This question was inspired by a ST:TNG repeat I just saw again today, The Outrageous Okuna. Okuna was telling Data that he could take on double the weight of Canaries his craft would hold if he kept half the Canaries flying always. Data said he didn’t think that was right. Anyways, this question isn’t meant for Café. This is just a segue into my actual question.
If you had a truck that could only hold 500 lbs. of Canaries and you put 1000 lbs. of Canaries in it but kept half of them always in flight, would that work? I know that the force exerted by their wings might equal their weight too , so let’s just say, for fun, the truck is extremely well ventilated. Air is constantly rushing in and out. Now it would seem to me that this just might work. But I realize I might still be wrong. So someone please correct me on this if I am .
It depends on if the truck is airtight of not. You can figure this out by looking at extreme situations.
If you want a system where it won’t affect the weight of the truck, just imagine the cargo area is just covered in a few pieces of crisscrossed yarn. Obviously birds won’t weigh on the truck if they dart in and out of the yarn. The will if they land on the bed but as they take off, it is all just open space except for rather arbitrary markings (yarn).
For a system where the weight of the birds would count, simply substitute a tanker truck with fish swimming inside it for the birds and air. Air is a fluid too and it has weight that counts as part of a closed container as well as anything flying in it. Just as swimming fish won’t lighten the load, neither will flying birds in a completely airtight system.
How big is the canary cage?
Are they flying forward, rearward, or sideways? Right sideways or left sideways?
Are you sure only 500 pounds are sitting on the truck bed? Is there enough room for 500 pounds of canaries?
Are you sure 500 pounds or more are flying at any time?
If there are 500 pounds of canaries sitting on the floor then the truck is carrying 500 pounds of canaries. The other 500 pounds of canaries are on their own (wings)!
I have that itchy feeling that this thread is going to be one of those that turns into a fountain of stupidity and misunderstanding; **Shagnasty
**'s fish in the tanker analogy is relevant to closed trucks and everyone should just keep that in mind for the rest of the thread. Please lets not have any nonsense about the fish only contributing weight to the total when they touch the bottom of the tanker; that’s just stupid.
Now, non-closed trucks… well… the weight (lets just call it weight) of an object suspended in fluid must be propagated downward to the bottom of the fluid, or else something would move- the object is resting on fluid molecules,which are resting on other moleculkes and so on, all the way to the bottom, but it isn’t going to be propagated downwards in a simple, tight column of force - it will be more like a cone or bell shape.
So the weight of a flock of canaries hovering over a flatbed truck is propagated downwards through the air as a superposition of cones; some of the footprint of this is going to be on the truck and some is going to spill over to the ground below.
The closer the birds are to the bed of the truck, the less distance the cone has to spread out and the greater proportion of the weight will be upon the truck; in the case of a bird hovering a mile over the bed of the truck, theoretically, some of the weight is still on the truck, but the cone is so dispersed that it will be impossible to measure.
I’ve seen this discussed several times before, and never seen an answer that convinced me either way.
In particular, the difference between a closed and non-closed truck. Does it make a difference? Just suppose you had a closed truck, the canaries are flying around. Now, you make a pin sized hole in the truck, allowing air to pass in and out a few molecules at a time. Does this make the truck suddenly non-closed? Does the weight of the truck change? What if there were two holes? Or twenty? What if the holes were 1mm wide? 1 cm? 10cm? Just how open does it have to be to make a difference?
And the bit about weight being propagated down through the fluid,lets try another thought experiment: instead of canaries suppose we suspend a weight from the roof by a thread. There is no air in the truck, it’s a vacuum. The thread breaks, and the weight falls. Does the truck weigh less for an instant? Since there’s no air to pr4opagate the weight through.
What if a fish swims 1 metre below the hull of a ship? Does the weight of the ship propagate down through the water and crush the fish? Or what if it’s a submarine?
Are they African or European canaries?
On the serious note, I’d side with those that say the truck would not be bearing the weight of the canaries in flight if it were airtight. If, say, the truck had walls consisting solely of chicken wire, the canaries would be no more a part of the weight of the truck than if a canary was flying above a pickup truck.
Not the same situation - the ship doesn’t have wings that support it by pushing water down - it supports itself by displacing water equal to its weight.
This is not an all-or-none proposition. If only a small amount of air can get out, only a small amount of force will be dissipated outside the system. The more air can get out, the greater the amount of force that will be dissipated.
So would the position of the hole in relation to the canary make a difference? Does it have to be within the cone of air underneath the canary? Or would any hole do?
Fluid dynamics is one of the most complicated math and physics topics there is. I suspect your question would require a supercomputer simulation to answer. The best we can do here, I suspect, is give the answers at the extreme ends of the scale. All we can say is that a closed system includes the weight of the air and everything else in it just like in water. As those systems become more open to any degree, our intuition simply isn’t capable of understanding it. I understand what you are getting at and I am not sure that anyone understands it fully today.
When the string breaks, the truck shows a loss in weight. When the falling weight reaches the bed of the truck, the truck’s weight goes back to what it was before.
In a truly open system the canaries would not really be in the truck but just flying along where the truck box should be. There is a wide range of solutions between completely open and completely closed and the answer becomes very complex.
How about another analogy for the closed truck. A water tank with a Da Vinci style helical screw flying machine. It has a density greater than 1 so it rests on the bottom . As the screw turns it produces a thrust vector with the reaction taking place against the bottom of the tank. Thrust will be subtracted from apparent weight but the net change will be zero. If thrust exceeds weight it will reach the top of the tank and stop. the upward force on the top of the tank will equal the reaction against the bottom.
The more open the truck, the more possibility for the canaries to be partly or wholly supported by air that is supported by air that is supported by air that is not supported by part of the truck; the total weight tends to decrease. It sounds like you’re trying to make this a boolean function when in fact it is nothing of the sort.
Well, you’re talking about a whole different problem now - the momentary weight of a system comprising one or more parts in free fall. If you had sufficiently sensitive scales, the truck would weigh less when the thread breaks, then when the falling object hits the floor of the truck, the total will momentarily appear to weigh more than the truck + the falling object, then it will settle back to just the weight of the truck plus the object.
The density/viscosity of the medium matters here; if the object falls through thick treacle rather than a vacuum, this acts to dampen the momentary change in weight; if it isn’t treacle, but water, the effect is dampened less and if it’s air, less again. If one of the canaries in the original sealed-truck problem suddenly dies, the (apparent) weight of the total system will decrease momentarily as the canary falls.
Probably not, because *again you’re describing a different problem; in fact, there’s no difference in pressure because the extra weight of the ship is equal to the amount of water it has displaced by floating there.
It’s probably worth mentioning at this point that we can only really talk about simplified vews of the problem; in reality, canary flight consists of up and down motion along with brief periods of controlled falling; the ‘cone’ thing I mentioned earlier is a vast oversmplification , but still a valid way of thinking about it.
How about another one?
I have an empty, airtight glass case on a very sensitive set of scales; I open it and add a balloon filled with helium, then seal it again; does the measured weight change? if so, how?
Or another:
On my scales now rests a container filled with water to the very brim; I weigh it, then add a model boat; what happens to the weight?
I think : You displace an amount of air, and replace with a lighter gas. The material of the balloon is heavier than air, and adds a little weight. If the balloon would float in air, the glass case becomes slightly lighter. If the balloon would sink in air, the case becomes slightly heavier.
An amount of water exactly equal to the boat’s weight would be spilt. The container would weigh the same.
, so let’s just say, for fun, the truck is extremely well ventilated. Air is constantly rushing in and out. Now it would seem to me that this just might work. But I realize I might still be wrong. So someone please correct me on this if I am 
