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I have a GPS on the middle of a plate as can be seen in attached picture. Clearly, the GPS gives me the Lon. and Lat of the center of the plate with no problem. My question is how can I estimate what is the Lon. and Lat at each end point (A,B,C,D) as the distance between end points and GPS is known? I believe there should be a formula for this purpose.
I appreciate any idea and suggestion in advance.

GPS and four points around it:

![][1]

[1]: https://i.stack.imgur.com/cwWVU.jpgenter image description here

1 Answers1

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Have a look at http://www.movable-type.co.uk/scripts/latlong.html ... specifically the section headed Destination point given distance and bearing from start point

John Machin
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  • Thanks John for the link. What does bearing mean here exactly? And I guess I can assume d/R is almost 0 right? Because d is just 1.5m –  Feb 23 '17 at 06:56
  • @Mechatronics "θ is the bearing (clockwise from north)" in your case for points A, B, C, D it's 0, 90, 180, 270 (degrees) respectively. Don't forget to convert it to radians. Yes, your d/R will be almost zero ... what's your use case for such a small d? – John Machin Feb 23 '17 at 20:48
  • I need this for designing a Robot. But there is a problem here. My GPS gives me Inches accuracy. But as I try on mentioned website, the coordination at all points are the same! Do you have any idea to solve this issue? –  Feb 23 '17 at 21:56
  • @Mechatronics Do you mean you tried using the calculator on the mentioned website? If so, please show what you entered and what the result was. Do you mean that you tried to write code following the formulas on the mentioned website? If so, please edit your question to show the code, plus the input and output. – John Machin Feb 23 '17 at 22:23
  • I checked on that website. Still don't have any code. I tried 90 as bearing and 0.001 as distance and didnt change the long lat as it doesnt matter to test –  Feb 23 '17 at 22:57
  • Have you noticed that the calculator doesn't show its output to any great precision? E.g it shows lat and lon to 1 second i.e. 1/3600 of a degree which corresponds to about 31 metres on the ground. Time to write some code? – John Machin Feb 23 '17 at 23:29
  • yep! that was the answer! works! Thanks John! I also found this for converting different format. Might be useful for someone in future: https://en.wikipedia.org/wiki/Decimal_degrees –  Feb 23 '17 at 23:51
  • Glad I could help. Please consider upvoting and/or accepting my answer. – John Machin Feb 24 '17 at 00:06
  • I got another question regarding this problem and I appreciate it if you could give me an idea. To use this formula we need both Lon Lat and also angle to the North. I'm wondering which NMEA message gives me these two information at the same time? I already use GGA but it doesn't give me that angle. As I did research RMC message might be the answer. Do you think 'Track angle in degrees True' in RMC message is the angle I should use (I mean the bearing? ) –  Feb 24 '17 at 21:09
  • @Mechatronics (1) ask a separate question (2) I have no idea what NMEA/GGA/RMC mean (3) the "angle to the North" (bearing) is supplied by you; to get your 4 points you use the "formula" 4 times with bearing 0, 90, 180, 270 – John Machin Feb 24 '17 at 21:34
  • I hope you still read this. I changed the picture in problem please have a look. In this configuration, what are d and bearing angle? I hope my drawing is clear and makes sense. –  Apr 16 '17 at 19:52
  • for me A and B are important in this configuration. But C and D are on the opposite side of A and B respectively. I was thinking maybe I can assume four points and GPS are on the same surface and Z dimension actualy doesn't affect Long and Lat, but elevation. What is your idea? Thanks –  Apr 20 '17 at 01:39
  • sorry, but your 2nd diagram less useful than the 1st diagram. – John Machin Apr 21 '17 at 06:11
  • It's not easy to explain what is the real life application for this configuration. But assume there is a GPS on top of a truck and we need to know the position of back wheels! –  Apr 21 '17 at 06:29