# Field Of View Calculation

I am shooting a commercial where every shot is a static from a crane looking directly down at the ground. Is there a simple formula for calculating the field of view of a given focal length lens so we can work out what dimensions our 'sets' need to be? For instance, at a height of 30 feet I would like to know what the dimensions of the space on the floor visible in a 16x9 super 35mm frame will be on a specific focal length.

Thanks

Jake Polonsky
DoP London

Jake Polonsky Wrote :

>Is there a simple formula for calculating the field of view of a given focal >length lens so we can work out what dimensions our 'sets' need to be?

Not sure of formulae, but there is a freeware FOV calculator out there that may be of use :

http://cameraassistant.sourceforge.net/

As long as you have Linux or something that can compile it (or an iPaq)". Looks like thereâ€™s other FOV calculators dotted around if you do a web search.

If you want formulas, you may have luck if you look in the ASC manual or similar?

Chay Donohoe
LD / UK

The easiest way I do this is by using the palm pilot program pCAM which does the calculations for any given image size.

The next easiest way is to use the formulas that can be found in the ASC manuals and in David Samuelson's Hands On Manual for Cinematographers.

Oddly enough, I cannot find the most useful formula in the latest (eighth or "blue" edition of the ASC manual.

This is in the earlier editions.

The basic formula is â€¦

O/A = D/f

where O is object size
A is image size
D is distance
f is focal length

To solve for one of them, do the math, so that...

O=D*A/O
D=O*F/A
F=D*A/O
A=F*O/D

You have to convert everything into the same units to do the math, so this is much easier with a calculator.

So, you need to know what the horizon and vert dimensions of your image are.

Find out from the camera rental house what the dimensions are of the 16:9 ground glass, which might be around24mmx13.5mm or maybe 23.5x 13.2mm and use those dimensions in the image size of the formula.

You have to do the math once for the vertical size and once for the horizon for each calculation.

Note : You CAN mix units if you know how and not convert everything, but you have to be careful - David Samuelson's book explains it the best.

If you don't own his book, you should buy it.

If you do own his book, than you should have looked in it before
posting this to the list - that's what reference books are for.

Good luck.

As always, any inaccuracies in reporting the formulas are to be expected - I am only one coffee into the day and a lousy typist. If you do the calculations and they come out weird, check against the standard horizon. field of view tables in various manuals for Super 35.

Mark Weingartner
LA based
Glad I paid attention in parts of math class

Hi Jake,

Seems to me you'd need not only field of view, but the Magnification of your lens as well.

Here are the calculations I use for field of view & magnification when shooting for compositing/miniatures. I'm sure I'm not alone....

To calculate the field of view, we first have to calculate the magnification of the lens (M), given its focal length and the distance to the subject :

f
M = ---
s-f

Where f is the focal length of the lens, and s is the distance to the subject (remember to keep apples with apples--if you start in millimetres, STAY IN MILLIMETERS... Your units of measure MUST remain consistent.

Given the magnification, the horizontal and vertical field of view are calculated :

Fw
Vh = ----
M

Fh
Vv = ----
M

where Fw is width of the gate aperture (in mm), Fh is height of the gate aperture and M is the subject magnification (derived from the above formula). This should get you on the way--I would check with the rental house for the exact dimensions of the gate they're using (in millimetres).

For good measure, here's the Angle of View Calculation, in case you find you need it...You'll need a scientific calculator or a good ol' look up table from high school Trig-- you may or may not be able to get the Angle of View from the rental house--I prefer to double check though.

H
2tan-1 ----
2*F

where H=image size (either the aperture height or width in mm) and F is the focal length of the lens (in mm).

You might need to massage the calculation a bit given that your gate won't likely be 16x9 (although the ground glass will be...).

I'd love to hear what others do...Just one guy's opinion...

Best,

David Waldman
Cinematographer
LA CA

Look in David Samuelson's "Hands-On Manual for Cinematographers." In the back there is a huge section of formulas in mathematical formulas. I have a number set up in an Excel spread sheet just for these questions.

Kind Regards,

Mark Woods Cinematographer
http://www.markwoods.com

Jake Polonsky wrote :

> I am shooting a commercial where every shot is a static

If you or your assistant have a Palm (even the Palm-based phones will run it) PCam does this easily and will even preview the size for you with a human figure. Link in the cml downloads section.

Rory Moles
1st AC London

>where f is the focal length of the lens, and s is the distance to the >subject (remember to keep apples with apples

Doesn't need to be this complicated for the specific question we have here.

>For instance, at a height of 30 feet I would like to know what the >dimensions of the space on the floor visible in a 16x9 super 35mm >frame will be on a specific focal length.

The ratio of subject distance to subject size is the same as the ratio of focal length to frame size. Keep units symmetrical. In this case, your distance is in feet, so you will get the floor dimensions in feet â€“ provided the other two dimensions (lens and frame size) are in the same dimensions as each other.

Frame width is .980" or 24.9mm. So using a 50mm lens for example, the width of floor area you cover will be 24.9/50 times the camera distance you say 30ft, so the floor will be a fraction under 15ft wide. For the other dimension, obviously it's 16x9 ratio, so 15*9/16, or 8.43ft. (8'5").

If it's a 12mm lens, you get 24.9/12 x the camera distance. ie 2.075x. So at 30 ft you cover 2.075*30, that is 62'3" wide, and 9/16ths of that (35ft) high.

if you want it, the formula is simply o/i = u/v
o = object distance
i = image distance (= focal length except for extreme close-up/macro)
u = object size (why u? - no idea, it just is)
v = image size (film frame dimension)

Dominic Case
Atlab Australia

Try O=D*I/f or D= O *I/f or f=D*I/0

Where D = Distance, I= Image (film frame or whatever) and f = Lens Focal length and * = multiplied by and / = divided by. (For macro work and short distances D has to be from the front entrance pupil of the lens, not the tape-hook.)

It's in my Hands-on Manual page 322, section 15.6.1.1

David Samuelson