Question

The forward kinematics for an industrial robot is given as: $x = d_3\sin(\theta)$, $y = d_3\cos(\theta)$, $z = 1 + d_2$ If you are given the end effector position as [6,3,3], after solving the inverse kinematics for this industrial robot, $\theta$ in degrees is equal to: 63.43 14.10 None of them 26.57 44.40

          The forward kinematics for an industrial robot is given as:
$x = d_3\sin(\theta)$, $y = d_3\cos(\theta)$, $z = 1 + d_2$
If you are given the end effector position as [6,3,3], after solving the inverse kinematics for
this industrial robot, $\theta$ in degrees is equal to:
63.43
14.10
None of them
26.57
44.40

The forward kinematics for an industrial robot is given as:
x = d3sin(θ), y = d3cos(θ), z = 1 + d2
If you are given the end effector position as [6,3,3], after solving the inverse kinematics for
this industrial robot, θ in degrees is equal to:
63.43
14.10
None of them
26.57
44.40

Added by Andrew O.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Eduard Sanchez

Step 1

Let's substitute these values into the equations: 6 = d * sin(θ) 3 = d * cos(θ) 1 + d = 3 From the third equation, we can solve for d: 1 + d = 3 d = 2 Now, let's substitute this value of d into the first equation: 6 = 2 * sin(θ) 3 = cos(θ) To solve for θ, we Show more…

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The forward kinematics for an industrial robot is given as x = d * sin(Î¸) * y = d * cos(Î¸) = 1 + d. If you are given the end effector position as [6,3,3], after solving the inverse kinematics for this industrial robot in degrees is equal to: 63.43 14.10 None of them 26.57 44.40