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I am trying to estimate the footprint and Ground Sampling Distance for aerial footage using specifications of the camera and flying height.

I tried two methods to get there, one based on the camera sensor size, image size (pixels), focal length and flight height:

GSD =(sensor*altitude*100)/(focal*imagew)
FP_width =((sensor*altitude*100)/(focal*imagew))*imagew/100
FP_height = ((sensorh*altitude*100)/(focal*imageh))*imageh/100

The other based on the camera field of view (let's call it a, in degrees), image aspect ratio and flight height:

Width <- (2 * height * tan((a*pi/180)/2))/ sqrt(1 + (16/9)^2)

Height <- 2 * (16/9) * height * tan((a*pi/180)/2)/ (sqrt(1 + (16/9)^2))
  • Both methods give me different results for footprint, any idea of why and what I might be missing ?

For an example with a camera or sensor size (1/2.8', I assumed corresponds to 5.6mm x 3.15mm), a focal length of 4.3mm, image size (1920x1080px), flying height of 400m and field of view of 63.7 degrees, I get 520x293m footprint with the first method, and 433x243m with the second.

  • Is one method more accurate ?
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1 Answer 1

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There are several approaches. I think this one is best. A ray trace of the light rays after traversing the lens, traces out a triangle.

Actually, we trace out two triangles, one for the length (5.6mm) and one for the width (of the sensor (3.15mm. The height of these two triangles is the focal length of camera lens (4.3mm).

Now divide to obtain the ratios of the two triangles. Thus 5.6 ÷ 4.3 = 1.3 and 3.15 ÷ 4.3 = 0.7326.

Now we trace the light rays arriving at the camera lens from the vista being imaged. We can trace two triangles. The height of both triangles is the altitude of the camera i.e. 400 X 1000 = 400,000mm.

You need to know, the image triangles and the object triangles are similar. In other words, the angles of all triangles are the same and the lengths of all sides have the same ratios. That being true, if the ratios are known, we can solve for the length of any size by multiplication.

The length of the base of the image triangle is 5.6mm. The ratio of its height to base is 1.3. Thus, the length of the corresponding object triangle is its height (400,000mm) multiplied by 1.3 = 520,000mm = 520m.

The width of the base of second image triangle is also solved by multiplication using its ratio (0.7326). Thus, its base is 400,000mm X 0.7326 = 293,040mm = 293m.

Hope this helps. Again, the lens traces out image triangles base on the size of the sensor and the focal length. The object triangles have the same angles and the sizes of both sets of triangles are in ratio.

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  • \$\begingroup\$ Thank you for your detailed explanations. I'm glad at least one of my approaches yielded similar results to yours :) Would you have any suggestions as to why the computation using the angle of view provided on the camera specs (63.7) does not work ? No problem if not ! \$\endgroup\$
    – user116447
    Commented May 3 at 9:56
  • \$\begingroup\$ Another approach: Find AOF (angle of view) AOF = 2ATAN(d)/2/2f AOF Horizontal = 66.1⁰ Known dementions of image triangle Height = 400m Apex angle 66.1⁰ Create right triangle by bisecting angle and unknown base length. Apex angle now 33.05⁰ base length still unknown Tan of this 1/2 angle is 0.6506 0.6506 X 400 = 260.26meters. This is ½ the base. Multiply by 2 = 520 meters the unknown base length. \$\endgroup\$ Commented May 3 at 18:26
  • \$\begingroup\$ Replace image triangle words above with object triangle. \$\endgroup\$ Commented May 3 at 18:36

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