The height of a equilateral tringle is the length of a side times the square root of 3 divided by 2(.866 if the origional side was a length of 1.) Picture for a second the pyramid resting on its base with one of the bases points facing you. If we were to travel around the 0 line of longtitude we would pass two height lengths and one side length. 360 divided by three would be 120 per length, but remember 2 sides are .866 the 120 degrees. So really we have two height lengths that span 103.92 degrees and one side length that spans 152.16 degrees. Plotting each light to 0,0,0 and then moving it from there the 4 lights ended up at X=52, Y=0, Z=0, X=-52, Y=0, Z=0, X=0, Y=128, Z=0, and X=0, Y=-128, Z=0. But because I hav'nt found a Light Set Rotator, I then need to rotate the X axis of my Main Camera and my M2 Model -141 degrees, to get the light set to shine as in the above render.

This image was rendered with all of M2's materials set to Middle gray, and with each light set to a Middle Gray Intensity Value of 120, and with much postwork done with Photoshop's Levels

I left a few decimal points out of this explanation, but Poser has a bad habit of not keeping these exactly where you left them, so thought the set may not Perfect it's reallly close...and you can see that in many of the highlights. BTW - The postion of each of these lights is supposed to be indicative of where the center of each of the planes of the pyramid would make contact with surface of a sphere.

= )