Calculate floor cover based on hight and lens

[norly]

hi. I have a mathematic question for all you bright guys.

For a while a go I needed to calculate how big surface on a floor the camera would cover (angle of view?) based on the lens and the distance from camera/object. I didn't succeed very well, so I thought that you people might have a better formula for it...

The format is 4x5" (10x12 cm).
The distance between the floor and the film is 8m in teh example.
The lens has a output of 18 degrees i think (I dont remember how I got that number but its not correct)

Lets for the examples say its a 90mm Nikkor SW taking a 4x5 NEG from 10 meters hight. My calculation should work if I can calculate how to decide the out-falling angle on the lens. Any tips?

Se the photo with my failed calculation... Good luck

http://www.rasmusnorlander.se/forums/angleofview.jpg

[Dan Fromm]

tangent = opposite/adjacent

sine = opposite/hypotenuse

Now, which function is appropriate for your problem?

And shouldn't you consider the film's size (hint, not 10x12 cm, measure the film holder's gate if you want to get it right) too?

[aduncanson]

....................................... Neg size: 10cm x12cm ....... 3.75in x 4.75in
FL (mm)........ magnification ....... Floor size: (meters)
90mm ........... 86.9x ...................... 8.69 ...... 10.43 .... 8.28 ..... 10.48
135mm .......... 57.2x ...................... 5.72 ....... 6.87 .... 5.45 ...... 6.91
150mm .......... 51.3x ...................... 5.13 ....... 6.16 .... 4.89 ...... 6.19
180mm .......... 42.4x ...................... 4.24 ....... 5.09 .... 4.04 ...... 5.12
210mm .......... 36.1x ...................... 3.61 ....... 4.33 .... 3.44 ...... 4.35
240mm .......... 31.3x ...................... 3.13 ....... 3.76 .... 2.98 ...... 3.78
300mm .......... 24.6x ...................... 2.46 ....... 2.96 .... 2.35 ...... 2.97

Above are a set of results for several common focal lengths. Please don't ask me to show my work. Rather than actually do the algebra to solve the equations, I used Excel's Goal Seek function to find solutions to (Object Distance) + (Image Distance) = 8 meters as well as the "Lens Equation": 1/ (Object Distance) + 1/ (Image Distance) = 1/ (Focal Length)

I then took Magnification = (Object Distance) / (Image Distance) and multiplied by the negative dimensions to get the projection of the negative on the floor. The hard part is getting the table to display correctly on the forum.

[Leonard Evens]

Multiply each of the film dimensions by the quantity

subject distance/focal length - 1

(which is the reciprocal of the magnification).

For 90 mm lens and subject distance 8 m - 8,000 mm, that quantity is

8000/90 - 1 = 87.9 (to one decimal place)

The dimensions of 4 x 5 film are about 96 x 120 mm. So the projection on a plane 8 m from the lens has dimensions

8,437.3 mm x 10,546.7 mm or about 8.47 m x 10.55 m.

The subject distance should be measured from the front principal point.

[norly]

perfect, thanks.