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AIRPLANE FLIGHT ANALOGY
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fig. 2 TWO DISK-BALLOONS, STACKED ADJACENTLY ____ _-- _____ / _-- --_ __| / \ __| . | | | \_ _/ --_____-- They do not touch each other. Both have platforms. If I jump onto the first platform, but then I immediately leap onto the next platform, I can stay up there for a tiny bit longer.
Next, suppose we have a row of these disk-balloons one KM long. It looks
like fig. 2 above, but with hundreds of hovering balloons. Now I can
run from platform to platform, and I will stay aloft until I run out of
balloons. Behind me I leave a trail of rotating, downward-moving
balloons. I can remain suspended against gravity because I am flinging
mass downwards. The mass takes the form of helium mass trapped inside the
balloons. I am also doing much more work than necessary, since the energy
I expend in rotating the balloons does not contribute to my fight against
gravity. (In truth, all my work is really not necessary, I could simply
walk along the Earth's surface with no need to move any massive gasbags!)
To make the situation more symmetrical, let me add a second row of
platform-bearing balloons in parallel to the first row: fig. 3 END VIEW OF TWO LONG ROWS OF DISK- BALLOONS _____ _____ _-- --_ _-- --_ / \ / \ | . |__ __| . | | | | | \_ _/ \_ _/ --_____-- --_____-- There's one platform for each of my feet. I can run forwards, leaving a trail of "wake turbulence" behind me. The "wake" is composed of rotating, descending balloons. Fig. 4 below shows an animated GIF of this process.
Fig. 4 Forcing the balloons downwards
Also see: Smoke Ring animation
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"DISK BALLOONS" BEHIND AIRPLANESA 3D aircraft does much the same thing as me and my balloons: it remains aloft by shedding vortices: by throwing down a spinning region of mass. This mass consists of two long, thin, vortex-threads and the tubular regions of air which are constrained to circulate around them. The balloons crudely represent the separatrix of a vortex-pair: the cylindrical parcels of air which must move with closed streamlines.
So, how do airplanes fly? Real aircraft shed vortices. They inject
momentum into air which as a result moves downwards. They employ
"invisible disk-balloons" to stay aloft. The two rows of invisible
balloons together form a single, very long, downwards-moving cylinder of
air. This single cylinder has significant mass and carries a large
momentum downwards. Airplane flight is vortex-shedding flight.
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The downward-moving "reaction exhaust" below a wing, made visible. The aircraft flys horizontally above the fog bank, while the vortex-pattern descends into the fog. (See Hyperphysics and other photos.) Scanned laser smoke downwash
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FLY FASTER FOR LESS DRAG?My forward speed makes a difference in how much work I perform. If I walk slowly along my rows of balloons, each platform sinks downwards significantly. I must always leap upwards to the next platform, and each balloon is thrown violently downward as I leap. I tire quickly. On the other hand, if I run very fast, my feet touch each platform briefly, the balloons barely move, and the situation resembles my running along the solid ground.
Similarly, if a real aircraft flies slowly, it must fling the vortex-pairs
violently downward. It performs extra work and experiences a very large
"induced drag." If it flies fast, it spreads
out the necessary momentum-shedding across a much larger volume of
decending air, and therefore it needs only barely
touch each mass-parcel (each "balloon.") Hence, faster flight is
desirable because it requires far less work to be performed in moving the
air downwards. And if a slowly-flying, heavily-loaded aircraft should fly
very low over you, its powerful wake vortices will blow you over and put
dust in your eyes.
All of my reasoning implies that modern aircraft actually remain aloft by
vortex-shedding; by launching a long chain of combined "smoke rings"
downwards. Imagine one of the
flying cars in the old 'Jetsons' cartoon, the ones with those little white
rings shooting down out of the underside. But rather than launching a
great number of individual rings, modern aircraft throw just one very long
ring downwards, and they are lifted by the upward reaction force.
L. PRANDTL, FATHER OF ...STUFFIn wondering why the entire aero community seems ignorant of these simple concepts, I stumbled upon something interesting. Klaus Weltner found that L. Prandtl was the initial source of the equal-transit-time fallacy. Prandtl's 1924 paper teaches that, parcels split by the leading edge, must rejoin at the trailing edge. |
Looking at more Prandtl work, I find that he was using two
simplifications, changes which make the math easier by erasing any moving
vortexes (no vorticity in translation.) This eliminates any propulsion
effects, any fuel use or induced-drag. He did two things: analyzed 2D
airfoils, and also analyzed 3D wings in flight, but where the
tip-vortices extended back perfectly horizontally. (So, a "Prandtl
Horse-shoe"
diagram, but without any tilt to the vortex-lines.) These two changes
have profound
effects: direct violations of Newton's laws.
A CRUDE PREDICTIONHow well does the "disk balloons" model correspond to the real world? Well, we can pull an equation out of the motions of the balloons, and use it to predict both aircraft energy-use and induced drag. If the equation is at all similar to the actual aerodynamics of a real-world airplane, then the "disk balloons" are a useful model. If my equation turns out to be faulty, then my model only has weak ties with reality.Suppose the "disk-balloons" contain air which rotates as a solid object, (or imagine radial membranes in the balloons. Or stuff them with aerogel.) If I then add together the work done in creating the circulatory flow, plus the work done in projecting the constrained air downwards, I arrive at a predicted aircraft power expenditure of: Power = 8 * (M * g)^2 / [ pi * span^2 * V * density ]M * g being aircraft weight, V is velocity of horizontal flight, and "density" is the density of air. Induced drag should then be power/V: Induced Drag = 8 * (M * g)^2 / [ pi * span^2 * V^2 * density ]What happens if I assume that the air within the disk-balloons is not "solid", but instead it's made to whirl faster near the center of the balloon, such that the tangential velocity of the air is constant, regardless of its distance from the center of the balloon? (Imagine a wing which produces a downward velocity of net downwash which is constant at each point along the whole span of the wing.) If the "downwash" is constant across the wingspan, then the modified "balloon equation" predicts a power expenditure of 2x that above.
How does this match reality? I'm looking for information on this at the
moment. I'm told that these two equations are identical to the equations
of real aircraft, except that the number "8" is replaced by a factor which
is dependent upon the particular geometry of the wing. So, calculated
from first principles, without prior instruction in fluid dynamics.
Pretty good for an "amateur aerodynamicist", eh?
One final note. The downwash vortices of real airplanes contains rapidly
rotating air. This represents wasted energy, since only the "shell" of
each "balloon" needs to rotate as the region of air moves downwards. Is
there a wing which can produce a downwash vortex-pair without any spinning
cores? Maybe it would use less fuel than modern wings.
(See also J. Denker's critique and my response, 8/99) |
Misc. thought experimentsIf a balloon inside a sealed chamber experiences an upwards force, well, Newton'd 3rd must be obeyed, so where does the down-force appear? Obviously it appears as a slight increase in static pressure on the floor of the chamber. The balloon accelerates upwards, and the chamber accelerates downward, conserving momentum. Now if our balloon carries a load, and load KG is adjusted so fhs Lloon hovers in level flight, what then? The down-force of the load is exactly countered by the buoyancy up-force of the balloon ...but the physics isn't local-only. Therefore, the floor of the chamber must experience a small pressure increase whenever a balloon lifts a small mass, a down-force exactly equal to the weight of the suspended mass. (And no, we cannot eliminate the weight of a truck's load upon the Earth by suspending the truck via balloons! If the chamber is sealed, the balloons cannot reduce the weight of the truck. If the balloons are outside the chamber, then the "footprint" of down-force wide across the ground, but it still nets to exactly the weight of the suspended mass.OK, how about this one. Suppose I'm high above the ground, and I have a huge piston and cylinder. I pull the piston out; forming an evacuated volume inside the cylinder. I pull it so far out that the buoyant force of the evacuated cylinder now lifts the weight of the cylinder, piston, and myself (I'm that strong.) I'm now suspended, hovering. So, what's the footprint of my weight upon the ground? :) And, what are the dynamic changes, if any, since the ground cannot experience force-changes except after speed-of-sound propagation delays. |