320 watts

[charlie b]

Re: 320 watts

Originally posted by physis: Rather than looking it up, what is that Charlie?

I can?t use math symbols with the Forum?s word processor. So I?ll try just using words.

The indefinite integral of the fourth power of the cosine of ?a? times ?t? (where t = time) is given by

(3/8) times ?t? Plus (1/4a) times the sine of (2 a t) Plus (1/32a) times sine of (4 a t)

When you evaluate the definite integral from 0 to 2pi, the only thing that remains is (3/8) times 2pi. When you are taking the average of this function over that time interval, you have to divide by the time interval. You are now left with just the (3/8). Finally, since I was calculating the RMS value, the last step is to extract the square root. So if you divide 3 by 8, and then take the square root, you get the 61% that I had cited earlier.

 

[allenwayne]

Re: 320 watts

Remeber the days when the industry boasted quadrophonic and there was an 8 track tape to serve it I still have an old sony 100 watt with a pair of cerwin vega speakers that blow away these new compact units.Nut I did buy a sony surround sound system a few months ago for my addition and I have to admit the 5:1 tiny speakers and the sub woffer sound great and for the money I spent it was well worth it.They bill it as 800 watts but no way if I can only get my old sony to spread the sound like this unit does I`d be so happy

 

[rattus]

Re: 320 watts

Charlie B.,

For "RMS Power" to make any sense, it has to have an application. So have you or anyone else ever computed and used RMS Power? And, how did you use it?

How about an example and/or a reference?

True, the term is sometimes mistakenly used to indicate average power computed from RMS values, but that does not count.

Rattus

 

[physis]

Re: 320 watts

Well Charlie, I did some learning about calculus and I see that indefinate integrals are just regular stuff.

When I figure out how to get functions and intergals to give me something that represents the interval I think I'll be able to operate these over done number lines.

My vote on RMS power is, it doesn't describe anything in the real world. One of the benefits of using RMS values for current and voltage is that it gives you the average or mean power in the same way as P=VI for DC. It's already the actual power under the curve.

 

[physis]

Re: 320 watts

I'm confused by something.

You guys are using intgral calculus to define the area under a sine wave.

You define the function, and evaluate for an interval.

I think I mostly get that part. You charted a curve or some curves.

So now you have the data that you basically already knew. The whole point is the integration. It seems like what I'm reading is describing the integration with mathmatical equations and no English. It could take years to resolve all those equations just to see what they say.

I'm getting the impression that you use predefined integrals to integrate the functions?

And then I'm curious why you would go through the process of defining the area under the curve with integral calculus and omit the integration. Isn't that the same thing as just multipling the peak of the sine by .707? In fact, that's more desciptive because you at least are using a factor.

I might be just not getting this cause I'm sort of at a loss when I get to the integration.

 

[coulter]

Re: 320 watts

Charlie

Again, I appoligize I'm not able to get back sooner. This damned working for a pay check is really taking up a lot of time :roll:

Originally posted by charlie b:
... But you are getting ?Average Power? confused with ?RMS Power,? at least as it pertains to my calculations. ...

Nope, not a bit.

I clearly understand the "Average" mathematical operation as it pertains to evaluating a periodic function.

I clearly understand the "RMS" mathematical calculation as it pertains to evaluating a periodic function.

I clearly understand the difference between the two.

This should have been clear from my thesis. But, it was long, and a bear to read. I?ll fix that this time.

Originally posted by charlie b:
... Taking the integral of this function over the period of ?0? to ?w,? and then dividing by ?w,? shows us that the average value of the ?power as a function of time? is indeed, as you calculated, half the product of Vm and Im.

Yes, I agree with you, that's absolutely true, as I stated in my thesis.

Originally posted by charlie b:
... But to calculate the RMS, you first square the function P(t)

P^2(t) = (Vm Im)^2 cos ^4(wt)

... Taking the integral of this function over the period of ?0? to ?w,? and then dividing by ?w,? and finally extracting the square root, shows us that the RMS value of power is indeed, as I had calculated, about 61% of the product of Vm and Im.

Yes, I agree with you, that's absolutely true, as I stated in my thesis.


Now to the heart on my point:

As accurately stated by rattus, in far better words that I, the RMS calculation applied to a power function has nothing to do with the power transfered to a load.

carl

 

[coulter]

Re: 320 watts

charlie b

rattus did a great job of explaining why I don't think the term RMS power means anything.

Originally posted by rattus
... I would argue that the term "RMS power" is undefined and meaningless. RMS currents and/or voltages are used to compute average power. Furthermore, I am unaware of any application of the RMS formula to anything other than voltage or current even if it is mathematically possible.

... For "RMS Power" to make any sense, it has to have an application. So have you or anyone else ever computed and used RMS Power? And, how did you use it?

How about an example and/or a reference?

True, the term is sometimes mistakenly used to indicate average power computed from RMS values, but that does not count.

Rattus is dead on target. Sure, you can apply an RMS calculation to a sine function power equation, the number you get just doesn't mean anything - that number doesn't have anything to do with the power transferred to a load.

For example, using your equation for power:

Originally posted by charlie b:
Let V(t) = Vm cos ( wt)
Let I(t) = Im cos ( wt - a)
Then P(t) = V(t) I(t) = Vm Im cos ( wt) cos ( wt - a)

Do an RMS calculation for the case of a pure inductive load. a = 90 deg (or pi/4 if you prefer)

I did the math in my thesis - Alas, again it is buried in the body of the text.

An RMS power calc gives the answer of .707VI. And that is the wrong answer for the power transfered to the load.

As you already know, for an inductive load, the voltage 90 deg out of phase with the current, and there is no power transfered to the load. Hummm, now what?

Well, do an average power calculation for the power function. The answer is "Zero". And amazingly, that matches the power transferred to an inductive load.

Hey, these two integrations is ought to be a piece of cake for anybody that can integrate a sin^4


carl

 

[coulter]

Re: 320 watts

charlie b

If you are interested in checking my math, there is an easier way that ingegrating a sin^4.

I gave the two trig identies (again in a long boring thesis) that resolve the sin^2 to a cos(2a) term. You end up with a cos(2a) term and a cos(4a) term. And both evaluate to zero.

Be okay with me if you did check, I've been wrong more that once in my life.

carl

 

[physis]

Re: 320 watts

Al right Carl. It's not like you've discovered an as yet unknown element. I personally haven't given this subject a moment's thought until this thread. It isn't hard to find commentary like this on the internet, anyone have that? I have that.

The History Of How It All Went Awry
Almost every manufacturer, distributor, advertiser in the world involved in selling domestic audio equipment - and more than a few hi-fi magazines! - uses the term 'watts RMS' (incorrectly, as explained above). How did this go wrong?
What happened is that they started to use it as a short hand method of saying that the amplifier's output conformed with a now-defunct US amplifier standard known as IHF A202, which was introduced in 1978.

The idea was that the words 'watts RMS' would serve to show the continuous average power output of an amplifier had been measured correctly according to IHF A202. That is, using the correct test signal (a sine wave), the correct period of time for measurement (more than five minutes), a properly calibrated, true RMS-reading voltmeter with an accuracy of better than 1% of reading, without exceeding a specific level of distortion (0.1%) into a defined load (usually 8-ohms) with the amplifier first having been pre-conditioned by means of driving all channels simultaneously with a 1kHz sinusoidal signal to a nominal power output into the rated load equal to 33% of the rated power output for at least hour (or more if protective circuitry interfered with continuous operation).

This was misguided thinking even at the time, and it's now time to fix it!

I certainly can't say that you don't handle your math well, because you do.

But then I look at this whole discussion and at least I'll admit that I hadn't really noticed that RMS power is essentially legally implemented disinformation. It wasn't until I started checking the math that I realized it.

I don't think it's inteligent to pretend that Charlie doesn't know what he's talking about. I'm pretty sure he gets it.

What would be cool would be to help me understand integration. If you want to point fingers, I'm accusing you of omitting the integration in your definition of the power function.



And I know I don't know what I'm talking about.

 

[coulter]

Re: 320 watts

Originally posted by physis:
All right Carl. It's not like you've discovered an as yet unknown element. ...

...I don't think it's inteligent to pretend that Charlie doesn't know what he's talking about. I'm pretty sure he gets it ...

Sam -

You sure slapped me back into place. I can deal with charlie b telling me my application specific technical knowledge and therefore my reasoning is flawed. That I can research and confirm or deny. But you are telling me I am being unnecessarily mean. And that's different. No amount of research can fix that. I'll take my whipping and be more careful in the future.

Originally posted by physis:
...What would be cool would be to help me understand integration. ...

I don't know that I am up to a class on differential and integral calculus. There are a lot of concepts where I understand the application, but not necessarily the derivation. I wouldn't be doing you any favors promoting my misconceptions. Which if I knew which ones were the misconceptions, I wouldn't promote them - but, alas, I don't always.

Originally posted by physis:
... And I know I don't know what I'm talking about.

I don't know that. You do as good or better than any of us on grasping at new concepts.

carl

 

[coulter]

Re: 320 watts

Originally posted by physis:

The History Of How It All Went Awry

Thanks Sam -

That's what I was looking for from the start. I figured something like that had to be out there. I just didn't know what it was.

carl

edited to remove excess verbage. I gotta quit doing that.

[ May 22, 2005, 05:34 PM: Message edited by: coulter ]

 

[physis]

Re: 320 watts

Carl, I was just jealous that Charlie was getting all the attention.

 

[physis]

Re: 320 watts

But you are telling me I am being unnecessarily mean. And that's different. No amount of research can fix that. I'll take my whipping and be more careful in the future.

Well, I think I might have interfered with some stuff that I didn't see was an issue.

There's no reason for you to pretend that you're somehow subordinant to me Carl.

Do I really have to quote Rodney King?

 

[rattus]

Re: 320 watts

The Effective Value Fudge Factor for Sam

To compute the effective value of a sinusoidal voltage or current, we proceed as follows:

First, understand that we need the average of the voltage or current squared over one period. To do this, we perform a definite integral of the function over one period then divide by the period.

Now let

v^2(wt) = Vp^2*cos^2(wt)

Integration of trig functions is messy, so we take the result from the handbook.

Int[Vp^2*cos^2(wt)d(wt)]= Vp^2*[wt/2 + sin(wt)*cos(wt)/2]

= Vp^2*wT/2 (value at wt = wT ? value at wt = 0)

Now divide by the period (radians) to obtain the average value of v^2(wt)

Veff^2 = Vp^2*wT/2wT = Vp^2/2

Now take the square root,

Veff = Vrms = Vp/sqrt(2) = 0.707Vp

That is all there is to it Sam.

 

[physis]

Re: 320 watts

That's all there is to it huh.

 

[charlie b]

Re: 320 watts

I have been away camping over the weekend, and spent most of yesterday drying everything out and putting it away. So I have missed a bit of this discussion.

Coulter: I do not take issue with any of your math. I think we both know how to do these analyses, and know how to interpret their results.

Rattus: Your recent discussion of ?effective value fudge factor? is right on the money.

Coulter and Rattus: You are both right in asserting that the calculation of an RMS value for power does not have a useful meaning in the electrical industry, and in particular it is not related to the measurable or ?effective? value of power delivered to a load. I think if you look back at my posts, you will not see any statement of mine that contradicts you.

Originally posted by coulter: charlie b: If you are interested in checking my math, there is an easier way that ingegrating a sin^4. I gave the two trig identies (again in a long boring thesis) that resolve the sin^2 to a cos(2a) term. You end up with a cos(2a) term and a cos(4a) term. And both evaluate to zero. Be okay with me if you did check, I've been wrong more that once in my life.

I knew that I could have resolved the cos^2 into first order terms, before calculating the RMS value of power. But if I had chosen that route, I would have ended up with integration of a set of three terms of first and second order. That is because I would have been squaring a set of the two terms that you list. Looking up the integral of cos^4 looked like a less-complicated path to take.

 

[physis]

Re: 320 watts

Let me start here Rattus:

v^2(wt) = Vp^2*cos^2(wt)

π = pi (as shameful as it is)

Using Vp=169 @ 60Hz.

ω = 2π?
t = 1/60 (for one period)
ωt = 2π (for one period)

I'm getting v = 59.23

these exponential trig functiona are new to me.

cos?(ωt) I must be doing incorrectly. I looked into it and what I found said:

cos?(ωt) = cos (ωt)? Since it could just as easily be written that way, and it comes out wrong. I figure that's not the correct operation.

So let me verify the variables:

ω = 2π?
t = 1/60 (for one period)
ωt = 2π (for one period)

What is T
What is d

You know Rattus from one day to the next how easy or interesting this stuff is can change dramatically. This is very much work today.

These are math characters you can copy from here if it makes notation easier. ?ωπ?√

And also, Thanks Rattus.

 

[rattus]

Re: 320 watts

Sam, don't bother plugging in real numbers; that serves no purpose and makes the problem much messier. Just go through the exercise using the general terms because the result will be the same for any peak voltage, any frequency.

d(wt) is the differential of the variable, (wt). It amounts, in this case, to an infinitesimally small slice of angle.

T is the period in seconds, so whatever the frequency, wT = 360 degrees = 2pi radians.

But before you try again, try this little trick:

Averaging a Function with Integral Calculus?for Sam

Sam, in your spare time perform this little mind exercise. A spreadsheet is handy for the calculations.

1) Plot sin^2(wt) at 5, 15, 25, 35, 45, 55, 65, 75, and 85 degrees..

2) Now draw vertical lines at 10, 20, 30, 40, 50, 60, 70, 80, & 90 degrees.

3) You should now see 9 almost rectangles each with a base of 9 degrees. The 9 degrees corresponds, sort of, to d(wt). We will call it delta(wt).

4) Now compute the area of each almost rectangle by multiplying the values found in the first step by delta(wt). Don?t worry that this is not an actual area.

5) Now add all these areas and divide by 90 degrees. This is the almost average of the sine squared.

6) Now take the square root of the above result. Should be about 0.707.

7) Now cut the interval size in half and repeat the exercise.

8) Now repeat the last step again, and again, and again, until delta(wt) is zero. Then the result should be exactly sqrt(2)/2.

Obviously, we can?t perform the last step, but with integral calculus we can. That is all there is to it.

P.S. Note also that we needed only 90 degrees to compute the RMS fudge factor due to the symmetry of the sine squared function, although with calculus we could just as easily integrate over 360 degrees.

 

[physis]

Re: 320 watts

Geeze. It looks like I have some work to do.

(wt) isn't (wt)?

I know Newton and the other guy went to a lot of trouble to come up a with a better notation.

But it really sucks if everybody's settled on conflicting notaions.

So anyway, appearantly that's one of my problems with seeing the integration. wt has a real value but it's also is being used to describe an infinitesimal sample.

So I think I'll need to identify the difference between the two identical notations.

Before you bother to get back to me with this I haven't really looked over your whole post, and basically, instead of reading it I'm being sort of frustrated.

Like I said before too, I'm having a bad math day.

I'll try to identify what I think my problems are and see if I can organize them.

Stupid math!

 

[rattus]

Re: 320 watts

Sam,

(wt) is simply 2*pi*f*t.

d(wt) is an infinitely small slice of (wt).

You must realize that learning calculus in the classroom is near impossible. Learning calculus through this forum is infinitely more difficult unless you are a _______ genius like Forrest Gump.

Rattus

[ May 24, 2005, 12:58 PM: Message edited by: rattus ]