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 table (13.416407865' based on Pythagorean triangle proportions) as the triangle base.

The tangent of 39° (0.80978403319 roughly) is equal to the opposite leg divided by the altitude of the formed triangle. Since the opposite leg is 6.7082039325', the altitude (unknown distance to the table) is 6.7082039325' / the tangent = 8.28394196175' or 99.407303541" or 252.494550994cm.
[![!\[!\[camera set-up diagram][1]][1]
The values I used are approximate to the closest angstrom since we're talking optics and among friends.

I'd say roughly 8'-4" off the table centre should do it.

Many thanx to comment from sweber regarding erroneous arithmetic


  [1]: https://i.sstatic.net/Hn2YX.gif