AtLarge said:Therefore, for two similar shots with just a slightly different cut angle, the player is likely to choose the same one of the 8 alignments. And if the CB/OB separation is the same for both shots, he will also use the same bridge distance for both shots. The OB will then travel on the same cut angles for the two shots, whereas slightly different cut angles are needed. For the two shots, your robot knows about the changed angles of the alignment lines to the OB/pocket line; the human does not.
To have a different cut angle, either the OB or CB or both must have moved from the original position(s). If the player doesn't move (realign) then the original visualized CTEL and A or B or whatever lines are no longer visually in the right place. He is forced to move to restore the correct visual alignment, which causes him to find new CTE, etc, lines. Below is a picture; it's not in perspective, but the principle holds. I haven't figured out how to type below an attachment, so I'll have to do it here.
In both cases, we have an outside CTE line (the red line) and a left-CB-edge to OB B line (the green line). The ball with the roughly dashed outline is the ghost ball. Clearly the shots go at different angles. I probably should draw this out in 3D where the orientation of the table top on the screen would indicate the player's alignment.
John, I don't see how your comments address my point. I'm postulating that all of the following are the same between two shot set-ups: distance between CB and OB, choice of alignments from among the menu provided by Stan's CTE (including the side from which the pivot is made), and bridge distance. The only thing different is the cut angle needed to the pocket. How does Stan's CTE achieve the different angle?
In your two-shot diagram, the angles to the corner of the diagram (the pocket) are very different -- approximately 32 degrees (not sure why this isn't 30) versus approximately 18 degrees. I imagine that such a large difference in cut angle would require either a different secondary alignment point or a different side for the pivot. But let's assume for now that "B" is correct for both shots and that the pivot is made from the same side for the two shots. Then why would the two shot lines converge to the same point at the pocket?
Suppose there is no pocket on your diagram, and no left rail (from the shooter's vantage point). You go through the same body-alignment steps for both shots (CB center to right OB edge and left CB edge to OB "B." You pivot from the same side for both shots and use the same bridge length for both shots. Now shoot the two shots. Do you not think the OB would strike the end cushion in two different places? If you agree that they would, then what changes when the pocket is put back in the picture? And remember, I am postulating two shots that are the same in all regards except for a slight difference in the cut angle needed to pocket them.