Differences

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--- calibration [2019/07/19 16:06] – visentin
+++ calibration [Unknown date] (current) – removed - external edit (Unknown date) 127.0.0.1
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-====== Calibration ======
-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.
-==== Mechanical Limits of the Axis Motors ====
-The axis motors have mechanical upper and lower limits.
-^Upper limit^Upper limit detail^Lower limit^
-|{{: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|}}|
-|Upper limit calculation shown below.| Upper limit calculation detail |Lower limit of the Axis Motor measured on the Delta robot, γ is about 5°.|
-L = 17.5 \\
-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)// \\
-\\ \\
-Hence, the maximum angle the axis can turn is // 90° + β ≅ 120°//.\\
-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°//.