Lens angle and large scale shoots

[jamesklowe]

Hi if anyone could help me out, or give some advice...

i'm currently doing a large scale shoot. photographing roughly 60 people and 22 cars in a grid 6x4 (as pictured) from 2 scaffolding rigs 5m up. there will be 2 5x4 cameras on each rig, each photographing one half of the shoot. (ie, left camera photographing 11 cars on the left, right camera photographing 11 cars on the right etc.)

the cars are able to start a max of roughly 10ms away from the cameras.

i was just trying to work out the angle of view for a 150mm lens on 5x4. I know this has been brought up a bit in the past, but i just can't work it out myself.
due to time and lack of funds to keep hiring equipment doing tests, i was trying to work it out on paper as to how far apart each camera should be and how close each camera has to be to it.

does anyone know what the angle of view is for 150mm and 135mm?





thats the general layout.

the lighting will be broken up into 8 sections. first by in half, then into each row.
lack of lights and power means i can't do it in one shot.
i'm familiar to large shoots, but nothing this big. i have 10 tech. helpers on board to do this and 30mins to setup and shoot each lighting setup.

if anyones got any advice, love to hear it.

thanks

[Leonard Evens]

Let's assume the effective frame for 4 x 5 is 95 x 120 mm and that you are focusing at or close to infinity. The general formula for angle of view is

2*arctan(L/(2f))

where f is the focal length, and L is an appropriate linear distance in the frame. This assumes the subject is far enough away so that you can ignore bellows extension, which would be true in your case since the subject will be at least 10 m from the camera. L is usually taken to be the diagonal of the film, but that would seem to be of little use in your case. For example, L could be the long dimension (120 mm) or the short dimension (95 mm), either measured along a line through the point where the lens axis intersects the frame. Which would be more appropriate would depend on whether your frame was in landscape or portrait orientation. Here are the firgures for those dimensions.

For 150 mm lens

Long dimension 43.6 degrees
Short dimension 35.14 degrees
Diagonal 54 degrees.

For 135 mm lens

Long dimension 47.92 degrees
Short dimension 38.77 degrees
Diagonal 59 degrees

But in your case, the calculation is complicated because the camera is elevated, may be pointed downward, and you may be employing a fall of the front or rise of the back (or possibly a shift to one side.) This complicates calculating various angles.

Let's assume the camera is plumb, which simplifies things significantly, but there may still be questions about which angles you want.

Fortunately, you don't need to know the angles. Think of it this way. The frame is projected outward from the lens as a solid with rectangular cross sections parallel to the frame, the size of which depend on the distance from the lens. What you want is the width of that frame at various distances from the lens.

The formula for that is just

D *L/f

where f is the focal length, L the horizontal width of the frame (95 mm or 120 mm, depending on the orientation) and D is the distance from the lens. D can be measured in whatever unit you want, say, in this case, meters, and the answer will be in that unit. L and f must be measured in the same units, say mm.

Here are the answers at different distances from the lens for L = 95 mm (portrait mode) and f = 150 mm (L/f = 6.33)

10 meters 6.33 meters
12.5 meters 7.92 meters
30 meters 19 meters

If you use landscape mode instead, multiply the above numbers by 120/95 which is just about 1.25.

10 meters 7.92 meters
12.5 meters 9.9 meters
30 meters 23.75 meters

For a 135 mm lens, multiply the previous numbers by 150/135 = 1.11.

thus for portrait mode you would get

10 meters 7.03 meters
12.5 meters 8.79 meters
30 meters 21.09 meters

and landscape mode

10 meters 7.04 meters
12.5 meters 11 meters
30 meters 26.39 meters

It might be worth double checking these results using the formula.

[cjbroadbent]

Make it simple. Use a 5" lens on a 4x5" sheet of film.
You get a triangle inside the camera where the perpendicular equals the base.
You get a similar triangle on the set.
Eg. At 10 yards the field of view is ten yards wide.
I use a 180mm on 13x18 film and mark out big sets with string, a nail and chalk.

[cjbroadbent]

Addendum.
5" is 127mm so with a 135mm lens the field of view is shorter than the distance by 0.95. Likewise with a 150mm lens the factor is 0.85
More, wherever you're pointing, it's the center of the image which counts.

[jamesklowe]

thankyou guys,
sorry so late to reply.

had the shoot 2 days ago. the first lot of transparencies should be back this afternoon

was quite a messy shoot. 70 or so people running round finding their place. lighting was trouble. etc.etc.etc...

here's a rough test from the digital