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I have the following information:

  • s: Image sensors dimensions: 6.00mm, 4.80mm, 3.60mm (diagonal, horizontal, and vertical)
  • f: focal length range: 5.1mm - 51mm
  • f = (s * WD / FOV), where WD is the working distance and the FOV is the object size.

I would like to compute the optimal f for a given WD = 10m and FOV = 1m x 1m.

Which sensor dimension should I choose (diagonal, horizontal, or vertical)? Since the results differ quite heavily depending on that.

Thanks

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    \$\begingroup\$ Is your problem a photographic problem, or a math/geometry/algebra problem? \$\endgroup\$
    – osullic
    Apr 15, 2021 at 14:18

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Since your field ov view is square, you only have to worry about the smaller dimension of the sensor, its height.

With H and D: height and distance of object, and h and f: height of sensor and focal length:

D/H = f/h

so

f = h * D/H

So f = 3.6 * 10/1 = 36mm.

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  • \$\begingroup\$ Thanks, makes sense. But if I calculate the same for the horizontal direction, resulting in f=48mm, would only a fraction of the object in the vertical direction be in the image? \$\endgroup\$
    – gab
    Apr 15, 2021 at 10:45
  • \$\begingroup\$ Yes, you'll miss some since your field is square while your sensor is rectangular. \$\endgroup\$
    – xenoid
    Apr 15, 2021 at 11:08
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    \$\begingroup\$ It would make more sense to choose the sensor dimension that correlates to the subject... e.g. sensor width if the subject is wide like a car, or sensor height if the subject is vertical like a person standing; or sensor width for a person standing with the camera in portrait orientation. And you would probably want to add a buffer of 10-20%; because you probably don't really want the subject to fill the frame right to the very edges. \$\endgroup\$
    – Steven Kersting
    Apr 15, 2021 at 12:18
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This is actually a simple ratio problem: We can trace imaginary lines from the boundaries of the object being imaged to the center of the lens. A triangle is traced. The height of this triangle is the object distance = 10 meters = 10,000mm. The base of this triangle is the object’s width = 1 meter =1000mm. The ratio of height to base is 10000 ÷ 1000 =10.0

Inside the camera we can trace a similar triangle. Its height will be the focal length. Its base will be the selected dimension of the imaging sensor.

Since the object being imaged is a square 1 meter by 1 meter, lets select the sensor height which is the vertical measurement.

I think this dimension should be reduced by 3% to gain some manipulation room. Thus, let’s use 3.6 X 0.97= 3.5mm.

The focal length that meets this reasoning is discovered by multiplying the base of the image triangle by ratio of height to base of the object triangle --- thus 3.5 X 10 = 35mm.

35mm will be the zoom setting for this problem.

This works because optically we have an image triangle and an object triangle. These are similar triangles in that all angles are the same and all sides are in ratio.

Detractors will tell you that this method yields only an approximate answer because object distance is measured from the front nodal and image distance from the rear nodal. These will be separated by an unknown amount. You might improve the accuracy of this math by using the center of the lens barrel to measure object distance. I think the resulting accuracy will be good whatever.

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