Subject:
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Math/Optics Problem
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Mon, 13 Aug 2001 23:27:43 GMT
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Viewed:
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99 times
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So. Let's assume I have a photograph of Bob standing in front of his house.
Knowing Bob the way I do, I know that Bob is 6 feet tall. And asking him
later, I know that his house is 30 feet tall. Now I want to find out how far
in front of his house Bob was when the photograph was taken. In the picture,
Bob is, say, 2 inches tall, and the house is, say, 4 inches tall. As far as
I know, I need one more piece of information to solve the puzzle: the
"fisheye" constant of the photograph. (I don't actually care about the
answer-- I want to know how to solve the problem)
#1 Will the fisheye constant be consistant for the same camera?
#2 If not, will it at least be constant for the same zoom level?
#3 are there any other "missing pieces" to the formula?
#4 what is the formula for figuring out the distance between the two objects?
#5 can I use that same formula to find out how far away from the *camera*
the objects are?
I *think* the answers to the above are:
#1: No
#2: Yes
#3: No
#4: dist. from camera = (fisheye const.)*(actual height)/(photo height)
#5: Yes
Am I right?
DaveE
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Message has 2 Replies:
 | | Re: Math/Optics Problem
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| David: (...) No. (...) No. (...) Yes: Focus (I think). (...) dist. from camera = (fisheye const.)*(actual height)/(photo height) + (small obscure camera dependent constant) Since you only ask about the relative distance, the "small doesn't matter. (...) (23 years ago, 14-Aug-01, to lugnet.off-topic.geek)
| | | Re: Math/Optics Problem
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| (...) Alrighty. Since I have been studying this stuff for the past month and am supposed to pass a test about it, let's see what I can do. (...) Depends. Does it "focus"? The "fisheye" constant you're referring to is, in technical terms, called the (...) (23 years ago, 14-Aug-01, to lugnet.off-topic.geek)
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