Simulated missions and contingencies

Example 1 A 6 m/s wind along the pipeline

The battery has 700 Wh usable, with 20% kept in reserve on arrival home. Cruise power is 1,200 W, airspeed 16 m/s, and the wind blows along the pipeline away from home at 6 m/s (assumed values).

import numpy as np

cap_wh = 700.0              # usable battery energy
reserve = 0.20              # minimum reserve on arrival home
p_w = 1200.0                # cruise power W
v_air = 16.0                # airspeed m/s
wind = 6.0                  # wind along the route away from home m/s (tailwind out, headwind back)

def energy_wh(dist_m, ground_speed):
    return p_w * dist_m / ground_speed / 3600

usable = cap_wh * (1 - reserve)
out_gs, back_gs = v_air + wind, v_air - wind
d = np.arange(0, 40001, 10)
need = energy_wh(d, out_gs) + energy_wh(d, back_gs)
pnr = d[need <= usable].max()
print(f"usable energy {usable:.0f} Wh")
print(f"point of no return with wind: {pnr/1000:.2f} km from home")
d0 = usable / (2 * p_w / v_air / 3600)
print(f"same calculation in calm air: {d0/1000:.2f} km")

usable energy 560 Wh
point of no return with wind: 11.55 km from home
same calculation in calm air: 13.44 km

A 6 m/s wind shrinks the point of no return from 13.4 km to 11.6 km. The 20 km pipeline therefore cannot be flown out and back in one flight from home; it must be split into two flights, use launch points at both ends, or be launched mid-route by vehicle. This calculation must be redone daily for the actual wind, using wind at flight height, not at the surface.

Example 2 Which stretch of the pipeline has nowhere to land?

There are four landing sites: home, a school field at 6 km, a temple yard at 12 km and the end depot at 20 km. In an emergency the drone must reach a site within 3 km (assumed from reserve energy). Check every 0.1 km.

import numpy as np

route_km = np.round(np.arange(0, 20.05, 0.1), 1)   # points along the pipeline every 0.1 km
sites = {"home": 0.0, "L1 school field": 6.0, "L2 temple yard": 12.0, "end depot": 20.0}
reach_km = 3.0                            # distance that must reach a landing site in an emergency (assumed from reserve energy)

gaps, best = [], []
for x in route_km:
    name, pos = min(sites.items(), key=lambda s: abs(s[1] - x))
    dist = abs(pos - x)
    best.append(dist)
    if dist > reach_km:
        gaps.append(x)
print(f"worst distance to a landing site: {max(best):.2f} km")
if gaps:
    print(f"uncovered stretch: {min(gaps):.1f} to {max(gaps):.1f} km")
    sites["L3 new field"] = 16.0
    worst = max(min(abs(p - x) for p in sites.values()) for x in route_km)
    print(f"after adding L3 at 16.0 km, worst distance {worst:.2f} km")

worst distance to a landing site: 4.00 km
uncovered stretch: 15.1 to 16.9 km
after adding L3 at 16.0 km, worst distance 3.00 km

The stretch from 15.1 to 16.9 km is more than 3 km from any site. Adding a field at 16 km closes the gap, but the worst distance becomes exactly 3.00 km, the limit, with no margin at all; more sites or more reserve energy are needed. Checking every 0.1 km also matters, because checking too coarsely can miss short gaps.