Mathematical lens formula?

[ShaneR]

Here's the dilemma - I need to fill the frame (vertical) with a 6 foot tall person who is 90 ft away from me. I don't have my 10D with me or I would just go outside and try it. I need to order a new, 2.8 lens for this purpose and need to know if I need something in the 28-70 range, or the 70-200 range.
Is there some mathmatical formula that will let me figure this out?

[stuartf287]

You need to determine the angle of coverage by using trigonometric functions. Compute the degrees in the angle of a right triangle that is opposite to a side which is 6' and the bottom side is 90'. Then you can select your lens for that angle of coverage (which is a more reliable measurement for lenses than focal length anyway). I can't do the math right now myself, but I would guess that you are going to need a rather long lens, at least 200mm and possibly even 300mm to 400mm.

[ShaneR]

Thanks, I didn't think about the angle of coverage. I guess that math degree will come in handy after all.

[robertwgross]

People who really have the math degree spell it *mathematical*.

---Bob Gross---

[john_houghton]

With the camera in portrait orientation, I make the focal length required approximately 337mm.

John

[Webster]

With the 10D's 1.6 focal length multiplier, a 200mm lens gives the coverage of a 320mm lens on 35mm film. Almost a perfect match. Or did John already take that into consideration in his calculations?

[ShaneR]

robertwgross wrote:
People who really have the math degree spell it *mathematical*.

---Bob Gross---

Bob - it's a typo. Get over it. I really have the math degree.

Everyone else - thanks for your help. Several possibilities are now available, such as the Sigma 70-200 EX 2.8 and the Tokina 80-200 AT-X 2.8. Any comments on either of these?

[john_houghton]

webster wrote:
With the 10D's 1.6 focal length multiplier, a 200mm lens gives the coverage of a 320mm lens on 35mm film. Almost a perfect match. Or did John already take that into consideration in his calculations?
The 1.6 factor was taken into consideration, i.e. 337mm actual, 540mm effective. This should give an angular field of view of 3.81 degrees = inv tan (6/90).

John

[ifurlong]

Bob, I have a MS in math and still quite often misspell it(and many other things) ;)

[ShaneR]

I have a feeling this is going to end up costing a lot more than I anticipated for a situation specific lens. Then again, I'll be making money with it (after I pay it off) and I'm sure it will be better than what I have now (Sigma 70-300 f/4-5.6).

Thanks for everyone's help.

[stuartf287]

Shouldn't your 75-300 zoom be good enough at the long end? You should be able to substantially fill the frame at that distance (90') with a 6' person. Or do you actually need to have his head tangent to the top of the frame with his feet touching the bottom? I would think you would be able to leave a small margin for framing errors, or you could simply crop the resulting shots without losing too many pixels. Good luck. This has been an interesting mathematical exercise. Kudos to john_houghton for coming up with the formula.

[ron chappel]

For those of us that will never learn the language of maths(to put it politely in honor of my maths teacher who at least tried,LOL),here is a handy web page.Halfway down is a simple program for doing exactly what is wanted here.

http://www.photo.net/making-photographs/lens

Don't forget the focal length crop when using a digital !

[ShaneR]

stuartf287 wrote:
Shouldn't your 75-300 zoom be good enough at the long end

Unfortunately, part of the shooting will take place under stadium lights at night with no flash allowed, so I need the f2.8. I was talking to another shooter last night (with the 12 mp Fuji) and he was at ISO1600, f2.8, and 1/125 and it was still pushing underexposure.