With 77% less wing area, how does the Opener BlackFly do a power off stall at less than 28mph (or fly with no power at 28mph)?
I understand the Lazair with 140 sq ft of wing area, was able to fly at 25mph ( some sources quote 18mph), with a MTOW of 450lbs.
With the roughly same MTOW of 450 lbs, in order to fly with just a single prop at 28mph, the blackfly would require a wing area of 180 sq ft, according to my calculations (not sure why that's higher than the Lazair above, but at least it's close), but it does this with only about 40 sqft, or 77% less wing area.
In fact the blackfly can fly a lot slower than 28mph. The blackfly can even land vertically.
It does this, like NASA's X-57, by using props along it's wing's leading edges to produce faster flowing air, thereby creating more lift. When flying slowly, it also uses those props to produce vertical thrust offsetting it's weight, so the wings don't have to produce lift equal to it's weight.
What about in a 28 mph power off stall, when those props are not powered?
The ability to stall under 28mph with power off, is required for it to fly in the US ultralight class.
How does it do this?
My gut feeling is that:
- Demonstrating that it doesn't stall above 28mph probably takes less then a minute . Maybe the props are still turning at thousands of rpm ( 8,000-12,000??) and probably still producing enough thrust to produce the lift required at 28mph even though they are unpowered.
- Maybe the props are producing lift in autorotation during a decent, just like helicopters do in a power off situation (in an ideal world), which provides enough lift with the wings to momentarily fly at 28mph with no power. When partially stalled, I understand wings can still produce 70% lift.
Is this correct thinking?