Sure. Happy to oblige.

Using trigonometry, the distance can be calculated based on the tangent of the FOV half-angle. I used the diagonal of the sensor aspect ratio that will include the table  (14.06' based on Pythagorean triangle proportions) as the triangle base.

First, you must fit the subject into the sensor's image field-of-view.<br>
The subject is a 6' x 12' Snooker Table.

[![!\[Sensor information][1]][1]

The image sensor image is given in angle-of-view
43.3° / 70.42° = 0.61 (Aspect ratio of 1: 1.64)

We know that we want to fit 12' into the sensor's Field of View.

0.61 = X / 12'

X = 0.61 x 12' = 7.32'

7.32' x 12' is the field size of the sensor when at the minimum height above the Snooker Table to fit.

The altitude can be determined by trigonometry using the TAN function for the triangle formed by the sensor image field diagonal.

[![Illustration for camera position details][2]][2]

Sensor diagonal = SQRT ((7.32')^2 + SQRT (12')^2) = <br>
SQRT ((53.58) + (144)) = <br>
SQRT (197.58) = <br>
14.06'

14.06' / 2 = 7.03' is the opposite (base) leg of the triangle.

TANGENT Angle = Opposite leg / Adjacent leg

0.71 = 7.03' / X

X = 9.9'

The values I used are approximate.

I'd say roughly 9'-11" off the table centre should do it.

Many thanx to comment from sweber regarding erroneous arithmetic


  [1]: https://i.sstatic.net/ebRGM.gif
  [2]: https://i.sstatic.net/19Pwk.gif