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If there is an object (tree) which gives back the below data measured from a given GPS location:

  1. Azimuth angle
  2. Slope distance to the tree or object
  3. Inclination angle to the tree or object.

How to calculate the GPS offset of this object?

For example, a man is standing at a given GPS point called GPS-A while the object (tree) gives back the above points back by a rangefinder.

What is the GPS-B (the GPS point of the tree or object)?

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Since distances are small you can simplify the calculation.

Use the following as the basis for the calculation:

  • At the equator, 1 minute of Latitude = 1 minute of Longitude = 1 Nautical Mile = 1852 M, but as you move away from the equator, lines of Longitude get closer together, lines of Latitude stay the same distance apart.
  • 1 Minute of Latitude = 1 Nautical Mile
  • 1 Minute of Longitude = 1 NM * cos(Latitude)

You can calculate the horizontal distance in M from GPS-A to the target using slope distance and inclination angle (simple trigonometry)

You can calculate the vertical distance in M from GPS-A to the target using slope distance and inclination angle (simple trigonometry)

You can calculate the Latitude offset in M from GPS-A to the target using horizontal distance and Azimuth angle (simple trigonometry)

You can calculate the Longitude offset in M from GPS-A to the target using horizontal distance and Azimuth angle (simple trigonometry)

Then using the equations for distance to minutes of arc, you can calculate your target position. And for target height it's just addition.

Note: Laser rangefinders typically give headings with an accuracy of at best 1 degree, often 2 to 5 degrees error, so if your distance to the target is significant, bearing error can contribute significant position error.

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  • thanks for your response. I have found this link igismap.com/… Can you please comment on it as per formula (Formula to find a lat lon point, when bearing, distance and another lat lon is given) where bearing is azimuth right ?. Please help in commenting on this link. thanks Trams Oct 18, 2023 at 6:37
  • @amedsvensson - that site looks reasonable. What they provide is a complicated formula which works at long distances, while my answer was only valid for short ranges (what you get from rangefinders.) What I tried to give is enough of an explanation that you can understand the calculation and YOU CAN WORK OUT A FORMULA FOR YOURSELF. I'm not trying to give the whole answer. By the way, I've been using a site for years which has a similar calculation to the site you linked. movable-type.co.uk/scripts/latlong.html
    – Trams
    Oct 19, 2023 at 7:29
  • thanks . Yes I need to do the work my self . I will go again through your guides and will look into the link you mentioned. Thanks Oct 19, 2023 at 14:22
  • Great link really. Oct 19, 2023 at 14:28

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