One for the camera and/or maths buffs

[Mark]

An object approximately 75 feet long by 50 feet wide is visible 81,000 feet away and occupies approximately 1/16th of the total frame on a TV camera (16:9 aspect ratio).

Approximately what zoom lens would this require?

PS - I realise this information is incomplete - your best guess will be fine as this is only to satisfy my curiosity. :)

[petemc]

Get close up and use a 10-20 :D

[Dr. Z]

I have had a think and I don't think its possible to calculate it without knowing more information. I shall ponder it a bit more - the thing is you need to know the size of the projected image.

If you assume a 24mm frame height (quick google turned up that) and you assume that the object is exactly vertical in the frame, you get a projected image size of 1.5mm for a 22.9m tall object.

A quick look at the lens equation should allow you to do the rest - draw a ray diagram! I need to go out else I will be late but if you havnt worked it out by the time I get back I will do it for you.

[Mark]

Actually 1/16th is wrong - it's lower than that, but let's say 1/4 of the frame height. 22.9m is fair enough, we'll stick with that one. Can't help with the projected image size I'm afraid as the camera used is most definitely not 'off-the-shelf'. It's also entirely possible the camera wasn't on the ground, though that would only introduce a 25% error and I'm not really bothered about that.

Don't spend too much time on it as it's not important, but I'd still like to know. Thanks. :)

[Pumpkinstew]

81,000 feet?

Have you hacked into a spy satellite or something?

[Darrin]

"81,000 feet?

Have you hacked into a spy satellite or something?"



Or at least a spy plane picture?