# Birds in a truck - I'm not getting it

1. Aug 5, 2013

### Graeme M

First up, I'm a very irregular visitor here. I should point out that I have no science background and only a high school education, so in my discussion here, I might misunderstand simple concepts and I almost certainly won't know the right terminology. I apologise in advance. :)

Now, I know this topic has probably been dealt with before on this forum, but I'm looking for some pretty pointed responses. I've not got much time spare for digging around trying to gather a broad cross section of explanations and my first few searches turned up a LOT of commentary some of which was way out of left field, some over my head, and some made sense. But I have a niggling problem with the 'standard' explanation. So I'm hoping there are some patient people here to help me out.

As far as I can tell, the agreed view of the answer to this problem is that a truck with sealed walls, floor and roof and which contains a flock of birds will weigh the same whether the birds are in flight or at rest. There are all sorts of variations on the theme regarding cages or meshed walls or whatever, but that seems to be the answer. The argument is that it is the flapping of the birds' wings that causes this - the downthrust bears on the floor of the truck with a thrust equal to their weight.

I struggle with this for a number of reasons, but before I dive into that have I got it right? To put it more simply:

1. A sealed box with the air inside at normal atmospheric pressure, placed on a scale, weighs X.
2. Add a bird of weight Y so that the total weight now equals X+Y.
3. If the bird takes to the air inside the box, flies around for awhile, and then resettles, the total weight of the box and bird 'system' is always X+Y.
4. The reason is that the birds flapping produces a downwards force that registers on the floor of the box as a force equal to the weight of the bird.

Is that a fair summary? If so, I have a couple of questions. If not... well.... :)

2. Aug 5, 2013

### Staff: Mentor

Yes. Fire away

3. Aug 5, 2013

### voko

The total MASS of the system stays constant.

The total WEIGHT may and will change as the birds roam in the box.

To see that, replace the birds with a cannon and a massive soft rubber ball. As the cannon fires, its recoil will force the floor down, thus the WEIGHT of the box registered by an external scale will increase. As the rubber ball collides with the ceiling, the box will jerk up, making its WEIGHT decrease.

4. Aug 5, 2013

### jbriggs444

If by "weight", one means the external support force (as measurable by a scale) required to keep the box in one place while its contents fly and jump around then the above is quite correct.

If by "weight", one means the force of gravity on the box+birds+cannon+rubber balls then that "weight" is constant.

If by "weight", one means the force of gravity minus any corrections for the rotation of the earth then that "weight" is constant as well.

5. Aug 5, 2013

### Staff: Mentor

However, the weight will still be constant and equal to $mg$ if you average around these fluctuations. Let's let OP ask his questions, see which departure from the idealized situation he's interested in.

6. Aug 5, 2013

### voko

@jbriggs444

The second and the third definitions of weight remove all the drama from the problem, because they are (A) defined to be fixed no matter what; (B) not measurable directly.

The intrigue in this problem has to do with measuring the weight of the truck as the birds fly around, which clearly uses the first definition.

7. Aug 5, 2013

### Graeme M

Yes, that's sort of where I was headed. As I understand it, there are two sorts of weight - operational weight ('OW') and gravitational weight ('GW'). Operational weight is the one we can measure with a scale while gravitational is I think a sort of potential weight. The mass ('M') times the gravity gives us the gravitational weight, but the weight expressed on a scale is the operational weight. Not sure if these are the right terms but the idea jibes with my own thoughts.

So in our box system, M and GW remain the same. However, OW must vary. If the birds are aloft, then their weight no longer bears on the floor of the box and hence the OW is less. Any downforce expressed through their flapping around would be dispersed around all 6 surfaces and should not have a significant effect on OW.

Of course, I assume that a force on the floor of the box not derived from mass being accelerated by gravity is not weight in terms of GW, but it does affect the reading of a scale and hence can be considered a component of OW.

Anyways, the end result is simply that whilst the birds are aloft, the OW of the system is reduced. Which is NOT what many of the answers I have found claim to be the case.

Here in Australia it's late, so I'll check back tomorrow my time for your thoughts.

8. Aug 5, 2013

### cabraham

In free fall the system weighs less. The original version of this question involved a truck driver who was at a weigh station. He wanted to register the lowest weight possible, so he knocked on the trailer walls to prompt the birds to fly, thinking that in flight, the bird's weight doesn't register. Of course the above answers are correct, birds in flight do not reduce the system weight.

But consider this. If the birds are sitting on a shelf high up near the ceiling, then just when the weight is about to be recorded, the trucker raps on the side and the birds jump off the shelf and do a free fall. For the 1.0 second or so they are falling, the system weight is less since the birds in free fall exert no force on the truck floor. Without flapping their wings there is no downward air motion. So I think it is safe to say that with birds in free fall, the system is a little lighter. Any comments?

9. Aug 5, 2013

### Staff: Mentor

This is not correct. The downforce is directed downwards. Any dispersed air does not provide lift. The downforce has a significant effect on OW which is equal to the gravitational weight of the birds assuming level flight.

You can analyze this situation very simply. If the center of mass (COM) of the box/bird system is unaccelerated then the OW is the same as the GW. If the COM accelerates upwards then the OW will be greater than the GW. If the COM accelerates downwards then the OW will be less than the GW. Since the box is stationary the COM only changes due to the motion of the birds. Since there isn't a lot of room in the box any acceleration of the birds is fairly transient, so if you average over time you get average OW equals GW.

10. Aug 5, 2013

### CWatters

Replace the birds with a hovercraft. Makes it easier to understand in my opinion.

11. Aug 5, 2013

### A.T.

No. If they are hummingbirds hovering in place then OW = GW.

Yes

Last edited: Aug 5, 2013
12. Aug 5, 2013

### cjl

This is where your intuition is leading you astray. A bird (or any other thing that flies due to aerodynamic effects) stays aloft by generating a net downwash from its wings. Yes, there are all kinds of complex flows, but the net overall flow off of the bird's wings is directed downwards.

Now, in the truck, the air obviously cannot have any net downwards motion, since it is restrained on all sides by the walls of the truck. Thus, the walls of the truck must stop this downwards flow of air. To do so requires that the truck exert a net upwards force on the air, which means (by Newton's third law) that the air exerts a downwards force on the truck. On average, if you measure for long enough to smooth out the fluctuations, this force from the air is equal to the weight of the bird. Thus, on average, the truck will weigh the same with the birds flying as it will with them stationary on the bottom of the truck.

As explained above, the OW (using your definition) is not reduced. It may fluctuate some, but on average, it will be identical with the birds aloft to what it is with the birds sitting on the bottom of the truck.

13. Aug 5, 2013

### cjl

Sure. Then the system will be heavier while the birds arrest their downwards motion (regardless of whether this is due to them flapping their wings, or hitting the bottom of the truck). The overall average weight will be unchanged though.

14. Aug 5, 2013

### cabraham

Of course the *average* weight is the same, but only if we understand that the time interval over which the weight is averaged exceeds the free fall time. If free fall time is 1.0 second, then averaging weight over a time span of 2.0 seconds or longer gives the same value. If the scale is a fast reading type taking only 0.5 seconds or less to acquire a reading, then it could happen that the reading is taken while the birds are in free fall, resulting in a lower weight, so that the driver pays less toll.

I agree that in any practical situation, the free fall time is less than the average weigh time so that we get the same value no matter what we can train the birds to do. We seem to have a consensus. Best regards.

Claude

15. Aug 5, 2013

### rcgldr

It might help to consider how the air inside the sealed truck exerts it's weight onto the truck. It does this by creating a pressure differential, lower pressure at the top, higher pressure at the bottom, so that the net downforce exerted by the air equals the weight of the air.

If you add one or more flying objects into the air inside the truck, and assuming the center of mass of the system is not accelerting vertically, then those objects end up increasing the pressure differential inside the truck so that the net downforce equals the weight of the air and any objects supported by the air.

16. Aug 5, 2013

### Graeme M

Now we are getting somewhere. This is the course of many arguments I've read. So let me summarise.

1. The downforce from the birds wings IS directed downwards.
2. That downforce is expressed on the floor of the box.
3. If the bird were in freefall, it's weight would NOT be expressed on the floor.
4. When it flies, the bird alternates between freefall and upwards movement. Thus the weight itself alternates but the average is the same.

Now, my view is that 4 is misleading. It's a red herring that originates with the idea of a bird. We could for example replace the bird with a helicopter. The question is, does an object that remains aloft by doing work cause the total box system to weigh more or less.

Now 3 too is misleading and seems to me to reflect some kind of misunderstanding by me or the people who make that claim.

To look at the problem from a slightly different angle, the bird flapping its wings is no different to a large flat object falling.

To explain. The wing sweeps downwards through the air and forces the air away. I know there's all sorts of aerodynamical stuff going on, but at a broad level that has to be the process - it is the wing moving against the air that creates the force that registers on the floor of the box. However, if I drop a large flat weight from the ceiling of the box, I am replicating that process. My flat weight may need to be larger than the bird's wings, and maybe it won't generate the same downforce until it reaches a certain speed, but on the whole it's the same process - something moving through the air generates a force away from itself that propogates through the air.

This means that our falling weight must exert the same sort of force on the floor of the box as the bird's wings flapping. Now, we know that a freefalling object has no weight, yet in this closed system this now weightless object continues to exert its weight on the scale. How can this be so?

In other words, the OW of the falling weight is zero, yet according to the argument in respect to the birds wings beating downwards, its motion causes the total system's OW to remain the same. On the other hand, some argue that the bird in freefall weighs zero and the box system is lighter, so our falling weight version should similarly weigh less..

Which is it?

Last edited: Aug 5, 2013
17. Aug 5, 2013

### Staff: Mentor

My analysis in post number 9 is completely general. It applies regardless of whether the contents of the box is birds or helicopters or weights or anything else. Ignore everything else besides the motion of the center of mass of the box and contents.

18. Aug 5, 2013

### cjl

If you have a large flat weight that is moving air downwards, it isn't in freefall. If it is in freefall, it isn't moving air downwards (and isn't equivalent to the bird). You can't have both.

19. Aug 5, 2013

### sophiecentaur

If, instead of the birds, you had a man hanging on a rope from the roof of the van. The weight of the man+van would be what registered on a weighbridge. If the man let go of the rope, the registered weight would be just that of the van - until he hit the floor. At this time, there would be an impulse on the weighbridge but the final weight reading would again be man + van. If the man pulled himself up on the rope, there would be a brief increase in registered total weight.

The situation with the birds would be essentially just the same but just a bit harder to analyse in detail. If the birds hover steadily then the measured weight will be van + birds.

Don't forget, every atom of the van and the contents is in constant agitation, with a massive number of varying forces vertical (much the same as with a flock of birds - only more so). The mean weight is due to the sum of all these random forces. The statistics smooths out any variation in what you measure.

20. Aug 5, 2013

### Staff: Mentor

When you drop the weight, it generates nowhere close to the amount of lift as the bird until it reaches terminal velocity. So no, the situations are not even close to the same.

So:
At the instant of being dropped, the weight generates zero lift and as a result, does not register on the scale. There is no contradiction.