Х 91 93 -Z F3 -Zo Figure 1: Geometry to derive inverse kinematics Use the geometry given in Figure 1 to derive equations for q1 and q2. Note: the triangle XYd is in the horizontal (x – y) plane and triangle -Zdk is in a vertical plane that is perpendicular to the (x – y) plane. Hint: q1 can be derived trigonometrically from the Cartesian position (x, y, z), and q2 can be derived in terms of the other angles o and y. 3n/2-q3

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Chapter1: Units, Trigonometry. And Vectors
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Xo
\91
-Z
O F3
-Zo
Figure 1: Geometry to derive inverse kinematics
Use the geometry given in Figure 1 to derive equations for q1 and q2. Note: the triangle XYd is in the
horizontal (x – y) plane and triangle -Zdk is in a vertical plane that is perpendicular to the (x – y) plane.
Hint: q1 can be derived trigonometrically from the Cartesian position (x, y, z), and q2 can be derived in
terms of the other angles v and y.
3n/2-93
Transcribed Image Text:Xo \91 -Z O F3 -Zo Figure 1: Geometry to derive inverse kinematics Use the geometry given in Figure 1 to derive equations for q1 and q2. Note: the triangle XYd is in the horizontal (x – y) plane and triangle -Zdk is in a vertical plane that is perpendicular to the (x – y) plane. Hint: q1 can be derived trigonometrically from the Cartesian position (x, y, z), and q2 can be derived in terms of the other angles v and y. 3n/2-93
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