Projector Throw Distance For SOS

Here are some notes on projector throw distances and lens calculations. In general, if you can, use the lens calculator for the projector (like at projectorcentral.com)

Here is how to calculate whether the lens for a given projector will work with SOS. The main numbers needed are the throw ratio.

In projector lens calculations, everything is proportional. The throw ratio is the the distance (or throw distance) from the center of the lens to the screen, divided by the width of the screen. For SOS, the screen is located is the distance to the center of the screen and the width that we care about is the diameter of the sphere. On zoom lens, the throw ratio is usually represented by two numbers which correspond to either end of the zoom stops. For example, the Sony VPL-FE40 standard zoom lens has a throw ratio (TR) of:

1.91 - 2.41 : 1

You obtain this TR from the manufacture. Sometimes its hard to find these numbers.

Using the throw ratio, you can calculate the size of the projected image at various distances. Our sphere is 72" in diameter, that's the minimum height the projected image needs to be. The projectors are 4:3 aspect ratio projectors (3-4-5 triangle), so the image width (W) we need is 72" * 4/3 = 96"

Using the throw ratio's, multiply 96 * (throw ratio) to calculate two distances for the projector.

SD - Sphere Diameter
TR - Throw Ratio
TD - Throw Distance
W - Width of projected Image @ TD distance
H - Height of projected Image @ TD distance

Use this formula:

SD * (4/3) * TR = TD

For example, using the Sony's throw ratios (tr) of 1.91 and then 2.41:

72" * 4/3 * 1.91 = 183.36" (or 15.28 feet)
72" * 4/3 * 2.41 = 231.36" ( or 19.28 feet)

The smaller number represents the closest the projector can be to the sphere and still illuminate the entire sphere. The larger number represents the optimal, farthest distance from the sphere that should be used to maximize the number of pixels used to display data. Any distance past that maximum, the projected image starts to become larger than the sphere and the actual data must become smaller to fit onto the 6' sphere.