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Math Help - Spherical Geometry Azimuth understanding

  1. #1
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    Spherical Geometry Azimuth understanding

    Hi,
    I need to calculate a GPS azimuth under the following conditions (diagram below). What I have is a known GPS location and bearing (azimuth of 145.56°). At this known position and orientation I also know that my field of view is centred on this azimuth and is 12° in total i.e. 6° of view either side of this azimuth. I want to calculate the azimuth if I changed my bearing to travel along the extremes of this field of view. What I have been doing is a really simple calculation of 145+6 and 145-6 to give me my requirements, however I am not at all confident that it is this simple. I think both degree systems being calculated may not be compatible as the azimuth is spherically related and the field of view if planar. Any advice, direction, help would be greatly appreciated.



    Thanks,

    Paul9038
    Last edited by paul9038; June 25th 2008 at 08:33 AM. Reason: Wrong Thread should be in College/University Maths Help
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  2. #2
    MHF Contributor ebaines's Avatar
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    Sorry - I'm not quite understanding your question. Are you asking whether the azimuth of an object in the sky changes if you change your location on the earth's surface? In general the answer is yes, but practically speaking it's only significant if you move a sufficient distance along the earth's surface, so that the apparent position of the object in the sky is changed. Consider this - the azimuth of the sun is the same for you and your neighbor across the street, but is clearly significantly different for someone several miles north or south of your location. If that's what you are trying to do here, you have quite a problem in spherical geometry. If you do a google search you should be able to come up with some formulas that convert Altitude and Azimuth to Right Ascension and Declination (akin to longitude an latitude) - this would help you solve your particular problem.
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  3. #3
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    Firstly, thanks for your reply and, secondly, for all users, sorry for the double post but I meant to post in the college thread and not the High School one. I tried deleting or moving my mistake but failed.

    To clarify, (or try to clarify) based on the sample image. I am travelling along a road with a GPS unit which is generating a gps track from which my azimuth is calculated as I travel (heavy black line with azimuth 145.56°), I know that the human Field of View is almost 180°, but for the sample image lets say its 12°. If I suddenly stop following the known azimuth (heavy black line) but move in a new direction of travel that is adjusted to either the left or right of this azimuth by the extremes of my field of view i.e. 6° either side, can I calculate my new azimuth. This will mean that either of the thin black lines will be my new direction of travel, but I want to calculate the azimuth of this new travel line. If I change my direction of travel to the edge of my original field of view, i.e. 6° to the left on the image, will my new azimuth be 145.56° - 6° = 139.56°?
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  4. #4
    MHF Contributor ebaines's Avatar
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    When you say your azimuth is 145.56 degrees, do you mean that you are traveling at a bearing of 145.56 from true north? I originally interpreted your use of the term "azimuth" to mean that the GPS satellite is positioned in the sky at 145.56 degrees. But it sounds like what you want to know is this: if you travel initially at a particular bearing, and then turn left or right 6 degrees, is your new bearing simply the addition (or subtraction) of 6 degrees? The answer to that is yes - it's that simple.
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  5. #5
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    The azimuth parameter I have is obtained from a GPS NMEA message string and as I understand it this parameter represents the bearing or course of travel. I'm just not confident that I fully understand its meaning, in navigation terms its reference plane is true north (wikipedia) with 0° being a direction of travel directly north, but is a field of view reference plane calculation of 12° really a simple addition or subraction to the azimuth reference plane or does a projection into another reference plane need to be done first?
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  6. #6
    MHF Contributor ebaines's Avatar
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    Yes - as I said earlier it's that simple! Reason is that the +/- 6 degrees that you're concerned about is as measured from your eyes on the ground to the horizon (0 degrees altitude), so you're +/- 6 degrees is in the same plane as the bearing measurement. If, on the other hand, you were looking up above the horizon (altitude > 0 degrees), then a 12 degree arc would translate to a larger change in bearing - but that doesn't seem to be the case here.
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  7. #7
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    Thanks for this Ebaines, really appreciate it. Given your knowledge, could you recommend specific areas of study that I could brush up on this stuff. While my example is assuming very simple parameters, this probably won't be the case as I do further work, so altitude and orientation (possibly roll, pitch and yaw of my field of view) I'm thinking may need to be considered. Is this pruely a spherical geometry area where my simple examples solution of easy addition/subtraction can be defined in the same terms as more difficult problems along the same lines.
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  8. #8
    MHF Contributor ebaines's Avatar
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    You could take a look at the following web site - specifically the discussion about astronomical coordinate systems. You are interested in the altitude-azimuth system. You can see how to convert from that system to the Right Ascension-Declination system (which is akin to latitude-longitude on earth) and vice-versa.

    http://ads.harvard.edu/books/1989fcm..book/Chapter2.pdf

    Here's an on-line calculator that automates this for you:

    Celestial to Horizon Co-ordinates Calculator - Converts celestial coordinates of right ascension and declination into horizon coordinates of altitude and azimuth
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