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Full Version: hp49g+ and vectors
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i am writing a survey progarm and want to use vectors so that i can easily change coordinate systems, by going from rectangular to polar it will put my vectors into an azimuth and a distance. problem vectors read 0 going east and turn counterclock wise, i want 0 to be north and turn clockwise?

From a mathematical view point i think if you rotate your vector through 90 degrees about the Z or yaw axis and then about 180 degrees about the x or roll axis you should then get the vector in the required axis system. The result is that the azimuth angle will
be arctan(originalX/originalY) or arctan (newY/newX).

Edited: 6 Apr 2006, 4:53 a.m.

Hi, Chelsey;

I have already written many survey programs to either RPN- or RPL-based calculators, and the same circunstance applies. You see, there is no matter how the calculator counts or positions angles because it doesn't have any reference to quadrants. You just have to worry about how to create these cardinal references and how to identify in which quadrant the coordinate is located. If you enter

`(-200,-200)`
and asks for its argument, the HP49G+ returns
`-135`
If you enter
`(200,200)`
the argumetn now is
`45`
If you consider that either (-200,-200) or (200,200) are coordinates and you have them already 'coded' as north, south, east or west by simply working with their signal, then it doesn't matter how the calculator interprets angles, you just need to work with the returned value as it is. Now, if you enter
`(200,-200)`
and asks for its argument, the HP49G+ returns
`-45`
and if you enter
`(-200,200)`
the returned argument is
`135`
I do not have the listings right here, but I remember that these outputs did not cause me much trouble computing the azymuth.

If you need further info, let us know.

Cheers.

Luiz (Brazil)

Edited: 5 Apr 2006, 4:45 p.m.