# Weight of the flying bees

1. May 14, 2007

### Dr. Nitrus Brio

We have bees flying in a jar, not touching it. Jar is closed and air in it has standard atmospheric pressure. Here is the question: if we put that jar on a weighing-machine, is ti going to show the mass of the bees?

I had some theories about it, but many people thought differently, trying to convince me that they are right. I don't believe them - I want to hear your opinion.

I apologize if I'm spamming. I'm new here and I searched through many different threads, but I didn't find anything similar.

2. May 14, 2007

### Staff: Mentor

F=ma

Force on the scale is weight. - mass of bees times g, the acceleration due to gravity.

While bees are flying around inside, the their wings are pushing air downward, equal to their acceleration due to gravity. This means the air is likewise "pushing" against the bottom of the container.

If the bees were parked on top of a vertical thin rod, the answer would be obvious. Because we cannot see air movement we don't make a connection between the bees and the container walls.

Yes, the mass will show.

Last edited: May 14, 2007
3. May 14, 2007

### Sojourner01

My thought is yes.

The bee must exert a continuous force downwards in order to maintain itself in midair. It might fly higher or lower, but it's confined within a fixed distance - between the bottom of the jar and the top - so the time-average of the force it exerts is constant. Now technically, the bee is pushing the air downwards, but because the air is itself confined in the jar it must at some point transfer the momentum given to it by the bee to the bottom of the jar. This exerts a force on the bottom of the jar, on average equal to the weight of the bee.

So yes, you can weigh the bee.

4. May 14, 2007

### Dr. Nitrus Brio

Ok. That sounds pretty logical. But, when bee moves her wings, she makes the air to flow. I was thinking that this flow of the fluid might be turbulent. Does this have anything to do with it? Maybe the bee keeps herself in the air thanks to Bernoulli's principle. Helicopters can fly above us, but we are not going to feel their weight.

5. May 14, 2007

### cesiumfrog

Brio, never seen helicopter-induced crop-circles on tv?

6. May 14, 2007

### DaveC426913

This is identical to a classic riddle involving a truck filled with birds crossing a bridge. (I wish I could find an example.) The owner bangs on the side of the truck to get the birds flying, hoping to reduce the weight of the truck. The answer is: no it doesn't work.

Last edited: May 14, 2007
7. May 14, 2007

### Dr. Nitrus Brio

Of course I did (although I don't watch the TV very much).

Nice example. I have something more to ask. When bee moves her wings, she transfers kinetic energy of the wings and impulse of force to the particles in air. This force impulse then transfers to the other particles, if the collisions between particles are perfectly elastic.
If the particles have same mass then we get: mv=mv1+mv2 => v=v1+v2 The more collisions happen, the particles will have smaller velocity, but there will be more and more of them and the starting impulse will be preserved. Correct me if I'm wrong. But what happens with the particles which don't hit the bottom of the jar?

8. May 14, 2007

### Sojourner01

Many of them won't - I don't believe the mean free path in air is that long. However, a volume of air molecules directly underneath the bee will have a net momentum downwards, which must be conserved in some fashion. If they don't reach the bottom of the jar, they will transfer some of their momentum to whichever air molecule they strike, which will then have a momentum with a statistically greater probability of being downwards. The downward momentum in this way spreads out, but it is definitely conserved and must eventually reach the bottom.

9. May 14, 2007

### pixel01

I like the explanation of Jim Mc and Sojourn.
My way to explain this is just as simple. The jar and the bee including all the gas inside is an isolated system. For an isolated system, the sum of momentum is zero. So no matter the bee is flying or standing, the jar will not exert any extra force on the scale.
In your example of momentum of the air particles, not all the particles will touch the jar's bottom wall, but those are impacted will transfer the impulses to the bellow and so on. Finally the momentum will get to the bottom.

10. May 14, 2007

### Dr. Nitrus Brio

But, no matter if the impulse is preserved, if all of the particles in air don't hit the bottom wall, then the whole weight of the bee(s) is not going to be "transfered" to scale. Of course, they eventually hit the bottom, but not at the same time.

11. May 14, 2007

### Staff: Mentor

Why not? What holds up the bees? The air! What holds up the air? The jar. What holds up the jar? The scale.

Even if the air particles being smacked down by the bee's wings do not hit the bottom, they will certainly increase the pressure underneath the bee enough to support the bee's weight. And that increased pressure will be transmitted to the bottom of the jar.

12. May 14, 2007

### pixel01

I think DocAl explain clearly enough, but I would like to show you the weird point in your reasoning. You said: "if all of the particles in air don't hit the bottom wall, then the whole weight of the bee(s) is not going to be transferd to scale". That's not true, even if they don't hit the bottom wall, they transfer the impulse to the layer bellow and so on. The force will immediately exert onto jar bottom and onto the scale in no time at all. Why don't you ask youself : what make some of the air particles float inside the jar?. Afterall, they must rely on each other and finally onto the bottom wall. The bee is just one special air particle that is so big that it has to flap itself to be float among other air particles.

13. May 14, 2007

### Dr. Nitrus Brio

Of course that the pressure will be transmitted to the particles underneath, but the bottom of the jar can only be hit by approximately same number of particles (pressure dependant). So, all particles that have the "extra" impulse cannot hit the bottom wall at the same time. Some of them must hit the side walls of jar before transferring their impulse to those below.

14. May 14, 2007

### Staff: Mentor

Not quite getting your point. Air will hit the sides as well as the bottom, but it's the bottom that will do the pushing up needed to support the bee.

15. May 14, 2007

### Dr. Nitrus Brio

Maybe I expressed myself a bit wrong. I realise that the impulse will be transfered to the layers below, but the direction is not only down. They also hit the side walls. What happens then? Do they lose their velocity if they hit the wall? They transfer the impulse they have to the walls, not only to each other. Am I right?

16. May 14, 2007

### Dr. Nitrus Brio

Doesn't the higher pressure below the bee cause the force which "neutralizes" her weight? I don't see what the bottom has with it. It's just being hit by these particles later.

17. May 14, 2007

### Staff: Mentor

Sure. And now that higher pressure must be supported by the layers below, as pixel01 explained.
I am assuming a steady state situation: the bee is hovering. The air is continually being pushed down by the flapping wings, creating a greater pressure beneath it, which in turn gets transmitted to the bottom of the jar.

18. May 14, 2007

### Sojourner01

That's right, but momentum is a vector quantity, so it must be conserved in direction and not just in magnitude. An air molecule hitting the side may or may not transfer a portion of its momentum downward; whether it does or does not, the quantity is conserved. The average momentum of the air in the jar is not zero - it has a net momentum downward, which since we know no air is escaping from the jar must be transferred away - hence a force downward.

19. May 14, 2007

### Dr. Nitrus Brio

I FINALLY realised. I wasn't sure about all of this by now. Particles don't lose anything when they hit the side walls. The walls are being hit approximately with the same force from all sides. So, the impulse from the bee goes straight to the bottom and scale shows the mass of the bee (of course, scale cannot measure mass, it measures weight). Maybe some of them can escape from the bottom, but I think that it itsn't the crucial thing.

20. May 14, 2007

### Ivan Seeking

Staff Emeritus
Just for fun, we tried to do this experiment after a chem lab class one day. The trouble was that insects don't like to fly when you put them in something. They would never stay aloft long enough for us to see anything but the impulse created when they took off and landed.

Last edited: May 14, 2007
21. May 15, 2007

### peaceharris

I think it will show a weight less than the combined weight of the jar and the bees. Using Bernoulli's principle, where the velocity of the air is more, the pressure will drop there.

If you have a aeroplane flying just above you, not touching you, obviously you wouldn't be squashed by the weight of the plane. Same thing here, there will be a local pressure drop due to the increased velocity of air just around the bee in accordance to Bernoulli's principle.

22. May 15, 2007

### Dr. Nitrus Brio

Yes, the static pressure will drop, the overall pressure is always the same. But what happens with the dynamic pressure? It increases. Does it have anything to do with this?

Maybe because the pressure is transferred all over the air, not just downward like in this bee case. And smaller number of particles will hit us. The bees may fly thanks to Bernoulli's principle, but not in the same way like planes.

What if I put some body in the water? Am I going to feel its weight, despite the lift force (if I put all of the water on the scale also)?

Last edited by a moderator: May 15, 2007
23. May 15, 2007

### peaceharris

I think you might be able to experimentally verify this if you have a sensitive weighing scale. Get some swimming fish, record the weight of the tank. Then kill the fish and reweigh the tank. I suspect that the weight of the tank with the dead fish should be more than with the swimming fish.

I did a internet search for someone who has done a similar experiment, but haven't found any.

24. May 16, 2007

### Sojourner01

Wrong. We've already 'proved' that this thought experiment has the opposite conclusion.

25. May 16, 2007

### peaceharris

I disagree with the 'proofs'. The proofs have all neglected that pressure will decrease when velocity increases in accordance with Bernoulli's principle.

Consider a huge box with an airplane flying inside, and a scale underneath the box. Will the weighing scale show the same reading when the plane has landed inside the box, and when the plane is flying inside the box?