

Dec 13, 2012, 06:59 AM  

Quote:
Trying to hover next to a ceiling would be very difficult ! Trying to hover right next to the ground is also problematic Ideally the rotor does not have to deal with air flow except for air which comes from directly above Everyone who has actually tried this  what do you think? 

Dec 13, 2012, 09:49 AM  

Quote:
EDIT: Actually, I have to make one more note here. ShoeDLG is right in that near the ground the effect is statically stabilizing, whereas near the ceiling it is destabilizing. However, near the ground you still run into dynamic stability issues because there is relatively little damping of those vertical oscillations in hover. 

Dec 14, 2012, 04:53 AM  

The above responses suggest proximity to the ground (or a ceiling) affects both the static and dynamic instability of a hovering helicopter. It has been noted that proximity to a floor promotes static stability, whereas proximity to a ceiling promotes static instability. The effects of turbulent recirculation in ground effect were brought up as well.
I'm sure I'll be accused of overcogitatin' the problem with book larnin', but I submit you can't learn everything there is to know about ground effect by kicking over a piece of plywood. Why try to confuse a tractable problem by oversimplifying it? 
Dec 17, 2012, 05:05 PM  
Ashtabula, OH USA
Joined May 1999
2,676 Posts

Quote:
This helps to accelerate the air downward, as the wing moves through the air. Along with the fact that the air is a fluid, and "sticks" to the top of the wing  we get a lot of air both accelerated, and moved in a downward direction, as it falls down the back/top of the wing as the wing moves forward. This is nothing less than Newton's third  action and reaction. The wing's action is moving air downward as the wing passes through the previously still air  and the reaction is a force called lift. So  it's good that your brain doesn't accept the observation you noted  it's all good. You are in sync with NASA's understanding of lift. Can't be bad. BTW  This force added to air is measurable, with a large scale, placed under a hovering quadcopter or helicopter. 

Latest blog entry: Vector FC on an F450 using e300 power


Dec 18, 2012, 10:18 AM  

So Crossplot, I think there’s value in trying to “tie up” the discussion about the connection between the normal acceleration (flow curvature) experienced by a fluid element and the pressure variation in the fluid. I think we are in complete agreement about how the radial pressure gradient at any point in the fluid is related to the acceleration normal to the flow streamlines:
dp = density *( v^2 / r) * dr At any instant in time, a fluid element moves along a path that can be described by a circle tangent to the path that has the same radius of curvature. The above equation says that if you move out from the center of the circle by a very small distance dr, the pressure will increase by a small amount equal to the fluid density times the velocity squared times the distance you moved divided by the radius of curvature. The remaining question is: how do you turn this expression into one that relates pressure changes in the fluid to velocity changes in the fluid? Your approach has been to multiply the right hand side by r/2, where r/2 represents an “integration factor” to account for the effects of pressure gradients elsewhere in the field. Multiplying the right hand side by r/2 and replacing the left hand side with the change in pressure gives the familiar Bernoulli equation: Change in pressure = (density/2) * change in (v^2) I don’t think you have any real support for this approach (multiplication by r/2) other than “it works”. There is an alternative approach that doesn’t involve any mysterious integration factors. If the fluid is irrotational, then you can relate changes in location (radius) within the fluid to changes in velocity: dr = (r/v)dv or equivalently: dv = (v/r)dr This equation says that if you move a very small radial distance dr in an irrotational flow, the velocity will decrease by a small amount equal to the fluid velocity times the distance you moved divided by the radius of curvature. If you combine the equation relating pressure changes to radius changes and the equation relating velocity changes, you end up with an equation relating pressure changes to velocity changes: dp = density *( v^2 / r) * dr = density *( v^2 / r) * (r/v)*dv This simplifies to: dp =  density * v * dv You can easily show that if the change in velocity (dv) is very small compared to v, then v*dv is almost exactly equal to (1/2)*[(v+dv)*(v+dv) – v*v]. In other words, the change in the value of (1/2)v^2 is equal to v times the change in v. If you need convincing, just compare these two expressions using a variety of values for v and dv (keeping dv<<v). This lets you write: dp = density * d[(1/2)v^2] where d[(1/2)v^2] represents the change in the quantity (1/2)v^2. If the density stays the same throughout the fluid, then you can also write this: dp = d[(1/2)*density*v^2] where d[(1/2)*density*v^2] represents the change in the quantity (1/2)*density*v^2. This equation tells you that the change in pressure as you move from one location to another is equal to the change in (1/2)*density*v^2 (with a minus sign). This is true whether the changes are big or small. This gets you to the steady, uniform density, irrotational form of the Bernoulli equation without having to resort to a mysterious and unsupportable r/2 “integration factor". Psst... post # 821 is total nonsense. 
Dec 18, 2012, 11:28 AM  
Ashtabula, OH USA
Joined May 1999
2,676 Posts


Latest blog entry: Vector FC on an F450 using e300 power


Dec 18, 2012, 11:30 AM  
Ashtabula, OH USA
Joined May 1999
2,676 Posts

Quote:
But I won't haul you in this time  I'll let NASA do it  from here ... http://www.grc.nasa.gov/WWW/K12/airplane/lift1.html Since they do it so much better than I. Here's a cutnpaste from that link (I added the underlines) ... HOW IS LIFT GENERATED? There are many explanations for the generation of lift found in encyclopedias, in basic physics textbooks, and on Web sites. Unfortunately, many of the explanations are misleading and incorrect. Theories on the generation of lift have become a source of great controversy and a topic for heated arguments. To help you understand lift and its origins, a series of pages will describe the various theories and how some of the popular theories fail. Lift occurs when a moving flow of gas is turned by a solid object. The flow is turned in one direction, and the lift is generated in the opposite direction, according to Newton's Third Law of action and reaction. Because air is a gas and the molecules are free to move about, any solid surface can deflect a flow. For an aircraft wing, both the upper and lower surfaces contribute to the flow turning. Neglecting the upper surface's part in turning the flow leads to an incorrect theory of lift. 

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