The funny part is, that gives me the right answer, but it cannot be accurate. With that formula, every shot of an object would be the same distance, since its dimensions are fixed. Pixel height rather than actual, but then how to account for actual dimensions?
Although a lover of math at an earlier time in my life, I believe that it would be much nicer if camera/lens manufacturers would just ad the focus distance that the lens reports in the EXIF data:) $0.02 Turbo
I used to love math problems when I was much younger, today I look for a simpler answer to the questions. Why don't the Lens/Camera manufacturers just add the focus distance of the lens in the EXIF data? Certainly the question of DOF comes into play, but the lenses today are pretty smart and accurate. At what distance did the lens focus? Manufacturer's please tell us! Turbo
Your geometry is a little off. The problem is that the triangle you are have used is not a right triangle. The right triangle in the frame is from the center of the image to the corner, not corner to corner. The formula is actually:
tan(t/2) = (a/2)/b
Also the angle of view specified is only at infinity focus. For a lens that focuses by extending the lens, the angle of view decreases as the focus distance decreases. Many modern lenses focus by moving internal elements relative to each other instead of by extending the lens. This internal focusing changes the focal length of the lens as focusing distance changes. All this means that the formula above applies only when the lens is focused at infinity.
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Perhaps this is getting a bit Geeky, but while Gary's formula is mathematically correct, for longer focal lengths (100mm and longer) your formula is considered a reasonable approximation.
If you crunch the numbers of your example using both formulas, and assuming I didn't err,the difference is only about 3 inches or so in the 14 foot coverage.
But Gary does give a pertinent bit of information that the specified focal length and FOV is only accurate at infinity focus. But I would think that at the distance of your example, the deviation would most likely be minimal and would be more of a concern at closer focus distances.
But it does provide a big wrinkle in your attempt at figuring a rule of thumb.
The back story and my conclusion. Walking in Pelican Valley in YNP, several clicks from the nearest road. I saw this:
I took this pic and enlarged enough to confirm it was a grizzly. My son was with me and I showed him what I was looking at. Of course, we started to stalk the little fellow to get a better look. I grew up in Wyoming and I know full well that there is a fine line between stalking and being stalked. I have no question who is the better at that game. I may be a carnivore but I buy my meat in packages.
To make a long story short. This was a close as I could get, without being caught spying.
The first pic he measures 85 pixels, top of hump to feet (best I can tell). The usual estimate is 1 meter. That yields a distance of 1210 meters. Certainly in the ball park as I estimate nearly a mile, and Joe guessed closer to a half. 1200 meters is about 3/4 mile.
The second pic is the rub. He measures 700 pixels (as near as I can tell). That puts him at 146 meters. That's just way off. He was definitely on the same soccer pitch as me, maybe on the same half.
Clearly, the focus is well short of infinity. I can only assume that is the problem. My conclusion is that distance estimates using this method are rough at best, and best when the lens is focused close to infinity.
>The second pic is the rub. He measures 700 pixels (as near as >I can tell). That puts him at 146 meters. That's just way off. >He was definitely on the same soccer pitch as me, maybe on the >same half.
If pixels had a fixed size, it would all be easy to figure out. I think that's where your problem lies. Pixels vary in size depending on display density (usually measured in pixels-per-inch or PPI) and print density (usually measured in DPI). Zoom in or crop or zoom out and any height or distance calculations go way off the rails. Even without zooming or cropping or magnifying, a pixel is not a size measurement but rather is used to describe the amount of data provided on a display or captured in an image.
For example, using ACDSee Pro, CaptureNX2, Lightroom or a number of other products, I can render your image at the exact same physical size (either on a collection of 21" monitors or printed at 16" x 20"), but at various resolutions, i.e., the exact same physical size with more display pixels or with fewer display pixels, or with more dots or fewer dots. You can make all the different versions of the same image look essentially exactly the same when viewed from a distance of one meter or so.
So it follows that anyone trying to use the number of pixels comprising the height of the bear as a factor in a distance calculation (or vice versa) based on either the angle of view at a particular focal length or the field of view at a particular distance would end up completely frustrated. A pixel is not a measurement factor unless you're in a mathematically compensated and tightly closed environment.
David is measuring his pixels based on the pixel density of the D800 sensor.
Regardless of the resolution of the output device a full frame image from the D800 will be 7360px wide by 4912px high. The only thing that changes with the resolution of the output device is how many of those pixels are displayed per square inch.
Using the fixed pixel density of the sensor – the image is always recorded (FX mode) as 4912 pixels high – out of those 4912 only 700 of those pixels covered the height of the bear.