Two drone paths that only cross on the screen
An air-traffic controller is tracking two drones. Drone A has position vector A(t) = (2 + t, 1 + t, 5) km at time t seconds, t >= 0. Drone B has position B(t) = (3t, 4 + t, 9) km at the same time t, because both are launched simultaneously. A two-dimensional display shows only the x- and y-coordinates. On that display the two tracks cross at (7.5, 6.5), and the controller immediately declares a collision risk. A second controller notices that the displayed crossing does not show altitude and asks for a three-dimensional check. The safety rule requires the separation between the drones to be greater than 5 km at all times; at exactly 5 km a warning is issued. A proposal is also made to delay drone B's launch by 30 seconds. The team must decide whether there is a collision, what the minimum separation is, and whether the warning is triggered.
1. Determine whether the two 3D paths intersect. Explain why the 2D display shows a crossing.
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To test whether the paths intersect in 3D, solve A(t)=B(s) for possibly different parameters t and s:
2+t=3s, 1+t=4+s, 5=9.
The third equation is impossible. Therefore the two lines do not intersect in 3D; they are skew. The display shows a crossing because it ignores z. At the displayed point (7.5,6.5), drone A is there at t=5.5 s with z=5 km, while drone B is there at s=2.5 s with z=9 km. Same x and y does not mean the same point in space, and the times are different as well. A 2D crossing is not evidence of a collision.
2. Assuming both drones fly simultaneously with the same parameter t, find the minimum separation and decide whether the 5 km warning is triggered.
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For simultaneous flight, use the same t. The relative position is
R(t)=B(t)-A(t)=(3t-(2+t), 4+t-(1+t), 9-5)=(2t-2, 3, 4).
The separation squared is |R|^2=(2t-2)^2+3^2+4^2=4(t-1)^2+25. This is minimized at t=1, where |R|=√25=5 km. At t=1, A=(3,2,5) and B=(3,5,9). Since the rule requires separation greater than 5 km and the minimum is exactly 5 km, a warning is triggered at t=1. There is no collision because the separation never reaches 0.
3. If drone B's launch is delayed by 30 seconds, its position becomes B(t-30) for t >= 30. Determine whether a collision is possible and find the new minimum separation after both are airborne.
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With the delay, B(t-30)=(3(t-30), 4+(t-30), 9)=(3t-90, t-26, 9). Equating to A(t) still gives z: 5=9, impossible, so no delay can produce a collision while the altitudes remain 5 km and 9 km. The new relative vector is
R(t)=B(t-30)-A(t)=(2t-92, -27, 4).
Then |R|^2=(2t-92)^2+729+16, minimized when 2t-92=0, i.e. t=46 s, when both are airborne. The minimum separation is √745≈27.3 km, which is greater than 5 km, so no warning is triggered after this delay. The delay removes even the 5 km close approach.