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calibration [2019/07/19 16:06] visentincalibration [2019/08/14 14:52] visentin
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-====== Calibration ====== +===== Calibrating the Delta Robot ===== 
-The ''Calibration Sequence'' can be started by triggering the safety event ''doCalibrating''. This can be done by pressing the red button until this lights up. Press and hold the red button for 2 seconds, this will trigger the Safety Event ''doCalibrating''. When finished with calibrating, press the red button once again. Press the green button to get back in normal operation mode. +If you need to calibrate the positions of the blocks for the Delta robot follow this steps: 
-==== Mechanical Limits of the Axis Motors ==== +  - The Delta robot has to be powered up and running 
-The axis motors have mechanical upper and lower limits.+  - If the Delta is not in it's homing position: 
 +     Press the blue button to stop the normal operating mode 
 +  - Else: 
 +     - Press the green and blue button together 
 +  - Remove all plates
  
-^Upper limit^Upper limit detail^Lower limit^ +You have to calibrate the Delta with each plateStart with plate 0 -> that means no plate is used, and position 0. 
-|{{:delta:zero_ref_ntb.jpg?400|}}|{{:delta:zero_ref_ntb_2.jpg?300|Upper limit calculation detail}}| {{:delta:motor-axis-calculation-lower-n.jpg?300|}}| +  Put the actual plate to the actual position and move the TCP to the center of this position. 
-|Upper limit calculation shown below.| Upper limit calculation detail |Lower limit of the Axis Motor measured on the Delta robot, γ is about 5°.|+  Hit the blue button. 
 +  Move the TCP inside the gap of the current block only in x and y direction. 
 +  Hit the blue button again
 +  - If the actual position is number 3 go to the next stepElse increment the position number and start again with step 1. 
 +  - If the actual plate is number 3 you are finished. Else use the plate with the next higher number and start again with step 1.
  
-L = 17.5 \\ +If you have calibrated all blocks on every position, put all blocks back and press the blue button. The Delta will move to it'homing position.
-B = 15   \\ +
-s = 4    \\ +
- +
-The upper limit can be calculated as follows: +
-\\ \\ +
-//(1) tan(β) = sin(β)/cos(β) = (B-y)/L// \\ +
-//(2) cos(β) = s/y// \\ +
-\\ +
-Substitute (2) into (1) to get: \\ +
-\\ +
-//(3) L*sin(β)-B*cos(β)=-s// \\ +
-\\ +
-Using following trigonometric formulas in (3):  +
-\\ \\ +
-//(4) t = tan(β/2) // \\ +
-//(5) sin(β) = 2*tan(β/2)/(1+tan^2(β/2) // \\ +
-//(6) sin(β) = (1-tan^2(β/2)/(1+tan^2(β/2) // \\ +
-\\ +
-Leads to: +
-\\ \\ +
-//(7) (s+B)*t^2 + 2*L*t + (s-B) = 0// \\ +
-\\  +
-Solutions are:  +
-\\  +
-// t1 = 0.2736 -> β = 30.6°// \\ +
-// t2 = -2.1157 (neg. solution)// \\ +
-\\ \\ +
- +
- +
- +
-OLD OLD OLD +
- +
-The subtraction of a cosine function from a sine function results in another sine function with phase shift. So we are looking for a function in the form of \\ +
-\\ +
-//(4) A*sin(β+φ) = L*sin(β)-B*cos(β)// \\ +
-\\  +
-By using the addition theorem we get \\ +
-\\ +
-//A*sin(β+φ) = A*sin(β)*cos(φ)+A*cos(β)*sin(φ) \\ +
-\\ +
-Now we can compare the coefficients from sin(β) and cos(β) \\ +
-\\ +
-//(5) L=A*cos(φ)// \\ +
-//(6) -B=A*sin(φ)// \\ +
-\\   +
-By dividing (6) by (5) we get \\ +
-\\ +
-// -B/L=sin(φ)/cos(φ)=tan(φ) -> φ = arctan(-B/L)=-0.70863 (rad)//\\ +
-\\ +
-Taking the square of each function and adding them together results in\\ +
-\\ +
-// L<sup>2</sup>+(-B)<sup>2</sup> = A<sup>2</sup>*(sin<sup>2</sup>(φ)+cos<sup>2</sup>(φ)) = A<sup>2</sup>// \\ +
-// A = sqrt(L<sup>2</sup>+(-B)<sup>2</sup>) = 23.048 //\\ +
-\\ +
-Inserting the values in (4)\\ +
-\\ +
-//23.048*sin(β-40.60°)=17.5*sin(β)-15*cos(β)=-4//\\ +
-\\ +
-we can now solve for β\\ +
-\\ +
-//β=arcsin(-4/23.048)+0.70863 = 0.53420 (rad) = 30.6°\\ +
-\\ +
-Hence, the maximum angle the axis can turn is // 90° + γ + β ≅ 120°//.\\ +
- +
-==== Mechanical Limits of the Tool Center Point Motor ==== +
-The TCP rotation is also limited by two metall pins. The maximum turn angle is calculated as follows:\\ +
-{{ :delta:tcp-limit.jpg?300 |TCP limit}} +
- +
-//sin(γ/2) = r/R//\\ +
-//γ=2*arcsin(r/R)= 0.0600 (rad) = 3.438°//\\ +
-\\ +
-That is the value for one side. The other side is identical, so the maximum turn angle is\\ +
-\\ +
-//2*PI - 2 * 0.06 = 6.16 (rad) ≅ 353°//.+