Give me the coordinates, the wind speed and direction vectors, and the drone’s performance envelope. I’ll run the regression and plot the optimal path on a heatmap—no intuition, just numbers.

First step: pick a desired ground track – say due north (0 °) for the shot. With a 25 m/s airspeed limit, the wind’s 6.2 m/s westward component reduces the northward ground speed to \(v_g = \sqrt{25^2 - 6.2^2} \approx 24 m/s\) But that leaves a 6.2 m/s west drift. To cancel it, turn the heading eastward by \(\theta = \arctan\!\left(\frac{6.2}{24}\right) \approx 14.6 °\) So fly 14.6 ° east of north. Fuel usage: at full load a 15‑minute flight at 25 m/s burns 1.25 kW, so a 10‑second shot consumes about 20 Wh—tiny compared to the battery. If you need a tighter path or a faster turn, adjust the heading proportionally to the wind vector and recalculate the required acceleration (≤2 m/s²). That’s the only data‑driven answer; intuition doesn’t help here.

The wind vector you gave is purely westward – no north component, so the tail‑wind you mentioned isn’t in the data. If you do have a northward element, add it to the vector, recompute the heading with the new horizontal components, and then recalculate the required acceleration. For battery safety, bump the reserve up to 30 % of total capacity; that gives you a cushion for unexpected wind shifts or a longer hold‑over. The revised heading remains roughly 14.7° east of north to null the westward drift. Double‑check the numbers, and we’ll hit the quad with precision.