


Joined Dec 2011
62 Posts

Thanks for the input. I had hoped that someone would take a look. It is holiday time so it may take a few days to figure out what I think fits.





Crossplot, beyond the connection between the flow curvature and the local pressure, there's another important consideration. From your webpage, it appears you are trying to connect the conditions at the surface of a cylinder (or airfoil) to the pressure there. While they may determine the pressure gradient, the conditions at a single point on an airfoil do not alone determine the pressure at the surface (as has been pointed out, deflecting a flap will influence not just the pressure on the flap, but over the whole airfoil).
According to classical Physics, the motion of the air over a cylinder or airfoil is governed by conservation laws (conservation of mass, momentum, and energy). By considering a single point in the flow, it is possible to use the conservation laws to derive a system of equations governing the fluid flow. In general, the conditions at any point within a continuous flow can be described by five quantities: the pressure, density and the three velocity components. In order to solve for the five "unknowns", you generally need five equations. You get one governing equation from conservation of mass, one equation from conservation of energy and three equations from conservation of momentum (one for each component). For the case of a Newtonian fluid (stress proportional to strain rate), these five equations are known as the NavierStokes equations. Under certain assumptions, the governing equations can be significantly simplified. For example, if the flow is steady, irrotational and uniform density, the field equations reduce to just a single equation (Laplace's equation). The flow over a circular cylinder discussed in "The Theory of Wing Sections" was determined by solving Laplace's equation. Solving Laplace's equation requires that you specify the conditions not only at the surface of the cylinder (or airfoil) but on every boundary of the fluid. In the case of an isolated cylinder, this can be done by requiring tangential flow at the surface of the cylinder and uniform flow far from the cylinder. If the cylinder is not isolated (such as when the ground is nearby), then you need to solve Laplace's equation subject to the appropriate boundary conditions at the surface of the cylinder, at the surface of the ground, and far from the cylinder. Changing the boundary conditions anywhere generally changes the pressure on every point on the surface of the cylinder (or airfoil). The presence of the ground influences the pressure on a cylinder passing overhead because it subjects Laplace's equation to a different set of boundary conditions. The bottom line remains that you can't determine the pressure at a point on the surface of a cylinder or airfoil purely from the boundary conditions there. You have to know the conditions over the entire fluid boundary, and then you have to solve the applicable governing equation(s). The discussion in "The Theory of Wing Sections" can be misleading in that it may appear that the flow is determined entirely by conditions at the cylinder boundary. A careful look at the derivation shows that both the governing field equation (Laplace's equation) as well as the boundary conditions far from the cylinder are needed to arrive at the solution. Edit: Understand you'll be be busy over the holidays, I just wanted to tie up some loose ends from our discussion before this thread sinks back to the "nanny nanny boo boo" level. 


Joined Oct 2004
3,349 Posts

Oh, I do, of course. I know several ways, all of them sharing a common basis. Let's consider a practical case: You have to airlift an airplane. Now, an airplane is a tricky underslung load. On one hand it's generally light enough, and it has a nice big fin on the back so once you picked up some speed it will align nicely with the airflow. On the other, it has those pesky wings, that keep wanting to lift it up in the air. You don't really want the plane to start flying on its own, the sight of your supposedly underslung load suddenly flying formation with you is, well, bad. On the other hand you don't want to add a ton of drag and weight to it either, and limiting your flying speed might mean you won't have enough fuel for the flight. So, how do you prepare an airplane for a safe airlift, if you can't remove the wings? Humor me.



Ashtabula, OH USA
Joined May 1999
4,282 Posts

Quote:
Your question is all over the place  if you're trying to test me, and you are, then ask something specific  something with a specific answer. BTW  the shuttle is carried on top, with little change to the carrier or the shuttle. The carrier gets the tail beefed up, as the shuttle messes with it  and the shuttle sometimes gets a temporary tail section added during a ferry to reduce the turbulence the shuttle puts on the carrier's tail  and also, the tail of the carrier gets more pitch control, to make up for the downwash coming off of the shuttle's wings. The shuttle, it turns out, has the least drag during a carry, when it's at 3 degrees AOA. And yes, the shuttle is lifting, a bit, during a ferry  but not enough to interfere with the carrier's flight. Now you know something you didn't know before. Awww  but you're a sly dog. Aren't you Brandano? 



Joined Oct 2004
3,349 Posts

Oh, I knew the shuttle carrier arrangements. And it's not the same thing as an underslung load at the end of a rope. It's not the same thing as towing a glider either. I just want you to try and use your logic and fail miserably by your own words. And I did chose a specific case I know well, but that can't be easily googled. You can get there by logic alone, but your understanding of lift will never get you to what is consolidated practice in these cases.



Ashtabula, OH USA
Joined May 1999
4,282 Posts

Quote:
So  flame on brother  flame on. NASA's got my back. This forum is a good example of mob science at work  and you  Brandano  are part of the mob. And you don't have a logical discussion with a mob  you stand back  and watch it work  and you can prod it, from time to time, to see what happens. 




To be clear, several posters have expressed disagreement with the content presented by some of NASA's educational websites. These websites serve a useful role in debunking some longstanding myths about lift generation, but to suggest they somehow represent NASA's depth of Aeronaurtics knowledge is a bit of a stretch. If you want to really understand the relationship between lift and momentum transfer to the air, you might read here (the author has worked with NASA for decades):
http://www.desktop.aero/appliedaero/...efftzlift.html I am not (and I don't think anyone else here is) suggesting a "Bernoulli theory of lift". Nor am I suggesting that a lifting wing doesn't exchange momentum with the air. I have simply been pointing out that when you go through the rigor of adding up all the momentum that a lifting wing transfers to the air, you discover that the rate of momentum transfer is not in general equal to the lift. It can further be shown that this result is entirely consistent with Newton's Laws. Your description of the fundamental mechanism behind lift, while more accurate than most popular descriptions, is not completely accurate. Local momentum exchange is certainly fundamental to lift, but the amount of vertical momentum imparted by a wing to the air depends on how the air is bounded (only under rare conditions is exactly equal to the lift). This reality detracts from the simplicity of a momentumbased explanation of lift, but careful analysis (see above link) shows that it is reality. 


Ashtabula, OH USA
Joined May 1999
4,282 Posts

Quote:
And in your analysis, is the momentum transferred to the air more, or less, than the lifting force? 




You cannot make a meaningful comparison between the lift force and the momentum transferred to the air as they don't share the same units. You can, however, compare the lift force to the rate of momentum transfer.
The relative magnitude of these quantities depends on how the air is bounded. In the limit where the vertical extent of the air influenced by the wing is very large compared to the horizontal extent, they are equal. In the limit where where the horizontal extent of the air influenced by the wing is very large compared to the vertical extent, the rate of momentum transfer is zero (the lift force is larger). The rate of momentum transfer is always somewhere between zero and the lift force. 


Ashtabula, OH USA
Joined May 1999
4,282 Posts

Quote:





Quote:
It's worth pointing out that rotating lifting surfaces (like helicopter rotor blades) have a completely different wake structure than that of a steadily lifting wing. The analysis used to determine the rate of momentum transfer for a wing does not apply to rotating wings. It shouldn't be hard to recognize that the momentum transfer should be different for a helicopter rotor than for than for a steadily lifting wing. 



Ashtabula, OH USA
Joined May 1999
4,282 Posts

Quote:





Just because something is moving doesn't mean that it has momentum. The (linear) momentum of an object is, by definition, equal to the sum of mass times (vector) velocity of each of the elements that make up the object. If some of an object's elements are moving one way and some the other, it is possible for some (or all) of the object's elements to be in motion despite the fact that the object has no net momentum.
The air passing through the rotor disk of a helicopter in ground effect cannot flow into the ground or collect below the helicopter. Instead, it is deflected outward and upward. The fact that there is significant upward motion of the air is well illustrated in the attached photo sequence. At any instant in time the air's downward momentum is equal to sum of all the momentum of the downward moving air minus the sum of all the momentum of the upward moving air. The $64,000 question is: how does the momentum of the downward moving air compare in magnitude to the momentum of the upward moving air? In the limiting case of ground effect, it is easily demonstrated that they are equal in magnitude (no net vertical momentum). To answer your question. In the limiting case of ground effect, the rotors are making the air move, but they aren't changing its net vertical momentum. 



"Net" may be a small word, but it's an important one. If you are going to make a connection between the lift generated by a wing and the momentum transferred to the air, there is no basis for neglecting the upward momentum it transfers to the air. From an action/reaction standpoint, the air deflected upward is of equal importance to the air deflected downward.

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