# Birds in a truck

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

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

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On (1) No. Not if they are steadily descending (zero vertical acceleration) and I think we both mean that by "gliding down."
(2) If flying without vertical acceleration their wings have a upward force on them equal (average over wing cycle) to their weight and all of them (their body) has a downward gravity force equal to the weight. I.e. net force F = 0 = ma so a = 0. AS STATED by "if condition" at start of this reply. If a (vertical) is not zero there is a change in the force on floor from when birds are just standing on it. There may be lots or no lateral force - it does not affect the scale weight. If there is net lateral force on the truck walls made by the birds, (not equal opposite on both) the birds have a horizontal acceleration.
(3) Definitely not agreed. Only if the bird is in "free fall" (has g acceleration towards the center of the Earth like an astronaut) does his weight disappear.
(4) No agreement. We only need consider the downward force flying bird applies to the air. IFF it is equal to his weight then bird has no vertical acceleration and his weight shows fully on the scales (with that of the truck) exactly as if he were standing on the floor (another zero vertical acceleration case.) Have some faith in Newton's laws especially the first: "A body at rest ... or in motion ..."
(5) While door is open (and flying bird not too deep inside the truck) the downward motion it gives to the air may be partially stopped by truck floor and partially stopped by ground behind (and under) the truck. As doors are shut more is stopped by truck floor.
(6) Quickly closing doors may compress some air inside the truck and make it heavier until the extra air leaks out. Even before the bird enters the truck some of the downward air flow is probably being stopped by truck floor. The full weight of level flying birds is stopped / supported by something all the time. As noted earlier if truck is very tall and bird is near ceiling that "stopping" could be by wall friction with the walls then pushing down on the floor. THESE DETAILS ARE NOT IMPORTANT, stop being concerned by them. Only the vertical F = ma matters and I have several times explained how. I will not again.

Last edited by a moderator: Mar 10, 2009

3. ### RJBeeryNatural PhilosopherValued Senior Member

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Syzygys, my final thought in the juggler thread applies here as well: put the truck in space, where the scale would read zero. If the "weight" of the truck could be affected from within then the birds would be able to maneuver through empty space. It should be obvious that this violates conservation laws.

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

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Alright, completely changing the set up.

First, where did you get this? Second, it doesn't even follow.

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

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The problem with this analogy is that air weights pretty much nothing, water does. Swimmers in the water doesn't leave the water and their weight is constantly in connection with the truck's floor, not like the bird's...

Again, if I keep opening and closing the truck's door the system's weight keeps fluctuating???

8. ### RJBeeryNatural PhilosopherValued Senior Member

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Umm, water and air are both weighted mediums. Air weighing "pretty much nothing" should not change the laws of Physics. What if we used a heavier gas? What if we filled the truck with water vapor and fog? At what density of the substance filling the truck is the phenomenon altered such that you agree that the average weight will not change?

Yes it does but not by 10 lbs "instantly". It will show very small perturbations as the downforce shifts from being distributed on the truck floor + the outside ground to being only distributed on the truck floor (and back again).

Ok try this one: replace the birds with a single, abnormally large bladed helicopter (it's a big truck). Now, make the truck bed circular with a radius exactly equal to the helicopter blade. Do you think that the truck weight will now change with the height of the helicopter?

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

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Quoting from the previously posted link:

"Fish in a tank is a terrible analogy. Fish don't rely on lift, they are neutrally buoyant through their swim bladders."

A fish with a swim bladder is analagous to a bird with a hot air balloon...

Shouldn't it? After all in one moment it is an open system, in the next it is a closed system. By closing the system one could argue that there is no air dissipation, although I am still not sure of that..

Here is an other one: Let' shave a 10 lbs ball hanging from the roof of the truck inside. Once we cut the string, the ball will fall. While the ball is falling does its weight added to the truck's weight or not?

If not, it is the same when birds are freefalling or gliding... (although gliding could be different if there is still downward pressure...)

Last edited: Mar 11, 2009
11. ### przyksquishyValued Senior Member

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For the record:
Actually, it wasn't. As Pete pointed out a few posts later in this necromanced thread, it's the (vertical) acceleration of the birds that is relevant. Explicitly digging out the math, the truck's effective "weight", as an idealized scale would measure, is its total mass multiplied by $g$, plus the (upward) acceleration of its centre of mass:

$F_{\text{w}} = M g + a_{\text{cm}}$​
(assuming an airtight truck, and not bothering with the bouyant force due to the surrounding atmosphere, which should be subtracted from the RHS)

If you assume the truck frame itself is rigidly fixed to the ground (probably not very realistic, since the thing has suspension and tyres and no object is perfectly rigid anyway), then the centre of mass acceleration is proportional to the acceleration of the combined centre of mass of any "loose matter" inside (eg. birds, air they're flying in, fuzzy dice . . .):

$a_{\text{cm}} = \frac{m_{\text{ls}}}{M} a_{\text{ls}}$​
("ls" = "loose")

so:
$F_{\text{w}} = M g + \frac{m_{\text{ls}}}{M} a_{\text{ls}}$​

For "birds in air", the acceleration that appears in the formula will lag behind what the birds are doing (by the time it takes the pressure waves and air currents to propagate to the sides of the truck) and be "smoothed out" due to dissipation. If you relax the assumption that the truck is rigidly fixed to the ground, then the truck itself will behave in some ways like an extension to the air inside it (trucks can conduct pressure waves too, and they have moving parts), increasing these effects.

EDIT: incidentally, all of the above is still valid if you substitute "birds" for "fish" and "air" for "water".

Last edited: Mar 11, 2009
12. ### SyzygysAs a mother, I am telling youValued Senior Member

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You keep saying if this and if that. If there is no vertical aceleration, if they are freefalling, etc.

Let's say the birds are flying completely randomly, thus in all kind of patterns. What is your answer to the original question THEN???
No more ifs, just complete randomness...

----------------

Here is what I think about closed system. What if the closed system is so big that it could be considered an open system?

Let's say the back of the truck is HUGE, 1 cubic mile and the birds are kolibris? Isn't that pretty much the same as a bunch of kolibris flying in the open air? After all we can agree that from a kolibri flying half mile above the truck's floor, no downward pressure reaches the floor.

so it seems to me that the truck's size : bird's size ratio can effect the answer.

13. ### przyksquishyValued Senior Member

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You're forgetting that a truck on the ground can exchange momentum with whatever it's in contact with. It's not an isolated system.

14. ### PeteIt's not rocket surgeryRegistered Senior Member

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HI Syzygys,
It doesn't matter how big it is. If the truck is enclosed, then the downward pressure from a hovering bird (even a hummingbird) will result in an increase in the force due to air pressure on the floor of the truck equal to the weight of the bird.

If the floor of the truck is very big and the weight of the bird is very small, then yes, that will be a very small increase in pressure. For example, a 1/6 oz bird hovering high over a 1 square mile floor (or flying around randomly around some average height), this means an average pressure increase of only 18 nanopascals (2.6 trillionths of a psi). But even that very small pressure is not the same as no pressure at all - and it happens to increase the weight of the truck by 1/6 of an ounce.

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

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You are becoming a troll. this has been answered completel already in points (5) and (6) of post 101. At least three different people have explained to you that ONLY WHEN THERE IS VERTICAL ACCELERATION does the weight deflect for the long term average while that acceleration exists.

Stop trolling. Read and try to undersand what has been well explained, even mathematically proven by several people.

"FREE FALL" is ONLY when the desent is accelerating downward with G. A steady descent hazs ZERO acceleration and thus makes no change in the scale display.

HOW many times must you be told, troll?

Last edited by a moderator: Mar 11, 2009
16. ### SyzygysAs a mother, I am telling youValued Senior Member

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You are coming again with your conditional. A simple YES would have done it.

(Not to mention you didn't even answer that particular question.)

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

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What question has not been answer at least twice by different people?

It is not "my conditional" - it is a law of physic, Newton proclaimed. When the acceleration is zero there is no net force applied or created. (A special case of F=ma where both F and a are zero.) When the bird's vertical acceleartion is not zero the scale will be deflected from the long term average. There are only two cases (a is or is not zero) everything else is irrelevant.

Last edited by a moderator: Mar 11, 2009
18. ### RJBeeryNatural PhilosopherValued Senior Member

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Yeah guys but what if you replace the swimmers and birds with SNAKES and the truck with a PLANE?

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

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yes but it all depends on whether or not the swimmer is juggling the snakes at time of replacement and what color they are, does it not?

good idea, me too.

20. ### geistkieselValued Senior Member

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Suspend a hundred balls from the ceiling of the enclosed truck which is weighed as, Wt (weight of truck), plus 100 units equal to the total weight of the balls.The beginning measured system weight is Wt + 100. A switch is engaged that releases the 100 balls simultaneously. Before the balls come to rest on the bed of the truck what is the total measured weight of the truck + ball system.?

Your example is equivalent to this one is it not?:shrug: ​

21. ### PeteIt's not rocket surgeryRegistered Senior Member

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Hi geistkeisel,
No, they are not equivalent. In your example, all the balls are accelerating downward.

22. ### RJBeeryNatural PhilosopherValued Senior Member

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...and as was discussed in the thread in many posts any accelerative forces would have a temporary impact on the weight of the truck system. I believe someone made the same point by picturing an elephant "jumping up and down" inside the enclosed truck. Over time, however, the average weight will remain the same.

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

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So as a summary, the weight of the system is going to FLUCTUATE sligthly according to the birds flying pattern!

From the mythbuster message board:

http://community.discovery.com/eve/forums/a/tpc/f/9401967776/m/7831944098

"The “Birds in a Truck” myth that was prematurely “Busted” is flawed due to a lack of precision in the scale used in the test. The registered weight of the truck should change, but OPPOSITE to common sense, and only in FLUX as the birds either rise or fall.

Birds gain loft when the normal force of the air flow they generate downward is GREATER than their weight, thus pushing them up. Since scales measure this normal force, birds taking flight in a truck should therefore INCREASE the weight of the truck. Now imagine that birds in flight suddenly stop flapping in midair and fall - they should therefore become “weightless” in freefall (they generate no normal force) and the truck should weigh LIGHTER. If the birds are at rest on a perch or maintaining a constant elevation, there should be no net weight change.

Or, think about it in terms of gravitational potential energy. Recalling that forces (like weight) are a derivative of energy with respect to distance, as the birds rise up in the truck, the truck must sink down onto the scale according to Newton’s third law, which registers as MORE WEIGHT. This explains why the weight should change only when the vertical position of the birds changes, and equal and opposite to their direction of change!

The above comments will only be observed in the most ideal conditions (for example, they ignore lateral forces), however they agree with theory, and anyone with a knowledge of Newtonian physics will agree - the truck will actually get HEAVIER as the birds FLY UP!

The problem with the Mythbusters method is that the scale used in the test was not precise enough to detect such a minute change in weight (they themselves noticed “noisy” data). The truck rig they used must have weighed at least 1000 lbs., and the total weight of birds was at most 1% of this. I imagine (but cannot confirm) that a difference of +/- 1% is difficult to detect on an industrial-sized scale used in this test. I recommend a much lighter rig and more birds before this myth can be confirmed or busted."