Focal length converted into actual distance to subject/motive?
Anyone have some rough numbers they could throw out there?..

Comments (8)

I'm not sure that I understand what you mean. But I'll assume that you want to estimate the effect of focal length on image size..

A subject of size x at distance d, with focal length f, will produce an image on the sensor with a size of approximately fx/d. So a subject 3 feet high photographed approximately 30 feet away with a 50mm lens (actual focal length, not "35mm equivalent") will project an image 5mm high onto the sensor..

Now what you need to know is the size of your sensor. For the most common sensor sizes, these are the *approximate* dimensions:1/2.5" digicam: 5.8 x 4.3 mm1/1.8" digicam: 7.2 x 5.3 mm4/3 DSLR: 18 x 13.5 mmAPS-C DSLR: 22.5 x 15 mm"full frame"/35mm: 36 x 24 mm..

Comment #1

Here are explanations:.

Focal Length -

Read about Normal lens (50mm), telephoto and wide angle lenses following links in the article.



Comment #2

OK, let's try this: Say you are to photograph a car with an array of different lenses, ranging from 12mm to 600mm (and everything between). At what distance from this car would you need to be standing in order to get the whole car featured within the picture?..

Comment #3

Krwl wrote:.

OK, let's try this: Say you are to photograph a car with an array ofdifferent lenses, ranging from 12mm to 600mm (and everythingbetween). At what distance from this car would you need to bestanding in order to get the whole car featured within the picture?.

What kind of camera? How big is the car?.

I gave you the formula in my first posting. It's a simple multiplication and division. Subject size divided by subject distance equals (approximately) projected image size divided by focal length..

Example. Assume an APS-C DSLR with a sensor that's 22.5mm wide, and we want the car to fill that 22.5mm width. Assume a 45mm lens. The ratio is therefore 0.5 (22.5/45). In that case, a car 16 feet long would need to be photographed from a distance of 32 feet (16/32 = 0.5). With a 90mm lens it'd be twice the distance, or 64 feet...

Comment #4

If I take your question without trying to interpret it in any way I would say that there is no relationship between focal length and subject distance..

However, the focal length does determine how much you can get in the viewfinder and picture because that is determined by the angle of view..

Your question makes me think you want to read a book about the basics of photography and lenses. Usuanlly such a book will contain examples to illustrate how useful some zoomlenses can be...

Comment #5

It was great reason to move away from PC. You got me out to parking lot..

Ford Thunderbird fit into frame of my Minolta 7D DSLR (1.5 crop body) from following distances:.

17mm from about 9 feet.

50mm from about 30 feet.

210mm from about 180 feet.


Comment #6

Here is THE answer with online calculator and formulas:.


Comment #7

I, like the others, couldn't make sense out of your subject. That last word is especially confusing..."motive"? Hmmm....

Krwl wrote:.

Anyone have some rough numbers they could throw out there?.

How rough do you want? I always thought "37" was pretty rough. But if you are looking for something smaller, "2" can be rough too, especially when the correct number is more like "2.73". "3.14159" is rough too, if you need something with more digits. Let me know which of these is closest to what you intended and I'm sure I can suggest more rough numbers....

Sorry, I just couldn't restrain myself!.

You might be interested in Barnack, if you are good with numbers (I can't tell until you reply and let me know about the above rough numbers). Barnack is the best lens simulator I know of. Unfortunately, it doesn't cost anything, making it less frustrating when you can't figure out how to drive it..

Get it here:.


Good luck....

Charlie DavisNikon 5700 & Sony R1HomePage: http://www.1derful.infoBridge Blog:

Comment #8

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This question was taken from a support group/message board and re-posted here so others can learn from it.


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