I'm gonna be "that guy", Mr. Pedantic, aka Mr. Argumentative. I am channeling my justnum.
Let's try some math. Let's make some assumptions as we go.
OK, let's say a stroke is medium speed, say, 10 MPH. I'll provide calculations for lower stroke speed too.
Let's assume that 1/5 of that forward speed can be transferred to a swooping lateral tip motion. Let's also assume that the miscue limit of a cue ball is where a striped ball's color meets the white. I measured a ball, and it's 5/8", or 0.625".
OK, so let's assume that with a side swoop of 1/5 the forward speed of the cue, laterally it is 2 MPH (the cue tip travels 2 MPH sideways at CB contact), on a medium forward stroke of 10 MPH (considering a pro shoots medium at half break speed), that means for the tip to travel 0.625", it takes 0.0178 to travel the sideways distance from CB center to miscue limit. (I can do the calculations of what side swoop speed would be at maximum given a 6" stroke, and a 0.625" lateral movement, which provides two sides of a triangle, but my brain would hurt looking up the appropriate sin/cos/tan/etc. tables for such a triangle. So I'll overestimate it as 1/5 forward speed.)
I fail to understand how anyone can reasonably control the cue contact point swooping within 0.0178 seconds.
Let's assume that the forward speed is even less; let's make it only 5 MPH, so the cue swoop is only 1 MPH laterally. Then the swoop window balloons to a still nearly instantaneous 0.0355 seconds.
Is it really possible a swoop, traveling at 1 MPH laterally, with a 0.0.355 second window to hit it perfectly at the miscue limit, can consistently add additional controllable spin, compared a straight stroke at the miscue limit? Or is is entirely possible that most if not all swoop happens after tip contacts the CB?
Please, someone correct my math. I genuinely want to be wrong because so many highly respected people here say it's possible to swoop to good effect.