Just wondering
I once saw an equation in MA that lists stall speed as a function of wing loading (among other things), But nowhere in that equation did wingspan have an effect on lift. I have also read that lift is not influenced by chord, but much more by wingspan, being a result of vortices. I have read that the bigger the wingspan. The more lift is generated I think (I might be wrong) That as lift gets larger, then stall speed decreases. Then how come the equation for stall speed does not have wingspan incorporated into it. If the equation is true, wouldn't the most aerodynamic airplane (Apart from turning very badly)have a very narrow wingspan and a big chord. There would a low stall speed from the wing area, but small frontal area. And what about flat bottom airfoils? The way I have read it is that the flat bottom is a built-in angle of attack. Doesn't that influence the stall speed somehow?
Unfortunately I'm still pretty uneducated in aerodynamics(I plan to be an aircraft engineer some day), so I beg for layman's terms
Thanks

You may be thinking about aspect ratio. Aspect ratio = span / chord

If the chord is not constant, then it is span^2/area

Higher aspect ratio wings are more efficient than those with lower aspect ratios. The Lift/Drag ratio is better. Either more lift, or less drag will improve the L/D ratio, but I think less drag is what is happening when you go with higher aspect ratios.

Also, the lift versus angle of attack curve is steeper, hence (.. or, and...) they stall earlier than wings with low aspect ratios but otherwise identical airfoils. This simply means a high aspect ratio wing reaches the max lift coeficient sooner (i.e. lower AoA) than a low aspect ratio wing using the same airfoil.

Stall speed is defined as the speed required to support the weight of the plane when the wing is operating at maximum lift coefficient. Buried in that max lift coefficient could very well be the factors you expect such as wing span. But the problem is not really tractable at such a level of detail so a single number "lift coefficient" for the entire wing (not 2D airfoil, *wing*) is used. One could say, *plane* not wing for that matter.

I think answer is drag. Low aspect ratio wings have poor lift/drag ratios. Low L/D = more fuel, or lesser glide ratio if you're a glider.

I think that answer is yes of course, and many other things. Not very satisfying... I know. Perhaps someone else can do better

.... you sure packed a lot of questions in there.

Best of luck with your endevours.

To add a little to Jim's excellent answer.

Yes airfoils with higher maximum lift coefficients will slow down further before stalling. That's one item that should have been in that formula. If you go back and read the article you may find that they assumed a max Cl of 1 or some other amount. Cl=0.9 to 1.2 is quite a common max lift coefficient range for model type airfoils so assuming the value equals 1 isn't that far out.

And flat bottomed airfoils do not have a built in angle of attack. Folks just draw them that way (like Jessica Rabbit ). Properley presented a ClarkY should be drawn with the chord line from the very peak of the leading edge to the point on the trailing edge in horizontal alignment. When done that way the flat lower surface will be angled. Don't be confused by common convention even if it IS common.

Here are some equations for you to play with. Understanding aerodynamics has a lot to do with understanding the math, so to really get a good understanding of the subject you need to start playing with the math.

1:Velocity/Glideslope=Sinkrate
2:Glideslope= Lift/Drag
3: Power for level flight=Sinkrate*weight

(There is a close relationship between Aspect ratio(AR) and Glideslope)
so substitute AR for glideslope....

Velocity/AR=Sinkrate
AR=lift/drag

Lets take two plane going 100 fps (feet per sec) AR=5 the other AR=10. Both weigh 10 pounds.

In the first case 100/5=20 fps sinkrate and Power would need to be 20*10=200 fp/s
In the second case 100/10=10 fps sinkrate...power would be 10*10=100 fp/s

So if the AR is equal to the glideslope then the power for level flight changes inversly with the aspect ratio. (if I double the Aspect ratio then the power needed is cut in half)

The next set of equation require a little bit of algabra: If you have not had algabra and earth science/physics maybe this will help you to take the subjects serious and see that they do relate to the real world.

Lift=Cl * .5 * q * V^2* area (remeber Lift/Drag=glideslope)
Drag=Cd * .5 * q * V^2 * area

Induced Drag= Cl^2/pi*AR*e

Cl is the coeff lift
Cd is the coeff drag
q is the mass density air
V is the velocity
Area is the area of the wing.

Now most likely your equation for Stall speed was the Lift equation rearanged:

V=sqrt( Lift/(Cl * .5 * q * area)) If a plane is flying level then Lift=Weight
V=sqrt(2 *weight/(Cl*q*area)) weight/area is just wingload so:
V=sqrt (2 *wingload/Cl * q)
V=sqrt(wingload)*sqrt (2/Cl*q) Now input q=.002378 at sea level
V=sqrt(wingload) * sqrt(841/Cl) Now change from pounds to onces
V=sqrt(wingload) * sqrt(52.565/Cl) Now convert to MPH
V = sqrt(wingload) * sqrt(25/Cl) Now set Cl = 1
V= sqrt(wingload) * 5 (in MPH)

You will notice that before we get to the last equation Cl is included.
and (induced drag=Cl^2/pi*AR*e )so:

Cl=sqrt(induced drag/pi*AR*e) so the aspect ratio is included in the equation as well as the induced drag.

Maybe that well give you the math prof.....

Here are some good websites you might want to look at:

[link]http://www.auf.asn.au/groundschool/contents.html[/link]
[link]http://142.26.194.131/aerodynamics1/Performance/Page3.html[/link]
[link]http://www.grc.nasa.gov/WWW/K-12/airplane/foil2.html[/link]


Steve

Thanks for the answers
And as to the stall speed equation, it was published in the article "Matching Propeller to Model Mission" by Don Brooks published in Model Aviation in the September 2002 issue
The Equation is as follows
V(sub stall)=(5.46)*((W/S)^.5)/d*C)^.5)
Where W=Weight
S=Wing Area
d=density of the air
and c is the lift coefficient
And I kinda see how it is derived from your equations, but I'm confused by the 5,46 in front.
And I actually understood the math.

I'm glad you understood the math. As far as the 5.46 number I have no idea.
The equation that your have may have been published but it looks like a garbage equation to me. The nice thing about the Lift/Drag equation is that they are
non-dimensional, then someone comes along and makes them dimensional, but even worse they don't even use a standard units of measurements....SI units (Newton, meters, seconds) English system(pound, feet, seconds). So, with the equation you have I need to know what to input....is W in ounces? Is area in feet? Is the result in MPH or fps? I haven't read the article so I don't know what they where attempting to do but I would not waste much time worrying about the equation, it is not profound. On, the other hand the Lift/Drag equation stated in there correct form are elegant. Let's look at them a little closer:

If you have had earth science then you have probably seen the equation for Kinetic Energy:

Ke= .5*mass*V^2

This look very similar to the Lift and Drag equations.....so lets say that (.5*q*V^2) is the energy in the system.

Lift=Cl*.5*q*V^2*area
So the amount of Lift that is generated is equal to some portion (Cl) of the energy in the system (.5*q*V^2) interacting with the wing (area).

So if you are going to play with equations....I would play with the non-dimensional equations...and get use to converting to the standard units of measurement before. Then after you have solved the equations you can convert it to what ever form you want.

the one in the magazines seems awful similar [sm=confused.gif]
To help
Stall speed is in mph
weight is in ounces
wing area is in square feet
and air density is in grams per liter
I think the 5.46 is a conversion factor
What are the units in your equation?
And it says that the quation was derived from MArtin Somon's Model Aircraft Aerodynamics

Try to get ahold of a book called Theory of Wing Sections by Abbott and Doenhoff. It will answer all of your questions about aerodynamics, plus it has airfoil data on just about every NACA airfoil ever tested. Another good book is The Illustrated Guide to
Aerodynamics by H.C. Smith... it is not as technical, and has a lot of illustrations to help visualize aerodynamics.

Ty

Mitek...The general form of the equation is good for either english or metric units, as long as you stick with the standard units (feet, seconds, and slugs and their derivatives in english, or meters, seconds, and kilograms and their derivatives in metric). Consistency is the key. What you have there is a form of the equation that's been de-standardized so that units familiar to and used commonly by r/c modelers can be directly plugged in without conversion. Unless the units for each variable are specified, that form of the equation is completely meaningless.

Indeed, that's quite a mishmash of units. English nonstandard mph, english nonstandard ounces, english standard square feet, and metric nonstandard g/L.

The units for the general form of the equation would call for, respectively, ft/s, slugs, square feet, and slugs per cubic foot; or m/s, kg, square meters, and kilograms per cubic meter.

Slugs?
What are those? I have deduced that they are some unit of mass, but the Standard unit I'm familiar with for mass is pounds.

Pounds is a unit of force (or weight). One pound is equal to ~32 slugs-ft/sec^2 on the surface of the earth, when dealing with the weight of an object.

Mitek

Maybe an example of how to use the equations will help.

Lets say we have the following:

Wing Span = 48 inches
Wing Chord = 12 inches
Weight = 4 pounds 8 ounces
Cl = 1

q= .002378 slugs per cubic foot at sea level and standard temp pressure

So:

First convert everthing to standard English system of measurements (foot, pound, sec)

wing span= 4 feet
wing chord= 1 foot
weight = 4.5 pounds

Wing area = span * chord= 4

Now the Equation:

Lift=Cl * .5 * q * V^2 * area

If we want to find out the Velocity at a given Cl then rearange the equation:

If we want to know the min speed that a plane can fly then we would need to know the max Cl. (note: generally a plane stalls at a little past the max Cl)
(lift=Weight)

V= sqrt(2 * weight)/(Cl*q*area)
V= sqrt((2* 4.5)/(1*.002378*4))
V = sqrt (9/.009512)
V = sqrt(946.17)
V = 30.75 feet per sec

Convert to MPH
3600 seconds in 1 hour
5280 feet in a mile so:

30.75 * (3600/5280)
MPH= 20.97

Now from what we have hear we can calculate the Induce Drag:

Induced drag= Cl^2/(pi*AR*e)
e is the eliptical lift distribution just set it to 1 for now (e=1)
AR=span^2/area so
AR= 16/4=4

Induced drag= 1/(3.14*4*1)
Induced drag=1/12.56
Induced drag=.0796 or about .08

Plug this into the Drag equation:

Drag=Drag_coeff * .5 * q * V^2 * area (keep every thing in standard units)

Lets say the only drag we have is the induced drag so Drag_coeff=Induced Drag
Drag= .08 * .5 *.002378 * 30.75 ^2* 4
Drag=.36 pounds

So the optimal lift/drag ratio would be 4.5/.36 = 12.5 this would be the best glideslope that the plane could reach at this speed and Angle of Attack. But planes don't only generate induced drag so the parasitic drag(drag not related to lift) is about double the induced drag. So the lift/drag ratio is closer to 4.5/.72 or 6.25.

Velocity/Glideslope=sinkrate

30.75/6.25= 4.92 fps

Power for level flight= sinkrate*weight
Power for level flight= 4.92*4.5
Power = 22 fp/s (note a .40 will put out about 100 fp/s at this speed)
( 1 Horse power is 550 fp/s)

Anyway hope that helps

I SHALL NOT BE USING EQUATIONS AS SOME PEOPLE SEEM TO MAKE IT A SYSTEM OF UNITS ISSUE.WHAT COUNTS IS REALLY WHETHER THE WING IS USED AT A SMALL OR WIDE ANGLE OF ATTACK RANGE,i.e. WHETHER YOU ARE CONSIDERING A GLIDER (large angle of ATTACK RANGE) OR A POWERED PLANE---FOR SMALL RANGES,INDUCED DRAG IS SMALL,THAT IS A SMALL SPAN,LARGE CHORD WING IS LIGHTER,LESS MOMENT OF INERTIAETC AT MARGINALLY HIGHER DRAG --ALWAYS CONSIDERING SAME SURFACE--A GOOD EXAMPLE FROM THE FULL SIZE WORLD BEING THE COSMIC WIND OR SIMILAR WITTMAN DESIGNS---MUCH APPRECIATE THESE FORUMS--ZERO AIRSPEED

Mitek,

Here is another way you can play with the equations:

Lift= Cl * .5 *q*V^2*area //again solving for V

V= sqrt(2Weight/(Cl*q*area) ///becomes the expression

So what happens if we Double the area ( keeping Cl, weight, q the same).

V= sqrt(1/2) = .707 so doubling the area will give us 70.7 percent of our origanal speed.

In the example before the speed was 30.75 fps so by doubling the area we would reduced the velocity to 21.74 fps, or in MPH it was 20.97 so (20.97*.707=14.8 MPH)

So what if I increased the weight by 2.5 times?

V=sqrt(2.5)
V=1.58 so if we increased the weight of our example plane we would go from
(30.75 *1.58) fps to 48.6 fps

So this is how you can play with the equations without units of measure and it tells you a lot more.....we learned that for a given set of conditions (cl,weight,q the same) doubling the area of the wing decrease the speed by 29.3 percent this would be for all planes that meet the conditions not just one plane.

Mitek,

As someone pointed out, when you consider the effect of span for a fixed area, you are considering the effect of aspect ratio. The equation you refer to derives stall speed from coefficient of lift, and the maximum coefficient of lift is not strongly affected by aspect ratio, except when the aspect ratio has an extremely large or small value. Overall, the equation you saw is correct not to include aspect ratio, as it does not have a primary effect.

banktoturn

Pardon me for going so far back
What I was meaning was that they are not at an angle in relation to the ground, but rather, that a flat bottom airfoil acts as a symmetrical one with a positive AOA
Or so my dad says....
And thanks for all the math
Keeps me from forgetting it over the weekend

Mitek,

I like to look at wings(foils) in terms of absolute AOA....and All planes have to fly with a positive absolute AOA (if it was zero then we would be generating no lift). At, 10 degree absolute AOA you get into the beginning of stall and at absolute AOA of about 15 degrees you are at max Cl and shortly there after you are stalled.

To change the foil cl to the Cl.... the three diminsionsal form, use the following equation:

Cl=cl/(1+(cl/(Pi*AR*e)))

ex:
AR=3
e=1
cl=1.5

Cl=1.5/(1+(1.5/9.42)
Cl=1.5/1+.16
Cl=1.29

Have you checked out
[link]http://www.grc.nasa.gov/WWW/K-12/airplane/foil2.html[/link]This has the AR correction in it and you can try out different shaped foils,speed,AR and see how they effect things.


Steve

HI DIPSTICK -GREETINGS FROM EUROPE-DO YOU DO MATHS AGAIN INSTEAD OF GOING FLYING?

What's up Duchy,

I'll try to get more flying in if that would make you happy.....right now the weather is cold and rainy so building is the next best thing...I admit that I'm a fair weather flyer.

Actually, some people do crossword puzzels, play trival persuit, etc I like equations. I blame this on my germantic heritage....my ansesters are from the Baden and the Rhine regions so just think we could be related...just thinking about that possiblility will probably really brighten your day, so...

Happy Flying

Steve

The nasa program is excellent! I have tried X-foil, but it was really confusing for me. The only fault I have found in this program is that it does not show the turbulence that is assotiated with a high AOA. And just wondering, what is the difference between cl and Cl?

Mitek,

The cl is the 2D lift coefficient (infinite wing model). CL is the 3D lift coefficient (finite model) and takes into account the air that moves from underneath the wing around the wing tips and dumps out on top of the wing (this is the wing tip vortices you may have heard of). The problem is the votices does two things: 1. creates drag. 2. The "span wise flow" reduces the lift that would normally be produce in this area of the wing which reduces the performance of the whole wing....so the CL will always be less than the cl. On the NASA sim page you have a AR correction BUTTON this is taking the model from the infinite to the finite model. In general the lower the Aspect ratio(AR) the larger percent of the wing that will be impacted by the vortices's.....building infinite wings or even very long slender wings are impractical, so AR from about 5 to 10 are the most practical to build (at least for large planes).

If you display the CL vs AOA plot graph on the NASA page and activate the STALL MODEL you will see a curve although it is not real noticeable at 10 degree absolute AOA the wing starts to stall. What this means is that the air (boundry layer), starting at the trailing edge and progressing to the leading edge, is separating from the wing. This causes drag but it also means that less of the wing is doing the work. At around 15 degrees AOA the amount of increased lift due to higher AOA is off set by the amount of lift being lost due to separation so this is the max CL (the highest point on the graph, the top of the hill). At around 19 degrees the air flow over the top of the wing become completely detached (STALL AOA). So the reason the CL vs AOA curve kind of looks like a hill is it is showing graphically the porgressive detachment of the boundary layer.

I will look around.... I use to have a very good link to a web-site that show the progressive stall of a wing with some pictures and some drawing. One thing that is a little confusing is the idea of smooth air flow(laminar), turbulent flow, and flow separation. Might be something to do a search on if your interested......have you ever seen the smoke coming from a cigar, etc....right after it leaves for a few inches the smoke is very uniform (laminar) at some point the smoke starts swirling and bubbling...turbulent flow. A foil can have either laminar flow or turbulent flow but not be stalled and sometimes foils are designed to force the flow to turbulent in order to keep the boundary layer from seperating(stalling).(see the princeton link with the tennis ball... the wire is creating a turbulant flow which allows the boundry layer to stay attached longer) Stall is caused from 'boundary layer separation' of the air flowing over the wing. I may be getting off track here but if you interested you might look for pictures that show drag....and flow...tuft picture(little pieces of string that indicate flow) Bluff body pictures etc.

Here are a few links to get you started:
[link]http://www.grc.nasa.gov/WWW/K-12/airplane/boundlay.html[/link]
[link]http://www.princeton.edu/~asmits/Bicycle_web/blunt.html[/link]


Happy Flying!!!!!

Thanks.
I have seen that the Graph on the nasa site has the parabola you described, but at at about 20 degrees AOA, the lift disappears entirely, meaning complete stall, or complete seperation of the boundary layer, right? Another question: How does this explain the harrier manouver during 3d-style flying? And does the nasa simulator show this seperation?
Another curious thing I noticed on the nasa simulator is how a ball or a cyllinder wil have lift only if its spinning. How come. And if according to the simulator an ellipse has more lift than a typical airfoil, why aren't airfoils ellipse shaped?

Mitek,

I'm not sure if I know enough to answer all of your question but I'll answer as best I can. Your right on the complete seperation at 20 AOA. If you take the AOA and mutiply it by .11 this will give you an ideal line any were that a Cl curve very from this it shows that something is going on to reduce lift...(spanwise flow, a bubble, boundry layer seperation).

Harrier manouvers are done with planes with Low AR.....the spanwise flow that is coming over the wing tips kind of energize the boundary layer creating turbulant flow and keeps the boundry layer from seperating (trapping the boundry layer).....this is a power "induced Angle of Attack" and harrier manouvers are only possible with a substantail amount of power and is not very effecient flight....the nasa sim does not show the induced AOA....and I don't know enough about it to quantify what the effect does. About all I can tell you is AR under 3 and closer to 2 or 1 can produce induced AOA. Remeber that your AR correct equation kind of tells you the amount of the wing that is being effected by the spanwise flow.

As far as the cyllinder or ball....lets say that they are spinning so that the top of the object is moving the same direction as the flow...this will create a low pressure area because the flow is being accelerated just like a foil. On the buttom the object would be spinning opposite the air flow so it would slow it down creating a high pressure area at the bottom of the object....very simular to what happens with a wing at some AOA. If a ball or cylinder is not spinning then the pressure is equal above and below so no lift is generated.... this would be like a symetric foil at zero AOA.

With the ellipse, lift is not the only thing that is important when considering what is the best airfoil....you must also consider drag....the two together (drag/lift) will give you the effecieny of a wing. I don't really know what the drag on an elliptical type foil looks like. Further the other issue that is important is were the center of pressure is on a foil and how it moves. In order to build a controllable plane this most be accounted for and the ellipse may be difficult to control...i'm not sure about that....would a parasail be an example of an elliptical wing? Maybe someone else would have a better explination?

Steve

Mitek,

I'm going to add to my answer a little. Lets talk about pressure, this is what cause a wing to lift. I assume you have used a straw to drink at some point. What your doing is creating a low pressure area inside your mouth. The atmosphere is at a higher pressure so it tries to equalize the pressure difference by pushing the fluid up the straw. Your lungs work the same way. Your diaphragm expands your lungs creating a low pressure area. Wings work just the same it's the pressure differances around a wing that cause lift...when the boundry layer seperates the pressure differance is reduced. If you notice that the nasa sim has a probe that will allow you to check the pressure around the wing. Using the probe is helpful but a better way to see this is with a picture. At the end of the following link there are some color pictures of the high and low pressure areas around a foil.

[link]http://www.mh-aerotools.de/airfoils/index.htm[/link] (check the end of the page for pictures)

Here's another link you that might help:

[link]http://www.eng.fsu.edu/~dommelen/research/airfoil/airfoil.html[/link]



Happy Flying!!!!
Steve

Great Links, DipStick
The pictures in #2 were absolutely marvelous and I could finally see boundary layer seperation. At the last picture, though, you can see that the boundary layer is still close to the wing, shouldn't the wind still have suffucient lift. And I presume that this stall is different from the one where the airplane is going too slow and it stalls. I have heard that in the stall described, the airplane can stall at any spped. Is this true? And I dpn't know where the pictures are on the second page,(the website got updated) but I found a great airfoil simulator.