0
$\begingroup$

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)?

Opener Blackfly top view

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:

  1. 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.

or ....

  1. 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?

$\endgroup$
10
  • $\begingroup$ It's the same voltage, just crazy amount of amps. Obviously you have to size the motors for this increase. I understand brushless DC motors can also withstand short increases in load. That's why Tesla plaid can boast 1,000 Hp, but only for 10 seconds. It's called the C rating of a battery. Pretty standard stuff in the MAV world. $\endgroup$
    – Fred
    Aug 30 at 19:52
  • $\begingroup$ Sorry, I disagree. Power is volts x amps. Some of these motors draw up to 180 amps. $\endgroup$
    – Fred
    Aug 30 at 20:04
  • 2
    $\begingroup$ Where in the website for the entire corporation does it mention this aircraft's power off stall speed, and what that term actually means for this aircraft? $\endgroup$ Aug 31 at 2:15
  • 7
    $\begingroup$ I'm a bit confused by this question. Shouldn't the question be: why does this helicopter have so much wing area? $\endgroup$ Aug 31 at 19:30
  • 3
    $\begingroup$ To put it simply: in a power off situation that thing has flight characteristics of a brick. $\endgroup$
    – Jpe61
    Aug 31 at 21:07
1
$\begingroup$

This machine is capable of vertical flight, and the 3 feet take off roll they talk about is just to maneuver the machine to the right attitude. The wing area (calculated) has to be around 112 sq ft, and wing loading 5lb sq ft. So clearly its clean foils wont b flying this machine at 28 mph, it simply passes through that zone in translative flight. With a 1,000 lbs of thrust from its 8 props this shouldn't be an issue even with some redundancy. For it is said to be able to fly and land safely with any 2 props out even in the same quadrant. Indeed the thrust almost equals 2x the UAW.

Since the props/rotors are fixed pitch I cant see the rather novel suggestion that it can autorotate, although its a neat idea. Perhaps I need to think about that.

Its emergency system is a ballistic chute.

I'll work out the Clmax at 28mph, and an approximate stall speed on a generous lift coefficient and get back to you. Problems, we don't know which data is right as there are 2 designs, one for US ultralight spec and one presumably for Canada. Also the data doesn't completely convince me as the wing area doesn't look right from images commonly found around the net.

I sourced the wing loading from this YouTube video. Other derived data are from the company website (screenshot below) and Wikipedia.

enter image description here

$\endgroup$
2
  • $\begingroup$ Welcome to aviation SE. This answer would be better if it had a source, but otherwise is a good answer. $\endgroup$
    – pirocks
    Oct 24 at 22:14
  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Oct 24 at 23:34
0
$\begingroup$

Well, since no one else is answering this question, I will take a stab at it.

My gut feeling is that the blackfly can still "fly" with power off at 28mph or lower is because it can descend slowly with its props in autorotation. See pic below.

I understand the Anotov AN-2 can also descend slowly to the ground by pulling the stick full back, with a forward motion of 14mph, and a descent of about 12mph.

Blackfly autorotation2

Blackfly Autorotation Thrust

Estimated Indicated Airspeed, IAS:28mph

Estimated Descent Angle: 21 deg

Estimated Descent rate = IAS x sin x radians(angle): 10mph

Estimated Descent, ft/min: 870 ft/min

Estimated Descent, ft/s: 14.5 ft/s

Estimated Prop dia: 1.5ft

Area, Single Prop: 1.8ft2

No. of props: 8

Total Prop Area: 14.1 ft2

Rho, imperial: 0.0765 lbs/ft3

Autorotation Thrust, T=2 x v2 x A x rho: 450lbs

Weight: 450 lbs

Conclusion: The thrust by the props in autorotation produced when descending at roughly a parachute descent rate of 10 mph or 870 ft/min is equal to its weight.

Therefore the blackfly could potentially "fly" under 28mph, as long as the descent rate is less than 870 ft/min, and an angle of descent of roughly 21 deg.

In actuality, this might even be slower due to fuselage drag. Also, you might want to have the props rotated at say 45 deg, not 90 deg, for a partially stalled wing so you can use your elevons for directional control.

Is this correct thinking?

$\endgroup$
2
  • $\begingroup$ Hmmm... good thinking, I completely disregarded the possibility of autorotation. Unfortunately I'm not mathy enough to verify if your calculations are correct. $\endgroup$
    – Jpe61
    Sep 15 at 19:51
  • $\begingroup$ Propellers with fixed pitch can only result in drag. You need variable pitch propellers that stores the energy in the rotor for release at a later time via variable pitch. For small to rotors with low inertia this idea doesn’t scale that well. $\endgroup$ Sep 20 at 16:08
-1
$\begingroup$

ok now I promised to do some work on the lift coefficient, which is work that requires gross weight, a wing area, and in doing so gives us a stall speed.

Preliminaries first, this is a tandem wing aircraft, therefore there is a difference required between the front wing and the rear for safety. The front wing must stall first, so that the machine dives away forward, picks up airspeed and recovers. Conventionally, we satisfy this by increasing the lift coefficient of the front wing or canard to 2.0. Given these are unflapped wings were we would expect a coefficient of 1.4 clean. This means the aft wing or mainplane has a working coefficient of 0.8.

Given all that we can gather values, the video tells us the wing loading is 24.5 kg/M2, which is 5 lb sq/ft. Given a gross weight of 563 lbs (from data), that gives us a wing area of 112.6 sq ft. Wing area each wing (which look identical) is therefore 56.3 sq ft. With each wing at 13' 7' (13.58') the wing chord of 4.14 ft satisfies the area.

So lets plug it all together, the dynamic 'slug' is set at sea level value, which is the weight of a cu ft of air / the dynamic of 32.

Vstall = the root of (2 x GW) / (slug x wing area x lift coefficient) which equates to the root of 1126/(.00238 x 112 x 1.4 = 54.9 ft sec, or 37.3 mph)

to satisfy a stall of 28 mph we rearrange the terms 2x GW) / (slug x airspeed ft sec x area) plugged in: (2 x 536) / (.00238 x 41.16 ft sec x 112)

the Cl averaged across the wings is therefore 9.8, which is quite unrealistic.

Just an observation, but it is hard not to notice that the CG seems well aft of what I think is desirable. The weight distribution must be biased to the canard so that the fuse or higher lift coefficient stall can promote the machine to fall forward and recover airspeed. Im nursing the summation that the machine is not being required to address conventional stall behaviour, which I think is dangerous. This is unless, the quite different configuration of this aircraft somewhat deceives the eye, in that to look from the wing plan view instead of the machine as it sits flat on the ground, the CG is then correctly located.

cheers

EDIT: I found some good photos that conceivably make a kind of 3 view by eye. I think I can place the CG as the CG of the pilot. Looking down on the aircraft it does seem the CG rests where I would expect it should. There is quite a difference between the side profile and the plan view. It really is a very clever design, but I enjoyed chasing the designers tail and working it out.

$\endgroup$
5
  • $\begingroup$ Did you mean to post this as a second answer? I think it would be better as an edit to your previous answer. $\endgroup$
    – Bianfable
    Oct 25 at 10:36
  • $\begingroup$ No I would appreciate leaving them both independent of each other, as one sets up the 'scene' and the remainder uses the data from #1 for the workings. $\endgroup$
    – Rob
    Oct 25 at 18:55
  • 1
  • 1
    $\begingroup$ This isn't really a separate answer (as in, a different answer from the same person); it's an addendum, and as such ought to be edited into your original answer. I'd suggest looking up how to format your posts, too, so it doesn't look like a wall of text even with good paragraph breaks. Headers and subheaders make an answer (or question) much more readable beyond 3-4 paragraphs. $\endgroup$
    – Zeiss Ikon
    Oct 25 at 19:06
  • $\begingroup$ I see theres a delete flag, well go ahead and have at it. Its not me that needs you or looking for answers. cheers $\endgroup$
    – Rob
    Nov 5 at 6:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.