Discussion in "Electronics" started by    euler314    Jul 15, 2007.
Sun Jul 15 2007, 01:50 PM
I am stuck on this problem>

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 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??

Thanks!!
Tags Incremental EncodersOptical Incremental encodersIncremental coder calculations
Sun Jul 15 2007, 06:11 PM
I am actually confused with these statements

1) Encoder A sends a home signal, 30
2) Both send out 500 pulses per revolution

If i take it as the number of turns By encoder A then, the solution is simple...
i.e. if Enc. A is 30, then
motor turned = 30 x 49 = 1470
and signal by Enc. B = 1470/50 = 29.4
the above answers are only if.. i am taking the statement as 30 = num of turns by encoder A
But if i am wrong... in understanding then,
Tell me if 30 is the number of pulses by A...?
and 500 pulses per rev of what? what about channels?
or the value 500 is just given to confuse the question?
Sun Jul 15 2007, 06:31 PM
The way I understand it is this::

After the system has been running “awhile” Encoder A sends a signal home.

Then, 30 Encoder B pulses later Encoder B sends a signal home. ( so after encoder A sends it home signal-- 30 encoder B pulses later; Encoder B sends a signal home)

The track for both encoders (A & B) produces 500 pulses per revolution.

I am assuming there is a channel A and B for both encoders (Encoder A and Encoder B)

Thanks for the help Rickey

Sun Jul 15 2007, 07:08 PM
Well now i understood the question, i actually took it the other way.. hey bro, would u mind if i answer the question tomorrow?
I was actually designing the new theme for site.. tomorrow, i will give the site a new look so.. surely.. will answer you tomorrow
Good night... (oops! its 5 am morning )
Sun Jul 15 2007, 07:16 PM
That will be great!
Thanks again!
Mon Jul 16 2007, 12:55 AM
I hope i am not wrong this time

Number of turns by motor = x (say)

Number of turns by enc. A = x/49
Number of turns by enc. B = x/50

Number of pulses by enc. ch. A = 500 * x/49
Number of Pulses by enc. ch. B = 500 * x/50

As per the equation:

500* x/49 - 500* x/50 = 30

50(x/49-x/50) = 3

50(x/[49*50]) = 3
x = 3*49
x = 147

What do you say?
Euler like this.
Mon Jul 16 2007, 09:10 AM
I say you are correct.

I think I was putting to much thought into the problem.

Thanks again!!

You have an excellent site!
Mon Jul 16 2007, 09:27 AM
It always happen like this.. i mean..we always make problems bigger.. but actually they are not. enjoy!
and thank you for the complement!

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