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How to compute focal length when images lack EXIF data? #669
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@zharvey-welltok , 1920 was supposed to be the width of your sensor in pixels and 1.2µ sensor's pixel size. |
Thanks @polto but how did @pmoulon know what my sensor was?!? My point being that he somehow found out what my sensor was (my smart phone) and was able to look up the width & pixel size. I'm wondering how he did that since I will be processing images from cameras and sources whose make/model info I will not have access to. So there has to be something in the EXIF data, or some other way to detect sensor make/model info. |
Here the explanation about the values:
Then according those you can approximate the focal length as focal_ratio * max(w,h) => 1.2 * 1920 This formula provides a good guess for SfM if your camera have a "classical fov" near than 45. |
Thanks @pmoulon, this is all starting to make sense! When you say:
...do you mean that my |
You can either look to the max size from the JSON or just open the image properties and look to the image size. |
As always, thanks @pmoulon 👍 |
You're welcome, I'm happy to help and explain things. |
good |
In a previous question I was using
openMVG_main_SfMInit_ImageListing
without the-f
argument. I was also using images that did not have focal length info in their EXIF data, and so the resultingsfm_data.json
had (essentially) UINTMAX set for each view'sid_intrinsic
value.In that question I was advised by @pmoulon to re-run
openMVG_main_SfMInit_ImageListing
with a focal length of-f 1.2*1920 = -f 2304
.I'm wondering what the two magic numbers (
1.2
and1920
) came from, and how to derive them on other image sets in the future. Any ideas? Thanks in advance!The text was updated successfully, but these errors were encountered: