Mar 11, 2012, 06:37 PM
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Thread OP
(Edit May 2012-- my thoughts on this matter have now changed significantly-- I'm still convinced that the earth "feels" a downward push from the wing of an aircraft in flight, equal in magnitude to the weight of the aircraft, which is also equal to the upward gravitational attraction that the aircraft exerts on the earth, but I no longer believe that this downward force need involve any specific amount of downward momentum of the air (downwash). For more, see posts 58, 61, and 72. End edit.)

Tom, I certainly must confess to not having read the entire (downwash) thread.

I am not sure what you are holding constant. Is area constant?

If not, if we are doubling both chord and area, with set span (or better yet, endplates), then it seems we are doubling the volume of the downwash.

You are thinking of downwash as an angle and a velocity but surely there is a sense in which we must think of downwash as a volume times an average downward velocity of that volume. If we scale up a wing 1000 times, or even if we just make the chord 1000 times longer, surely the downwash volume changes, as the total lift force in pounds increases, even though the lift coefficient stays the same and all the velocities and angles stay the same.

As I read your question you seem to suggest that the aerodynamics of the flow behind the wing are staying exactly the same-- no change in downwash volume-- even though by adding more wing area we have more surface area to feel the pressure differences and create lift. But surely this is not really true-- surely if we are making more lift, then there is in fact a real change in the flow behind the wing. Am I mistaken on this?

Can someone else give Tom a better answer?

Is it possible that the aerodynamics of the flow behind the wing are unchanged by the addition of more chord (and creation of more lift) in the idealized infinite-span case, but not in the real-world case, and the action of tip vortices etc must be considered to understand how the downwash volume increases as the chord and area of the wing are increased?

In the real world, adding more chord and more area and getting more lift without changing the aerodynamics of the flow behind the wing sounds an awful lot like getting something for nothing...

For one thing, if the span is finite, and we increase chord and area, then we are decreasing the aspect ratio-- doesn't that increase the total downwash in terms of volume times time?

Steve

Quote:
 Originally Posted by Tom Harper Steve, OK - so let's go back to an earlier example: The downwash angle is determined by the coefficient of lift Cl The Cl is determined by the angle of attack Lift then is proportional to velocity squared v Using a flat plate section at an angle of attack A and some area S set v so the lift is 20 lbs Now double the chord of the wing at the same angle of attack and velocity The downwash angle and velocity will remain the same because the Cl and v have not changed However twice the chord will produce twice the lift from the same amount and angle of downwash If lift was in anyway dependent on downwash the downwash would have to have doubled or changed it's angle Tom
Last edited by aeronaut999; May 10, 2012 at 05:42 PM.
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Mar 11, 2012, 06:55 PM
Registered User
Thread OP
(Edit May 2012-- my thoughts on this matter have now changed significantly-- I'm still convinced that the earth "feels" a downward push from the wing of an aircraft in flight, equal in magnitude to the weight of the aircraft, which is also equal to the upward gravitational attraction that the aircraft exerts on the earth, but I no longer believe that this downward force need involve any specific amount of downward momentum of the air (downwash). For more, see posts 58, 61, and 72. End edit.)

Quote:
 Originally Posted by HerkS Momentum is conserved in a closed system - so if you fly your helicopter inside a box there is no net force on the system. Not so in an open system - an aircraft in flight is an open system,
The purpose of this particular thread is to consider an aircraft in flight as a closed system. The planet and the atmosphere are part of the closed system. We may admit solar radiation into the system to provide power to over come drag. That doesn't change the fundamental conservation of (vertical) momentum issues. I think that there is a legitimate subject to discuss here.

Helicopter in a closed box-- gravity is pulling down on the helicopter from outside the box. If the box is on a scale, the weight of the helicopter will register on a scale even when the helicopter is in flight within the box. This can only happen if the helicopter's downwash is hitting the bottom of the box with a force equal to the helicopter's weight. Surely we all agree on that.

This thread argues that the same is true of an airplane in flight over the earth. The earth "feels" the aircraft's weight. That must be due to some sort of downward momentum reaching the earth's surface. Which could be measured by means of a (large enough) scale. Unlike the pressure increase from a balloon in flight-- this occurs on a global not a local scale, and could be measured with a (sensitive enough) barometer at any point on the earth's surface, but not with a scale located somewhere under the balloon no matter how large the scale is. I gave a more detailed argument for this in an earlier post, based on global mass distribution-- the center of mass of the earth/ balloon system is located at the center of the earth, but the center of mass of the earth/ airplane system is not located at the center of the earth. See items 4) and 6) in post #1.

Steve
Last edited by aeronaut999; May 10, 2012 at 05:42 PM.
Mar 11, 2012, 07:19 PM
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Thread OP
(Edit May 2012-- my thoughts on this matter have now changed significantly-- I'm still convinced that the earth "feels" a downward push from the wing of an aircraft in flight, equal in magnitude to the weight of the aircraft, which is also equal to the upward gravitational attraction that the aircraft exerts on the earth, but I no longer believe that this downward force need involve any specific amount of downward momentum of the air (downwash). For more, see posts 58, 61, and 72. End edit.)

Quote:
 Originally Posted by HerkS Momentum is conserved in a closed system - so if you fly your helicopter inside a box there is no net force on the system. Not so in an open system - an aircraft in flight is an open system
PS we would reach the same conclusions as I am suggesting, if we just considered in airplane in a closed box. Feel free to make the box as large as you want. Naturally there is a scale under the box. How does the airplanes weight get transmitted through the bottom of the box to the scale? By means of a nebulous pressure increase due to the kinetic stirring-up of the air by the wing, that is pressing equally on all sides of the box, pressing up on the top of the box just as hard as it is pressing down on the bottom of the box? That couldn't work. Or by means of the momentum of the downwash pushing down on the bottom of the box?

If the airplane pushes over into zero-G flight inside the box, does the box get lighter, as measured by the scale under the box? I say it does. Then the box gets heavier during the pull-out that follows.

Steve
Last edited by aeronaut999; May 10, 2012 at 05:42 PM.
 Mar 11, 2012, 07:21 PM Registered User Thread OP Ok enough for today for me. I would like to see some more input from others regarding posts # 29 and #31 as I'm not sure I fully know the answer to this. Steve Last edited by aeronaut999; Mar 11, 2012 at 07:34 PM.
 Mar 11, 2012, 07:22 PM Registered User The mass and gravity of a single aircraft in comparison to the entire planet is negligible at best. It's like comparing the mass of a grain of sand to that of a car. 747-8's mass is approx. 442,000 kg. Earth is 5.9722 × 10^24 kg. Don't have time/will to build a complex argument here but I agree with Pat, Bruce and Herk.
Mar 11, 2012, 07:29 PM
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Thread OP
Quote:
 Originally Posted by slow2by2z The mass and gravity of a single aircraft in comparison to the entire planet is negligible at best. It's like comparing the mass of a grain of sand to that of a car. 747-8's mass is approx. 442,000 kg. Earth is 5.9722 × 10^24 kg. Don't have time/will to build a complex argument here but I agree with Pat, Bruce and Herk.
I think this type of argument misses the point of this thread. Remember, the plane is going to fly in circles around a point on the earth's surface for a long, long time, so it has lots of time to accelerate the earth upward. Think of sunlight hitting a solar sail as it sets of on a voyage across the solar system. But if that's your view, how do you handle the case of the plane in the box as detailed in post #33?

Steve
Last edited by aeronaut999; Mar 11, 2012 at 07:39 PM.
 Mar 11, 2012, 07:47 PM B for Bruce Herkm I am SOOOOO in agreement with you on the aspect of the downwash coming off the trailing edge being what is left after the airfoil does its job. To my mind it was the reason why I told the originator of that last thread why I was totally done more then once. I guess I just can't say "no" to a good mind puzzle.... Aeronaut, the whole conservation of momentum seems like a good principle. But since the airplane develops its lift by reacting with the air mass and not neccesarily by pushing against the earth I'm not sure the whole argument holds water, so to speak. Is it really a crime if the earth is "falling" towards a plane in flight? It's not like it's enough to make the earth shift to any significant degree in it's orbit. And in truth the earth does not revolve around the true geometrical center anyway. The distribution of the plates that make up our landmasses both above and below the water of our oceans ensure that the center of mass of this ball of dirt is anything but centered on the overall average spherical center. Tossing in a few planes in flight wouldn't cause more than an undetectable blip in the gravitational "noise" level. Balloons and fish at their balance point, in other words neither rising nor falling, don't add or subtract to the whole picture as their volumes at that point are of the same density as the air around them. They simply become part of the homogenious mass of the water or air of the globe.... unless you want to look at them as providing an increase to the overall volume of water or atmosphere and by being part of the water or air mass raising the height of the oceans or atmosphere and thus contributing to some change and effect of such a change. Ya know... thinking along those lines we're getting back dangerously close to the age old mystery of how many angels are able to dance on the head of a pin..... The whole conservation of momentum certainly applies to someone that jumps off the earth and then they suck themselves back together and "impact" to a landing. But I'm not so certain that it applies to an airplane which is able to develop lift in various directions (but mostly "up") by interacting with the air around it to re-direct air in such a way as to create reaction like thrust through the action of wing surfaces. But I'll freely admit that it's all a feeling and that I don't know enough to say either way with authourity.
Mar 11, 2012, 08:16 PM
I don't want to "Switch Now"
Quote:
 Originally Posted by aeronaut999 PS we would reach the same conclusions as I am suggesting, if we just considered in airplane in a closed box. Feel free to make the box as large as you want. Naturally there is a scale under the box. How does the airplanes weight get transmitted through the bottom of the box to the scale? By means of a nebulous pressure increase due to the kinetic stirring-up of the air by the wing, that is pressing equally on all sides of the box, pressing up on the top of the box just as hard as it is pressing down on the bottom of the box? That couldn't work. Or by means of the momentum of the downwash pushing down on the bottom of the box? If the airplane pushes over into zero-G flight inside the box, does the box get lighter, as measured by the scale under the box? I say it does. Then the box gets heavier during the pull-out that follows. Steve
Even in the static case the pressure in a box is not equal on all sides.
There is more pressure on the bottom than the top because the pressure is higher there. Bernoulli explains this
It might be hard to visualise this with air, but it is clearly true if the box is filled with water.

Pat MacKenzie
Mar 11, 2012, 08:19 PM
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Thread OP
So, earth is 1.35117647 × 10^19 times as heavy as 747. Earth is attracted toward 747 at rate of g / ( 1.35117647 × 10^19 ) =7.25293862 × 10^(-19) m/s/s . So, if the 747 were not pressing down on the earth, then how long does it take for a 747 circling above a point on the earth to accelerate the earth to the speed of light? Even using the simple Newtonian formulae.

(Edit-- I dropped an exponent sign in my earlier calculation, so that
10^(-19) was transformed into (10-19). I'm now getting
something like 1.3 x 10^19 years. But the point remains the same, )

Of course the details are not the point of this thread. Rather, the point is that momentum must be conserved and the earth cannot be attracted upwards at all. Something must push the earth down.

Steve

Quote:
 Originally Posted by slow2by2z The mass and gravity of a single aircraft in comparison to the entire planet is negligible at best. It's like comparing the mass of a grain of sand to that of a car. 747-8's mass is approx. 442,000 kg. Earth is 5.9722 × 10^24 kg. Don't have time/will to build a complex argument here but I agree with Pat, Bruce and Herk.
Last edited by aeronaut999; May 08, 2012 at 03:30 PM.
Mar 11, 2012, 08:22 PM
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Thread OP
Quote:
 Originally Posted by pmackenzie Even in the static case the pressure in a box is not equal on all sides. There is more pressure on the bottom than the top because the pressure is higher there. Bernoulli explains this It might be hard to visualise this with air, but it is clearly true if the box is filled with water. Pat MacKenzie
Granted. Still-- consider-- does this static pressure difference between top and bottom of box (as measured by barometers) change when the plane is in flight? Does the scale under the box read differently during the zero-G pushover and the 2-G pullout, than when the plane is in 1G flight? I say no to the first and yes to the second.
 Mar 11, 2012, 09:44 PM Recreational Engineer Steve, You are looking for fly turds in the elephant pen. I withdraw. Thanks, Tom
 Mar 11, 2012, 09:52 PM Registered User Thread OP (Edit May 2012-- my thoughts on this matter have now changed significantly-- I'm still convinced that the earth "feels" a downward push from the wing of an aircraft in flight, equal in magnitude to the weight of the aircraft, which is also equal to the upward gravitational attraction that the aircraft exerts on the earth, but I no longer believe that this downward force need involve any specific amount of downward momentum of the air (downwash). For more, see posts 58, 61, and 72. End edit.) Ladies and gentlemen this is a sincere thread. Motivated by the philosophy that we can approach the truth by many different converging roads-- i.e. it is entirely valid to seek non-aerodynamic evidence for the relationship between downwash and lift. That's the way the world (cosmos) is put together. As far as the aerodynamics side of the argument goes, I remain interested in getting better insight as to the questions posed in posts # 29 and # 31. The answer appears to lie in the role of the tip vortices? Not sure this is it though-- tip vortices may relate to the (drag) cost of producing downwash rather than the (volume* velocity) of the downwash field-- even with end plates (no tip vortices), in a wind tunnel, surely increasing the wing chord has some effect on the flow behind the wing? Scaling up the chord must also scale up the "picture" of what is happening behind the wing in some way, so that the (volume* velocity) of the downwash field is increased, even if the downwash angle as measured at some particular scale-independent point is unchanged? No? Thanks very much. Steve Last edited by aeronaut999; May 10, 2012 at 05:43 PM.
 Mar 11, 2012, 10:22 PM Grad student in aeronautics Conservation of linear momentum is a rewording of Newton's first law. Of course, this is just a special case of his second law where the net external force is zero. When we look at a system of particles (such as Earth+atmosphere+airplane) we are actually using Euler's laws of motion. Euler's first law states: For a system of particles, S, the inertial acceleration of the center of mass is equal to the sum of all external forces. Inertial acceleration just means that the acceleration is with respect to an inertial frame. It will be easier to apply Steve's though experiment with the box. So consider this box sitting on a scale with air and a bumble bee. Well, maybe we should pick an insect that can actually fly... To simply the numbers, zero out the scale before putting the bee in. Then when the bee sits on the bottom, the scale will read the weight of the bee. If the bee jumps up, the scale will: 1. momentarilly read more than his weight as he pushes off, 2. read zero as he goes through the air (not flapping), 3. read more than his weight as he lands, 4. and finally, after he comes to rest, it will read his weight again. Now if he used his wings instead of his legs you will have the exact same story. Similarly, if he hovers, the scale will read his weight. The increase in pressure on the bottom will equal his weight. If he hovers too long (thinking like a helicopter), the air will start to circulate in a toroidal fashion and it will become increasingly difficult to hover.
 Mar 11, 2012, 10:28 PM Registered User Thread OP Not intending to go to far in using a popular work as a definitive source, but it seems well written, and I do find this sentence "There is downward momentum in any air column behind the wing. There is zero momentum in any air column ahead of the wing, outboard of the trailing vortices, or aft of the starting vortex." In this summary http://www.av8n.com/how/htm/airfoils...rfoils-summary In what appears to be a fairly well thought-out presentation of lift http://www.av8n.com/how/htm/airfoils.html (I know, it doesn't address all the different nuances we've been talking about but thought it worth pointing out.) Steve
Mar 11, 2012, 10:35 PM
I don't want to "Switch Now"
Quote:
 Originally Posted by aeronaut999 Granted. Still-- consider-- does this static pressure difference between top and bottom of box (as measured by barometers) change when the plane is in flight? Does the scale under the box read differently during the zero-G pushover and the 2-G pullout, than when the plane is in 1G flight? I say no to the first and yes to the second.

I would say it has to be different, since the total "weight" of the box is the same if the plane is in your hand or gliding along.
It would also vary as the plane did pushovers and pullouts, just as it would if a motorcycle was doing loops on a vertical circular track. Or if you are throwing a ball up and down. Different mechanisms in each case but the principle is the same.

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