Rotor timing and adjustments

Somewhere in this section I was already writing about this. I tried to find it and after 5 minutes I could find it I found only something as “in this section there is an explanation”.

Making adjustments for rotor is important task, without it is almost impossible to play. There is many ways to do it with better or worst results but I like the most my way.

 Some people use stop watch to measure time rotor makes. Without going in to details by my opinion it is not very good way, converting rotor movement to time then converting time back to rotor movement differences is not natural.

Some people may use reference time  let’s say 2 seconds and observe how much rotor made then map it on cards using it in future spins.  It may be rotor made 25 pockets after 5 rotations the ball will be at number xx. It is a slow method it needs to be mapped for every wheel , every bal combination and every moment of prediction. In addition  if timer is used the timer doesn’t always start at particular diamond therefore noticing if the rotor made 20 or 27 pockets can be a problem.

 When I play I observe only the difference rotor makes. The difference is in pockets, same as adjustment should be,  so there is no conversation to time or any more complicated calculation required.

For example on the wheel I play within 2 sec reference time the rotor makes mostly 18 pockets (1/2 wheel).
When timer gives first vibration I notice the position of #0 against diamonds, and then transfer my sight to opposite side waiting for zero to arrive and for 2 sec vibration to happen. The rotor may make 3 additional pockets. So it makes 18 +3 =21. I do not care for 21, all I need is 3 pockets difference.

If my prediction is about 10 seconds before then the ball drops, it means it is 5 times longer than the reference time I use to observe for rotor change (10s/2s=5s). For that reason my multiplying factor is 5.

Applying it to spotted difference it makes 3 x 5=15 pockets.
It means the ball will drop on a number which is 15 pockets further in ball direction from number on which the ball drops if the rotor was making ½ of rotation.

This approach doesn’t require anything to remember, it is very flexible and adaptable to all kind of wheels or balls. If the player wants to play 12 sec before then the ball drops all he does is to use multiplying factor 6.

What if instead of 5 he uses multiplying factor 6?

Then he will have error of difference the rotor made.
If in 2 sec the rotor made 4 extra pockets then he will have 4 pockets error.
4 x 5 =20
4 x 6 = 24
The difference is 24-20=4

Forester, I do something very similar to what you describe.

I found it easy to just count the pocket difference from one spin to the next, over 2 sec and just multiple that by 5 or 6, depending on how fast the ball was snapped. If my rotor speed index number was 5 on the first spin (ACW rotor), then 16 on the next (2 pockets faster over 2 sec), I instantly project my ball onto the rotor 2 pkts x 5 = 10 pkts further on rotor in CW direction.

Actually the correct calculation would be 2pkts/2sec x 10sec left = 10 pkts, with the time remaining (10 sec) assumed to be close. If the ball is launched with a higher speed, I may have to use a 6 or even 7 multiple to adjust.

If the rotor speed is +/- 6 pockets or more over 2 sec, I find it necessary to start over and recalibrate (hoping the dealer will stay close to this new rotor speed) instead of trying to adjust my prediction.

I appreciate you sharing your experiences, :slight_smile:

If the rotor speed is +/- 6 pockets or more over 2 sec, I find it necessary to start over and recalibrate (hoping the dealer will stay close to this new rotor speed) instead of trying to adjust my prediction.

hahah that’s exactly what I do, I mean if there is a few spins I try to adjust but if it is continue I completely readjust.

I found it easy to just count the pocket difference from one spin to the next, over 2 sec and just multiple that by 5 or 6, depending on how fast the ball was snapped.

Not sure why it matters how the ball was snapped.

[quote author=forester link=topic=1510.msg12646#msg12646 date=1340690293]

Not sure why it matters how the ball was snapped.

Forester, you are right. - for 1sec/rev ball in upper track method you described, it would be the same time to ball drop from that ref point (gravity, ball mass, etc all constant).

I was thinking of L.S. crossover method where a faster ball launch and rotor speed could cause predictions to occur earlier in spin, leaving ball to circle upper track for more revolutions until it slows sufficiently to exit - but this does not happen too often, and I would end up recalibrating anyway. Then at the faster speeds, I might just let the ball circle around 1 or 2 more times just after the x-over pattern before making my observation, instead of adding time.

Sorry for the confusion ???

Trying to understand :smiley:

Sometimes I try to predict earlier to make dealers happy at least for few spins, regardless if I get it wrong or right.

Hi Forester,

I’ve read many of your posts (although not all 1400+). I can see that you already understand more than 99.9% of all the so-called experts out there. Yet, you still keep an open mind - a very admirable quality!

In fact I’ve found several knowledgeable players at this site. I would not hesitate to recommend this forum to anyone who is thirsty for more knowledge!

If i know any better place, i would join it :smiley:

[quote=“forester, post:1, topic:827”]

For example on the wheel I play within 2 sec reference time the rotor makes mostly 18 pockets (1/2 wheel).
When timer gives first vibration I notice the position of #0 against diamonds, and then transfer my sight to opposite side waiting for zero to arrive and for 2 sec vibration to happen. The rotor may make 3 additional pockets. So it makes 18 +3 =21. I do not care for 21, all I need is 3 pockets difference.[/quote]

What if in 2 sec the rotor make 3 pocket less ? you make 15 pocket backward from the droping ball point ?

Exactly ! 8)

I couldn´t figure out why when we have let´s say 2 pockets diference, multiply by 5 it will be 10 pockets LESS that the rotor will do in the next 10 seconds. It will be SHORTER 10 pockets. Why we add this pockets ACW from the DD if those pockets are missing?

If the rotor have more speed, we add CW.

Try to think out of the box… make a drow… ball direction, wheel direction, or take couple of spins on avidemux soft… its so simple, that nobody gonna bother to explane. It already been explaned.

Thanks, I will try hard.

I´m working with avidemux and the topics and all together, making the “soup”

If you observe rotor in 2 seconds time and the difference is minus 2 pockets (-1 pocket per second [-1x2=-2]), it means that in 10 seconds until the ball drops the difference would be - 10 pockets (-1 pocket per second [-1x1=-10]).

Feel free to ask any question.

Sergiy, some people here are new members and it may be confusing for them searching for everything.

Very nice topic Forester :slight_smile:

Thank you Forester

I´m reading a lot and searching but sometimes you have specific questions in the beginning…

Yes, I get that and I agree that is -10 pockets.

Why if that is the number of pockets that the rotor didn´t travel. The rotor is moving ACW so

Lets supose that we have the 0 at 12 o clock and DD at 12 o clock. If we timer from that point and we figure out that the rotor had a diference after 2 sec in minus 1 pocket and we spect the finish in let´s said 2 second so -1x1=-1 x 2 = -2. Then the 0 never pass the 12 o clock, instead it will be 2 pockets CW away from that point!

If the speed is faster it pass more times by a specific hour.

So, why we add ACW the minus difference from the DD if that pockets the rotor never did

I want to understand everything before to move forward

Ill try to reply it, with the permision of others… what we are calculating exactly is wich number gonna be ander the diamond in the moument of strike. We assume that time till the strike is constant. So ball gonna hitt the diamond(another assumption) after this time passed. When we take our visual read (during setup/preparation) we note the rotor speed. When we play/predict, we estimate rotor speed first. If , during estimation, the difference in pokets is observed, we multiply this differense by following formula: time till the end of the spin (in seconds) divided per time used to estimate rotor speed (in sec). So if our resoult is 5 and difference in pokets is 3pok. , we add 15 pok to the direction of shift. Hope it is anderstandable.

I understand the formula and how to apply it. What do you mean the direction of shift?

I only want to know why we add ACW the negative difference and add CW the possitive difference, why? Because the ACW and CW inversal relation between ball and rotor?

I simply can do the formula and make the adjustment as you said, but I refused to leave this unsolved in my head.

Forget ahout ccw. Think of moving back to compensate. In other words because the rotor is say 2 pockets slower in two sec than before, so the next two sec it will be another 2 pockets and the next 2 sec 2 more and so on.

In your example the rotor was slower by 1 pocket in 2 seconds.
It is 0.5 pockets slower per second. In remaining 2 sec time it will not be short 2 pockets but only 1 pocket, since -0.5 x 2sec = -1.

Don’t get confused with CW or ACW ball directions. It is always adding or deducting to the direction in which the ball is traveling. If the rotor was faster you adding to the balls direction.
For example if you need to add 2 pockets for a faster rotor and the ball is CW instead of 0 it would be 15.
If the ball was traveling ACW instead of zero it would be number 3.
The ball direction is your positive side and reversed of ball direction is the negative.

Ok, I think that´s my key point.

Thank for all the replies it would be very helpfull