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I have a GoPro7 that I will mount on a pickup vehicle to capture ground images with video capture.

The camera isn't top down but has about 20 (or 30) degrees inclination from the vertical axis. I want to capture the road that is 4m in width.

Other givens would be:

  • Vertical Field of View: 71.0deg
  • Horizontal FoV: 86.7deg
  • Diagonal FoV: 100deg
  • Sensor size: 1/2.3” (6.17×4.55mm)
  • Aperture: f/2.8
  • Zoom: 0%
  • Frame width: 4:3

I would like to know if it is possible (and how) to calculate the needed height of the GoPro for this.

bluewander
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  • The bottom of the frame will be narrower than the top of the frame because it is closer to you. Do you want to capture the 4m road at the bottom of the frame, the middle, or the top? Can the camera be turned toward the center of the road or must it face straight ahead? If straight ahead, how much of the 4m is left of centerline? The easiest way is to mark some spots on the road with chalk and move the camera until you can see what you want. Yes, it is quite possible if you describe exactly what you want. – Ross Millikan Feb 05 '21 at 16:15
  • Don't you remember SOHCAHTOA from high school trigonometry? – osullic Feb 05 '21 at 17:26
  • Related: How do I calculate the ground footprint of an aerial camera? – scottbb Feb 05 '21 at 21:02
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    What does this have to do with photography? – Michael C Feb 05 '21 at 23:13
  • I think this is an interesting question, but it might be better on a site such as Physics or Mathematics. – jng224 Feb 06 '21 at 15:52
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    @Jonas, I don't want to sound like a grump, but I don't think it's an interesting question. It's a basic trigonometry question that I would have expected to solve easily in high school. There is simply 2 triangles involved, a few measurements given, and the unknowns easily calculable using the formulae that I linked to earlier. – osullic Feb 06 '21 at 22:02
  • @Jonas that isn't as simple as that. you think that whatever the camera captures doesn't have magnification on it? – bluewander Feb 07 '21 at 06:25
  • my bad, yes. it should be @osullic – bluewander Feb 08 '21 at 08:32
  • @bluewander magnification is entirely irrelevant. The only figures you need are the horizontal field of view (86.7°), the width of road you want to include (4 m), and the angle the camera is pointed at (20° or 30°). You are left with 2 triangles, and you can solve for the unknown values using Tangent = Opposite ÷ Adjacent followed by Sine = Opposite ÷ Hypotenuse. Now, get out your pencil and paper... – osullic Feb 08 '21 at 09:34

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I will make a bunch of assumptions along the way along the lines of my comment. If I guess wrong, I hope you can adapt the calculation. Let us assume the camera is pivoted to the left so the 4m you want to see is equally split across the centerline of the frame. I will also assume the 4m is across the middle of the frame vertically. If the height of the camera is H the point on the ground will be H/tan(20) ahead of you. The half width of the frame is 2m, so you want 2=(H/tan(20))*tan(86.7/2), which gives H=0.77 meters or about 30 inches.

Ross Millikan
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