Calibration Question, Parallelogram

Shop Forum Makelangelo Polargraph Art Robot Calibration Question, Parallelogram

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  • #10048
    Anonymous
    Inactive

    Hi,

    I’m trying to fine tune things on my system and make sure I have everything measured and entered correctly.

    I am noticing the entire image is in the shape of a slight Parallelogram.

    The bottom left, and top right corners are acute, and the bottom right and top left are obtuse. I’ve tried shifting the home position in all directions, and it doesn’t seem to help. I also checked that the pulleys are the exact same size (I laser cut them).

    How far down should the home position be? one of the other posts said “about 16cm”. Is that from the motor spindles?

    #10049
    Dan
    Keymaster

    Can you show a picture? Is your pulley diameter ~1.27 cm?

    #10052
    Anonymous
    Inactive

    Sure, it is a little hard to see in the photo, I am putting a square up to it to measure the angles.

    My pulley has a diameter closer to 36.2 mm, but I had to enter 18.1 mm in the software. Initially things were all over the place.

    I drew a 150mm square, and the measured length was 146mm, so that is pretty close right now with the current settings.

    I am using the adafruit steppers. I changed the steps per revolution to 200. I’m guessing that I missed something along the way.

    Square Prints

    #10053
    Anonymous
    Inactive

    These are some better examples of the issue.

    Bottom Left Corner

    Bottom Right

    #10063
    Anonymous
    Inactive

    I had a thought on calibration that I am hoping will fix my problem.
    If I can get the length the bot thinks the belts are, I can measure and adjust the pulley size until it is exact. I can also do spot checking at different locations.

    In the where() function, I added the code below.
    I believe this is giving me the lengths, but I’m not sure the units (and Thread Per Step is 0).
    My results are.. X-0.00 Y-0.06 Z55.00 F20000.00 L1 20321 L2 20322 Thread Per Step 0.00

    If L1 and L2 are in steps, I should be able to do
    (Pulley Diameter * Pi) / Steps per revolution * L(value)

    Any thoughts on this method, other things I need to consider?

    Thanks
    Robin

      long l1, l2;
      IK(posx, posy, l1, l2);
    
      Serial.print(F(" L1 "));
      Serial.print(l1);
      Serial.print(' ');
    
      Serial.print(F(" L2 "));
      Serial.print(l2);
      Serial.print(' ');
    
      Serial.print(F(" Thread Per Step "));
      Serial.print(THREAD_PER_STEP);
      Serial.print(' ');
    #10064
    Anonymous
    Inactive

    I’m doing something wrong, or I don’t understand something completely.
    My home position comes out to 57.04 for l1, and l2
    It can’t be mm, if the spool diameter was converted to cm, that would be close, but still too short.

    
      Serial.print(F(" L1"));
      Serial.print(l1*SPOOL_DIAMETER*PI/STEPS_PER_TURN);
      Serial.print(' ');
    
      Serial.print(F(" L2"));
      Serial.print(l2*SPOOL_DIAMETER*PI/STEPS_PER_TURN);
      Serial.print(' ');
    
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