I am setting up cameras for an experiment filming animals. In a experimental protocol paper that I'm following, it states that:

Based on a pinhole camera model, the bound x_min on the pixel span of the animal in the image can only be guaranteed if the observation distance between animals and each camera is at most:

enter image description here

where X is the length of the animal, f is the focal length of the camera and p is the physical width of a pixel. For our studies, in the interest of observing the flight paths of the animals over a large distance, we chose to allow a small image size.

My question is what is the physical width of a pixel, and how do I obtain that number? In the paper, it was stated that their equipment was:

three thermal infrared cameras (FLIR SC8000, FLIR Systems, Inc., Wilsonville, OR, USA) with variable-focus 25 mm lenses and a pixel width of 18 μm, providing a 40.5 deg field of view. The 14 bit grayscale−1 video has a frame size of 1024×1024 pixels and frame rate of 131.5 Hz.

It wasn't explained how the pixel width was obtained. The number seemed strange to me as well. If the physical pixel width is 18 μm, then the physical length captured by the whole image would be only 1024 * 18μm = 18mm, while their subject are birds flying in the sky in the scale of several tens of meters across. I'm probably understanding the concept incorrectly.


1 Answer 1


"three thermal infrared cameras"... "and a pixel width of 18 μm"

Pixel width (size, pitch) is a sensor design specification... it is indeed 18um; and the sensor size is indeed 18.4mm wide. The three interrelated characteristics are pixel size, sensor size, and sensor resolution... with any two known the third can be calculated. But if you can find two of the specs, the third is also usually available.

I'm probably understanding the concept incorrectly.

Perhaps another formula would help... the focal length (magnification, mm) required to fill the sensor axis (width, mm) is equal to subject distance (m) divided by subject size (m) multiplied by sensor width (mm); or FL=(D/S)Sa

Given what you have provided I'll fill in the formula making some assumption. The first being that "the subject" is something like a starling murmuration and I'll give it the size of 30m as described, and a distance from the camera as being 100m. So FL=(100/30)18.4 = a 61mm lens is required to fill the sensor width. But your lens is only 25mm and the subject will therefore only cover ~ 40% of the sensor width and 410 pixels wide.

Of course the formula can also be rewritten as D=(FL/Sa)S to determine how close the camera/lens must be to fill the sensor; which is essentially the same as the formula you provided. Only your formula seems to be concerned with pixels on subject rather than filling the sensor per-se.

  • \$\begingroup\$ Thanks for the explanation, I see. So to obtain the pixel width, I would need to find out the sensor size and divide that by the resolution \$\endgroup\$ Commented Jun 14, 2022 at 13:00
  • \$\begingroup\$ Yes, that is correct. \$\endgroup\$ Commented Jun 14, 2022 at 13:34
  • \$\begingroup\$ @TianxunZhou Technically that will get you the pixel pitch - the distance between pixel centers, and not the pixel width (pixel width can be smaller than its pitch). But that doesn't change the answer. \$\endgroup\$
    – Peter M
    Commented Jun 14, 2022 at 14:57
  • \$\begingroup\$ And the 18um pixel width given is actually the pixel pitch specification. It seems that to get the exact pixel width (if it's different) you would need to contact the sensor manufacturer... sometimes it is specified (and the pitch/spacing isn't). Generally, they are close enough to the same as to be interchangeable. flirmedia.com/MMC/THG/Brochures/RND_018/RND_018_US.pdf \$\endgroup\$ Commented Jun 14, 2022 at 15:27
  • \$\begingroup\$ @StevenKersting thank you! Your explanations have been very helpful \$\endgroup\$ Commented Jun 15, 2022 at 1:01

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