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Autonomous Drone LiDAR Obstacle Clearance (Computer Engineering) Solution

Formulas & Definitions

Problem Overview

An autonomous quadcopter following the trajectory y=0.5x2y = 0.5x^2y=0.5x2 on x∈[−4,4]x \in [-4, 4]x∈[−4,4] detects a stationary transmission pylon at (0,4)(0, 4)(0,4) (coordinates in decameters, 1 dam=10 m1\text{ dam} = 10\text{ m}1 dam=10 m). We determine the closest point of approach (CPA) and minimum clearance distance using the Squared-Distance Optimization Technique and the Second Derivative Test, verifying compliance with the 2.0-dam2.0\text{-dam}2.0-dam (20-m20\text{-m}20-m) collision-avoidance safety margin.


Step-by-Step Mathematical Solution

Step 1: Formulate the Euclidean Distance Function

The Euclidean distance between the pylon at (x0,y0)=(0,4)(x_0, y_0) = (0, 4)(x0​,y0​)=(0,4) and any point (x,y)(x, y)(x,y) along the flight path is:

d(x)=(x−x0)2+(y−y0)2=(x−0)2+(0.5x2−4)2d(x) = \sqrt{(x - x_0)^2 + (y - y_0)^2} = \sqrt{(x - 0)^2 + (0.5x^2 - 4)^2}d(x)=(x−x0​)2+(y−y0​)2​=(x−0)2+(0.5x2−4)2​
(Governing equations defined in Applied Optimization Modeling Framework.)

Step 2: Formulate the Squared-Distance Objective Function

Applying the Squared-Distance Optimization Technique, we differentiate the squared distance D(x)=[d(x)]2D(x) = [d(x)]^2D(x)=[d(x)]2 to eliminate the radical chain rule:

D(x)=x2+(0.5x2−4)2D(x) = x^2 + (0.5x^2 - 4)^2D(x)=x2+(0.5x2−4)2
Expand the quadratic term:
D(x)=x2+(0.25x4−4x2+16)=0.25x4−3x2+16D(x) = x^2 + \left( 0.25x^4 - 4x^2 + 16 \right) = 0.25x^4 - 3x^2 + 16D(x)=x2+(0.25x4−4x2+16)=0.25x4−3x2+16
Feasible domain: x∈[−4,4] damx \in [-4, 4]\text{ dam}x∈[−4,4] dam.

Step 3: Differentiate and Find Critical Numbers

Differentiate D(x)D(x)D(x) with respect to xxx:

D′(x)=ddx[0.25x4−3x2+16]=x3−6xD'(x) = \frac{d}{dx}\left[ 0.25x^4 - 3x^2 + 16 \right] = x^3 - 6xD′(x)=dxd​[0.25x4−3x2+16]=x3−6x
Set D′(x)=0D'(x) = 0D′(x)=0 to identify stationary points:
x(x2−6)=0  ⟹  x=0orx=±6≈±2.449 damx(x^2 - 6) = 0 \implies x = 0 \quad \text{or} \quad x = \pm\sqrt{6} \approx \pm 2.449\text{ dam}x(x2−6)=0⟹x=0orx=±6​≈±2.449 dam
All three critical numbers lie strictly within the operational flight corridor [−4,4][-4, 4][−4,4].

Step 4: Classify Critical Points via the Second Derivative Test

Compute the second derivative:

D′′(x)=3x2−6D''(x) = 3x^2 - 6D′′(x)=3x2−6
Evaluate D′′(x)D''(x)D′′(x) at each critical number:

  1. At x=0x = 0x=0:
    D′′(0)=3(0)2−6=−6<0D''(0) = 3(0)^2 - 6 = -6 < 0D′′(0)=3(0)2−6=−6<0
    Because D′′(0)<0D''(0) < 0D′′(0)<0, the flight path directly beneath the pylon at (0,0)(0, 0)(0,0) represents a local maximum of the distance function among the stationary points (d=16=4 damd = \sqrt{16} = 4\text{ dam}d=16​=4 dam).
  2. At x=±6x = \pm\sqrt{6}x=±6​:
    D′′(±6)=3(6)−6=18−6=12>0D''(\pm\sqrt{6}) = 3(6) - 6 = 18 - 6 = 12 > 0D′′(±6​)=3(6)−6=18−6=12>0
    Because D′′(±6)>0D''(\pm\sqrt{6}) > 0D′′(±6​)>0, both symmetrical points represent local minima.

Step 5: Evaluate Coordinates, Endpoints, and Safety Margin

Evaluate the trajectory altitude y=0.5x2y = 0.5x^2y=0.5x2 at the critical numbers:

y=0.5(±6)2=0.5(6)=3 damy = 0.5(\pm\sqrt{6})^2 = 0.5(6) = 3\text{ dam}y=0.5(±6​)2=0.5(6)=3 dam
The two closest points of approach on the drone trajectory are:
P1(−6,3)≈(−2.45,3.00) damandP2(6,3)≈(2.45,3.00) damP_1\left(-\sqrt{6}, 3\right) \approx (-2.45, 3.00)\text{ dam} \quad \text{and} \quad P_2\left(\sqrt{6}, 3\right) \approx (2.45, 3.00)\text{ dam}P1​(−6​,3)≈(−2.45,3.00) damandP2​(6​,3)≈(2.45,3.00) dam

Calculate the minimum squared distance:

D(±6)=0.25(6)2−3(6)+16=0.25(36)−18+16=9−18+16=7 dam2D(\pm\sqrt{6}) = 0.25(6)^2 - 3(6) + 16 = 0.25(36) - 18 + 16 = 9 - 18 + 16 = 7\text{ dam}^2D(±6​)=0.25(6)2−3(6)+16=0.25(36)−18+16=9−18+16=7 dam2
Taking the square root gives the minimum clearance:
dmin⁡=7≈2.646 dam=26.46 metersd_{\min} = \sqrt{7} \approx 2.646\text{ dam} = 26.46\text{ meters}dmin​=7​≈2.646 dam=26.46 meters

Evaluate the domain boundary endpoints (x=±4x = \pm 4x=±4):

D(±4)=0.25(4)4−3(4)2+16=0.25(256)−48+16=64−48+16=32 dam2D(\pm 4) = 0.25(4)^4 - 3(4)^2 + 16 = 0.25(256) - 48 + 16 = 64 - 48 + 16 = 32\text{ dam}^2D(±4)=0.25(4)4−3(4)2+16=0.25(256)−48+16=64−48+16=32 dam2
d(±4)=32≈5.657 dam=56.57 metersd(\pm 4) = \sqrt{32} \approx 5.657\text{ dam} = 56.57\text{ meters}d(±4)=32​≈5.657 dam=56.57 meters

Comparing outputs:

  • Endpoints: d(±4)≈26.46 m<56.57 md(\pm 4) \approx 26.46\text{ m} < 56.57\text{ m}d(±4)≈26.46 m<56.57 m.
  • Central stationary point: d(0)=40.0 md(0) = 40.0\text{ m}d(0)=40.0 m.
  • Closest points: d(±6)=26.46 md(\pm\sqrt{6}) = 26.46\text{ m}d(±6​)=26.46 m.

Engineering Significance & Conclusion

  1. Safety Clearance: The minimum physical distance between the UAV and the high-voltage transmission pylon is 26.46 meters26.46\text{ meters}26.46 meters (2.646 dam2.646\text{ dam}2.646 dam). Because this clearance exceeds the required minimum safety buffer of 20.0 meters20.0\text{ meters}20.0 meters (2.0 dam2.0\text{ dam}2.0 dam), the navigation flight computer does not trigger an emergency trajectory abort.
  2. LiDAR Firmware Optimization: Computing D(x)=[d(x)]2D(x) = [d(x)]^2D(x)=[d(x)]2 rather than d(x)d(x)d(x) allows embedded flight microcontrollers to execute real-time collision checks in simple polynomial arithmetic, saving processor cycles and reducing latency in obstacle avoidance pipelines.