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I'm trying to figure out the focal length of my camera. I'm trying to figure out the distance from my camera to an object using the equation listed here (How do I calculate the distance of an object in a photo?) and I had this camera taking a video recording (http://www.bhphotovideo.com/c/product/1022655-REG/sony_hdrcx240_b_hdr_cx240_full_hd_handycam.html) which says the following under Focal Length:

35mm Equivalent Focal Length 29.8 - 804 @ Aspect Ratio: 16:9 Video mode 29.8 - 804 @ Aspect Ratio: 16:9 Photo Mode 36.4 - 984 @ Aspect Ratio: 4:3 Photo mode

I'm simplifying my assumptions as much as possible and am trying to assume the video was not zoomed in at all. At this point I do not have any stills, which I understand contain more information than a video. Does that mean I should be using the "29.8" value for the Focal Length in the equation? (which seemed appropriate based on this post, How to get focal length from a camera specification which gives a value like "4.5-22.5"?)

When I do that my distance is outrageously small, and the Focal Length is the only thing that I don't have a good estimate for. Thanks!

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Manner
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1 Answers1

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Your B&H link gives the link to the camera user manual, and its spec sections says:

Lens: ZEISS Vario-Tessar Lens 27× (Optical), 54× (Clear Image Zoom, while recording movies) 320× (Digital) F1.9 - F4.0 Focal length: f=2.1 mm - 57.0 mm (3/32 in. - 2 1/4 in.) When converted to a 35 mm still camera For movies: 29.8 mm - 1609.2mm (1 3/16 in. - 63 3/8 in.) (16:9) For photos: 29.8 mm - 804.0 mm (1 3/16 in. - 31 3/4 in.) (16:9)

So the focal length is 2.1mm to 57mm (actual). It also says the field of view that it shows is the same (equivalent) view of a 35mm camera with a 29.8 - 804 mm lens (which is only handy if you are familiar with using 35mm cameras, which we used for decades.)

But the focal length is 2.1mm to 57mm. You only know those two extreme end points. There are many possible zoom points in between these.

WayneF
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  • Thanks for the response. So which of them do I take if I assume it was NOT zoomed in at all? The smaller value? – Manner Aug 11 '15 at 18:17
  • It is always zoomed to some value. Guessing that such camcorders when turned on will surely have a default zoom position, but which is not specified. Wildly guessing, perhaps about 30% more than its minimum 2.1, perhaps like 2.7 mm default? From there, it will still zoom a little wider, and probably a lot longer, but it turns on at this one place. And you are saying zoom is default. ----- We also know crop factor is 29.8 / 2.1 = 14.2 ----- Message is too long, so two parts: – WayneF Aug 11 '15 at 20:37
  • And I have a Field Of View calculator at http://www.scantips.com/lights/fieldofview.html --- It offers extra features, like aspect ration and crop factor. In Option 3 there, enter Crop 14.2 and "16:9 aspect ratio, camcorder". It computes sensor size to be Width=2.66 mm, Height=1.49 mm, Diagonal=3.05 mm. The Sony manual says 3.1 mm, possibly rounded, seems very close. ---- So now these methods can tell you default focal length and sensor size. ---- That is as far as my calculator goes, perhaps I need to consider adding another feature? – WayneF Aug 11 '15 at 20:43
  • Oops, struggling with this tiny length issue, I forgot to add the second part... Take any still picture at your default zoom. A good Exif viewer will tell you that focal length used (from Exif data in still image JPG file). It has to be a still picture, movie files do not show variable Exif data. – WayneF Aug 11 '15 at 20:48
  • Awesome, I will track down the camera tomorrow and try getting a still from the default zoom and retry my calculations. Thank you for the link and extra information. – Manner Aug 11 '15 at 22:05
  • OK, I whipped out a calculator for determining Subject Distance in an Image, which is at http://scantips.com/lights/subjectdistance.html --- You still have to know the numbers, but I hope it is useful. I took a picture of my front door from 18 feet, and it seems accurate, at least within a percent or two, which is my own error framing it in the viewfinder. :) My style is that the page probably gets edited for a couple of days. – WayneF Aug 12 '15 at 03:32
  • Thanks for the extra link, WayneF. I got the camera, took stills from it, and at default zoom it was 2.1 mm focal length (I used http://exif-viewer.com/ on the stillshot during a movie). I think now I need to figure out the mm size and pixel size of the sensor of the camera but only found one measurement, <Imaging Sensor 1/5.8'' (4.6mm) back-illuminated Exmor R® CMOS Sensor>. Do you think this is a vertical measurement? I found equation SensorSize = pixelSize * resolution, but plugging in 4.6 mm = ? * 2272 gives me .002024.. pixels height. Which doesn't make sense. Any ideas? – Manner Aug 20 '15 at 19:43
  • The 1/5.8" number is a very crude number, only about old TV camera video tubes, and is about 1/3 larger than the sensor, and not necessarily very meaningful (long story, but that number is just a confusion). But my previous comment here computed sensor Width=2.66 mm, Height=1.49 mm, Diagonal=3.05 mm. The Sony manual says 3.1 mm, which could be round off, so it seems very close. I guess we have to trust the Exif for focal length, but that is minimum zoom in specs, and I would have assumed it would always zoom wider than the power on default. – WayneF Aug 20 '15 at 23:13
  • Ugh, sorry, my last comment didn't make sense, and I hadn't reread your calculations for the sensor width/height/diagonal in mm. – Manner Aug 21 '15 at 21:55
  • Took too long editing the last comment. Anyway, I still want to calculate the sensor size in pixels, though -- is this just the actual picture dimensions? E.g., my photo is 4032 pix wide by 2272 pix tall. – Manner Aug 21 '15 at 22:04
  • 4032 x 2272 (16:9) is the sensor specifications from the camera specs, so that's the right number for sensor size. You also need sensor size in mm. If for example, the object you seek distance to is say 1000 pixels tall in that image, then its size is (1000/2272) * sensor height in mm, so you know object image size in mm. Then estimating object's real life height, then it is just similar triangles to get distance (from focal length). The calculator does this if you can supply the numbers. – WayneF Aug 22 '15 at 02:24
  • Awesome, thank you so much. I have some funky numbers somewhere (I know the object is 6 feet away, but it's calculated at about 5 feet away) but this is nearly there. – Manner Aug 25 '15 at 21:02
  • That is nearly 20% error, and my guess is it could be focal length. Your specs say 2.1mm is the maximum wide angle, so I doubt it is also the default power on focal length. If we said it was shorter than it was, that would make the distance also be shorter. Also the focal length number applies to infinity focus, and it becomes longer when focused closer. 6 feet should be hardly different, but say 2 feet would, and macro especially so. – WayneF Aug 25 '15 at 22:40
  • Yeah, I know it's a fairly outrageous error, but I got the focal length from starting up the camera, starting a video, taking a still shot during the recording, and running that jpg through the exif viewer. I didn't change any of the zoom settings or touch any other buttons. I took a few other videos & photos when my object was different distances away (e.g. 11 feet) without changing zoom and the still from that video said the same thing. I can always get the camera and try again though. – Manner Aug 27 '15 at 22:22
  • Don't mind me, I am naturally suspicious, but if it said 2.1mm (which maximum wide angle does not seem a normal value for power up default), you might try zooming it to be something a little different (will it go wider then?) and repeating the exercise, and see that you do get a different number? – WayneF Aug 28 '15 at 00:16
  • I took a video that included "default", all zoomed in, and all zoomed out. There is no default zoom- it turns out the camera just uses the last zoom that was set. The Exif data gave me "21/10" for max zoom out, "570/10" for max zoom in, and "61/10" when I just sort of put the zoom in the middle. So I've at least verified the focal length is changing and the pesky default setting. Now I'm going to get some easier-to-measure objects in front of the camera and see how exact I can get the measurements. The item I was using is curved and sort of weird so I'll switch to a cardboard box. – Manner Aug 28 '15 at 17:04
  • Sounds reasonable then, sorry. If interested, I added an option 8 at http://www.scantips.com/lights/fieldofview.html to compute sensor size from a measured field of view (knowing focal length). You already know both sensor size and focal length, but suspect an error. So one thing it will do is to correlate sensor size and focal length. The numbers should agree if all are accurate. Or it can confirm your error degree. Only has two variables, plus your measurements. Trial and error can suggest what one should be if the other is known. – WayneF Aug 28 '15 at 18:53
  • Your new option says 10 - 12 feet is even better; all of my data will be closer than that, more like 7 - 9 feet for the interesting stuff. How much more problematic will this be? I measured out the boxes at 7, 9, 11 feet (11 feet two different box positions), at min zoom and mid zoom, and it's still off. I'm using the calculation I think you use on your website, not your actual website, so I could use Excel instead of enter all those numbers by hand. My closest calculations were at min zoom (2.1 FL) and the mid-zoom was terrible. It could be the way I measure pixel size of object? – Manner Aug 28 '15 at 19:00
  • Focal length is marked for infinity focus, and so farther is better. This 6 to 12 feet includes 7-9 feet, but is just for computing and knowing sensor size. Since it won't change, it can confirm focal length... which does change slightly with focused distance (more so up close), but sensor size will not change. Six feet should not change much, but six inches changes a lot. Part of the point was if you can confirm focal length more distant, maybe you can determine how much it is changing up closer. On the equations, I feel sure they are all the same equation, it is just similar triangles. – WayneF Aug 28 '15 at 20:27