Tracking Actuator Angular Velocity (Robotics & Computer Engineering) Solution

Problem Overview

We evaluate the required angular sweep speed $\frac{d\theta}{dt}$ of an automated radar tracking dish following an approaching aircraft at fixed altitude:

\theta(z) = \tan^{-1}\left(\frac{3}{z}\right), \qquad \frac{dz}{dt} = -0.5\text{ km/s}

Step-by-Step Mathematical Solution

Step 1: Differentiate Inverse Tangent with Respect to Spatial Distance $z$

Let $u = \frac{3}{z} = 3z^{-1}$ . By the Inverse Tangent Rule $\frac{d}{dz}[\tan^{-1}(u)] = \frac{u'}{1+u^2}$ :

u' = \frac{d}{dz}[3z^{-1}] = -3z^{-2} = -\frac{3}{z^2}

\frac{d\theta}{dz} = \frac{1}{1 + \left(\frac{3}{z}\right)^2} \cdot \left(-\frac{3}{z^2}\right) = \frac{1}{1 + \frac{9}{z^2}} \cdot \left(-\frac{3}{z^2}\right)

Step 2: Clear Compound Fractions Algebraically

Multiply $z^2$ into the denominator sum:

\frac{d\theta}{dz} = \frac{-3}{z^2\left(1 + \frac{9}{z^2}\right)} = \frac{-3}{z^2 + 9}\text{ rad/km}

Step 3: Link to Time via the Chain Rule

\frac{d\theta}{dt} = \frac{d\theta}{dz} \cdot \frac{dz}{dt} = \left(\frac{-3}{z^2 + 9}\right) \cdot \left(\frac{dz}{dt}\right)

Step 4: Evaluate at Horizontal Distance $z = 3\text{ km}$

Substitute $z = 3$ and $\frac{dz}{dt} = -0.5\text{ km/s} = -\frac{1}{2}\text{ km/s}$ :

\frac{d\theta}{dt} = \left(\frac{-3}{(3)^2 + 9}\right) \cdot \left(-\frac{1}{2}\right) = \left(\frac{-3}{9 + 9}\right) \cdot \left(-\frac{1}{2}\right) = \left(-\frac{3}{18}\right) \cdot \left(-\frac{1}{2}\right)

\frac{d\theta}{dt} = \left(-\frac{1}{6}\right) \cdot \left(-\frac{1}{2}\right) = \frac{1}{12} \approx 0.0833\text{ rad/s}

Engineering Conclusion

When the aircraft is at $z = 3\text{ km}$ , the dish motor must rotate at $\frac{1}{12}\text{ rad/s} \approx 4.77^\circ\text{/s}$ . Robotics and mechatronics control engineers utilize this inverse trigonometric rate formula to determine whether the mechanical servo-motor torque and gearbox gear ratio can achieve the required slew rate without losing radar tracking lock.