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3D Coordinate Mapping Solution

In robotic systems and autonomous navigation, depth cameras (like RGB-D sensors) represent the surroundings as a 3D point cloud. A robot arm is situated at the origin (0,0,0)(0,0,0), and its software needs to calculate distances between detected objects to prevent collisions.

A depth sensor detects a cylindrical obstacle with a safety envelope modeled as a sphere centered at C(3,−4,2) metersC(3, -4, 2)\text{ meters} with a radius of r=1.5 metersr = 1.5\text{ meters}. A point obstacle is detected at P(4,−2.5,3) metersP(4, -2.5, 3)\text{ meters}.

  1. Find the distance between the robot arm at the origin O(0,0,0)O(0,0,0) and the center of the spherical safety envelope CC.
  2. Determine the standard equation of the spherical safety envelope.
  3. Calculate the distance from the point obstacle PP to the center of the safety envelope CC. Determine if the point obstacle has breached the safety envelope (i.e., is it inside the sphere?).

Step-by-Step Solution:

  1. Find the Distance from the Origin to CC: Using the 3D distance formula between O(0,0,0)O(0,0,0) and C(3,−4,2)C(3, -4, 2):

    dOC=(3−0)2+(−4−0)2+(2−0)2=32+(−4)2+22d_{OC} = \sqrt{(3-0)^2 + (-4-0)^2 + (2-0)^2} = \sqrt{3^2 + (-4)^2 + 2^2}
    dOC=9+16+4=29≈5.39 metersd_{OC} = \sqrt{9 + 16 + 4} = \sqrt{29} \approx 5.39\text{ meters}

  2. Determine the Equation of the Sphere: The standard equation of a sphere is (x−h)2+(y−k)2+(z−l)2=r2(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2. Substituting the center (h,k,l)=(3,−4,2)(h,k,l) = (3, -4, 2) and radius r=1.5r = 1.5:

    (x−3)2+(y+4)2+(z−2)2=(1.5)2(x-3)^2 + (y+4)^2 + (z-2)^2 = (1.5)^2
    (x−3)2+(y+4)2+(z−2)2=2.25(x-3)^2 + (y+4)^2 + (z-2)^2 = 2.25

  3. Check if the Safety Envelope is Breached: Calculate the distance from the point obstacle P(4,−2.5,3)P(4, -2.5, 3) to the center C(3,−4,2)C(3, -4, 2):

    dPC=(4−3)2+(−2.5−(−4))2+(3−2)2d_{PC} = \sqrt{(4-3)^2 + (-2.5 - (-4))^2 + (3-2)^2}
    dPC=12+(1.5)2+12=1+2.25+1=4.25≈2.06 metersd_{PC} = \sqrt{1^2 + (1.5)^2 + 1^2} = \sqrt{1 + 2.25 + 1} = \sqrt{4.25} \approx 2.06\text{ meters}
    Compare the distance dPCd_{PC} to the radius rr: Since dPC≈2.06 m>1.5 md_{PC} \approx 2.06\text{ m} > 1.5\text{ m}, the point obstacle is outside the safety envelope. The safety envelope has not been breached.

Meaning of the Answer & Real-Life Application:

Robotic collision-avoidance algorithms use these simple geometric checks in real time. By comparing distance vectors between spatial coordinates and spherical safety envelopes, robots can instantly halt or reroute their paths when a breach is detected, avoiding mechanical damage or injury.