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Tracking Actuator Angular Velocity (Robotics & Computer Engineering) Solution

Formulas & Definitions

Problem Overview

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

ΞΈ(z)=tanβ‘βˆ’1(3z),dzdt=βˆ’0.5Β km/s\theta(z) = \tan^{-1}\left(\frac{3}{z}\right), \qquad \frac{dz}{dt} = -0.5\text{ km/s}ΞΈ(z)=tanβˆ’1(z3​),dtdz​=βˆ’0.5Β km/s


Step-by-Step Mathematical Solution

Step 1: Differentiate Inverse Tangent with Respect to Spatial Distance zzz

Let u=3z=3zβˆ’1u = \frac{3}{z} = 3z^{-1}u=z3​=3zβˆ’1. By the Inverse Tangent Rule ddz[tanβ‘βˆ’1(u)]=uβ€²1+u2\frac{d}{dz}[\tan^{-1}(u)] = \frac{u'}{1+u^2}dzd​[tanβˆ’1(u)]=1+u2u′​:

uβ€²=ddz[3zβˆ’1]=βˆ’3zβˆ’2=βˆ’3z2u' = \frac{d}{dz}[3z^{-1}] = -3z^{-2} = -\frac{3}{z^2}uβ€²=dzd​[3zβˆ’1]=βˆ’3zβˆ’2=βˆ’z23​
dΞΈdz=11+(3z)2β‹…(βˆ’3z2)=11+9z2β‹…(βˆ’3z2)\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)dzdθ​=1+(z3​)21​⋅(βˆ’z23​)=1+z29​1​⋅(βˆ’z23​)

Step 2: Clear Compound Fractions Algebraically

Multiply z2z^2z2 into the denominator sum:

dΞΈdz=βˆ’3z2(1+9z2)=βˆ’3z2+9Β rad/km\frac{d\theta}{dz} = \frac{-3}{z^2\left(1 + \frac{9}{z^2}\right)} = \frac{-3}{z^2 + 9}\text{ rad/km}dzdθ​=z2(1+z29​)βˆ’3​=z2+9βˆ’3​ rad/km

Step 3: Link to Time via the Chain Rule

dΞΈdt=dΞΈdzβ‹…dzdt=(βˆ’3z2+9)β‹…(dzdt)\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)dtdθ​=dzdθ​⋅dtdz​=(z2+9βˆ’3​)β‹…(dtdz​)

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

Substitute z=3z = 3z=3 and dzdt=βˆ’0.5Β km/s=βˆ’12Β km/s\frac{dz}{dt} = -0.5\text{ km/s} = -\frac{1}{2}\text{ km/s}dtdz​=βˆ’0.5Β km/s=βˆ’21​ km/s:

dΞΈdt=(βˆ’3(3)2+9)β‹…(βˆ’12)=(βˆ’39+9)β‹…(βˆ’12)=(βˆ’318)β‹…(βˆ’12)\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)dtdθ​=((3)2+9βˆ’3​)β‹…(βˆ’21​)=(9+9βˆ’3​)β‹…(βˆ’21​)=(βˆ’183​)β‹…(βˆ’21​)
dΞΈdt=(βˆ’16)β‹…(βˆ’12)=112β‰ˆ0.0833Β rad/s\frac{d\theta}{dt} = \left(-\frac{1}{6}\right) \cdot \left(-\frac{1}{2}\right) = \frac{1}{12} \approx 0.0833\text{ rad/s}dtdθ​=(βˆ’61​)β‹…(βˆ’21​)=121β€‹β‰ˆ0.0833Β rad/s


Engineering Conclusion

When the aircraft is at z=3Β kmz = 3\text{ km}z=3Β km, the dish motor must rotate at 112Β rad/sβ‰ˆ4.77∘/s\frac{1}{12}\text{ rad/s} \approx 4.77^\circ\text{/s}121​ rad/sβ‰ˆ4.77∘/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.