Help with a home cinema design

[Neal Scanlan]

Hi,

I am just beginning to put together the idea of a home cinema, a bespoke building outside. I have seen some inspiring home cinema’s owned by members.

I don’t know if this is possible but is there a calculation that can be made that will give the size of the screen, at any given length, using a given lens. I would imagine that an anamorphic film, projected (being the widest aspect) would be the datum?
Rocket science to me!!
Any help would be much appreciated.

[Mark Gulbrandsen]

Hi Neal,

Give this one a try. It looks as though it may do what you need...

[Randy Stankey]

... is there a calculation that can be made that will give the size of the screen[?]...

Any measurement (dimension) on the film, times the focal length of the lens is equal to the same dimension, projected on the screen, times the throw from the projector to the screen. (Measured from the primary nodal point of the lens.) (Often marked with a "ø" or similar symbol on the lens body.)

For example, if one frame of film is 25 mm (≈1 in.) tall, projected through a 100 mm lens with a throw of 30 meters, your image will be just under 8 meters tall.

* Film = 25 mm
* Lens = 100 mm
*Throw = 30,480 mm (100 m)
*Picture = 7,620 mm (7.62 m)

Film x Lens = Throw x Picture

Just remember to keep your units consistent. It doesn't matter whether you use inches, millimeters, or cubits as long as you use the same units, throughout.

If you have an anamorphic lens, multiply (or divide) by the anamorphic factor. (Usually x2.) In our example, if we were measuring the film horizontally, that 25 mm dimension would appear to be 15.24 meters wide on the screen.

It's all just high school algebra. Set it up as a cross-multiplication problem. ("Means-Extremes Product Property") If you know three of the terms, you can solve for the fourth. You can do it with a pencil and paper or you can plug it into a computer spreadsheet.

[Neal Scanlan]

Hello Mark and Randy,

Thank you for taking the time to reply, really very helpful.

[Martin Brooks]

Or, instead of height, you can use width where FW=DA
F=focal length of lens in inches
W=width off the screen in feet
D=Distance from the screen in feet
A=Aperture of the film (width)

So F=DA/W

[Randy Stankey]

Yup! That works.

The way I solved it, any measurable dimension on the film can be used. You could measure the height of a character, as seen on the film, and predict the height of that character's image as projected on the screen.

Several years ago, a guy I knew had a 35mm photograph that he wanted to know the location where it was taken. He recognized the people in the photo and he knew the general area where the photo was taken but he didn't recognize any of the background. I scanned the negative at some insanely high resolution then used Photoshop to blow up the image and measure the heights of various objects in the scene. With a little bit of geometry and some help from Google Earth we were able to locate the photographer's position within a 10 m circle and the compass direction he should have been pointing.

If you measure the height of a person's image on film and you know that that image was shot through a 50 mm lens, you know how far away from the camera the person was. (Assuming that an average man stands 6 ft [1.8 m] tall.) Calculate the same for another object in the scene, such as a fence post, and you have three corners of a triangle. You can use that as a base to calculate the distance between landmarks in the scene such as a hill or a road.

I was able to e-mail that guy a link to Google Earth that showed where the photographer was standing and he said something like, "Oh! That was my Uncle So-and-So's house! I haven't seen him since I was a little kid!"

It's all just high school math and basic map reading skills like you should have learned in the Boy Scouts.