# How and why does the VB2 system work

Hello everybody!

In this topic I would like to explain the VB2 system as I understand it. Reading the forum I’ve noticed that many people can’t understand why does it work because the system doesn’t require particular starting point for observation to begin. Questions at forum are like this: “Yes, I understand that you may start your observation in about 5 sec. time range, but I don’t understand why it works? How can you be sure that the number you get in the end of observation has a correlation with a final outcome number?”

So let’s start from this. Imagine you’re doing traditional VB. Here you try to identify particular ball rotation after which the ball travels the same amount of rotations Let’s say rotor speed in this spin is 4.6 s/rot.; ball is CW; ball deceleration (Forester calls it ACC) is 220 ms in our targeted rotation; we’re trying to identify ball rotation when it is 7 rotations left before then ball hits DD. That moment ball speed is about 1 sec. per rotation. We identify this moment and read number 10 under the DD where the ball is passing right now. After about 10 seconds ball hits DD when number 29 is under the ball. For the next spin we add an offset of 12 numbers and continue playing. If the next rotation the ball is above the number 34 for example (wheel speed is the same, 7 rotations before then ball drops) - we predict number 34+12=16, ball hits number 16 when it is above the DD and we are happy.

What’s about VB2 in this case? According to the data we have - time “T” (which we need to calculate for VB2 T=1000ms/4.6s/220ms)is approx. 1 second. In VB2 we start our observations when the ball is above the DD. Imagine we start our calculated time (1 second) in the same spin, like in previous example of traditional VB. So we start it when there are 7 rotations left to the end, number under the ball is 10 (but we don’t care about it in this case) and when it finishes number 31 is under the ball. Then the ball hits DD at number 29 and we add an offset of 4 numbers for next rotations. The next rotation again we start timer when 7 rotations are left, we start at DD when number under the ball is 34 (we don’t care, we just start timer when ball is at DD) and time of 1 second finishes when there is number 23 under the ball. We add an offset of 4 and get number 23+4=16.

Until this I think everybody should understand that there is no difference in between the first and the second method. In all the upcoming spins we will get the same results. The ONLY difference is an OFFSET. In the first case it is 12 numbers while in the second 4 numbers. That’s all. Of course for different wheel speeds offsets will vary but the final results (predicted numbers + offsets) will be the same.
I.E. the rotor is faster by 1 pocket per second (it means that in 10 seconds which are left for the ball before it drops the rotor will travel 10 additional pockets) - therefore in the first case VB offset will be 22 (12+10), and in the second case VB2 offset will be 14 (4+10).

But where is the point of VB2 system? That’s where all the interesting things begin :). Forester have explained in a pretty good way, what is time “T” in VB2. I’ll just repeat that it is time during which the wheel travels the same amount of pockets as the ball slows down. What does it mean practically?
Imagine number 0 zero is at 12 o’clock. Ball is there too. Our T=2 sec. Ball speed 500 ms/rot. Rotor speed 4 sec/rot. We start time; when we finish we read number. The ball is at 6 o’clock and it is number 0 again. It is so because during 2 seconds ball slowed down and didn’t reach full 4 rotations, but only 3.5; on the other hand rotor moved for 18 pockets CCW, so number “0” meets with the ball again. The difference between start and final position is 6 hours (o’clock). We start timer again (when the ball is at DD at 12 o’clock and number 0 is at 5:30 o’clock) During next 2 seconds ball travels let’s say 1.9 rotations, so it is at 11:30 o’clock. The rotors moves 6 o’clock, so number 0 is at 11:30 o’clock too.
That’s it! During the time range in which the linearity occurs we can start our timer at DD at any rotation. For example linearity occurs from 13 to 7 rotations before then ball drops. It means that we can start our timer at DD when 13, 12, 11, 10, 9, 8, 7 rotations are left to the end and we will still read the same number!

Now you can connect this information with one I wrote above. If we target 7 ball rotation till the end - I proved that there will be no difference between traditional VB and VB2 (despite an offset). With VB2 we have read number 31. BUT! If we started timer at DD when 8 rotations left till the end (one rotation earlier)- guess what would be? Yes, you’re right! We will read number 31 again when the time ends. 9 rotations till the end - again 31. 10 rotations - and again 31!

Summary:

1. If you target particular rotation (i.e. 7 before the end), read number (first) when ball is at DD and add constant time difference after reading the number and then read number under the ball again (second number) - you’ll always get constant offset of numbers between the first and the second number.
2. If added constant time equals to time T calculated by VB2 formula -you will get second number not only when 7 rotations left before ball hits DD but when 9, 10, 6 or 11 rotations left before the ball hits DD (number of rotations suitable for VB2 application depends on time during which linearity occurs).
3. Therefore here is a constant correlation between a number predicted by VB2 and the number under the ball when it hits DD (same result as in traditional VB, but in less complicated way).

It is more convenient and practical to use VB2 to compare with methods which imply certain rotation recognition! If this post makes it clearer for someone - I will be very glad! Best wishes, Shotman.

Imagine number 0 zero is at 12 o’clock. Ball is there too. Our T=2 sec. Ball speed 500 ms/rot. Rotor speed 4 sec/rot. We start time; when we finish we read number. The ball is at 6 o’clock and it is number 0 again. It is so because during 2 seconds ball slowed down and didn’t reach full 4 rotations, but only 3.5; on the other hand rotor moved for 18 pockets CCW, so number “0” meets with the ball again.
You cant get such T on that conditions like You write because ball deceleration when ball speed is about 500 ms per rot can be in max 50-80 ms

I know, I just tried to explain principles of how VB2 works, therefore I said “imagine”. Of course in real life there will be other numbers but principles will be same. ball deceleration per rotation ball deceleration per second

on the green graph you can see reasonable linearity from 3-9 sec

LET’S SAY ROTOR IS 9POCKETS PER SEC.
In 4 sec it moves about one rotation.
All we need is “reference time” where;

When we apply it at 14 sec before end the ball makes 3 rotations
And if we apply it 10 sec to the end it makes 2 rotations
That was for easy understanding but it could be any amount of rotations as long as 4 sec. later it makes on less rotation.
It could be 2.5 then 1.5
If we apply reference time at 14 sec
The ball makes 2.5 rotations
If we applied it at 13 sec to end the ball makes 2.5 rotation minus 9 pockets
12 its minus another 9 …etc
How else to predict so early in time? : 