I know how to figure the lift of a Cessna wing, 1/2 rho V squared and
all that.  How would I figure the lift on  a large cambered wing with
approximately 200 knots over the top and zero airspeed (ambient, 14.7
psi) underneath?  I know it sounds goofy but humor me, I've got a
plan.  I expect this will require more exlpanation....

bummer-ga,

Here’s why this is a dumb idea… (this answer is meant to be facetious,
not condescending! I’m the devil’s advocate you were asking for!)

I had read through the discussion in the comments about four times to
be sure I had your setup right.

From what I can tell, what you basically have is a helicopter without
the helicopter. Basically something that looks like a frisbee or an
upside down saucer with a ducted fan built in the top that blows air
all around the airfoil-shaped surface.

When you get this device airborne are you going to get an upward
force? Yes. Are you going to get a downward force? Yes, and it will be
much larger than the upward force. But lets concentrate on the
mechanics.

For the incompressible case that you are interested in you can figure
out the pressure of the air over the top by using bernoulli’s
equation.  The pressure on the bottom is simply the ambient air. The
pressure on the top goes as the speed across the top surface, and the
pressure on the bottom is simply ambient. Whala... Lift!

What I don’t think you are accounting for is the momentum of the air
coming down on the disk. Momentum can be calculated as 1/2 emm vee
squared. Or if you are talking about fluids, it’s easier to think in
forces. 1/2 rho volume/second vee^squared (momentum transfer per
second... a force).  And it’s all coming –down- on your wing! If you
calculate this for your particular configuration you will likely see
that the momentum transfer on the top of the wing plus the pressure
over the top is likely higher than that on the bottom.  What you end
up with is essentially negative lift, not lift.  The suction force on
top of the ducted fan (producing an upward force) is likely to be very
small because the air going into the fan is travelling at very low
speed.

Here’s a Demon... don’t think up, think sideways.  Your configuration
is basically the front of a fuselage without the rest of the fuselage.
 Air coming straight on the disk, no fuse or wings behind it. LOTS of
drag. There is no simple way to calculate this, either stick it in a
wind tunnel or use computational fluid dynamics (that takes turbulence
into account). I doubt there is a way to get very highspeed airflow
over the top surface of your device without getting very turbulent,
circulating air.

One last demon... When the airflow coming off of the disk meets the
edge it will likely try to curl around it.  This is bad news because
when this happens the air will separate (become spoiled) and reduce
the “high” pressure on the bottom. In other words, the air on the
bottom is actually moving... reducing the ambient pressure quite a
bit.

I think that about kills it.  It’s probably worth the time to take the
numbers that you have for your proposed configuration and run them
through this model just to be sure. If I've glossed over a detail you
need, just request it and I'll be happy to provide it.

If I’ve screwed something up in the above explanation, please ask for
a clarification... Also, if my conception of your device is flawed, I
can take a second try... this was a very neat question!

Best Regards,

Krobert-ga


FYI... You keep referring to camber. A wing (or airfoil) has camber
for two reasons. 1) to help keep flow attached, 2) to get the maximum
pressure differential between the top and bottom of the wing
(somewhat, but not completely related to 1). In this device's case,
neither one nor two will apply because your device does not satisfy
the Kutta condition that your average wing does. The Kutta condition
states that the flow meets somewhere near the tail end of an airfoil.
Hopefully (for drag's sake) at the trailing edge of the airfoil. For
this device, in a real world configuration, the flow will meet
somewhere on the -bottom- of the disk, because the flow is bending
around the edge. It will also be unstable, because the flow will be
turbulent.

---

Request for Answer Clarification by bummer-ga on 22 Dec 2002 18:20 PST

krobert - Thank you for being the Devil.  I asked for a $50.00 answer
and have gotten much more than $50 worth of info and enjoyment.  But,
you're point about the momentum of the air pushing on the airfoil is
incorrect.  The second stage of the counter rotating ducted fan
re-accelerates and re-directs the airflow (about 60 degrees) parallel
to the lifting surface.

It's going to be just like blowing air over the top of a sheet of
paper, at 200 kts.

The Kutta thing might be the Achilles heal, and the vortices, and God
knows what else.  So, I've got an appointment with the wind tunnel
guys and UW, we're going to try to jury rig some of his gear so the
airflow is directed parallel to the lifing surface.  I did that with a
leaf blower and lifted the shell off the ground.

Thank you for your time, and pls respond to the momentum thing.  I'll
consider that an answer to we can take this thing off the Google
books.

I'd be happy to report the wind tunnel results, if I know to whom to
report them...

Thank you for your time

---

Clarification of Answer by krobert-ga on 22 Dec 2002 21:55 PST

Despite the redirection (actually because of it) there is still a
downward force involved. Your air is moving in a different direction
than whence it originally came, your problem is that a force is needed
to do this.

Think of the air as a series of baseballs. Your up to bat and the
pitcher is tossing balls at you.  You have a bat that you can
accurately use to bunt them (no swinging involved) along the third
base line of a baseball field. Once the balls hit your bat they now
have momentum pointing in the direction of third base. They have
roughly the same magnitude of momentum that they did when they were
hurtling toward you. 1=1 right?

Newton says: a body in motion stays in motion unless acted upon by a
force. What does this mean for you? You're deflecting the air 60
degrees from the vertical.  There is going to be a point somewhere in
your apparatus where a force is pushing the flow sideways. Because
this is a mechanical flow, there has to be a surface force (in
contrast to a body force like gravity) that is applied approximately
at 60 degrees to the flow. Think of the baseball.

<tt>
                             o
                             |
                             v


               <-o           /
</tt>

In this diagram (I hope it comes out alright), the "/" is your
baseball bat. The force required to change the momentum of the
baseball points down and to the right. This force can be broken down
into two components that -add up- to get the force that points -down
and to the right-... a "down" component and a "right" component.

The "down" component is what will be pushing down on your device...
"anti-lift".
The "right" component you could care less about because it's balanced
symetrically about the disk (just build your device out of sturdy
material).

(Professor steps down at this point)

I have to ask... did the shell hop when it was lifted? Or did it
actually stay airborne? If it hopped it may have been a temporary
pressure differential between the top surface and bottom surface. The
ambient air below would push the shell up if the pressure on the top
surface was lowered significantly. If you just lifted up an edge by
placing the blower end so that the airflow passes over a small portion
of the shell, the momentum discussion above doesn't apply because all
of the forces caused by momentum transfer are happening inside the
blower. You would have caused lower pressure on the surface of the
disk and a higher pressure below due to ground effect. The proximity
of the flow to the ground would cause greater localized swirling
effect and a "pushing" on the bottom of the disk because the airflow
has nowhere else to go.

Best Regards,

Krobert-ga

---

Request for Answer Clarification by bummer-ga on 28 Dec 2002 19:45 PST

Krobert:    Assuming your question was not rhetoric, "no" the shell
did not "hop," it "flew" up into the diffuser and stayed there untill
I turned off the airflow.

If your "downward force" argument held water helicopters could not
hover.  We already have PACV's (Air cushion vehicles) that "fly" on
the force of a single stage ducted fan.  The first stage of my ducted
fan will "pull" air straight down into the second (counter-rotating)
stage which will re-accelerate and re-direct the airflow
(approximately) 60 degrees, parallel to the lifing surface. The "lift"
from the second stage will be marginal because the lift vector will be
deflected 60 degrees from vertical.  I'm discounting the lift from the
fans altogether,  hoping to get significant lift from the difference
between the low pressure, high speed, turbulent air over the top of
the lifting surface and the zero airspeed, ambient, 14.7 psi air
underneath.

I've obviously explained this thing badly and apologise for taking up
everyone's  time.

I'm going to leave this discussion now.  Am happy to pay the $50.00
because of the the thoughtful, evocative, responses.  Still, my
original question: "how wouold I figure the lift on a large cambered
wing with approx. 200 kts over the top and zero airspeed underneath?"
was never addressed.

On to the wind tunnel...

---

Clarification of Answer by krobert-ga on 29 Dec 2002 07:20 PST

bummer-ga,

Sorry about the dissapointment with the answer bummer-ga. I'm glad you
feel that the discussion was worth it though.

krobert-ga

My question was not answered, but the discussion surrounding it was
well worth the price.

I'm not exactly sure how this whole process works (i.e. - Google
questions/answers), but just for your info, the lifting force on a
wing is 2/3 suction (from the top) and 1/3 "pushing" (from the
bottom).  Therefore, it's my belief that if any part of your equation
has a zero in it, there won't be much lift.  One way of imagining it
is to look at the top of modern cars or a race car.  The roof is
cambered to improve aerodynamics, but there is no lift (as far as I
can forsee) from the roof.  But then, what do I know...

Dan Dalton,
A.E.

Dan, I'd like to know your source for the "2/3 suction, 1/3 push"
formula.

The conventional lift formula applies the coefficient of lift to the
dynamic pressure (q) and multiplies that times the sq. ft. in the
wing.  I can do that.
What I can't do is figure the coefficient of lift when I've got lots
of air passing over the top of a (nearly) conventional wing, and zero
airspeed underneath.  Like.....  blowing over the top of a sheet of
paper...

There's got to be a way.

Kent

There's a neat applet called FoilSim on Nasa's website that can be
used to calculate lift on a number of different objects, speeds,
cambers, etc.
http://www.grc.nasa.gov/WWW/K-12/airplane/foil2.html

I doubt it would be of much help to your current question as airspeed
beneath the wing would be the same. Still you might find it useful for
other applications.

Just out of curiousity, are you planning on rotating a wing over a
fixed surface? I can't comprehend how to get 200 knots over the top
and 0 underneath?

Visualize a flattenned bell shape with the "skirt" forming a large
cambered, (Infinite Aspect Ratio) wing.  Now picture a counterrotating
(to zero out the torque) ducted fan mounted over it;  the first stage
brings air straight down, the second, counterrotating stage
reaccelerates and redirects the air (approx 60 degrees)parallel to the
lifting surface.  Compare all that turbulant, high speed, low pressure
air on top to the zero airspeed, stable, high pressure ambient air
underneath.  That has to produce massive amounts of lift, right?

I've been working on this thing for a few years now.  Have reached the
point where I need a Devel's Advocate to explain what a really stupid
idea it is, before I invest even more time and money in a project
destined to fail.  Then I can move on to the next dumb project.

Thank you for your time
Kent Vandervelde

Pardon my ASCII art, but something like this?

                |////////| (Ducted Fan)
                |   ||   |
                |\\\\\\\\|
          /----------------------\    
    /-----------------------------------\ (Airfoil)

That's it, the "shell" has an aerodynamic shape.  I bench checked it
with a leaf blower and lifted it off the floor.  Airspeed was 20 kts
coming out of the nozzle and 15 off the bottom of the the skirt. 
Understanding that lift varies with the square of the airspeed, 200
kts should generate a lot of lift.

I've missed something I'm sure, but damned if I can figure out what.

Kent

Do you have any drawing/photo/mockup of the device in question? I ask
because I'm at the cusp of understanding what you're talking about,
but I'm still slightly behind the point of comprehension. Post links
if you do... this question intrigues me.

Jickster-ga

Yes I have drawings/pictures/mockups. Don't understand "post links if
you do."
I scanned a couple drawings onto a floppy, would like to email  them
to you but am not sure how...   will  see if "post links" helps.

Let's not let this question "time out."  I badly need some definitive
answers.

My apologies for being obtuse... I was hoping that you might have had
the photos available on a website, or could make them available and
list an web address for where they could be seen. Unfortunately, if I
remember correctly, the rules that govern Google Answers researchers
prohibit us from providing direct contact info for ourselves. If you
can make your pictures or drawings available on a website, please
place a comment or clarification with their address. If not, I'll just
have to squint at that ASCII art a bit harder :)

jickster

This discussion is as old as aerodynamics itself, What about a wing
that is equally curved top and bottom like a fighter plane and a stunt
plane. It has always been an argument between Newton principal and
Barnulies principals. Newton says for every action there is an
opposite reaction. Lets look at an aircraft on a takeoff roll with
cemmetrical wings, and zero wind conditions. As the speed increases
the air moving over the top and the bottom of the wing is traveling
the same speed. When the pilot raises the nose of the aircraft it
increases the angle of attack, and the air on the bottom of the wing
has more pressure than the top of the wing, due to the relative wind,
thereby lowering the pressure on the top of the wing. The air on top
of the wing is being diverted due to the increased angle of attack and
has to travel further and faster to met the air on the bottom of the
wing. This is where Barnulies principal comes to play. I don't think I
spelled his name right, but I think you know who I am talking about.

I do indeed know what you're talking about.  Still I believe raising a
sheet of paper by blowing over it successfully marries Newton and
Bernoulli (sp?).  All of which begs the question - how can I predict
the lift of a wing that has lots of air over the top and zero airspeed
underneath?

I'm coming to the realization that nobody can - so - I've been talking
to the "Kirston Wind Tunnel" people at the U. of WA.  Am going to take
a couple of my airfoils in to see if they can jury rig the airflow
over (parallel to) the lifting surface on my 360 degree circular wing.

It'll cost $500 to "rent" their small, slow wind tunnel equipment for
a day.  Hopefully that will prove or disprove the viability of the
idea.  It will be worth $500 to get this monkey off my back.  Am
anxious to either fly this unmanned lifing platform or trash it and
get on with the next dumb project.

Please weigh in if you comments, criticism (constructive or otherwise)
questions, better ideas or whatever.  I love Google a lot but this may
not be the best venue for this discussion.

Thank you for your time