# Incremental Encoders

#### eulerpi

Joined Jul 15, 2007
1
Two incremental encoders are set up as a Vernier. Both send out 500 pulses per revolution (no quadrature). Encoder A is geared such that 49 turns of the motor result in one turn of Encoder A.

Encoder B is similar, only geared 50:1. Initially, the two index (AKA home) signals are synchronized. After the system has been running awhile, Encoder A sends a home signal, 30 Encoder B pulses later (constant direction), Encoder B sends a home signal.

How far has the MOTOR turned??

I came up with 120 Rev. but that seems like to many.

[360deg/rot] / [3deg/count] = 120

I do not think this is correct but I gave it a shot.

Thanks in advance for any help !!!

#### hgmjr

Joined Jan 28, 2005
9,027
hgmjr said:
After considering the problem as stated, It would seem that the difference in the number of encoder pulses between the home pulses from each of the two motors would always be an integer multiple of 500.

Is it possible that the problem is mistated?

hgmjr
See subsequent posting for my self-correction to the above statement.

hgmjr

#### JoeJester

Joined Apr 26, 2005
4,390
The question ... How far has the motor turned? ... is that relative since the begining or relative since the encoder A index signal?

120 revolutions is way too small if it's just from receipt of A's index to receipt of B's index.

#### hgmjr

Joined Jan 28, 2005
9,027
Ooopppsss. I see where I went off the rails. I had the gear ratio reversed on each of the motors. It is 49 motor revolutions per encoder revolution and 50 motor revolutions per encoder revolution on motor A and B respectively.

With that cleared up, the question boils down to how many encoder pulses does motor B lose each time motor A generates a home pulse.

hgmjr