Spindle orient 1/2 ratio advice

   Hello
I am retrofiting a mill with 21 tool ATC ( CINCINNATI MILACRON 500 ARROW), after a while is almost finished ,a lot of things work ,PB is very nice , relative easy to install and configuring for beginers like me.Retrofit  it s with 7i92+7i77+7i84 , +/- 10 V , same for spindle , keep the original drivers.
 For ATC ,I use parameters 1189+1190 to 2010 in .var file , they are volatile , but become persistent after  activate them(after  restart the tools remain in corousel)
The head is 1/2 ratio from spindle motor , where the encoder is located, and the M19 , orient spindle is work fine , but stops sometimes in good position O degree ,or with 180 degree plus position(the reason is the 1/2 gear) The position must be always O degree for the mechanics of toolholders , so I need some advice how to manage it.
I mentioned that the head has a inductive sensor , wich is active on 180 degree side of shaft and inactive for the rest of 180 degrees side of the head shaft.
I make a few routines in the orient.ngc file that is atached to spindle orient button from ui  , but still not work.
I apreciate if somebody can help me , with advices
Thanx a lot
Denis

the spindle itself needs an index point that repeats in the correct location everytime.. if you have a ratio of 2:1 and the encoder is on the motor, that means that the index pulse of the motor is consistent, but the spindle will always be 50/50 it repeats. for indexing use a index from the spindle itself and configure to find that index and then use the next motor encoder index. this will ensure the ratio is on the correct 50% side everytime!

Thanx a lot for the answer
This is what it already tryed , but without succes.this is the reason why I asked for help.
I will try again , hope will be better
Something like this :

o<orientspindle> sub
S50
G4 P2
one of this two commands
M66 P7 L3 Q2 # P7 is the pin from the spindle that is active on the half side of rotation , atached to motion and hal
or M66 P7 L1 Q2
o100 if [#5399 LT 1]
(abort , spindle not oriented )
M5
o100 endif
M19 R0 P1
o<orientspindle> endsub [1]
M2

Thanx again.

Denis

Perhaps I misunderstand, what is the encoder counting at a 1:1 ratio?

Hello
At 1/1 ratio , start rotation =0 , end one rotation=1.
i try
o<orientspindle> sub
M3 S5
M66 P7 L2 Q6 # P7 is the pin from the spindle that is active on the half side of rotation , atached to motion and hal , when the pin is from hi to low , then
M19 R0 P1
o<orientspindle> endsub [1]
M2

it is much better , but still have 180 degrees error sometimes.
thanx for help , i apriciated your implication
Denis