Birds in a truck - I'm not getting it

[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... :)

 

[Nugatory]

Is that a fair summary? If so, I have a couple of questions.

Yes. Fire away

 

[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.

 

[jbriggs444]

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.

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.

 

[Nugatory]

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.

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.

 

[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.

 

[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.

 

[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?

 

[Dale]

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.

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.

 

[CWatters]

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

 

[A.T.]

Anyways, the end result is simply that whilst the birds are aloft, the OW of the system is reduced.

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

So I think it is safe to say that with birds in free fall, the system is a little lighter.

Yes

 

[cjl]

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.

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.

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.

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.

 

[cjl]

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?

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.

 

[cabraham]

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.

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

 

[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.

 

[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?

 

[Dale]

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.

 

[cjl]

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?

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.

 

[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.

 

[russ_watters]

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.

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:

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?

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.

 

[Graeme M]

My apologies for the disjointed nature of my posts as again, this gets fitted into very small spare moments and I just don't get the time to read or think in detail.

Mentor, the COM argument makes sense, but I will assume the acceleration must be only where work is done in an opposing direction to local gravity. For example, if the box contains a vacuum, the acceleration of an object in the box must produce a force on the floor ONLY when the object is accelerated directly away from the gravitational field. If the object is in freefall and accelerated by gravity, then there is no force applied to the box floor and hence the system is lighter. If however the object is accelerated by doing some work perpendicular to the gravitational field, then a force may push against a wall of the box but that cannot contribute to OW can it?

So, the COM argument therefore informs us that it is only when the thrust of acceleration is applied to the floor of the box, or at least, to the surface on which weight bears.

Consider the nature of OW. It must be a sum of forces, some of which are not 'weight' at all. Weight itself, it seems to me, is best described as the 'extent to which an object is prevented from being accelerated by gravity'.

For example, a lead ball resting on the Earth's surface cannot be accelerated but the force expressed by the unrealised gravitational acceleration can be measured as operational weight. Similarly, if the ball is in freefall and has been accelerated by gravity it has no weight - its GW remains the same but its OW is zero (or close to it). I'll assume then that if we could place the ball on a scale that is allowed to fall at exactly half the speed of local gravitational acceleration, the ball would express half its GW on the scale - that is, its OW would be GW/2. Or have I misunderstood how this works?

The downforce of accelerating an object in a box containing a vacuum is not actually weight by that definition. It is an additional force produced by doing work, let's say the rocket engine we use. So OW in that context is a mixture of forces, one of which is thrust and not weight as such.

How close am I with this? I have more to go, but I'll wait until we have clarified this point, because the direction of my thought will depend on the accuracy of that notion.

 

[russ_watters]

For example, if the box contains a vacuum, the acceleration of an object in the box must produce a force on the floor ONLY when the object is accelerated directly away from the gravitational field.

When it hits the floor, yes.

If the object is in freefall and accelerated by gravity, then there is no force applied to the box floor and hence the system is lighter.

Correct.

If however the object is accelerated by doing some work perpendicular to the gravitational field, then a force may push against a wall of the box but that cannot contribute to OW can it?

By definition, the work is done in the direction of motion.

So, the COM argument therefore informs us that it is only when the thrust of acceleration is applied to the floor of the box, or at least, to the surface on which weight bears.

When the object hits the floor, yes.

Consider the nature of OW. It must be a sum of forces, some of which are not 'weight' at all. Weight itself, it seems to me, is best described as the 'extent to which an object is prevented from being accelerated by gravity'.

Yes.

For example, a lead ball resting on the Earth's surface cannot be accelerated but the force expressed by the unrealised gravitational acceleration can be measured as operational weight.

Yes.

Similarly, if the ball is in freefall and has been accelerated by gravity it has no weight - its GW remains the same but its OW is zero (or close to it). I'll assume then that if we could place the ball on a scale that is allowed to fall at exactly half the speed of local gravitational acceleration, the ball would express half its GW on the scale - that is, its OW would be GW/2.

Yes.

The downforce of accelerating an object in a box containing a vacuum is not actually weight by that definition. It is an additional force produced by doing work, let's say the rocket engine we use. So OW in that context is a mixture of forces, one of which is thrust and not weight as such.

How close am I with this? I have more to go, but I'll wait until we have clarified this point, because the direction of my thought will depend on the accuracy of that notion.

Sounds like you are trying to argue your way out of the impact of the object on the floor registering on the scale.

 

[Graeme M]

Not at all, just trying to get clear in my head all of the various forces and effects. And of course, thanks for the indulgence. I realize this is old hat to you guys.

To summarise then.

In a vacuum:

1. A box+object system will express an OW of X when resting on a scale that is not itself in motion.
2. If the object is accelerated perpendicularly away from the gravitational attraction, the OW of the system is unchanged as the thrust on the floor equals the OW of the object.
3. OW can therefore be composed of different forces.
4. Thrust used to accelerate an object inside our box can only register as a component of the systems OW equal to the objects OW at rest when it is applied exactly perpendicular and opposite to the local direction of gravity.

 

[Dale]

Mentor, the COM argument makes sense, but I will assume the acceleration must be only where work is done in an opposing direction to local gravity. For example, if the box contains a vacuum, the acceleration of an object in the box must produce a force on the floor ONLY when the object is accelerated directly away from the gravitational field. If the object is in freefall and accelerated by gravity, then there is no force applied to the box floor and hence the system is lighter. If however the object is accelerated by doing some work perpendicular to the gravitational field, then a force may push against a wall of the box but that cannot contribute to OW can it?

So, the COM argument therefore informs us that it is only when the thrust of acceleration is applied to the floor of the box, or at least, to the surface on which weight bears.

The COM argument works always, regardless of the contents of the box, including vacuum. As the contents fall through vacuum the COM is accelerating downwards and therefore the OW is less than the GW.

 

[Graeme M]

Well, yes, but that's a slightly different thing and I think you are glossing over an important detail.

1. If the COM is accelerated upwards, the thrust required bears on the floor and increases total OW.
2. If the COM is accelerated downwards by gravity the total OW is reduced by an amount equal to the OW of the falling object.
3. If the COM is accelerated downwards by the same thrust we had in 1 above, then the the total OW is reduced by an amount equal to the OW of the falling object PLUS the force of thrust.
4. If the COM is accelerated sideways, then we have a thrust force on a wall of the box and in effect the COM has no effect on OW.

Have I got that right?

 

[A.T.]

1. If the COM is accelerated upwards, the thrust required bears on the floor and increases total OW.

Yes

2. If the COM is accelerated downwards by gravity the total OW is reduced by an amount equal to the OW of the falling object.

Only if the object is truly free falling, with no air resistance and just gravity acting.

3. If the COM is accelerated downwards by the same thrust we had in 1 above, then the the total OW is reduced by an amount equal to the OW of the falling object PLUS the force of thrust.

Yes but the thrust adds to gravity, so the object accelerates more than in 1 and 2.

4. If the COM is accelerated sideways, then we have a thrust force on a wall of the box and in effect the COM has no effect on OW.

Yes

 

[Graeme M]

Thanks AT. And thanks to all who have contributed - I think I have a clearer idea now of the problem. Sadly this medium is not really conducive to a fuller discussion. I don't have the time and some of the concepts need a lot of thought and various restatements of what I am thinking. So I think I'll have to leave it here. I've pretty much come to the conclusion that only in some very limited circumstances will the bird+truck system weigh the same when the bird is flying as when the bird is at rest, and even then there are some questions around how that might happen. But no way can I express what gives me to think this in the space of a few comments.

 

[sophiecentaur]

I have a problem with the term "operational weight". It is a home-made term and not very relevant. "Weight" is well enough defined and the other forces which are involved with any problem like this are quite adequate to cope. What are you planning to include in the "operational" bit, which could change its value from the "weight".

 

[A.T.]

I've pretty much come to the conclusion that only in some very limited circumstances will the bird+truck system weigh the same when the bird is flying as when the bird is at rest

The measured weight (OW) is not perfectly constant even if the birds are not flying. They move their heads, they breathe so the OW is fluctuating. With flying birds those fluctuations might get larger, but they still fluctuate around the same mean value. So for all practical purposes the OW is the same.

 

[Dale]

I've pretty much come to the conclusion that only in some very limited circumstances will the bird+truck system weigh the same when the bird is flying as when the bird is at rest

It isn't a "very" limited circumstance. It is merely when the COM is not accelerating, which is always true on average.

You should really watch the Mythbusters episode on this.

 

[Graeme M]

What I meant is that the subject becomes incredibly complex as you dig and that leads to two things: one is that I don't have the time to dig and comment and this thread would become a rambling affair if I did, and two that some of the wrinkles are beyond my ability to grasp without doing some serious study.

SophieCentaur, by operational I am simply using the Wikipedia idea of gravitational definition and operational definition. I may have misunderstood exactly what is meant there, but in my mind it made a quite suitable distinction.

The gravitational definition is a mathematical construction about a relationship between mass and gravitational acceleration, while the operational definition reflects what I see as an obvious distinction - that the weight an object expresses on a scale can be composed of other forces. If we know enough we can filter out the additional forces (eg my finger on the scale is an obvious additional force which is not related to weight, but can be measured as weight).

DaleSpam you are probably right but to explore that in detail requires a LOT of thought and sicussion. I get it that you have the learning behind you, but I didn't even do basic stuff at school so it'd be a long road to get my head around it.

But just to help me see something about your most recent comment, what would be the total system OW in the event that I replace my bird with a small remote controlled powered aeroplane. Let me place that aircraft into a tight circle with wings vertical and held at a constant speed? As far as I can see, the COM is on average static in the vertical frame which is the one we care about, and it is not accelerating.

My understanding of aerodynamic forces suggests that lift is expressed horizontally in this example, so which force now applies to the floor of the box? Remember that earlier, we agreed that forces applied to the walls of the box do not contribute to OW.

 

[Nugatory]

But just to help me see something about your most recent comment, what would be the total system OW in the event that I replace my bird with a small remote controlled powered aeroplane. Let me place that aircraft into a tight circle with wings vertical and held at a constant speed? As far as I can see, the COM is on average static in the vertical frame which is the one we care about, and it is not accelerating.

If the wings are completely vertical, the COM isn't going to be on average static because the plane won't be able to maintain altitude - there won't be any lift to oppose the downwards pull of gravity, the plane will fall, the COM will move downwards, and the scale won't register the weight of the plane until the COM stops moving when the plane comes to rest on the floor.

If the wings aren't completely vertical and the aircraft is moving fast enough to that the vertical component of the aerodynamic forces on the wings is sufficient to exactly balance the downwards pull of gravity then the center of mass will be on average static in the vertical direction. This is the situation that CWatters mentioned way back near teh beginning of the thread: "Replace the birds with a hovercraft".

 

[russ_watters]

I've pretty much come to the conclusion that only in some very limited circumstances will the bird+truck system weigh the same when the bird is flying as when the bird is at rest, and even then there are some questions around how that might happen.

It will very rarely have the same apparent weight, but the average apparent weight will be the same: the apparent weight will oscillate as the wings flap.

But I suspect that's not what you've decided to believe...

My understanding of aerodynamic forces suggests that lift is expressed horizontally in this example...

Lift is a vertical force. If it weren't, it wouldn't be called "lift" and the plane would fall!

 

[Graeme M]

It's not a matter of what I believe Russ, and that probably doesn't matter much anyway. Look, I can take at face value the fact that a bird flying in a box still contributes its weight to the total system. It's really neither here nor there. But I want to KNOW why that is so, but to explain why I have doubts is to wander down the rabbit hole of my mind which I can't really do here.

I disagree with what you say about lift. I think that lift is a force expressed in relation to the direction of flow of a fluid and is nothing to do with gravity at all.

 

[russ_watters]

I disagree with what you say about lift. I think that lift is a force expressed in relation to the direction of flow of a fluid and is nothing to do with gravity at all.

Sorry, I misread your statement, though I think the statement is probably missing something:

You are correct that lift is measured perpendicular to the wings. That's "up" with respect to where the pilot is sitting.

But in your scenario of a plane with wings held vertical, do you recognize that the plane would spiral into the ground?

 

[Dale]

But just to help me see something about your most recent comment, what would be the total system OW in the event that I replace my bird with a small remote controlled powered aeroplane. Let me place that aircraft into a tight circle with wings vertical and held at a constant speed?

Is the circle in the horizontal plane or in a vertical plane?

As far as I can see, the COM is on average static in the vertical frame which is the one we care about, and it is not accelerating.

Yes, the COM is static on average in the vertical direction. If the circle is in the horizontal plane then the COM is always static (not just on average) in the vertical direction.

My understanding of aerodynamic forces suggests that lift is expressed horizontally in this example, so which force now applies to the floor of the box? Remember that earlier, we agreed that forces applied to the walls of the box do not contribute to OW.

Putting an airplane into a vertical orientation basically turns it into a helicopter with the propeller as the rotor. If the propeller cannot generate enough thrust to climb vertically then it will crash, as has been mentioned by others. The force applied to the floor is the wash from the propeller.

 

[russ_watters]

Dale, the way I read it (the second time) was that it is turning in tight, horizontal circles, with the wings vertical...

 

[Dale]

That is kind of what I thought too, but I wasn't sure. If so, it is basically either a helicopter or a crash.

 

[Graeme M]

I disagree. I am no aeronaut, but an aircraft in a tight turn maintains altitude through the work of the engine providing thrust via the agency of the propellor and the use of rudder to maintain nose-up. The process of flying is effectively an exercise in fluid dynamics I suspect. The fluid, in this case the air, provides the aerodynamic forces to offer lift. That lift is expressed in relation to the direction of fluid flow and the plane of the airfoil.

In a vertical turn (ie wings vertical in relation to the ground or floor), the thrust is directed behind the aircraft, ie horizontally. Lift is provided in this case horizontally as well. The only 'vertical' force would be that of the rudder but again that's largely an aerodynamic force (though I might be quite wrong in how I think an aeroplane flies).

It's the engine doing work that provides the thrust to allow the aircraft to create a fluid flow and that flow proffers lift in whatever plane the aircraft is in. Naturally, a tight turn without more power or rudder will result in the steady deterioration of the circle, but providing we have enough lift and power, the circle should not deteriorate.

In effect, the force of lift is cancelling the attraction of gravity, and it's by doing work that we can control the amount of lift.

It's a little like firing a rocket across the face of the Earth - with enough thrust we can counter gravity and create a largely straight path. With too little power, gravity will take over and decay the path toward the centre of gravity of the earth,

That's my take, anyway.

 

[Graeme M]

And yes, turning horizontally with wings vertical.

 

[russ_watters]

I disagree. I am no aeronaut, but an aircraft in a tight turn maintains altitude through the work of the engine providing thrust via the agency of the propellor and the use of rudder to maintain nose-up.

Oh, c'mon! [Dale wins]

If the propeller is providing the force that keeps the plane aloft, then it is just a complicated helicopter and the force to keep the plane aloft is still directed to the bottom of the truck.

Unimportant, but what you describe isn't how it typically works. Planes usually like to fly coordinated turns where the lift force vector is perpendicular to the wings. A plane in the configuration you are describing is side-slipping.

http://en.wikipedia.org/wiki/Coordinated_flight

The only 'vertical' force would be that of the rudder but again that's largely an aerodynamic force (though I might be quite wrong in how I think an aeroplane flies).

Er, you just said [correctly] in the quote above that the propeller provides the vertical force.

The rudder actually provides a downward vertical force.

It's a little like firing a rocket across the face of the Earth - with enough thrust we can counter gravity and create a largely straight path. With too little power, gravity will take over and decay the path toward the centre of gravity of the earth...

Only if the rocket is propelled at 25,000 mph. At any slower speed, it must pitch up and use the engine to provide lift.

 

[Dale]

providing we have enough lift and power, the circle should not deteriorate.

Regardless of how you orient the plane, in order to maintain altitude you must push the air down with a force equal to the weight of the plane, and this force in turn pushes on the box. The COM is stationary (vertically) so the OW is equal to the GW.

 

[Graeme M]

I should apologise if my tone is sometimes not appropriate, I make these posts in snatches between my normal stuff. So sometimes I miss things that have been said or I say things in a less than considered way.

Now, I did not at any stage say that an aircraft's propellor provides a vertical thrust when it is turning. What I am suggesting is that an aircraft flying in air uses aerodynamic forces to obtain lift and those forces are not necessarily directly relevant to gravity.

The propellor provides the thrust to make the plane move through the air - all that is doing is creating a situation analagous to the airfoil remaining stationary in a moving flow. It's only the flow of air over the airfoil that creates lift. The propellor in this case is giving us movement of the aircraft through the air and allowing the airfoil to generate lift in an otherwise static fluid (using our box model).

You'll need to show me somewhere how an aircraft doing a tight turn is transferring any force at all to the ground.

You said "Regardless of how you orient the plane, in order to maintain altitude you must push the air down with a force equal to the weight of the plane, and this force in turn pushes on the box. " I'm saying that the aerodynamic force of lift is relative to the plane of the aircraft and that the forces are expressed horizontally in the main. Yes gravity is still pulling down, but the forces of thrust and lift are sufficient to overcome gravity.

The rocket example demonstrates this. Yes you may need to go at 25000 mph because the rocket is using very little aerodynamic lift, but the thrust of the engine is what allows it to go faster enough for that force to overcome gravity. A rocket zooming across the face of the moon would exert no force on the moon's surface even though there is a gravitational attraction and hence the rocket has a gravitational weight.

 

[Graeme M]

PS when I get a moment I'll read your wikipedia reference to coordinated turns! :)

 

[russ_watters]

Now, I did not at any stage say that an aircraft's propellor provides a vertical thrust when it is turning.

You said (basically correctly):

[previous post]
...an aircraft in a tight turn maintains altitude through the work of the engine providing thrust via the agency of the propellor and the use of rudder to maintain nose-up.

So to put it another way, if the airplane is angled upward, the propeller directs a force downward, which is what holds-up the plane (if it can).

What I am suggesting is that an aircraft flying in air uses aerodynamic forces to obtain lift and those forces are not necessarily directly relevant to gravity.

Aerodynamic forces on what? You said the wings are vertical, so they can't provide an aerodynamic force to keep the plane aloft. You said clearly that the force that keeps the plane aloft comes form the engine. Yeah, the propeller uses aerodynamic forces.

The propellor provides the thrust to make the plane move through the air...

In which direction? Before, it had both horizontal and vertical components. But it sounds like you are hedging/backtracking on that.

It's only the flow of air over the airfoil that creates lift.

As you pointed out, if the wings are vertical, the lift is horizontal, so it can't be keeping the plane aloft.

You'll need to show me somewhere how an aircraft doing a tight turn is transferring any force at all to the ground.

It sounds to me like you are changing the scenario. You very clearly said before that it is angled upwards, so the engine is directing a force downward. If you want the plane (fuselage) horizontal, and the wings vertical, then there is nothing keeping it aloft: it death-spirals into the ground.

I'm saying that the aerodynamic force of lift is relative to the plane of the aircraft and that the forces are expressed horizontally in the main. Yes gravity is still pulling down, but the forces of thrust and lift are sufficient to overcome gravity.

You really white-washed that. If the plane is not dropping, there must be a force directed downward. Before, you said it came from the propeller. If now you are saying that the propeller isn't providing that force, then clearly you have a self-contradiction.

The rocket example demonstrates this. Yes you may need to go at 25000 mph because the rocket is using very little aerodynamic lift, but the thrust of the engine is what allows it to go faster enough for that force to overcome gravity.

No. That situation is a very specific one: that's escape velocity (actually, it could be a little lower and be in orbit). The rocket is still falling, but it is traveling along the Earth's surface and covering enough distance that the Earth drops away from the rocket faster than the rocket drops toward the earth.

 

[Simon Bridge]

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... :)

I would have said "yes and no" ... this is fair for the situation in static equilibrium - which is an implicit assumption.
The questions immediately following do not involve a static equilibrium.

My vet has a large scale for weighing large dogs, small horses etc.

I stand on a platform and it reads my weight - quite accurately I think since all other scales have me a half-kg heavier.
If I jump up and down on that platform, the scale wobbles about - and the vet tells me off.
If I stand still, the scale is still.

Drive an RC car around the scale-platform and it stays the same - so the distribution of weight across the scale does not matter.

However, there are lots of different ways to move around a closed box - they are not all equivalent.

 

[Graeme M]

Russ, great answer and I really want to take issue with you over most of it. Just no time right now. But we are getting closer to the crux of the matter from my point of view.

Bear in mind I am just going with my thoughts here, but this discussion is great in that it's making me THINK and I love that. So thanks for the patience so far.

But I'm going to rip into your last comment just as soon as I can. :)

Simon, agreed, but that too is getting close to the crux of the matter. What is weight, how is it expressed, and what part does an atmosphere play in the equation.

 

[berkeman]

Be careful how you "rip". Other Mentors are watching this thread now.

EDIT -- let me clarify. Other less patient Mentors than Russ. Please be careful to base your questions on actual science and aerodynamics, and please explain your questions fully, so others do not have to guess at what you are asking or saying. Thank you.

 

[rcgldr]

You'll need to show me somewhere how an aircraft doing a tight turn is transferring any force at all to the ground. ... I'm saying that the aerodynamic force of lift is relative to the plane of the aircraft and that the forces are expressed horizontally in the main. Yes gravity is still pulling down, but the forces of thrust and lift are sufficient to overcome gravity.

If the aircraft's wings are vertical, then the plane is generating lift in "knife edge" fllight by using the fuselage of the aircraft as an airfoil, diverting the relative air flow downwards. The prop axis is angled (yawed) upwards, so some component of thrust is also downwards. In order for the aircraft's COM to not be accelerating downwards, the aircraft has to generate enough vertical component of aerodynamic force to keep the aircraft from accelerating downwards.

 

[Graeme M]

Berkeman, not sure quite what you mean regarding the mentoring thing - I AM 'basing' my questions on science and aerodynamics, as best I understand them. But I freely admit my understanding is weak and my grasp of terminology similarly uncertain. What I am doing is as much aimed at querying my own ideas as anyone elses. So while I might engage in a somewhat intensive manner, I am not trying to 'prove' I'm right. But equally I want to test the claims others make. By the way, is that you on the bike in your avatar?

rcgldr, I will need to think about what you've posted. First up, a curved airfoil is needed to divert the airstream in a way that generates lift. I thought that a perfectly symetrical airfoil will not generate lift. If the aircraft is canted with wings vertical, I believe the fuselage cross section is perfectly symetrical and therefore unlikely to contribute much 'lift' in the vertical plane. Secondly, how is the prop axis angled up? Are you saying that in such a manoeuveur, the aircrafts actual longitudinal axis is slightly 'up' to the horizontal as it balances forces with control surfaces? If so, I guess that may explain the notion that the fuselage is contributing to lift as then we'd have a deflected airstream. But I suspect that the vertical lift component in this case is substantially less than the vertical component of level flight.