# polar rectangular conversions

Discussion in 'Math' started by gcp, Jan 25, 2009.

1. ### gcp Thread Starter New Member

Jan 25, 2009
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0
Im trying to revive my memory, its been awhile since I had to work with polar and rectangular. I want to add -1@210 deg plus 0@210deg. I converted to rectang and got .866+j.5. when going back to polar I end up with 5@30deg? What am I missing here? I was doing fine working nums up thru 180deg, but after that I am missing something here.

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I think you should have gotten the answer 1@30. The sign of the magnitude has changed from negative to positive and the angle has been decreased by 180 degrees. The change in angle offsets the change in sign.

-1@210 equals 1@30

hgmjr

3. ### gcp Thread Starter New Member

Jan 25, 2009
6
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Your right on it being 1@30, I typed the wrong number. Now how do I get from that 1@30 to knowing it should read -1@210?

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
1@30 is equivalent to -1@210. Adding 180 degrees to the angular component of the polar coordinates is the same as multiplying the magnitude by -1.

hgmjr

5. ### gcp Thread Starter New Member

Jan 25, 2009
6
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Thanks for the help...I thought adding 180 back in had something to do there but I couldnt remember gettting the 1 to neg 1. Heres another that has me stumped. Im adding -.68@200and .69@200 I get 0@0? why the )deg?

6. ### gcp Thread Starter New Member

Jan 25, 2009
6
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make that a 0deg

7. ### hgmjr Moderator

Jan 28, 2005
9,030
214

What do you mean by ")deg"?

hgmjr

8. ### gcp Thread Starter New Member

Jan 25, 2009
6
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Thats suppose to say 0 @ zero deg. My keyboard doesnt type what i want it to...lol

9. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I suspect that the reason you get 0 degrees is due to round-off error.

hgmjr

10. ### gcp Thread Starter New Member

Jan 25, 2009
6
0
Thanks again..now back to the grind.