calibration
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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 '' | ||
- | ==== Mechanical Limits of the Axis Motors ==== | ||
- | The axis motors have mechanical upper and lower limits. | ||
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- | ^Upper limit^Upper limit detail^Lower limit^ | ||
- | |{{: | ||
- | |Upper limit calculation shown below.| Upper limit calculation detail |Lower limit of the Axis Motor measured on the Delta robot, γ is about 5°.| | ||
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- | L = 17.5 \\ | ||
- | B = 15 \\ | ||
- | s = 4 \\ | ||
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- | The upper limit can be calculated as follows: | ||
- | \\ \\ | ||
- | //(1) tan(β) = sin(β)/ | ||
- | //(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/ | ||
- | //(6) sin(β) = (1-tan^2(β/ | ||
- | \\ | ||
- | Leads to: | ||
- | \\ \\ | ||
- | //(7) (s+B)*t^2 + 2*L*t + (s-B) = 0// \\ | ||
- | \\ | ||
- | Where: | ||
- | \\ | ||
- | // t1 = 0.2736 -> β = 30.6°// \\ | ||
- | // t2 = -2.1157 (neg. solution)// \\ | ||
- | \\ \\ | ||
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- | OLD OLD OLD | ||
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- | 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 \\ | ||
- | \\ | ||
- | // | ||
- | \\ | ||
- | 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/ | ||
- | \\ | ||
- | Taking the square of each function and adding them together results in\\ | ||
- | \\ | ||
- | // L< | ||
- | // A = sqrt(L< | ||
- | \\ | ||
- | Inserting the values in (4)\\ | ||
- | \\ | ||
- | // | ||
- | \\ | ||
- | we can now solve for β\\ | ||
- | \\ | ||
- | // | ||
- | \\ | ||
- | 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:\\ | ||
- | {{ : | ||
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- | //sin(γ/2) = r/R//\\ | ||
- | // | ||
- | \\ | ||
- | 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°//. | ||