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I'm not sure who the "we" is in your statement, but it certainly doesn't
include people who understand the concepts and mathematics involved. The problem with your approach is that there *is* such as thing as RMS power (as I described in an earlier posting), and it is *not* the amount that gives the same heating effect as the same amount of DC power. So while you may think of RMS power in this way, your thinking is wrong. I can calculate my RMS speed in going from point A to point B. But it isn't equal to the distance I went divided by the time it took to get there (unless, of course, I went the whole way at a constant speed) -- for exactly the same reason that the RMS power isn't the equivalent heating power. You can think of the RMS speed as being the distance divided by the time, but it isn't, and your thinking doesn't make it so. And that also doesn't make the RMS speed a useful value. Roy Lewallen, W7EL Paul Burridge wrote: I guess you're right, but when we speak of RMS power (sic) what we actually envisage is the amount of AC power dissipated in a resistive load that gives rise to the same heating effect as the equivalent amount of DC power. Sorry if that's not well put, but you no doubt get my drift... So it may be a misnomer, but that doesn't make it useless. |
While the discussion is interesting, one needs to consider the purpose. For a
pure sine wave, certain relationships are stable and hold true. For a complex waveform, like voice, the relationship between peak power and average/RMS power not a straightforward relationship. 73s, Evan |
Bill Turner wrote:
On Mon, 18 Oct 2004 11:15:43 -0700, Roy Lewallen wrote: The problem with your approach is that there *is* such as thing as RMS power (as I described in an earlier posting), and it is *not* the amount that gives the same heating effect as the same amount of DC power. So while you may think of RMS power in this way, your thinking is wrong. __________________________________________________ _______ Let me ask one more question and then I'll shut up. Actually, three. Suppose you have a 100 ohm resistor and you apply a 100 VRMS sine wave to it. Will 1 amp of RMS current flow? I would think yes, but after the previous discussions, I'm not taking anything for granted. :-) Yes. So if you have 100 VRMS and 1 ARMS, how much power is the resistor dissipating? Just pure power in terms of heat generation, forget whether it's RMS or anything else. How many watts? 100 watts. This is the average power. (The RMS power is about 122.5 watts, if I did the integration right -- corrections are welcome. RMS power is calculated from the instantaneous power exactly like RMS voltage or current is calculated from instantaneous voltage or current.) And now the real question: How much DC voltage would you have to apply to the resistor to get exactly the same power dissipation? 100 volts. The dissipation would be 100 watts. Breathlessly awaiting, I would hope that any Technician or higher class amateur could immediately answer those questions, no waiting required. The problem people seem to be having is the feeling (or, I'm afraid in some cases, certainty) that an RMS value of a quantity times an RMS value of another quantity HAS TO BE the RMS value of the product of the quantities. But it isn't. Here's something you can check on your calculator: Take sin(5 degrees) times sin(10 degrees). Is the product of the two equal to sin(50 degrees)? While the product assumption works for some functions, such as square and square root for example, it doesn't work for others, like RMS and sine. Unfortunately, if you (any reader, not Bill personally) aren't able to follow the mathematics involved in calculating RMS and average values (which I posted in brief form a short while ago), this is bound to remain something of a mystery to you, and it looks like you're left with four choices: 1. Hang tightly to a mistaken idea about the meaning of RMS and average, and ignore or resist any explanation that contradicts them. This will include explanations in any textbook. 2. Learn enough math to understand the definitions of average and RMS, which should make the explanations understandable. 3. Assume that I and the textbook authors know what we're talking about and believe us, without really understanding why. 4. Find a textbook oriented toward people without a very strong math background, which hopefully can explain it in terms you can understand. The fourth choice is probably the best for people who are really interested in learning and can't devote the time necessary to learn the math. The main problem with the first choice is that you'll be likely to pass the misconceptions on to others. But the choice is yours. Roy Lewallen, W7EL |
Let me add one more comment that will hopefully help in understanding this.
I think I might have identified another misconception that might be contributing to the confusion. That misconception is that the RMS value of a waveform is the "equivalent" or "heating" value. This leads to the mistaken notion that the RMS value of the power must be the "equivalent" or "heating" value. Voltage and current, by themselves, don't do any heating. Heating takes power. And the amount of heating is the determined by the average power. If two different power waveforms have the same average value, they'll create the same amount of heat and otherwise do the same amount of work, even if their RMS values are different. So what's the whole thing about RMS? The only importance of RMS is that when you multiply the RMS values of two waveforms together (such as V * V, I * I, or V * I), you get the average of the product of the two. That is, RMS(i) * RMS(v) = Avg(i * v), where i and v are the instantaneous values of the current and voltage, and i * v is the instantaneous power. Likewise, RMS(v) * RMS(v) / R = Avg(v*v/R) and RMS(i) * RMS(i) * R = Avg(i*i*R). In fact, RMS(x) * RMS(y) = Avg(x * y) where x and y can be any periodic waveforms or quantities. (I derived these mathematically in an earlier posting.) So we calculate the RMS values of voltage and current only so we can use them to calculate the average power -- not because the RMS value of any waveform is its "equivalent" or "heating" value -- which it isn't. Roy Lewallen, W7EL |
Bill Turner wrote:
On Tue, 19 Oct 2004 01:39:57 -0700, Roy Lewallen wrote: Voltage and current, by themselves, don't do any heating. __________________________________________________ _______ I don't follow you on this. Voltage can be present without any current, but current can only flow if voltage is present, so what is meant by "current by itself"? -- Bill W6WRT An electron beam in a vacuum can flow current without any voltage drop, as can a superconductor. Granted, these are not everyday occurrences, but they _do_ occur. The real point that Roy is talking about is that whenever a voltage drop occurs simultaneously with a current there is power being consumed -- and if it isn't coming out of the system as some other form of power then it's being dissipated as heat. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com |
On Tue, 19 Oct 2004 06:32:55 -0700, Bill Turner
wrote: Well. The answer you gave is exactly the answer I would have given, but you say my answer is wrong. I understand about deriving the RMS power from the instantaneous power in the same way that RMS voltage or current is derived, I just don't accept the definition. It's been taught the other way all my life, even though you say it's incorrect. I see your point, I just don't accept it. I agree. On the one had we've had a pillar we'd always accepted knocked away by Roy; on the other hand, it was Roy who knocked it away. Had it been anyone else I'd have dismissed them as a nutter. I will QRT for now, but thanks for taking the time to explain. I mean that sincerely and I do respect your point of view. As I'm sure we all do. If this is a misconception it must be an extremely widespread one. I've found a reference to RMS power and how it's calculated in 'Practical Radio Frequency Test & Measurement' (Newnes) and have posted the relevant page he http://www.burridge8333.fsbusiness.co.uk/cvcvcv.gif If we are wrong, it appears we're not alone... -- "What is now proved was once only imagin'd." - William Blake, 1793. |
Sure, current can be present without voltage (in a short circuit) just
as easily as voltage can be present without current (in an open circuit). But what I meant was that the value of voltage doesn't determine heating, nor does the value of current. It's the value of power that does. Roy Lewallen, W7EL Bill Turner wrote: On Tue, 19 Oct 2004 01:39:57 -0700, Roy Lewallen wrote: Voltage and current, by themselves, don't do any heating. __________________________________________________ _______ I don't follow you on this. Voltage can be present without any current, but current can only flow if voltage is present, so what is meant by "current by itself"? -- Bill W6WRT |
Bill Turner wrote:
On Tue, 19 Oct 2004 00:02:50 -0700, Roy Lewallen wrote: And now the real question: How much DC voltage would you have to apply to the resistor to get exactly the same power dissipation? 100 volts. The dissipation would be 100 watts. __________________________________________________ _______ Well. The answer you gave is exactly the answer I would have given, but you say my answer is wrong. Where have I said that's wrong? Of course it's not. I understand about deriving the RMS power from the instantaneous power in the same way that RMS voltage or current is derived, I just don't accept the definition. It's been taught the other way all my life, even though you say it's incorrect. I see your point, I just don't accept it. I have no idea who taught that to you, since the definition of RMS is in its very name (the square Root of the Mean of the Square of the function). It's your choice to ignore the accepted definition. I can only hope you don't teach your mistaken idea to others, who will then someday say the same thing. I will QRT for now, but thanks for taking the time to explain. I mean that sincerely and I do respect your point of view. You're welcome. The only reason I've taken the time for these postings is in the hope that it will help people understand and learn. Even if it hasn't worked for you, I hope some other readers have benefitted. Roy Lewallen, W7EL |
I really appreciate the compliment, and will do my best to try and
deserve it. Please look very carefully at the diagram at the URL you've posted, and notice that it's a voltage waveform (see the labeling of the vertical axis). Then read the text very carefully. Neither the diagram nor the text contradict what I've said. If you think it does, post the reason why, and I'll try to clear it up. Roy Lewallen, W7EL Paul Burridge wrote: On Tue, 19 Oct 2004 06:32:55 -0700, Bill Turner wrote: Well. The answer you gave is exactly the answer I would have given, but you say my answer is wrong. I understand about deriving the RMS power from the instantaneous power in the same way that RMS voltage or current is derived, I just don't accept the definition. It's been taught the other way all my life, even though you say it's incorrect. I see your point, I just don't accept it. I agree. On the one had we've had a pillar we'd always accepted knocked away by Roy; on the other hand, it was Roy who knocked it away. Had it been anyone else I'd have dismissed them as a nutter. I will QRT for now, but thanks for taking the time to explain. I mean that sincerely and I do respect your point of view. As I'm sure we all do. If this is a misconception it must be an extremely widespread one. I've found a reference to RMS power and how it's calculated in 'Practical Radio Frequency Test & Measurement' (Newnes) and have posted the relevant page he http://www.burridge8333.fsbusiness.co.uk/cvcvcv.gif If we are wrong, it appears we're not alone... |
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