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enord
no swatters please!
enord
& as the fly loses weight by burning fat & flying, the plane loses mpg?
adoucette
How about this thought experiment:

Take a cellophane box.

Put a fly in it.

Attach to the box a helium balloon, which displaces just enough air such that the box is lifted off the ground.

Wait till the fly lands and then let helium out of the balloon, atom by atom, until the box finally settles.

If the fly takes off, will the box rise?

Arthur
boit
QUOTE (adoucette+Nov 23 2010, 05:34 AM)
How about this thought experiment:

Take a cellophane box.

Put a fly in it.

Attach to the box a helium balloon, which displaces just enough air such that the box is lifted off the ground.

Wait till the fly lands and then let helium out of the balloon, atom by atom, until the box finally settles.

If the fly takes off, will the box rise?

Arthur

No, the balloon won't lift off. I believe it still won't lift off even if the fly stopped flapping its wings and flew level. This is my understanding although a seasoned pilot in this forum will disagree. He says as long as there is no flapping there is no extra weight generated. How do we make the set-up loose some weight even momentarily? If the fly takes a nose-dive (wings flapping or not).
AlexG
QUOTE
He says as long as there is no flapping there is no extra weight generated.


I said if there is no flapping, there is no downward force generated.

Let us imagine a large, hanger sized building, open at both ends, set on a scale. A small plane flies in through one open end, flies the length of the hanger in level flight, and exits the other end.

What does the scale read during and after the transit?
adoucette
QUOTE (AlexG+Nov 22 2010, 11:30 PM)
What does the scale read during and after the transit?

The scale would register the increased weight of the plane passing through it because the downwash behind a plane is vertical and if it is impacting on the floor of the the hanger and because of Newton's 3rd law, the mass X velocity of the downwash is equal to the weight of the plane.

I suspect if you did this in the real world, the increased weight on the scale would be a bit less then the weight of the plane due to not all the mass of the downwash is impacting the scale.
mudderrunner
what adoucette said sounds right to me. And I think the weight of the plane would decrease from the horizontal speed. I'm not sure by how much though.
adoucette
In level flight the mass X velocity of the air displaced downward is always equal to the weight of the aircraft, and that doesn't vary based on speed.

At lower speeds the wing has to "bend" the airflow displacing a higher volume of air but at lower velocity. Increasing the span and curvature of the wing by use of devices like leading edge slats and flaps on the wings help to do this at low speeds.

As a plane accelerates the volume of air displaced downward is less, but the velocity that it is displaced at increases.

Arthur
mudderrunner
horizontal speed decreases weight just like with an orbiting space station, or the moon. The speed it would take to reach weightlessness at ground level is probably very fast but a fraction of this speed would still cut a fraction of the weight.
boit
QUOTE (mudderrunner+Nov 23 2010, 08:34 AM)
horizontal speed decreases weight just like with an orbiting space station, or the moon. The speed it would take to reach weightlessness at ground level is probably very fast but a fraction of this speed would still cut a fraction of the weight.

I think i understand you correctly. The plane you're talking about is probably wingless. To maintain a level flat will be tricky near ground level. The speed you'll have to get at will surely take you to orbiting altitude. You won't fall fast enough to keep up with the curvature of the earth. That will essentially mean you're in free fall and therefore weightless. A spear thrown through the hanger is practicaly weightless in that setup.
enord
The fly in order to fly has to burn its flyfat. The fly flies up into the ceiling of the plane with as much force as it delivers to the floor of the plane as it drops after hitting the ceiling. The fly gets lighter as it converts flyfat to energy...the fly is flying against the force of gravity?
Capracus
As long as an object(the fly) is moving through a given volume of fluid(the aircraft atmosphere) it will transmit force to that fluid. Given that the aircraft containing the atmosphere which contains the fly are all subject to the same gravitational acceleration, the fly will transmit its gravitational force via the atmosphere to the aircraft, regardless of its velocity within the aircraft. Therefore whether the fly is airborne or at rest, the weight of the aircraft will remain unchanged.
NoCleverName
The Mythbusters did this a few years back with a closed truck full of birds. If I remember correctly, the total system weight did not change (within reason) when the birds became airborne.
enord
QUOTE (NoCleverName+Nov 23 2010, 07:35 AM)
The Mythbusters did this a few years back with a closed truck full of birds. If I remember correctly, the total system weight did not change (within reason) when the birds became airborne.

put me in the back of the truck & i will jump & land. the forces of jumping & landing will affect the springs? Can this effect be dismissed as a force & not a temporary change of weight of the truck? Also. I'll be burning calories to do the jumping which will lighten my weight
adoucette
QUOTE (mudderrunner+Nov 23 2010, 12:34 AM)
horizontal speed decreases weight just like with an orbiting space station, or the moon. The speed it would take to reach weightlessness at ground level is probably very fast but a fraction of this speed would still cut a fraction of the weight.

We aren't talking about orbital speeds.
Horizontal speed, in the atmosphere at aircraft speeds, has insignificant effects on the mass and velocity of the air that is displaced, only the relative values.
Arthur

adoucette
QUOTE (boit+Nov 22 2010, 10:52 PM)
No, the balloon won't lift off.

Correct, and Capracus explained why.

Now let's change the experiment:

Take a box.

Put a frog in it.

Attach to the box a helium balloon, which displaces just enough air such that the box is lifted off the ground.

Let helium out of the balloon, atom by atom, until the box finally settles.

If the frog jumps will the box rise?

Arthur


mudderrunner
QUOTE (adoucette+Nov 23 2010, 08:27 AM)
We aren't talking about orbital speeds.
Horizontal speed, in the atmosphere at aircraft speeds, has insignificant effects on the mass and velocity of the air that is displaced, only the relative values.
Arthur

I'm not talking about mass, I'm talking about weight.

Orbital speed at the surface of the earth is about 17700 miles/h. If the plane were traveling 400 miles/h then it would lose approximately 2% of its weight.

boit
QUOTE (adoucette+Nov 23 2010, 05:34 PM)
Correct, and Capracus explained why.

Now let's change the experiment:

Take a box.

Put a frog in it.

Attach to the box a helium balloon, which displaces just enough air such that the box is lifted off the ground.

Let helium out of the balloon, atom by atom, until the box finally settles.

If the frog jumps will the box rise?

Arthur

Yes! It will momentarily rise when the frog is airborne. It will be in free fall in its ascent and descent, even in the fraction of a second that it will be still in the air. If the box is set on top of a scale it will register increased weight during the jump and landing. Balloon comes back down. smile.gif
adoucette
QUOTE (mudderrunner+Nov 23 2010, 09:50 AM)
I'm not talking about mass, I'm talking about weight.

Orbital speed at the surface of the earth is about 17700 miles/h. If the plane were traveling 400 miles/h then it would lose approximately 2% of its weight.

Which is why I'm saying the speed of the aircraft is insignificant to the discussion about the effects of the weight of the plane on the hanger it flys thru.
mudderrunner
QUOTE (adoucette+Nov 23 2010, 10:29 AM)
Which is why I'm saying the speed of the aircraft is insignificant to the discussion about the effects of the weight of the plane on the hanger it flys thru.

first of all you said 'mass' before. And secondly I never said it would be significant, although 2% is not exactly insignificant. I'm just pointing out an additional factor in the scenario.
adoucette
Yes, the plane in level flight displaces a mass of air which times the velocity of that air is equal to the mass of the aircraft x its gravitational acceleration.

Arthur
boit
QUOTE (boit+Nov 23 2010, 06:10 PM)
Yes! It will momentarily rise when the frog is airborne. The frog will be in free fall in its ascent and descent, even in the fraction of a second that it will be still in the air. If the box is set on top of a scale it(the box) will register increased weight during the jump and landing. Balloon comes back down. smile.gif

I felt I had to clarify all those "it''s. It is the frog that will be in free fall. smile.gif
enord
QUOTE (adoucette+Nov 23 2010, 09:34 AM)
Correct, and Capracus explained why.

Now let's change the experiment:

Take a box.

Put a frog in it.

Attach to the box a helium balloon, which displaces just enough air such that the box is lifted off the ground.

Let helium out of the balloon, atom by atom, until the box finally settles.

If the frog jumps will the box rise?

Arthur

whaen the frog jumps the balloon will descend ? when the frog lands , the balloon will descend? the frog will be burning frogfat to jump & the frog will be lighter due to the bburnt frogfat.......
adoucette
If the box is sitting on the ground when the frog jumps the balloon will rise.
When the frog lands the balloon will settle to the ground.

Presume the box is sealed, so though the frog is lighter the air is heavier with all that exhaled CO2 and H2O.

Arthur
boit
I believe this thought experiment shows where proponents of gyroscope populsion get their erroneous idea that their contraption generate lift. When the rotating weights precesses ever so slowly, like a spinning top towards the end of its spin, the slow falling (but accelerating) will take some of the weight out of their system. This will all be returned when it reaches it's final descent.
enord
QUOTE (adoucette+Nov 23 2010, 02:19 PM)
If the box is sitting on the ground when the frog jumps the balloon will rise.
When the frog lands the balloon will settle to the ground.

Presume the box is sealed, so though the frog is lighter the air is heavier with all that exhaled CO2 and H2O.

Arthur

the fat burning of the jumping frog may result in a warmer air [rising like a balloon] & thus raising the thingy?
boit
QUOTE (enord+Nov 23 2010, 10:37 PM)
the fat burning of the jumping frog may result in a warmer air [rising like a balloon] & thus raising the thingy?

No. Remember the box is sealed and likely does not expand. If you convert the heat generated by the frog to mass using the famous equation E=mcc, the loss of mass x 9.8m/s/s will give a negligible weight loss as the heat escapes the system. a hydrogen ballon flying inside a helium balloon does not lessen the overal weight of the system nor increase bouyancy. much less warm air from the frog.
enord
QUOTE (boit+Nov 23 2010, 02:50 PM)
No. Remember the box is sealed and likely does not expand. If you convert the heat generated by the frog to mass using the famous equation E=mcc, the loss of mass x 9.8m/s/s will give a negligible weight loss as the heat escapes the system. a hydrogen ballon flying inside a helium balloon does not lessen the overal weight of the system nor increase bouyancy. much less warm air from the frog.

duh, the frog gets hot from jumping & heats the air & hotter air rises?
boit
QUOTE (enord+Nov 23 2010, 11:08 PM)
duh, the frog gets hot from jumping & heats the air & hotter air rises?

Yes. Hotter air rises and cooler air is displaced downwards to take its space. Say you're raising a flag aboard a ship, there is no decrease of weight in that ship despite the fact that the flag is being raised. Thanks for discussing these issues mate. We'll clear it together.
adoucette
QUOTE (enord+Nov 23 2010, 03:08 PM)
duh, the frog gets hot from jumping & heats the air & hotter air rises?

Not if the mass of the air is held constant, as in this box.
Capracus
QUOTE (NoCleverName+Nov 23 2010, 12:35 PM)
The Mythbusters did this a few years back with a closed truck full of birds. If I remember correctly, the total system weight did not change (within reason) when the birds became airborne.

QUOTE
A truck carrying birds will be lighter if the birds fly around than if the birds sit/stand.
busted

Adam and Jamie constructed a large box and placed it on top of scale and then filled it with captured pigeons. Then, the Mythbusters activated a special contraption that would force the pigeons to fly into the air, but they could not detect any discernible difference in the weight of the box. They then placed a model helicopter inside the box and had it hover above the ground, but this method also failed to produce any results. The Mythbusters theorized that the air being displaced by the birdsí wings and the helicopter rotors was pressing down the box, which is why there was no change in the overall weight.
http://mythbustersresults.com/episode77
boit
That pretty much closes the case. Substituting spring suspension and pneumatic tyres with wings does not take the weight off the ground. Just like a table will transfer my weight through its four legs, so does a plane through wings and air. Mythbusters should now bust gyroscope propulsion.
enord
1- frog in box, balloon suspended jumps so it almost reaches the ceiling of the box
2- the frog jumps higher so it impacts a force on the ceiling of the box.

assuming the frog can hit the ceiling of the box [can it?], will this make a difference?
adoucette
It does, but it also depends on what the frog pushes against.
If the box is sitting on the ground than the force of the frog jumping is absorbed by the earth, not the box and so any velocity the frog has when it hits the top of the box would be counter to the force of gravity and would allow the box to rise even more.

If the box wasn't in contact with the earth though, then to jump the frog would push the box down as it jumped, and then while the frog is in the air, the box would start rising again. The net of the two movements would essentially cancel each other out.
enord
not complaining but just realized that since the frog showed up , the fly disappeared! was this a planned conspiracy related to thanxgiving? fly gets eaten by frog which gets served at a french dinner that "tastes like chicken" but much better than turkey
adoucette
Sorry, but we had to use the fly to get the frog to jump!

Happy Thanksgiving.....
boit
QUOTE (adoucette+Nov 24 2010, 07:22 PM)
Sorry, but we had to use the fly to get the frog to jump!

Happy Thanksgiving.....

Hahaha. Suppose I short circuit that long process and just eat the fly. Was it a lake-fly? biggrin.gif
We can have other parallels, A hydrogen balloon inside a helium balloon, a fish in an aquarium. A stone reaching its terminal velocity in a jar of water on a weighing scale...
58beerman
@mudderrunner
QUOTE
horizontal speed decreases weight just like with an orbiting space station, or the moon. The speed it would take to reach weightlessness at ground level is probably very fast but a fraction of this speed would still cut a fraction of the weight... ...Orbital speed at the surface of the earth is about 17700 miles/h. If the plane were traveling 400 miles/h then it would lose approximately 2% of its weight.

I believe you are missing the real source of the 'decrease' in weight for fast-moving orbital bodies like the moon or the space station. It has nothing to do with the 'horizontal speed' you mention (at least not directly), but the centripetal acceleration/force.

Think of it this way, assuming a circular path (such as the surface of the earth), the inward force needed to 'curve' the path of the object (which would 'rather' move in a perfect strait line through space) is always [mass]*[velocity]^2/[radius]. Yet gravity always provides a force of [mass]*[accel of gravity, g] which happens to usually be way, way beyond the centripetal force needed to keep an object on the curved surface of the earth. So we usually need something (air, roads, railroad tracks, the ground, etc.) to provide a 'normal' force to counter-act the 'extra' gravitational force beyond that needed to curve the path and it is this contribution we perceive as 'weight'. As velocity increases, more centripetal force is needed (proportional to v^2, not v as you inferred in your calculation above) and is readily supplied by gravity, and there is less 'left over' gravitational force to counteract with a normal force. At the orbit velocity for a given radius and gravitational acceleration, [velocity]^2/[radius] exactly equals [accel of gravity], there is no 'left over' gravitational force, and no supporting normal force is needed: the object is in a perpetual weightless "free fall," to use orbital terminology I'm sure you've heard before.

All of this being said, the fact the the weight decreases at all is dependent on the presence of a curvature in the object's path. In the original thought experiment you referenced (an airplane flying through an open hangar) it is entirely possible and, I would say, most likely, that the building (and more importantly the flight path of the airplane) are actual true, linear, Cartesian strait-line entities without any curvature (ok, technically infinite radius of curvature) which, incidentally as a consequence, actually vary slightly in altitude along their length when compared to the curved (sea-level) surface of the earth. So your "additional factor" is likely nil, and in any case adds unneeded complication.
mudderrunner
centripetal force is not involved. Gravity is not a force.

2% is not nil.
58beerman
@mudderrunner
QUOTE
centripetal force is not involved. Gravity is not a force. 2% is not nil.
Obviously I am referring only to the Newtonian mechanics which are fully adequate for this problem (though admittedly approximate in light of Relativity and such). In this model, gravity is a force, and it serves also as the centripetal force required to curve any moving object through space. See [en.wikipedia.org/wiki/Classical_mechanics] and [en.wikipedia.org/wiki/Centripetal_force]. Even if you assume the hangar and the flight path have the same curvature as the surface of the Earth (which you must assume in order to have any reduction of weight as you suggest), then at 400 mph or 586.67 ft/s, the centripetal acceleration (assuming the radius of the earth is 20,925,524.9 ft as provided by Google Calculator) is 0.01645 ft/s^2 which when compared to the gravitational acceleration (32.2 ft/s^2) is only a 0.05% reduction in weight.

Remember that the felt decrease in weight is proportional to velocity squared, as previously explained. Therefore equivalently, to use your numbers of 400 mph for the airplane and 17700 mph for the velocity needed for orbit, we have: [400 mph]^2/[17700 mph]^2==0.0005==0.05% .

Again the method (and your data... congrats) is confirmed. I will definitely keep an open mind to any objections to my logic you have, as I enjoy the discussion. If you can provide any outside references to your position, that would definitely help your case.
PatGhosh
Well, according to my understanding, we can calculate this by change of momentum. Let us assume that the airplane has a mass 'M', and the fly has mass 'm', and they are both stationary at the beginning, and the fly is sitting on the floor of the plane.

Here we should consider the outside frame of reference. That is the air plane is not stationary in the plane of reference (for a little logical understanding).

So, at the start of the experiment, the fly+plane combination has (m+M)g in weight. And, as it moves to a steady speed v, the initial momentum, is (m+M)v. Any change in momentum is due to a net external force. The fly remains within the plane during the thought experiment.

Then the fly steadily starts to rise in the plane. During this, it is flapping its wings, and going up. Did the momentum of the plane+fly change?? YES! It did, but the total momentum must remain constant. Therefore, the plane must show a increase in momentum (negative momentum) on the exactly opposite side.

Mathematically, mu+Mz=0
where, u>>>> velocity of the fly, and z>>>>> velocity of the plane. w.r.t the FoR

Say if the fly is going up, the plane has to go down, in order to conserve momentum. The only way a plane is going to go down, is by changing its weight. Or, W>B
The vectors are shown in bold face.

Therefore, we can conclude that the fly indeed adds something to the plane, but it is only a false addition. The mass doesn't increase of anyone. So, apparently the (m+M)g will not change. And as our dear bios would say, they burn fat. So apparently, the wight goes on decreasing. But a moving fly does actually induce movement in the plane.

Beware of the fly's might!
PatGhosh
QUOTE (adoucette+Nov 23 2010, 05:24 AM)
In level flight the mass X velocity of the air displaced downward is always equal to the weight of the aircraft, and that doesn't vary based on speed.


But, how can you equate the momentum with force??????????
58beerman
QUOTE
They then placed a model helicopter inside the box and had it hover above the ground, but this method also failed to produce any results. The Mythbusters theorized that the air being displaced by the birdsí wings and the helicopter rotors was pressing down the box, which is why there was no change in the overall weight.
QUOTE (->
QUOTE
They then placed a model helicopter inside the box and had it hover above the ground, but this method also failed to produce any results. The Mythbusters theorized that the air being displaced by the birdsí wings and the helicopter rotors was pressing down the box, which is why there was no change in the overall weight.
Substituting spring suspension and pneumatic tyres with wings does not take the weight off the ground. Just like a table will transfer my weight through its four legs, so does a plane through wings and air.


Huh, interesting... I find these results very puzzling because (to use the second part of the MythBusters experiment) if the weight of the hovering helicopter was felt by the box floor (via the air, or whatever), it would indicate that the box floor directly supports the hovering of the helicopter: i.e. no floor would mean that the helicopter (or it's air cushion) had no support and therefore could not hover. This conclusion seems invalid to me. After all, why would an airplane or a helicopter need the ground for support?

I hypothesize (although I didn't see that episode) that the MythBusters' assessment is correct... for a 'small' box where pressure differences created by the down-wash of air that has nowhere to go (rather than acceleration of that air) are what primarily act to keep the helicopter airborne, a phenomenon known as "ground effect" in aircraft.
(see en.wikipedia.org/wiki/Ground_effect_(aircraft); sorry apparently I'm too new to link)

However, back to the original thread, I don't see a tiny fly hovering high in a large, spacious airplane cabin benefiting from ground effects. I would still have to say that the airplane is in fact lighter when the fly is airborne, with no added weight from the fly if hovering at a sufficient height; viscous dissipation preventing any change in pressure or velocity adjacent to the high-flying fly from reaching the cabin floor.

An equivalent question: could an airplane fly if there were no ground under it? (Although not as efficiently if flying very low using ground effects...) I say yes, it could. Therefore, the ground plays no role in keeping the airplane up and should not feel any effect from its presence (again, assuming it is high enough). If you disagree with me, please explain two points:
a) Exactly what do you see wrong with the strait and level flight of an airplane that flies in a theoretical infinite universe of air (and 'downward' directed constant gravity, the direction of which "strait and level" can be referenced to)?
b) How does the fly's weight get communicated to the plane cabin floor assuming it is 'far' from any objects?
PatGhosh
QUOTE (58beerman+Nov 30 2010, 06:06 AM)
missing the real source of the 'decrease' in weight for fast-moving orbital bodies like the moon or the space station. It has nothing to do with the 'horizontal speed' you mention (at least not directly), but the centripetal acceleration/force.


@58beerman,
Totally agree, with you, and your second post too. the idea of calculating the drop in acceleration is correct in case of bending frames like the earth.

In my model, I drop the fact that the plane with the fly passenger is flying in a curved path, and hence discontinue circular motion. Whatever the case, the flying fly will hardly cost a dime.

tongue.gif
mudderrunner
QUOTE (58beerman+Nov 30 2010, 02:49 AM)
@mudderrunner
Obviously I am referring only to the Newtonian mechanics which are fully adequate for this problem (though admittedly approximate in light of Relativity and such). In this model, gravity is a force, and it serves also as the centripetal force required to curve any moving object through space. See [en.wikipedia.org/wiki/Classical_mechanics] and [en.wikipedia.org/wiki/Centripetal_force]. Even if you assume the hangar and the flight path have the same curvature as the surface of the Earth (which you must assume in order to have any reduction of weight as you suggest), then at 400 mph or 586.67 ft/s, the centripetal acceleration (assuming the radius of the earth is 20,925,524.9 ft as provided by Google Calculator) is 0.01645 ft/s^2 which when compared to the gravitational acceleration (32.2 ft/s^2) is only a 0.05% reduction in weight.

Remember that the felt decrease in weight is proportional to velocity squared, as previously explained. Therefore equivalently, to use your numbers of 400 mph for the airplane and 17700 mph for the velocity needed for orbit, we have: [400 mph]^2/[17700 mph]^2==0.0005==0.05% .

Again the method (and your data... congrats) is confirmed. I will definitely keep an open mind to any objections to my logic you have, as I enjoy the discussion. If you can provide any outside references to your position, that would definitely help your case.

my objection to your last post came mainly from this:

QUOTE
I believe you are missing the real source of the 'decrease' in weight for fast-moving orbital bodies like the moon or the space station. It has nothing to do with the 'horizontal speed' you mention (at least not directly), but the centripetal acceleration/force.


specifically when you said, "real". Since gravity is not really a force, I disagreed. But I may have misunderstood.

Besides that I would agree with your conclusion. I shy away from the math so I didn't try to attain the exact result. I just googled the orbit speed, picked a reasonable plane speed, and made a rough estimate. My point was to reveal an additional factor effecting the posed problem. I wasn't suggesting it was a significant factor. However, 2% would have been much more significant then .05%.

Thanks for the help.
PatGhosh
QUOTE (58beerman+Nov 30 2010, 07:58 AM)
could an airplane fly if there were no ground under it?

I agree with you here too beerman. A plane can fly if there is a medium, with sufficient density, and minimal Bulk Modulus. Something like the air, or the water is good. But it is wrong to assume that a medium does not transfer a floating bodies' weight. In fact, we know the result as the famous Archimedes Principle.

following it, we can take a case when a giant aquarium is taken on a very sensitive scale, and the air is not allowed to escape. If by some Magic Spell we could transfer a fly inside it, we would see that the scale equivalently shows an increase in the weight. Ofcourse, a little increase in pressure does not harm the fly, does it???
Capracus
QUOTE (58beerman+Nov 30 2010, 07:58 AM)
However, back to the original thread, I don't see a tiny fly hovering high in a large, spacious airplane cabin benefiting from ground effects.  I would still have to say that the airplane is in fact lighter when the fly is airborne, with no added weight from the fly if hovering at a sufficient height; viscous dissipation preventing any change in pressure or velocity adjacent to the high-flying fly from reaching the cabin floor.
The downward acceleration of the air displaced by the motion of the fly's wings will be directly proportional to the vertical acceleration of its mass, and that force will be transmitted to the body of the aircraft.

In flight, the fly is nothing more than a constituent of the aircraft's atmosphere, its motion has no more effect on the weight of the aircraft than the motion the cabin air itself. Does the downward deflection of circulated cabin air give the plane lift?
PatGhosh
But Capracus, surely, in purely theoretical sense, a moving fly must provide change in momentum, if initially it was sitting there on my seat. So, if the plane is moving and the fly moves up, the plane must have a small velocity downwards??? Albeit only for nanometers/second, but surely there will be a velocity!!
58beerman
QUOTE
following it, we can take a case when a giant aquarium is taken on a very sensitive scale, and the air is not allowed to escape. If by some Magic Spell we could transfer a fly inside it, we would see that the scale equivalently shows an increase in the weight.
Ahh... yes, I agree. Very good example! So the fly's weight is transferred to the airplane not necessarily through aerodynamic forces such as downwash, but though buoyancy considerations (an increase in static pressure) which can be viewed statically, and are independent of whatever method the fly (or airplane in the other example) actually uses to stay in the air (this line of thinking even works for a hot air balloon).

So that answers my question b) for any objections I previously posed, and to answer my own question a) about the airplane in a theoretical universe of air and downward gravity: this imaginary universe is itself not consistent with our universe's physical laws for a single reason... what keeps the air (which itself has mass and weight) from accelerating downward under the influence of this universal 'downward' gravitational field? Something has to, and I wrongly didn't address this in my air-only universe.

Back in reality on Earth, the ground serves this purpose. So the combination of ground and gravity (or something similar to contain the air, such as your giant sealed aquarium or a 747 with a pressurized cabin) is required for an airplane to fly at any altitude, no matter how far away from ground effects.

Interestingly though, does this mean that whatever air container is used to hold the suspended flying object (fly or airplane) must be airtight in order for the weight to be transferred? "Airtight" meaning a container able to withstand an increase in pressure needed for buoyancy. (Of course, I mean when using the buoyancy explanation PatGhosh came up with and neglecting dynamic considerations such as ground effects and downwash.) In this case, the suspended weight and volume (buoyancy) of the fly in an unpressurized cabin would result in an (minuscule) increase in pressure in all of the world's atmosphere (which would be the only container that is 'airtight' in this thought experiment). Put another way: does a unsealed box weigh more with a hot-air balloon inside of it? A sealed box would weigh more, as PatGhosh explained with buoyancy (repped), but it seems to me an unsealed box would simply loose the weight-equivalent amount of displaced air to the atmosphere, so the box would not feel the difference if there was, say 8 oz of balloon compared to if 8 oz of air occupied that same space.

So to summarize my current stance on the original question: the fly would add weight to the airplane only if it had a pressurized cabin, neglecting dynamic considerations such as the ground effect or whether or not the cabin is big enough to let the air move around freely (as likely wasn't the case with the MythBusters experiment).

Awesome question by the way, enord. Thoughts?
boit
Put a normal sized aquarium on a scale. Drop a fish (dead or alive) in. Extra weight will be registered no matter if the thing is sealed or not. Another thing, a balloon with air inside weighs more than an equal balloon with no air inside. An aquarium will register less weight if the fish somehow jumps out but extra weight will be momentarily registered in the process. Even dropping a coin into the aquarium will slowly register increase weight reaching maximum when the coin reaches terminal velocity due to viscousity. It does not have to hit the bottom to achieve that.
boit
QUOTE (PatGhosh+Nov 30 2010, 10:55 AM)
But, how can you equate the momentum with force??????????

Remember that the mass of air started with zero velocity? So it is actually the rate of change of momentum that is important here. That was corrected some posts below. Just check. smile.gif
58beerman
@boit
QUOTE
Put a normal sized aquarium on a scale. Drop a fish (dead or alive) in. Extra weight will be registered no matter if the thing is sealed or not.
This is exactly as predicted with my previous explanation because gravity keeps the displaced fluid (in this case, water) from escaping; the aquarium is in fact "sealed" because gravity provides the seal just as in our own atmosphere. (Okay... air would escape the aquarium, but the displaced air's contribution to buoyancy is negligible compared to the displacement of the much more dense water.) In my explanation, "sealed" or "airtight" simply means
QUOTE (->
QUOTE
Put a normal sized aquarium on a scale. Drop a fish (dead or alive) in. Extra weight will be registered no matter if the thing is sealed or not.
This is exactly as predicted with my previous explanation because gravity keeps the displaced fluid (in this case, water) from escaping; the aquarium is in fact "sealed" because gravity provides the seal just as in our own atmosphere. (Okay... air would escape the aquarium, but the displaced air's contribution to buoyancy is negligible compared to the displacement of the much more dense water.) In my explanation, "sealed" or "airtight" simply means ...a container able to withstand an increase in pressure needed for buoyancy.
In your aquarium, the increase in pressure comes from the displaced water (which cannot escape) which has no where to go but up (the water surface will get higher if measured along the side of the container). Therefore, the bottom of the aquarium is slightly deeper under the surface of the water (which has slightly risen) and thereby feels a higher pressure due to the increased depth. The increase in overall force due to the pressure on the bottom of the aquarium will exactly equal the weight of the fish (which itself is equal to the weight of the displaced water, thanks to Archimedes Principle). So the weight of the fish is transferred to the water, then to the aquarium through increased pressure, then to whatever holds the aquarium (such as a scale).

An example of what I see as an "unsealed" aquarium in your example is one filled to the brim such that any displaced water actually falls out of the system being weighed to the floor. Thus, for the aquarium, the decreased weight of the escaped water equals the increased weight of dropped fish and, indeed, there is no change in weight, as predicted.

I totally agree with the jumping fish/falling coin scenario, but I am trying to keep my attention focused on comparing two systems that are static and don't change in time: airplane A without a fly, and airplane B with a fly (that has, if "sealed" before the fly is deposited within, the same mass of air as airplane A).

@PatGhosh or Anyone
But reading my own words brings to mind a question for PatGhosh: Are we not supposed to be comparing airplane A with a fly that is landed, and airplane B with a fly that is flying rather than the empty plane A? So (neglecting hydrostatic changes in air density as a function of height in the cabin) the landed fly would have the exact same (very small) buoyancy as the airborne fly and would therefore exert the same force on the cabin walls through Archimedes Principle regardless of Plane A or Plane B. So where does that leave the buoyancy talk? Assuming the aerodynamic ground effects and downwash of the fly's wings can not push on the cabin walls because the fly is too far away from them ("too far" meaning that viscous forces dampen out any air movement at a sufficient distance away from the fly's flapping wings), how do you account for the extra weight (beyond its small buoyancy) of the (not standing) fly in Plane B making it to the cabin walls? For Airplane A this extra weight is transferred through the fly's legs to the cabin wall. It seems buoyancy effects can't be the answer because they are identical in both cases.

All this flip flopping I am doing is making me dizzy... my bad. I just want a definitive answer that doesn't have any holes.
Capracus
QUOTE (58beerman+Nov 30 2010, 09:49 PM)
Assuming the aerodynamic ground effects and downwash of the fly's wings can not push on the cabin walls because the fly is too far away from them ("too far" meaning that viscous forces dampen out any air movement at a sufficient distance away from the fly's flapping wings), how do you account for the extra weight (beyond its small buoyancy) of the (not standing) fly in Plane B making it to the cabin walls?
The damping, or dissipation of motion does not translate to negation. The mass and net gravitational acceleration of the cabin air remains unchanged.
Capracus
QUOTE (PatGhosh+Nov 30 2010, 09:40 AM)
But Capracus, surely, in purely theoretical sense, a moving fly must provide change in momentum, if initially it was sitting there on my seat. So, if the plane is moving and the fly moves up, the plane must have a small velocity downwards??? Albeit only for nanometers/second, but surely there will be a velocity!!
Any upward acceleration of the fly is balanced by a proportional downward acceleration by the plane, the net acceleration of the fly/plane system does not change.
boit
QUOTE (Capracus+Dec 1 2010, 03:44 AM)
Any upward acceleration of the fly is balanced by a proportional downward acceleration by the plane, the net acceleration of the fly/plane system does not change.

Precisely! The 'plane's motion will on longer be visibly level if a jovian frog keep leaping to catch the fly. It will thread a wavy course along it's path. The same of course will be observed with the frog. This question is similar to the old riddle: ''What weighs more? A ton of metal or a ton of air? Assuming of course on the same gravitational field, the earth's surface.
PatGhosh
Without quoting any quotes, and considering a non-airtight system, I will proceed to explain my point of view.

Of course, negate the jumping frog and the hovering fly, as well as any change caused thereafter by the Coriolis force and any other kind of astronomical phenomenon. Also, let us consider a curved path of flight of this airplane, so the gravitational acceleration is slightly reduced to a value 'g'.

Conclusively, Weight of the fly+plane = (M+m)g. [which will be because we consider it as a system]

Question arises: will the downwash from the wings of a fly add force to the weight, so that a scale placed beneath it would notice the increase??

YES. Consider an alternate case: a bomb going off in the plane. It will surely amount to a huge pressure wave which ruptures the hull. Come on. I mean, there will be some pressure changes. I cannot tell you anything about the damping , cause I am too bad at the mathematics involved in the damping due to viscous force.

Taking an analogy of a vessel filled with water, one can guess what happens when pressure waves start from the bottom of the vessel: the water spills over. But, considering the fact that air cabins have compressed air, it is difficult to believe that air will spill out of the plane.

ANOTHER QUESTION: will the plane have to carry the fly? Of course, i don't see any reason why it should not. Not only it brings disrepute to the stewards, it has no option, as the air doesn't leak out, or even if the air does, I don't see why the fly needs to head that way.

IF the plane is accelerating, then it no longer has to support the fly. The fly will get squashed at the back wall. But thereafter, the fly has to work.

WHEN THE AIRPLANE IS NOT ACCELERATING, AND IS USING ITS fuel to merely cancel out the drag produced, the airplane is not carrying anyone's weight on board. [newton's second law].

aNY HEAT PRODUCED by the passengers, the fly will be dissipated by the plane, so yes the weight will decrease, but by 2*10^-3 grams. Considering the rate of fuel loss per second, the balance has to be pretty sharp!

So that is my take on it. For the more complicated stuff, I have no idea. I am only a student, not a professor.
boit
QUOTE
WHEN THE AIRPLANE IS NOT ACCELERATING. . . It is not  carrying anyones weight on board, . . .Newtons second law.
We are mixing stuff here. The weight we were talking about is the one from acceleration due to gravity, the vertical one that lift has to overcome. Not the horizontal one that the propeller's 'lift' has to create. If I lie on bed in the east west direction with my bathroom scale at my feet, it is true I won't register any weight even at the equator. Either direction.
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