# Birds in a truck

Discussion in 'Physics & Math' started by BenTheMan, May 20, 2007.

1. ### D HSome other guyValued Senior Member

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I think it's a bad analogy for many reasons.
• First and foremost, birds do not leap a mile up into the air. Birds lift themselves off the ground with their wings, not with their feet. This alone invalidates the analogy.
• The gun generates all of the force needed to propel the shell a mile upward in a short period of time, and does so with a rigid connection to the truck floor. For a mortar with a 1 meter barrel (the 177.7 m/s muzzle velocity does not need a long barrel) and a smallish 8 kg round, the gun expends about 250,000 newtons of force over a 10 millisecond interval, nearly an order of magnitude larger than the weight of a typical 16 foot truck. Birds generate lift by flapping continuously. The force applied by birds to loft themselves extends over a much longer period of time. The medium through which the birds apply the force to loft themselves is loosely coupled with the truck floor.
• The gun operates in a vacuum. Birds do not. The downward deflection of air is critical in explaining how birds fly.
• The "birds in a truck problem" implicitly asks for the steady-state solution. The steady state in the gun situation is the ballistic trajectory. The truck will weigh less while the shell is flying than it did before the gun is fired. Once the birds achieve level flight the truck will weigh more-or-less the same that it did before the birds took off, with some small deviations that result from the birds' flapping.

When you change too many aspects of the problem the analogy becomes a false analogy.

3. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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I used a rail gun to avoid introducing any gas in the truck. As noted before, it is the fact the reaction force is transmitted by invisible air that obscures understanding. A good analogy should clarify understanding (usually by removal of the obscuring aspects, as I did.)

Another useful tool that often facilitates understanding is to exaggerate some aspects. For example: When a weight is thrown down on the short arm of an asymmetric, frictionless, mass less, perfectly rigid, "sea-saw" to throw a smaller weight, resting on the longer arm up and conservation of momentum is (too simply) applied it appears possible to have net gain of energy. Replacing the real Earth with, tiny earth, the conflict with conservation of energy is understood. Thus, I made the rail gun very tall so the upward acceleration could be minutes long. The projectile never comes to end of the rails". When it falls it back it causes the scale to read less, until the drop into sand pile on floor makes huge brief increase in scale reading.
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*Someone else made a better (than mine) analogy that immediately made clear that the scale would deflect with every vertical acceleration (via the "go to extreme" approach) I.e. they suggested that the bird be replaced by an Elephant that jumped up.

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5. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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It does appear that the focus on details has caused you to not understand the problem. As stated in my prior posts, only the vertical acceleration is important (for reason already given - Newton's 3d law etc.)

Thus if the bird is in "level flight" (defined as bird's center of gravity remaining a constant distance from the floor) there is no vertical acceleration so there is no deflection of the scale – wing flapping does not matter - does not produce "small diviations" if bird is in level flight.*

If you must look at details; note that wings accelerate downward (to accelerate air downward also). The downward impulse the wings give to the air striking the floor is offset by this downward acceleration of the wings as far as scale is concerned. (Air is incompressible at these subsonic velocities, so changes in the vertical column air mass’s action on floor can be neglected.) Upward accelerations increases the scale reading (weight) and downward accelerations decreases the scale reading.

Stated another way:

For there to be an increment of extra force downward on the floor, the must be the 3d law reaction force upward on the bird. If there is a net upward force on the bird, then it is accelerating upward, not in "level flight."

Again you need to reduce the problem to its essential aspects, not focus on the details such as wings vs feet, air vs. vacuum etc. if you want understanding of it.

P.S. Most readers (including me) did not assume that the bird must remain in level flight. Also I think birds rarely are in "level flight" but do normally have at least small oscillatory vertical accelerations and these are reflected in "small diviations" of the scale readings.

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*This is similar to fact that skipper of sail boat cannot use an electric fan (on the boat) blowing into the sail to move the boat forward when the wind dies down.

Last edited by a moderator: Jul 7, 2008

7. ### D HSome other guyValued Senior Member

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No. You are the one who has focused to much on details. The transient behavior while the birds take wing and rise is largely irrelevant. Everyone else is talking about steady-state behavior; your analogy doesn't even have a steady-state.

8. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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No, you are wrong about that too. Most (ALL but you as far as I tested) do NOT assume the bird is only in “level flight.” ONLY the non level flight is relivant. -i.e. makes the scale reading change. Working back from recent posters:

Hypewaders, Post 71: “in our imagined sealed box with birds flying up and down within”
Post 69: “in our imagined sealed box with birds flying up and down within”

Enmos, Post 68: “it's moving up and down the whole system, not some tiny part of it.”

James R , Post 61: “imagine an enclosed truck completely filled with water, and fish swimming around in the water, … when a fish swims upwards, what happens to the weight? When it swims downwards, what happens? …”

SIDE NOTE: This is one of the rare cases where James is wrong. His analogy with fish is exactly the opposite of a trick I often use (“going to extreme”) to aid understanding. Many post back someone did that well by replacing the bird with an elephant vertically accelerating. James’s slowing accelerating fish are trying to defend his wrong view by making the acceleration forces smaller, and certainly less obvious than the jumping elephant! , which as James notes: “There's no fundamental difference - just a different {animal} …”

Kalster Post 60: “As BillyT said, the initial acceleration of the bird launching would create a temporary increase in weight… The only things that might lessen the recorded weight would be radiated thermal energy, the absorption of momentum from the turbulent air by the box and the decrease in weight of the birds from being farther from the earth’s centre of mass. IMHO …” He is correct about the weight decrease with the bird moving farther from the center of the Earth, but wrong on the “momentum effect” (It will of course be conserved, despite spread by the turbulence). Going to the extreme on the radiation effect to understand it: Imagine the bird holds and fires a million of the world’s most powerful lasers. If pointed down, the recoil of launching all that momentum in photons will accelerate the bird upward (and conversely if pointed at the ceiling). Not wanting to be slammed into the ceiling, the bird reduces the wing activity and can hover with CoG at constant altitude with less wing effort so long the laser beam is being ejected at the floor. Again the details of how “level flight” is achieved are not important. Either Kalster misquotes me or I spoke too loosely. Weight is the gravitational force a planet applies to an object. It is a variable with planet mass and distance from the center of the planet, mass is not. Impulse is a time varying force. A scale measures the sum of the weight and down impulse applied to it, not weight.

Kevinalm Post 57: “In order for a bird to fly in a moving sealed truck, it would have to hover wrt to air…” {not quite true but cieling and floor do limit the vbertical range} May assume level flight here, but he is replying and stating what is needed for level flight, not necessarily assuming it. I doubt that he actually assumes the bird neither rises nor falls.

I quickly skimmed back to post 1 trying to learn what Kevinalm was assuming about the bird’s distance from the floor (constant or not?) but 57 is his only post. In doing so noted most clearly indicated their understanding that it is fundamentally the vertical acceleration, Pete especially clear on this, that casue the scale to change it readings

I could not find even one other poster who clearly assumed that there was never any vertical acceleration. Thus clearly you assertion that "Everyone else is talking about steady-state behavior" is wrong. Only you assume this and think the vertical accelerations are "irrelevant."

Last edited by a moderator: Jul 7, 2008
9. ### szabo23Registered Member

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7
My original problem with the birds in a truck was that I couldn't see a mechanism whereby the flying birds transmitted their 'weight' to the floor. D H has now explained where I was going wrong and elucidated the actual mechanism --

Originally Posted by szabo23
To elaborate somewhat: as far as I understand it, when a bird flies, air passes over its wings such that the air above the wing is moving faster than the air below the wing. This causes a localized pressure differential -- lower pressure above the wing, higher pressure below -- which results in a net lifting force on the bird's wings.

Originally Posted by szabo23
My question is: how does this wake of disturbed air then manage to produce a net downward force on the floor of the closed system? Or is there some other mechanism that I'm completely overlooking?

All right! Now I get it! Just to make sure I've understood this, I'm going to restate this in my own words -- 'The downward deflection of air molecules' is a better (ie more fundamental and more complete) way of describing what what the birds must do in order to fly. And when these downwardly deflected molecules (which can also be considered as a downwardly-moving pressure wave) hit the floor, the cumulative effect of all these tiny impulses sums over time to a more or less constant force that is equivalent to the weight of the birds.

@Enmos: I think this also means that you were right in saying it doesn't matter whether it's heavier-than-air flight (birds, helicopters, etc) or near neutral-density floating (helium balloons etc) because despite the differences (eg in force magnitude, in the amount of work that needs to be done to maintain equilibrium) in both cases, the forces involved are mediated by the air molecules in much the same way.

Anyway, thanks guys. Now I can start to consider what would happen if all the pigeons were attached to the roof of the truck with their wings tied to render them flightless. If the truck driver released them all from the roof so that they fell in ('weightless') free fall, would that affect the weight of the truck? (I figure it would: the weight of the pigeons would 'vanish' from the measuring device and then reappear with a vengeance a short time later when the pigeons hit the floor.

Assuming that's true, then here's a math question: If the truck was traveling at 60km/hr and the bridge was 7m long, how high would the truck have to be for the truck to travel the entire length of the bridge at its 'unladen' weight? Assume that the truck's front and back axles are 3m apart, and that the driver releases the pigeons (or cabbages if you're feeling humane) just as the front wheels of the truck reach the bridge.

10. ### KALSTERRegistered Senior Member

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I am thinking that some (very small amount translated into heat) of the momentum would be absorbed by the walls and ceiling of the box as a result of the turbulence, the rest would indeed even out in an overall downward directed force.

Yes, I guess I did not explain myself fully. I am assuming that the bird hovering would exert a constant downward force, but in reality each flap of the wings will be registered by a precise scale (impulse). When the bird is a bit higher up the amount of force necessary for it to stay afloat would be a little less, as its weight would be less. So if one were to draw a graph and get the average of when the birds are flapping around, it should be a little less than the stationary graph, but only very slightly so, only a very small fraction of the weight of the birds.

11. ### SyzygysAs a mother, I am telling youValued Senior Member

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Exactly. Thus if there is 1 bird and it flies downward pushing the air upward its pushed air actually makes the truck lighter.

12. ### SyzygysAs a mother, I am telling youValued Senior Member

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This is the correct answer. A bird flying sideways is puching air to the side of the truck, thus not towards the floor, so it adds way less if anything to the weight of the truck/system. A bird falling or gliding downward doesn't push ANY air to the floor.

Also another thing to consider the side of the truck and the birds. Pidgeons are relative big and create quite a lot of airpush. But if it is a moving truck with 8 feet height and the birds are small, their airflow won't even reach the floor of the truck but dissipates in all direction.

I think the mythbuster experiment failed to measure the difference because they were using pidgeons and the box wasn't big enough, like a moving truck...

13. ### RJBeeryNatural PhilosopherValued Senior Member

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Syzygys, this is why I made the bird/truck reference in the other thread. It was the average weight of the juggler + balls that we were discussing. Note that the enclosure on the truck is critical! A truck with mesh sides would in fact be lighter while the birds were in flight because the downward force would escape to the ground. In an enclosed truck, all downward force eventually pushes against the truck bed regardless of the moment-to-moment antics of the birds!

14. ### SyzygysAs a mother, I am telling youValued Senior Member

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My point is that when birds are flying, they are creating sideways, upward or no force at all too, not just downward force. Tell me what kind of downward airforce is created by a gliding bird...

Another huge difference between the 2 cases is that the balls are kept up by the juggler (outside force for the ball) but the birds create their own force to keep themselves up...

15. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Except for the very small effect of slightly moving the center of mass more or less distant from the center of the Earth, the STEADY FLIGHT up or down does not change the weight of the truck and bird(s) Only if they are accelerating up or down, does the weight total change.

Imagine there is a diagonal pole inside the truck. If a bird is steadily walking up (or down) it THE TOTAL WEIGHT DOES NOT CHANGE. Can you not see that the force of his feet on the pole does not change? (Is the same as if only standing on the floor as his body has acceleration a = 0 and F= ma still applies. There is no force on bird's feet applied by the pole except the one equal to his weight pushing up.)

Well exactly the same is true if there were no pole and he were in STEADY vertical flight (except it is now his wings, instead of his feed thru which his weight is counter acted (to keep the every present force of gravity from accelerating the bird). The total force on the bird is zero if it is not accelerating..

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16. ### RJBeeryNatural PhilosopherValued Senior Member

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It is true that less force is placed on the truck bed while the bird is gliding, but the bird is losing altitude. The only way to gain the altitude back is to create an extra down force which will be felt by the truck bed. The birds being kept up "on their own" is irrelevant because the truck feels the pressure whether the birds are flying, or are thrown by a juggler, or are using stilts, or jump into the air, etc.

Trust me, the average enclosed truck's weight remains the same. The enclosure is what makes this scenario precisely analogous to the juggler puzzle.

17. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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No it is not if he is not vertically accelerating - see post 92. Imagine he is steadly walking down a pole on exacly the same trajectory, at the same rate as the glide to see this after reading post 92.

18. ### SyzygysAs a mother, I am telling youValued Senior Member

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OK, so you are:

1. Changing the situation saying that the birds have to fly STEADILY.

2. If they don't, you agree that the total weight DOES change.

I can imagine it, but it is totally irrelevant, because the bird's weight at any given moment still pushing the pole down. But when it is flying and specially freefalling, there is no downward force.

Also, you didn't address my point of the created airforce dissipating in all direction not just downward. If it pushes the side of the truck, that doesn't add to the system's weight...

19. ### SyzygysAs a mother, I am telling youValued Senior Member

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Now one could make a point that in the case of more birds the accelerating and falling birds kind of even out each other, thus no change in the weight. But here we still have to assume that an accelerating bird makes more force on the floor of the truck...

20. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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I have ALWAYS said that vertical acceleration up, while it last makes the weight indicated increase (and conversely) but the long term average is same as if birds just sit on the floor (Or flys constant altitude circles inside the truck).

No not changing anything. Only taking "gliding" to typically mean a steady rate of sinking inside the truck (not flapping wings). Then so long as that zero vertical acceleration state exist, the weight indicated is the long term average. (Except as I carefully noted, to be 100% correct, the scale indicated weight is slightly increasing as the truck/bird center of mass moves closer to the center of the Earth with the bird's steady glide down.)
And you think air is not being pushed down? What is irrelevant is the means (pole or air) by which birds weight is tranferred to the truck floor. Likewise the bird could suport his weight at constant altitude or steady vertical motion with a powerful laser pointed down at the floor and that means of support would also be irrelevant.

That is true of "free fall" as then by definition the bird is accelerating downward at acceleration g. Again, when there is no vetical accelration the scale indicated the long tem average (weight of truck + birds) except for CoM shift effects.

There is no need to - lateral forces do not effect the weight, not even lateral force the bird might make accelearting in level flight towards one side of the truck. Except in free fall, the air is accelerated downward by the bird's wings. It could come to rest again via friction with the walls in a very tall truck, instead of impact with the floor. Then the walls push down on the floor with more than their weight. You are failing to see what is relevant and see lots which is not.

I did not address the air in the tires either, but that is more important than lateral forces as that does effect the weight.

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21. ### SyzygysAs a mother, I am telling youValued Senior Member

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Let's assume all of the birds are gliding for 2 seconds at the same time. Shouldn't their weight be subtracted from the system's weight???

I think air is pushed all around, not just down. And as you note later on, lateral push DOESN'T add to the weight.

So we agree that falling and gliding birds' weight don't add to the system's.

So we agree again, thus if the air is pushed all over not just down, the birds push way less force downward then their own weight.

Here is a scenario. You are in a big rental truck used for moving. The overall weight with you is 2 tons. The backdoor is open and you are getting ready to load, you are inside. Suddenly 5 birds (weighting 10 pounds) fly in. What is the weight of the system?

I would say still 2 tons. Now you are saying that if you quickly close the doors, the weight becomes 2 tons + 10 pounds. What you are claiming is that depending on when you open and close the backdoor, the truck's weight is getting 10 pounds lighter or heavier.

Is that so??? I don't think so...

Last edited: Mar 10, 2009
22. ### RJBeeryNatural PhilosopherValued Senior Member

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Syzygys, put 96 lbs of sand and 4 pieces of lumber on a scale. Now make a small square box out of the lumber, and fill the box with the sand. It is clear that the sand exerts pressure upon the insides of the box, but do you expect the sand + lumber to now read something other than 100 lbs? It works the same with air in an enclosed truck - the lateral force exertion does nothing to the weight, and (maybe this is a point of confusion) the birds themselves cannot use lateral force exertion to gain altitude anyway.

I think we're all in agreement, for the most part, that the average weight of the truck will not change (whether the birds are flying or not) although it may fluctuate. Is that fair to say?

23. ### RJBeeryNatural PhilosopherValued Senior Member

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Replace the air in the rental truck with water, and replace the birds with swimmers. Are you claiming that if the swimmers are not touching the bottom of the truck then it is lighter? The same phenomenon that you claim is reducing weight in the air (lateral for exertion) exists in water.