Question about rotor speed changes

Lets asume that with X speed of rotor the observation point in 3 revolution out is 12 ocklock(cw ball)

and that with that X speed plus 1 more pocket per sec(faster) the observation point changes to 3 ocklock in 3 revolutions out again.

SO WE HAVE 1 MORE POCKET PER SEC = 3 MORE HOURS CHANGE POINT(FOR THIS EXAMPLE)

[b]now the question is this

with 4 revolution out…will this 3 hours change be the same in +1 pocket per sec again…or it will be deferent???
[/b]
eg if in 4 revolutios out the observation point with X rotor speed is at 9 ocklock …in the next spin with rotor speed X + 1 pocket per sec will the observation point be at 12 ocklock???(same change as in 3 revolutions) ???

I am not sure if I understand your question.

Are you saying if you predicting 3 revolutions before then ball drops, rotor of 1 p/s faster makes 9 pockets more.

If so, then 3 ball revolutions together in your example are 9 sec long. (not realistic)

If happens that you predicting 4 revolutions before then the ball drops then rotor change will be for 9 sec + time length of 4th revolution. So instead of 9 pockets you may have 9-12 pockets change. If 4th rev was 2.25 sec then it would be 11.25 pockets.

It would also mean that your rotor speed x is 3 s/r.

From 4th to third rev it shifted ¾ of rotation (9 to 12 o’clock ACW).
And in 9 sec it made 3 rotations.

Well i think the question is how you adjust offset to different rotor speeds.

Lets assume you have CW and 3.0 3.3 3.5 3.0 3.5…

Cheers

Well I did some explanations with my vb how I do it.
I actually never know rotors speed.
Since my VB predicts in any ball rotation I do not have mistakes of predicting wrong number just by picking wrong ball rotation.

Rotor speed I adjust by observing rotor movement change in reference time.

If I was playing 3.5 sec rotor I can see how far from prediction the ball really drops on rotor.
3.5 sec rotor also makes X amount of pockets when observed in my reference time.

If the rotor changes to 3 sec. then I would notice that such rotor makes X+Y pockets during reference time. But also the point where the ball will drop on rotor changes by Z amount of pockets.
Therefore Z/Y is multiplication factor. For example if it is 5
Then 2 pockets change of rotor traveling in reference time would make 2x5 =10 pockets change to final result.

Hello ewerybody
I am new in VB. End I want to ask someone who understend in the topic. Whose books I need to read to understend better VB?I play european wheel end I understood from forums that Laurences books are good but not for europeans wheels.
thanks