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This manual on page 101 states that the camera has a sensor size of 6.3mm (1/2.9 type) I found this on sensor sizes ant tough that I could aply the 1" 16mm rule to this sensor.

1" = 16mm 1/2.9" = 5.517241379

1/2.9" * 25.4 = 8.75862069

But both methods don't give me 6.3mm What am I doing wrong and how would I get from the 6.3mm diagonal to the horizontal and vertical size.

Some more details are described in this post.

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    \$\begingroup\$ No, please don't cross-post. Do split up the posts into multiple questions. How to find out what the sensor size is of a camera? How to understand crop-factor? That fits on this SE (after you've done research by searching the web and this SE). How to calibrate a computer vision model? That's more on topic on other SE's. But currently your question would've been a better fit on meta. \$\endgroup\$
    – Saaru Lindestøkke
    Sep 22, 2020 at 10:12
  • \$\begingroup\$ Okay, thank you for your reply I am going to change this question soon, just give me some time. \$\endgroup\$
    – Phönix 64
    Sep 22, 2020 at 10:35
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    \$\begingroup\$ Does this answer your question? Why is a 1" sensor actually 13.2 × 8.8mm? \$\endgroup\$
    – scottbb ♦
    Sep 22, 2020 at 16:11
  • \$\begingroup\$ @scottbb the OP mentions that they tried to use the "the rule of 16" of the answer to that question to convert from the type size to a mm diagonal (that's what they do in the first calculation), but the result is different from the diagonal in the specs. So I guess the answer to that question does not apply in this case? \$\endgroup\$
    – Saaru Lindestøkke
    Sep 22, 2020 at 17:15
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    \$\begingroup\$ @SaaruLindestøkke Hmm. I mean, the answer is basically the same: so-called 1" (or related 1/2.3", 1/2.9", etc.) are scaled by 1":16mm. In this particular case, it comes to "nominal" or 1/2.9" class, as opposed to precisely 1/2.9". But the answers to the proposed dupe make it clear that 1", 1/2.9", etc, are just names, not actual dimensions. \$\endgroup\$
    – scottbb ♦
    Sep 22, 2020 at 18:12

2 Answers 2

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As the answer you linked to describes, the 1/2.9" is not actual inches and the 16mm conversion factor is an approximation. This dpreview article has some more information.

Wikipedia lists the 1/2.9" Sony EXMOR IMX322 sensor with a 4:3 aspect ratio size as:

diagonal: 6.23 mm
width: 4.98 mm
height: 3.74 mm

Given the same type description, aspect ratio and very close diagonal (I suspect 6.3 is a rounded value), these values could perhaps be used for your case.

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  • \$\begingroup\$ okay, thank you for clarifing. \$\endgroup\$
    – Phönix 64
    Sep 22, 2020 at 11:37
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The nomenclature that describes todays imaging chips as to their size was established in the 1950’s before the advent of digital imaging chips. This information is just a footnote in history however we still use this sizing method. In the era of the 1950’s TV stations with their TV cameras were popping up all over the world. These TV cameras used an imaging tube that had a short lifespan. This vacuum tube was called a Vidicon. Various size camera’s meant that the shelfs were stocked with various size Vidicon Tubes. These were sold by their diameter. This value was most commonly expressed in inches as most were manufactured in the USA. While the Vidicon Tube had a round active image surface, the camera masked this to a rectangle so that its aspect ratio fit the size of TV sets. This nomenclature, now obsolete remains as the way imaging chips are titled. The size of a 2/2.9 inch digital imaging chip, if a Vidicon, would fit a camera that accepted a tube 2.9 inches in diameter.

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