# Incremental Encoders

Discussion in 'Homework Help' started by eulerpi, Jul 15, 2007.

1. ### eulerpi Thread Starter New Member

Jul 15, 2007
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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 !!!

2. ### hgmjr Moderator

Jan 28, 2005
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See subsequent posting for my self-correction to the above statement.

hgmjr

3. ### JoeJester AAC Fanatic!

Apr 26, 2005
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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.

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
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