Proposal for hovering flight.

I was watching a Science Channel show discussing how birds fly. It
discussed the rotation a birds wing undergoes at the end of its
The description was of the vortices this produces and how this
contributes to lift. No doubt this does occur, but I was reminded of
the way that a swimmer rotates his hand on the return stroke while his
hand is still in the water so as not to counteract the thrust he
already produced.
I thought then this same idea could be used to produce craft capable
of hovering, as helicopters do. You would have a cupped surface moved
vertically downward producing a downward thrust, but on the return
stroke the surface would be collapsed so as not to counteract the
thrust already produced. I'll refer to the cupped surface as a canopy
whether of not it is stiff during the propulsion phase or flexible as
an actual parachute.
I'll use the formula for drag to calculate the thrust that would be
Drag (physics)
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It is the drag is 1/2 times the density, times a reference area,
times the drag coefficient, times the velocity squared. Note that
usually the reference area in this formula would be the cross-
sectional area, as for instance for an airfoil. However, for a
parachute the reference area that goes in this formula is the surface
The power required to overcome drag is also given on this page as the
velocity times the drag. From this we see the ratio of thrust to power
required is 1/v, where v is the velocity. This means it would be more
efficient to move the canopy at lower velocities.
For thrust given in terms of newtons, and power in watts, this thrust
to power ratio would be in newtons per watt. This lifting capability
for the power required is sometimes given in terms of kilos lifted to
kilowatts of power or grams per watt. For this scenario and for the
velocity given in meters/sec, the numerical value of grams to watt
power would be about 100/v.
The efficiency of helicopters is in the range of perhaps 10 grams
lifted per watt. Then this proposed method would exceed the efficiency
of helicopters if the downward velocity of the canopies was less than
10 m/s, 36 km/hour.
If we made the downward velocity say 5 m/s the efficiency would be 20
gram/watt. Say we wanted to make a personal hovering device capable of
lifting a person. Let's suppose the total weight was 200 kg, including
the person, engine, canopies, support structure etc. Then we would
have to have the thrust be at least 2000 newtons. The density of air
at sea level is 1.22 kg/m^3. For a parachute, the drag coefficient Cd
is perhaps 1.5 to 2. The surface area of a hemispherical canopy is
2*Pi*radius^2. This would result in a radius for the canopy of about
3.2 meters or a diameter of 6.4 m. You could of course break the
surface area up into more than one canopy. Note that to reduce the
volume of the craft, it may also work to have several of the canopies
in a single column.
For an efficiency of 20 gram/watt and lifting capability 200 kg this
would require a power of 10,000 watts, about 13 horsepower, quite low
for a power plant capable of lifting a person.
I originally was thinking of this for heavy lift applications
however. A common heavy lift helicopter is the S-64 Skycrane. It can
lift a gross weight of 20,000 kg using 7,000 kwatts of power with a 22
meter diameter rotor. This is actually a weight to power ratio of only
3 to 1.
Suppose we wanted to get a 5 to 1 ratio using the same power. This
would require a velocity of 20 m/s in our scenario and a lifting
capability of 35,000 kg, or a thrust of 350,000 N. Then to get the
350,000 N thrust would require diameter of about the same size as the
helicopter rotor. But this would be able to lift almost twice as much.
To allow further lift during forward flight and reduce the forward
drag we may make the canopies in the shape of parafoils. Note that
parafoils create most of the reduction in descent speed from the fact
they produce lift, with a relatively small amount coming from the drag
they produce. Then it might be advantageous to use these moving in a
horizontal direction to generate the hovering effect. We might also
want to change the shape of the canopies depending on the direction of
flight, vertical, horizontal or a combination.
The idea might be usable for supersonic or hypersonic propulsion as
well. In this instance the canopies would not have to be moving at
supersonic velocity with respect to the surrounding air. Their
velocities would be at or a little more than that at which the ambient
air is moving rearward (with respect to the craft) in order to get the
highest efficiency.
Bob Clark
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Robert Clark
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The problem is that this involves a reciprocating or flapping structure. On each stroke the inertia of the structure itself has to be overcome. This is THE problem with all such proposals.
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The pectoral muscle of a flapping bird acts much as a spring, storing energy as it is stretched - on the upstroke, as the lift on is wings, coupled with the weight on its body applies a torque to the muscle through a rotation relative to the roll axis. This energy is re- released on the start of the downstroke as the pectoral muscle starts to contract again.
The mass of the wing and the "spring" of muscle constitute an oscillatory system, albeit damped. Once the oscillation is established, the bird expends only the energy required to compensate for losses due to drag and less-than-perfect efficiency (internal friction within the bird that leads to elevated body temperature, for example). Hooke''s Law is sufficient to overcome the inertia of the moving wing.
The changing angle of attack in the bird's wing is an evolutionary adaptation necessary to reduce energy losses in a flapping structure my maintaining an optimal angle of incidence of the wing against the air as the relative velocity of the wing through the air changes during the cycle. This takes the form of a wing surface that pivots relative to the leading edge. All evolved flapping-wing systems (flying birds, bats, insects exhibit the same structure and behavior. Animals with non-flapping wings ("flying squirrels", lizards of the genus Draco) can only glide.
Tom Davidson Richmond, VA
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A fair point. The main purpose of this idea is get high lift at low power. Note then the best efficiency is obtained when the canopy is moving at quite low speeds, say at 5 m/s or 10 m/s. Also, using advanced materials we can now probably markedly reduce the weight by using microscopically thin canopies. As it is, a 10 meter diameter parachute may only weigh 15 kg. For the idea of using this for supersonic or hypersonic propulsion, we might have the canopies move in a circular path.
Bob Clark
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Robert Clark
I can see how you might make parachute might work on the "down stroke" but the up stroke?
It's interesting to take a look at how man powered flight has been achieved - bear in mind that racing cyclists can only put out about a third of a horsepower continuously...
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I believe the prize for a man powered helicopter is still unclaimed although some have managed to get off the ground on less than one horsepower!
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You might have more than one canopy directly above one another so that as one is returning for another stroke there is another one still headed downward. Thanks for those links on human powered flight. I found this link that suggested top cyclists could generate 500 watts continuously and 900 watts peak (though perhaps with some chemical augmentation ;-) )
Olympic Human Powered Flight Human Powered Flight Undergoes Renaissance With Olympic Competition. "//humans can produce about 100 watts continuously, with peaks of 400 watts or more// That may be so on average, but for a trained cyclist those stats are quite low. I'm at 300 watts continuous / over 900 peak with moderate training. Lance Armstrong somehow manages 500 continuous. It's probably safe to say Olympic cyclists can maintain at least 400 watts."
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The prize discussed on the human powered helicopters site only requires that the craft hover at 3 meters under human power only for one minute. A fit male probably could do leg presses of twice his body weight for several reps and bench presses of his weight for several reps. These would only have to be done for 60 seconds. To see these are reasonable capabilities note that in climbing stairs we're actually supporting our weight most of the time on one leg. We're using both legs only for the short time when we're switching from one leg to the other. So just climbing stairs we're doing the equivalent of supporting twice our body weight. A normal fit person climbing stairs certainly could go up one stair in one second, with the stairs being about 8 inches, 20 centimeters high. Then a good athlete could do the equivalent of 100 centimeters, one meter, per second up the stairs if only for one minute. For the bench press, a normal fit male could do several reps of push- ups, which is supporting your weight with your arm and chest muscles, once each second. The distance would be about 2 feet. Then a good athlete should be able to raise his weight in a bench press 1 meter once per second for several reps for a minute. Note for both of these I'm assuming that both the positive and negative directions count as a single repetition, not a full cycle, as in normal exercising. This corresponds to how this would be useful for our scenario since both the positive and negative directions could be used for driving the craft. However, there is a question of whether both the upper and lower body could be used at this intensity at the same time. I think for the top athletes this could be possible at least for one minute. Then we may suppose a top athlete raising and lowering 3 times his weight 1 meter once per second, that is, a full up-down cycle taking 2 seconds, for 60 seconds. For a small 60 kg athlete, this would be 180 kg raised 1 meter per second or 1800 N over 1 m per second, or 1800 watts. For the weight of the person of 60 kg, this is a weight to power ratio of 33 to 1. I will assume that the canopies and the shell holding the craft can be made quite light, and will take simply the total weight to be 60 kg. Remember now that the weight to power ratio is 100/v. So v is 3 m/s, for the speed at which the canopies are brought down. Then plug this into the equation for the drag produced to see how large the canopy has to be raised 60 kg. We need to produce 600 N of force. This requires a canopy diameter of about 6.6 meters.
Bob Clark
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Robert Clark
As I recall several of the man powered flight projects looked at how best to get the power out of a human. For long flights they concluded it didn't matter how as the limit was about the same as just cycling. I believe it's to do with oxygen take up of something. In the end they went for the simplest scheme which was cycling. I guess it might be different for short bursts though.
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You might be right about long flights. But this may not be the case for a test flight of only 1 minute duration. I found this site that gave some values of maximal human power output for different activities:
Energy Data & Calculations.
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Look at the table on this page labeled "Power Output for Human Walking and Running." It gives the power output for a 60 kg person at 15 mph, 403 meters/min, as 1670 watts. The world record for the 400 meter dash is under 45 seconds for men and under 50 seconds for women. Moreover, I looked up some winning times for university track meets and found that the best times for the 400 meters are routinely less than 60 seconds for both men and women. So being able to put out close to 1700 watts over 1 minute is routine for competitive mid distance sprinters. Note that since the power output is dependent on the weight of the runner this amount would actually be higher than 1670 watts for the male runners since most male sprinters weigh more than 60 kg. But we're concerned with getting this high power at low weight so it might actually be a woman sprinter who could be meet this goal. Also, look at the table labeled "Human Power Limits". It gives a maximum of 4500 watts for a single vigorous jump or lift over a time of less than 1 second, and a maximal power of 2200 watts for a sprint over a 20 second duration. Then it is quite conceivable that you could get a maximal power of 1500 to 1800 watts over a duration of 60 seconds.
Bob Clark
Reply to
Robert Clark
This page discusses the fact that wind acting on a parachute can contribute to lift:
Parachute Design and Construction.
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Then we can perhaps move the canopies at an angle to use both drag and lift. See Fig.1 on this page. This passage describes the effect of the mode of descent on the descent rate: "The gliding nature of a parachute is another reason that using Cd as a measure of the effectiveness of a parachute can be misleading (Fig. 1). When a parachute descends, it may have both a downward component of velocity as well as a horizontal component (in other words, rather than descending straight down, it will descend at an angle). Air flowing around the parachute at a certain velocity (V) generates both lift and drag forces -- the drag (D) acting opposite to its line of motion, and the lift (L)acting perpendicular to this, tending to reduce the descent rate, Therefore, the drag coefficient measured from free fall "drop" tests may indicate a significantly higher Cd (than, say, that measured in a wind tunnel), as a result of this gliding phenomenon. # The flow of air around and over the canopy of a parachute may produce an oscillating (spiralling), or coning, pattern to the state of motion of the descending parachute, as flow separation and suction forces alternate in direction. Therefore, a parachute may be considered to be capable of descending in either a gliding mode, or an oscillating mode, or a combination of both. Gliding tends to prevail at lower descent rates, and oscillating at intermediate rates of descent. The resulting Cd can vary significantly, depending on the mode of descent, as indicated by the following data for a typical full sized parachute:
Descent velocity Descent mode Cd 23 fps restrained 1.26 20 fps oscillating 1.60 16 fps gliding 2.40"
Bob Clark
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Robert Clark

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