Plant Transfer Function Derivation
Initially, we modeled the dynamics of a gantry crane and analyzed the forces acting on it to obtain a plant transfer function. From the diagram, we can solve for the acceleration of the payload using the vector sum of the acceleration of the trolley and the acceleration of the payload with respect to the trolley.
Then, we break up each of these accelerations into their normal and tangential components, and then recombine them.
The force due to gravity can be broken up into tangential and normal directions as well:
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The tension on the pendulum cable can only exert a force in the normal direction. Therefore we can apply Newton’s second law to the tangential acceleration and equate it to the tangential force due to gravity.
We can cancel the mass term on both sides, and using the small angle approximation we obtain:
We can then take the Laplace Transform of this second order differential equation to find the transfer function:
To write this in terms of the applied motor velocity, we first write the acceleration in terms of the force and then plug this into the equation above.
Plugging this into the above transfer function gives:
However, this transfer function does not account for wind resistance. After adding in the force due to wind resistance and using a coefficient b to account for the geometry of the payload, the transfer function becomes:
This equation is non-linear, and the variable theta cannot be isolated. We chose to approximate this equation by assuming that theta will not exceed +45 degrees , meaning the angle varies between 0 and +/- 0.785 radians. We will make the assumption that the overall effect of the angle theta on the term can be approximated by a proportionality constant P:
If we define
Then the plant transfer function with wind resistance becomes: