At 200 mph you will find the body provides all the lift needed.
Note. For cruising with or without a wing suit, the propellers do not need to angle upwards. Any angle required will be only to assist in maintaining the correct pitch angle of the flyer/pilot.
Manolis. Supporting the mass of the flyer cantilevered beyond the pilots shoulders in horizontal flight is not a trivial matter. The engine and propellers will tend to "flop" downwards under their own weight.
You write: “A wing suit, flown very well, has a glide ratio of about 2.5:1 This is NOT an effective alternative to a wing for generating lift. Almost all your lift will have to be from engine power.”
The glide ratio depends heavily on the weight the wing (or wingsuit) carries.
Take the best (in generating lift) wing of the world,
load it with, say, 50Kg/m2 (10lb/sq ft) - i.e. about as much as the wingsuit of Visa Parviainen – and calculate its glide ratio.
In the following video:
Visa Parviainen proves in practice that with his “NOT effective in generating lift” wing (i.e. his wingsuit) he needs only 32Kp (70lb) of horizontal force in order to sustain horizontal flight.
From 1:06 to 1:14 one can see the “fuel tank” used, and the way it was “secured”.
What Parviainen proves is that with only 20bhp mechanical power, a wingsuiter can fly horizontally for as long as there is fuel in the tank.
And he flies at 100mph or so.
Compare the 20bhp that actually “push” Parviainen, with the 1,000+ bhp power output of the state-of-the-art JetPacks to get an idea of what “inefficiency in generating thrust force” means.
You also write: “From your drawing of 'high speed' the inclined angle is about 10° from horizontal; you would need about 6 times hover power to fly at this attitude, neglecting lift from the body.”
Quiz for all:
You are hanged –with your hands – from a bar, and a 200mph (300Km/h) horizontal wind is falling on you. What is the angle your body will lean from vertical?
Well, good luck Manolis. I'm not sure where this magic flying energy comes from, but I suggest you not volunteer to be the first pilot. It's all good and well to make verbal arguments and make nice drawings with bodies hanging off a propeller but really, I wouldn't try this at home. Seriously, the math does not support this and a non-rigid body being pulled along at 200 mph is not a wing; think about the moments being applied to the motor unit with a trailing lifting body. As an added thought, how does the pilot, at 200 mph, raise his/her arms to grab the hand bars and control the throttle et al? I like your idea for the engine, but the rest is a pipe dream, especially the control system with flailing limbs at 200 mph. Out of curiosity what is your calculation for the energy needed to hover?
The propeller axes look “slightly” upwards to take the weight of the Portable Flyer.
If the pilot feels “nose down”, he redirects the propeller axes more upwards.
From Visa Parvilainen flight presentation:
“Recovering from the stall is easy because of the agility of the human body to change flight profile easily” (the opposite is the case for airplanes).
Limiting the tip speed (of the 39’’ diameter propeller) at only 150m/sec (45% of sound velocity) for “quiet” take off, the resulting propeller rpm is 2,900rpm.
With 28’’ pitch and 3 blades per propeller, the static thrust at 2,900rpm is calculated (with the http://www.godolloairport.hu/calc/strc_eng/index.htm propeller thrust calculator) at ~75lb (~35Kp, 350N), while the power absorbed by each propeller is calculated at ~15bhp.
At the “quiet” take off, the total upwards thrust is 4*75lb=300lb (136Kp, 1360N); with a total weight of 250lb (114Kg, 1140N) this means ~0.3g upwards acceleration; the required power per engine is 2*15bhp=30bhp.
With 2.4:1 “crankshaft to propeller” reduction ratio, the 2,900rpm of the propellers at the above “quiet” take-off, translates into 7,000rpm for the engines.
In order a 21.5ci (350cc) 2-stroke engine to provide 30bhp at 7,000rpm, it needs to make 22 lb*ft (30mN) of torque at 7,000rpm (this is 86mN/lt specific torque, which is easily attainable even for naturally aspirating 4-stroke engines; i.e. the engines run at partial load at take-off: the low tip speed, the partial load with lean burn and the lightweight is the recipe for low noise).
End of Quote.
The above are for a vertical take-off with 0.3g (i.e. 3m/sec2) upwards acceleration.
For pure hovering the propellers rotate at ~2,500rpm with the engines revving at ~6,000rpm providing ~40bhp of power output.
The total power required for hovering is 40bhp because the pitch of the propellers is tailored to be power efficient at 100+ mph cruising.
If it was desired to just loiter hovering around, then a substantially lower pitch for the propellers would be sufficient, which substantially reduces the required power to even less than 20bhp in total.
So how does the pilot recover if they invert at 200 mph? Seems to me your machine might be stable in a vertical power dive.
Edited to add: A quick calculation shows the power to overcome the drag of a supine human in horizontal flight is approximately 55 horsepower @ 200 mph, assuming Cd = 0.3 and an area of 0.3 m².
What is the longest flight duration undertaken with wing suits? Human anatomy doesn’t seem very well suited to supporting its weight, plus the weight of the powertrain, on arms and legs spreadeagled. At least not for periods over a few minutes.
Fortune favours the prepared; she has no favourites and takes no sides.
Truth is confirmed by inspection and delay; falsehood by haste and uncertainty : Tacitus
The propeller axes look “slightly” upwards to take the weight of the Portable Flyer.
If the pilot feels “nose down”, he redirects the propeller axes more upwards.
The problem is the effort required to "redirect the propeller axes more upwards". Looking at a FBD of the flyer, the weight of the flyer (Mg) is vertical-down, the "lift" is provided by the pilot and with drag added gives the resultant force vector at the shoulders - "F". The thrust from the propellers is vector "T". The lower diagram shows the three forces cancel but because they do not intersect at a common point there is a net moment M. If you sum the moments about the pilot's shoulders you see the "T" and "F" vectors produce zero moment but "Mg" is trying to tilt the flyer down and produces a moment about the pilot's shoulders. The effect is no different than having the pilot lie face-down on a table with his shoulders and the flyer beyond the edge of the table.
This moment would be impossible to resist without handlebars and even then would eventually cause fatigue. One solution would be a rigid "spine" extending down from the flyer frame and attached to the pilot's waist or hips with a belt. The handlebars might even become redundant.
You write: “So how does the pilot recover if they invert at 200 mph?”
The skydivers, who at free fall achieve higher speeds, recover easily; and the only thing they have is their body weight (which permanently “looks” downwards) and their body posture (following video, from 0:42”):
The Portable Flyer pilot besides his weight and his body pose, has also a strong thrust force (from the powered propellers) that can instantly and effortlessly be vectored at any direction.
“What is the longest flight duration undertaken with wing suits? Human anatomy doesn’t seem very well suited to supporting its weight, plus the weight of the powertrain, on arms and legs spreadeagled. At least not for periods over a few minutes.”
According Wikipedia the longest flight duration of a wingsuiter is 9’:06”.
If you think the Portable Flyer pilot as lying (supine pose) on an air mattress, his weight cannot be better supported.
Compare the case with a Harley Davidson driver cruising at 100mph:
In the FDB plot the Mg should be near (or at) the center of the propeller, and the aerodynamic drag force acting on the engines of the Portable Flyer is missing (worth to mention: the frontal area of the engines is not too smaller that the frontal area of the pilot who – at supine pose - has a lower coefficient of aerodynamic friction).
For the torque:
Take the case wherein at horizontal flight (as in the FDB plot) the pilot releases the Portable Flyer which, alone, will accelerate forwards and slightly upwards (from the vertical component of the thrust force T it is subtracted the weight Mg of the Portable Flyer).
I.e. the moment from pilot's shoulders / torso (if any) is not for keeping the engines from falling downwards, but from moving upwards.
Well, that's all good and well to have verbal descriptions of how you wish things worked, but until you do some analysis, determine the center of gravity and the aerodynamic center, see if it is vastly different from the center of pressure, and look at longitudinal stability to see if it's positive or negative I really don't think you have any data to support your assumptions about control via moving legs; if the center of gravity is forward of the center of pressure you will need a corresponding torque to offset the tendency of the machine simply nosing over (airplane tail). [edited to add; Conversely, if the c.g. is far aft of the c.p. it will be difficult to get out of hover]. Transition from hover to forward speed will be very tricky with a constantly changing center of pressure and gravity as the body flexes. I have no idea how you propose to transition from forward high speed to brake mode....
I especially have concerns about the ability of a human body, with an engine cantilevered off its shoulders, being able to exert controlling torque to guide this thing. The body is not a rigid structure and especially at speed I seriously doubt that moving legs around will give control. This is not sky diving or wing suit flying. You claim lift from the body and lift from the motor; how do you reconcile these? Also, the body would seem to be more hanging and causing drag than providing lift. Most lift from the body, I suspect, would be kite lift thus requiring even more power. As I have mentioned previously, in a vertical dive your device looks to be stable, which would be an ugly situation.
If you've analyzed all this stuff and come to good conclusions I would be interested in seeing your work. I thought about doing it, but lack data, time, and motivation. Your motor is interesting and kudos for building a running model.
Last edited by Rodak on 11 Nov 2019, 06:42, edited 1 time in total.
In the FDB plot the Mg should be near (or at) the center of the propeller, and the aerodynamic drag force acting on the engines of the Portable Flyer is missing (worth to mention: the frontal area of the engines is not too smaller that the frontal area of the pilot who – at supine pose - has a lower coefficient of aerodynamic friction).
For the torque:
Take the case wherein at horizontal flight (as in the FDB plot) the pilot releases the Portable Flyer which, alone, will accelerate forwards and slightly upwards (from the vertical component of the thrust force T it is subtracted the weight Mg of the Portable Flyer).
I.e. the moment from pilot's shoulders / torso (if any) is not for keeping the engines from falling downwards, but from moving upwards.
Thanks
Manolis Pattakos
Agreed, my FBD does not account for aerodynamic lift acting on the (inclined) flyer. This is the only reason I can see for inclining the flyer upward. Conventional aircraft do not do this, the thrust from the propeller is used solely to overcome drag.
If the flyer is released from the position shown in the FBD it loses a force vector "F" whose line of action passes below the CG and creates a moment tilting the flyer downwards so the "released flyer" should not be compared to the "flying system" as represented in the FBD.
The Portable Flyer pilot besides his weight and his body pose, has also a strong thrust force (from the powered propellers) that can instantly and effortlessly be vectored at any direction.
Instantly and effortlessly? How is the thrust force vectored? What moment arms are applying torque to change the vector? The counter torques must ultimately come from aerodynamic forces and these seem to be legs in an air stream on a flexible body.
