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### #1 tuxster

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Posted 02 September 2008 - 06:11 PM

I am new to the Classpad 330 and am having some difficulty entering polar calculations.
example would be impedance calcualtions for single phase circuits:
(3+j7)+(6-j5) = take the answer and divide it by voltage which is expressed as a polar value ie (28kv<38)

This involves quick on the fly Pol to Rec conversions.

I can enter the rectangular values but am having difficulty converting to polar and entering polar values.

Any help would be appreciated.

### #2 tuxster

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Posted 03 September 2008 - 04:28 PM

Anyone help me out here?

Is there another forum that I should be asking?

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Posted 03 September 2008 - 05:34 PM

I am new to the Classpad 330 and am having some difficulty entering polar calculations.
example would be impedance calcualtions for single phase circuits:
(3+j7)+(6-j5) = take the answer and divide it by voltage which is expressed as a polar value ie (28kv<38)

This involves quick on the fly Pol to Rec conversions.

I can enter the rectangular values but am having difficulty converting to polar and entering polar values.

Any help would be appreciated.

I don't know if I can help - but I'd like to figure this out. It's been 12+ years since I've done stuff like this and I'm pretty rusty.

CP seems to distinguish between complex-polar and real-polar values. For instance, if you use the expression comToPol(1+j) you get sqrt(2)e^(pi/4*j) -- whereas if you use the rectangular expression toPol([1,1]) you get [sqrt(2), <pi/4] (in radian mode)

The only way CP converts polar back to rectangular (that I can find) is toRect( ) and that insists on a vector argument, so (as far as I could tell) you can't convert from complex-polar to complex-rectangular.

I don't know if this will give you the type of answer you're looking for, but you could do the following (I'm in Std/Cplx/Rad):
compToPol((3+j7)+(6-j5)) ---> sqrt(85)e^(arctan(2/9)*j)

ans/(28e^(38*pi/180*j)) ---> (9/28 + j*1/14)*e^((-38*pi/180)*j)

from which you can read your result in polar form ( (9/28 + j*1/14), <-38)... though I confess I can't tell if it makes any sense.
If it doesn't, perhaps you can describe the calculations to me more so I can dig deeper. What I've been able to find via the web hasn't been terribly helpful.

### #4 tuxster

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Posted 04 September 2008 - 01:39 AM

I purchased a 330 because it can manipulate a graph real easy so I can take a non linear plot of current vs time for distribution fuse coordination. I have tables which I can lookup the impedance of a cable XL and the real resistance R and use this complex number together with a polar voltage, 44kv @ 23Deg to find the current at a specific spot and to not be over the 50b (low set instantanuous) I to trip the LV breaker.

I can use my HP50G+ for an answer but you do not have the ability to move the graph with the wisp of your pencil or easily plot table values. I had a TI-89 that I purchased which can also enter the values in and multiply on the fly between rectangular and polar values but it went missing, hence the 330 purchase.

In a nutshell I need to work with rectangular values I=V/(r+xl). v is expressed as a ang b theta (polar) number and denominator is in rectangular (a+bi) format.

This is a good calculator, I really do not see why this has to be hard? Casio must have thought this calculator may be used for electrical engineering?

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Posted 04 September 2008 - 07:21 PM

In a nutshell I need to work with rectangular values I=V/(r+xl). v is expressed as a ang b theta (polar) number and denominator is in rectangular (a+bi) format.

This is a good calculator, I really do not see why this has to be hard?

You're right, it shouldn't have to be hard. It seems like it can't do exactly what you're wanting - though I tried to do what you suggested on an HP50G and was irritated because it wouldn't accept my angle in radians (even in radian mode).

Anyhow, I talked it over with a friend and we came up with a (possible) workaround. Take a look at it and see if it helps.

the theory here is that you call I( [44,<23] / ((3+j7) + (6-j5)) )

where I is input in the CP as follows:
define I(z)=re(z[1,1]+z[1,2]*i)+im(z[1,1]+z[1,2]*i)*i

it'll perform the division, remove it from the resulting vector, then separate the values into real and complex parts. And after you define it once you should be good to go from then on.

is it the same as having Casio take care of it for you? Nope... but hopefully it'll at lease solve your problem between now and whenever they release the next update (and we can hope that it'll be addressed in that update).

If, for some reason, that doesn't do what you need I'd be happy to keep trying - just describe the problem in detail. I'm enjoying the challenge!

### #6 tuxster

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Posted 05 September 2008 - 03:39 AM

Well that seems like an interesting fix, although it works it seems a little over kill for what this calculator should be able to do on the fly. I still have the ti-89 emulator and performed a calculation to show you what I am looking for.

http://www.flickr.co...864458/sizes/o/

Let me know if you are able to access the photo.

Thanks,

Mike

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Posted 05 September 2008 - 05:22 PM

I checked out the photo - thanks! it gives me a better idea of exactly what you're typing in.

You know, I think you should be able to use complex-exponential form (r*e^(theta*j)) to do the same thing and that would (I think) solve your problem. But I admit it is possible that I don't fully understand the use of that form. I've also found some results that are confusing me. I'll write more when I have a better idea of what I'm talking about.

in the meantime you might try using compToPol( ) -- that will convert complex-rectangular into polar form (exponential, but you can easily read r and theta). But I think the CP may have a bug in the way it simplifies polar (exponential) back to rectangular... so just be aware.

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Posted 05 September 2008 - 05:49 PM

figured out my problem -

if you're willing to use the polar form r*e^(theta*j) then you should be able to do all the calculations you need. and you can convert back into this polar form by using the function compToPol( ).

BUT - I just figured out that that particular form is only defined for radians and therefore does funny things when you plug in degrees. So, if you use it, you need to convert your degree values to radians.

Sadly, another work-around. but perhaps one you would be somewhat comfortable with...

hopefully, Casio will tweak this in a future software release.

### #9 tuxster

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Posted 06 September 2008 - 12:27 AM

I just bought the 330, do you know if an update is in the works? or maybe Casio going to introduce another Calculator? Haveyou heard of anything.

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Posted 09 September 2008 - 04:37 PM

Sorry, haven't heard any details of a future release. I'd imagine, just given the nature of technology these days, that they'll have to update the software (in the near future) and the hardware (probably will take longer) - but that's true of any company, just to stay competitive. And timing is anybody's guess.

### #11 Behnoud

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Posted 01 October 2008 - 08:10 AM

In most of companies, major releases of new product, take place before Christmas and before college and school new term start.
I don't know if any forthcoming exhibition we have in next few months. Is there any?!

### #12 Guest_Guest_John_*_*

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Posted 10 June 2010 - 02:09 AM

Hello Guys! Do you have any updates ? How can we input ( 56 ∠ 27Â°) * ( 10 ∠ 10Â°) and get correct result?

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