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Old Dec 13, 2012, 04:13 AM
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Originally Posted by richard hanson View Post
Ground effect defined- Just tip over a piece of plywood - note the slow contact with the floor.
Simple explanations are usually better than complicated ones, but good ones accurately capture behavior over a wide range of conditions. The above explanation suggests ground effect results from air having to “move out of the way” of a wing in the same the way it has to move out of the way of a piece of falling plywood.

I suppose this analogy might be used to explain why a plane experiences less drag (and a higher lift curve slope) in ground effect, or why a helicopter needs less power to hover in ground effect.

When you play around with an RC helicopter, you quickly notice that it’s easier to find a power setting for a steady (constant altitude) hover when you are within a rotor diameter or two of the ground (compared to the middle of the room). As the helicopter settles toward the ground, the upward force on the rotor increases, contributing to altitude stability.

If you try to find a power setting for a steady hover within a rotor diameter or two of the ceiling, you will find it MUCH more difficult compared to the middle of the room (try it). As the helicopter rises toward the ceiling, the upward force on the rotor doesn’t decrease, it increases. This contributes to altitude instability.

If you were to try to rapidly push a piece of plywood against a ceiling, you would find that the air provides resistance to your effort (it certainly doesn’t try to suck the plywood up toward the ceiling).

Bottom line: the falling plywood analogy would suggest a helicopter would be easier to hover near the ceiling, not more difficult. While it may be visually compelling, this analogy completely fails to capture behavior with just a slight modification to the conditions.
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Old Dec 13, 2012, 06:59 AM
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Originally Posted by ShoeDLG View Post
the falling plywood analogy would suggest a helicopter would be easier to hover near the ceiling, not more difficult. While it may be visually compelling, this analogy completely fails to capture behavior with just a slight modification to the conditions.
- I think you made a bit of a leap on the comparison.
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?
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Old Dec 13, 2012, 09:49 AM
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Quote:
Originally Posted by richard hanson View Post
- I think you made a bit of a leap on the comparison.
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?
I have to agree with Richard here. Hovering near the ground is difficult for the exact same reason hovering near the ceiling is difficult - thrust changes as a function of altitude. That doesn't mean the aerodynamic phenomena causing the stability issues are the same, though.

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.
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Old Dec 13, 2012, 10:11 AM
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They get a bit skittish--
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Old Dec 13, 2012, 10:46 AM
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Actually turbolent recirculation makes hovering harder in ground effect as well. It might help with very widely spaced multirotors, perhaps. Looks like it on the human powered helicopters.
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Old Dec 14, 2012, 04:53 AM
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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 over-cogitatin' 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?
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Old Dec 14, 2012, 07:55 AM
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Now that's just downright wrong-
Just because I offered a simple example about how pressure changes as an object approaches the ground

You may have a lot of book larnin- '
I have little - and I do have the patents to proove it.
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Old Dec 15, 2012, 12:41 AM
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Got it. It's OK for you to admonish others when you perceive they are making a discussion too complicated, but not OK when you are admonished for making it too simple.
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Old Dec 15, 2012, 08:29 AM
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There you have it!
To be candid - I do like to see good explanations - but I don't take myself too seriously - The info swapped should be both interesting and fun.
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Old Dec 17, 2012, 05:03 PM
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whoopsie boopsie
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Last edited by Tim Green; Dec 17, 2012 at 05:08 PM. Reason: somehow created two msgs - so deleted this one -
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Old Dec 17, 2012, 05:05 PM
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Originally Posted by eflightray View Post
Low pressure on top, higher pressure underneath. Yet the low pressure air causes downwash into the higher pressure air ?

Still can't get my head around that.
Maybe not - but you hit the nail on the head, so to speak. Low pressure on top is caused by the fact that the wing is moving forward through the air - that creates a slight vacuum (low pressure zone) behind the wing. Which is also the top of the wing.

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.
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Old Dec 18, 2012, 01:47 AM
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Tim, how do you disrupt lift on a wing?
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Old Dec 18, 2012, 10:18 AM
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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.
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Old Dec 18, 2012, 11:28 AM
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Originally Posted by Brandano View Post
Tim, how do you disrupt lift on a wing?
Do you really not know this?
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Old Dec 18, 2012, 11:30 AM
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Quote:
Originally Posted by ShoeDLG View Post
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.
Yeehaw - I got a nibble. Let's see what's on the line now ... it feels like a Bernoulliest, from the way the line's wiggling around.

But I won't haul you in this time - I'll let NASA do it - from here ...

http://www.grc.nasa.gov/WWW/K-12/airplane/lift1.html

Since they do it so much better than I.

Here's a cut-n-paste 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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Last edited by Tim Green; Dec 18, 2012 at 11:36 AM.
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