Incremental rotary encoder motor help

AI Thread Summary
The discussion focuses on calculating the angle change and no-load speed RPM for an incremental rotary encoder with 128 counts per revolution. The initial method of dividing encoder counts by 128 to find revolutions and multiplying by 360 for degrees is confirmed as correct. To find RPM, the sample frequency must be known; using a sample time of 5.25 seconds, the calculations yield approximately 567 RPM. Clarifications are provided on how to derive RPM from revolutions per second based on the time taken for the encoder counts. The conversation concludes with an understanding of the correct approach to these calculations.
AC130
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Homework Statement


Hi please see the image that I have attached. The text colored in red is the data I need to calculate.

Homework Equations

The Attempt at a Solution


The encoder has 128 counts per revolution
So to calculate corresponding angle change this I what I did
I divided encoder count in decimal number by 128 which gave the number of revolutions, I then multiplied it by 360 to find the angle change

So for example I did
6350/128 = 49.61 revolutions
49.61 * 360 = 17859.34 degrees

Is this method right?

To calculate no load speed RPM, I divided the revolutions by (time*60) to get it to RPM

So for the 6350 encoder count which gave 49.61 revolutions

I did

49.64/(5.25*60) = 0.1575 RPM

Is this RPM calculation right?

For me it looks as if the RPM is well below and so I must have made a mistake somewhere but I don't know specifically where
 

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AC130 said:
So for example I did
6350/128 = 49.61 revolutions
49.61 * 360 = 17859.34 degrees
That's correct.

To calculate the speed, you must know the sample frequency. Say it is 1Hz. Then you will have 6350 counts per second.
Convert to RPM. ( 128 counts per rev, 60 sec/min ).

The result is close to 3000 RPM.

If the sample frequency were 2Hz, the motor speed would be the double ( ≈ 6000 RPM ).
 
Last edited:
Hesch said:
That's correct.

To calculate the speed, you must know the sample frequency. Say it is 1Hz. Then you will have 6350 counts per second.
Convert to RPM. ( 128 counts per rev, 60 sec/min ).

The result is close to 3000 RPM.

If the sample frequency were 2Hz, the motor speed would be the double ( ≈ 6000 RPM ).

Hi can you please tell me where RPM came from?
I know that it took 5.25 sec for the angle change and so frequency would be 1/T = 1/5.25 = 0.1905Hz

But I don't really know how to find RPM
 
AC130 said:
So for the 6350 encoder count which gave 49.61 revolutions
AC130 said:
I know that it took 5.25 sec for the angle change
So 49.61 rev. within 5.25 sec, which gives 9.449 rev./sec ≈ 567 rev/min. = 567 RPM. ( 1 min = 60 sec ).

The RPM's in #2 is just an example, not knowing your mentioned 5.25 sec.
 
Ok thanks I understand it now
 
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