Here is the remaining battery capacity vs. time for the previous 7 km out and back.. I have marked the start, turn around and end times.
The rate (slope) consumption is fairly constant as a function of time in the upwind and downwind directions. If air speed is governed to ~12 m/s power consumption should be fairly constant. If you were to plot remaining battery vs. distance traveled then you calculate:
For 6000 mAhr
upwind rate = 42%/7km = 6%/km
downwind = 26%/7km = 3.7%/km
So using 14km to turn around: 6%/km * 14km = 84% would be used of 6000 mAhr to get there
and 3.7% /km * 14km = 52% of 6000 mAhr to return.
Which looks possible with 12,000 mAhr and no climbing.
We could refine this and have a two part equation by testing consumption in a sustained climb with 12000 mAhr of batteries. Based on the above you have about 30% of 6000 mAhr to climb about 900 m over 14 km. Climbing into the wind will definitely help so we are on the edge of plausible. I would like to include wind speed into this estimation at some point.
Making some assumptions here but that's better than just hoping.
We do need to do more testing with climbs to ascertain the most efficient climb angle/speed. The difference between the upwind and downwind consumption reinforces the fact that we need to watch the weather closely and choose our mission day with as close to zero windspeed as possible.
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