 570419"] I'm going to break this done for you one last time. First, I will state that you do not really understand as much as you would like to think you do. Second, we can all see this and you're not fooling anybody. Type what you mean next time. If you know what inductance is, don't type this: "i think it was pretty safe to say (without knowing wat inductance is) that..." That is what you typed but you edited to make yourself seem more educated than you really are. Let us read something I have already typed: [I]"In a DC environment, your current is constant (hence the Direct Current) and thus the part of the voice coil that is acting like an inductor is now acting like a short (think of the resistance and inductance of the sub as if they're in series with each other). If the inductance is a short, you can say that its value of resistance is 0."[/I] Yep, inductance is equal to 0 in DC, just like helotaxi also said. I never stated this. I DID state multiple times that your amp will never see a load lower than your DCR. Read and understand; don't put words in my mouth. In a DC environment, it will be. This is the context I was talking about. When you're working with a DC source, inductance will be 0 so DCR + 0 = impedance = DCR. Just like 1 + 0 = 1. Read in context and understand. And that is similar to what I said here: [I]"With this flux, we're going to have inductance added to the DCR . . . So we now have the inductance (which will vary over frequency) added with the DC resistance of the coil. This value, over the usable range of the sub while free-air, will be roughly 2 ohms per voice coil."[/I] The problem is that impedance is going to change because inductance changes over frequency and it will change based on the environment the sub is in. This value of inductance will only increase though and will never become negative in a sub. Also, it's not equal to the nominal impedance as you will read here later. Nominal does not equal lowest... sorry. That value is also taken while free-air. When you load down the sub in an enclosure, your nominal impedance will be different from it's rated value; this new value will almost always be higher. Actually, that "i" value is used to describe the phase angle of the impedance when in polar form. "i" is used as the 90* phase angle portion of the impedance. Like I have said before, impedance is the summation of the purely resistive load and purely inductive load. This is not a normal sum like you would think it to be though. Because we have an imaginary value, we switch to a polar coordinate system instead of our standard geometric system. This form of addition is done by taking the square root of the sum of each value squared. This will give you the magnitude of the impedance. The phase angle of the impedance is going to be the tan^1 of the imaginary value over the resistive (real) value. Example: Something has a DCR of 2ohms and an inductance of 2i ohms at some random frequency. To find the actual impedance, we solve this in our calculator: sqrt[2^2 (real value) + 2^2 (imaginary value)]. This is equal to the sqrt(8) which equals ~2.83ohms. This is the magnitude or total value of resistance. Now, the phase angle is found by taking the tan^-1(2/2) which is equal to 45*. If your amp was worth a ****, it will handle a load slightly lower than it is rated for. I have a shitty profile amp from the early 90s that can handle a 2 ohm load bridged for hours and it is fine. It sounds like you have wired your amp to .5 ohms. [IMG]//content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif[/IMG] I'm not taking it as arguing but rather you trying to defend your ignorance. Trust me, you're failing hard at this. You're young so I wouldn't expect you to know this stuff. It's honorable that you're trying but there comes a point when you are not fooling anybody and you just need to stop. You are way past this point and you're too far over your head. Oh, and when you sum DCR and inductance, you don't get the nominal impedance but the magnitude and phase of the impedance at that frequency. There IS a difference but again, you don't understand it. [/QUOTE]
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