Helium Balloon: Newsletter Challenge (07/06/10)

Weight transfer. I'm not an engineer but I do know that when a vehicle accelerates from a "standing start" the back end hunkers down while the front end lifts. In a really hot car and extreme circumstances this results in a wheelie. The front of the cab is now higher than the rear and so the balloon must move to the front. The same would be true when cornering. When making a left turn, for example, the right side of the vehicle would be lower than the left and, again, in extremes the result would be a rollover. Now the left side is MUCH higher than the right. Seems simple enough to me but I could be missing something. I'm no engineer.....

Good thoughts...

So, if you tether the balloon so that it is held in the center of the car's cab from a single string, what would happen?

You don't have to be an engineer, but it's good to be a critical thinker. So, that is why I am posing the question to you.

Also, after you have though about the answer and proposed what will happen, can you try to create an experiment to test your theory? How would you do it (safely, of course) and what factors or forces might complicate the experiment or spoil the results?

Not having a car would be one. ;-) But I just wanted to get you thinking and have some fun with it.

If the balloon were to be tethered, then the pivot point would cause the balloon to arc downward as the vehicle pulled away until the motion of the balloon caught up with the motion of the vehicle.

A simple experiment would be to turn this upside down basically, and use a pendulum. A potatoe on a string! Hold this still from the string, a twelve inch height perhaps, and then walk briskly. The potatoe or weight should hold still for a moment under its own motionless weight, be drawn by the the motion and then swing into action.

the descrivbed situation results in very, very small effect to the balloon so it is possibly not possible to test the effects; what is if the ceiling is not a straigth line or theres an angel to the horizontal?

all these little things effects to the result!

I can't remember the full 'dramatics', but owing to Einstein's and Newton's Theorem, because the balloon is resting within its own gravitational 'seating', it should roll back-over as the car moves forwards. The balloon will remain in place, basically, vertically above the road.

If it is touching the roof of the car's innards, the the friction will cause it to roll against the material.

The same principal sits as when a ball or object rests upon a floor of a box and is dropped from a height, then the ball with it's own weight/force will leave the surface of the innards of the box as gravitation within the movement alters.

Assuming the ballon is floating i free air within the car, it (skin and helium) must have less density than air, and the movement described above is true. This means a very light ballon skin.

Also, being flexible, the ballon skin will transfer any outside pressure to the inside of the ballon. The ballon moves forward just because the density is less.

The Answer will be posted right here on CR4 on August 3rd.

Great! that will give me time to get a helium filled baloon and try it.

If a window is open, the balloon will be blown any old where. This is known as an uncontrolled experiment.

As the car moves forward, the friction between the balloon and the car roof will not be enough to accelerate the balloon, the difference between then will cause the balloon to rotate, first rolling back, then when the balloon hits the rear the energy stored in the balloons rotating action will cause it to bounce forward, and its anyones guess after that.

Regards JD.

Assuming the ceiling is flat, the balloon will move forward.

Air is not compressible under these conditions* in any meaningful sense. However, the G vector direction changes, and the balloon floats accordingly. If the car were accelerating at a rate of one G, the balloon would move forward at roughly the same rate as it would otherwise float upward in still air (ignoring surface friction, static charge, etc., etc.).

In turns, helium balloons go to the inside of the turn for the same reason.

*i.e., no more than is seen in the density of air at 1 foot above ground vs and 7 feet up, for example.

"*i.e., no more than is seen in the density of air at 1 foot above ground vs and 7 feet up, for example."

Now think about that. If the pressure gradient was zero over the horizontal span of the car's cabin (as you suppose), what are you proposing is the mechanism for the ballon to move forward in the cab?

Ummmm? It floats.

The explanation is not nearly complete. So, why does it float?

For example, imagine a tank of water. Now put something buoyant into the tank at 1/2 the depth of the tank and midway between the walls. Which way will the buoyant object go?

If you follow the logic stated by post #5 it could take any direction; up, down, or sideways. Why would the object not got to the bottom of the tank?

Post #5 assigns some form of anti-gravity property to the buoyant object. However, post #5 is wrong in this respect.

A buoyant object rises to the surface of the tank because there is a pressure gradient at work. Gravity's influence simply creates that pressure gradient in the medium that the buoyant object is surrounded in. In the case of the water tank, the fluid at the bottom of the tank is slightly denser than that of the top of the tank with a linear gradient in between top and bottom of the tank.

In the case of the ballon in the car acceleration, not gravity, creates a pressure gradient inside the car in the horizontal direction. The forward movement of the car creates a gradient where the density of air is slightly greater at the rear of the car's cabin than the front. Earth's gravity creates a gradient that is denser at the floor of the car. The buoyant object, which is the balloon, now strives to reach equilibrium in the media that it is surrounded in and seeks the lowest density of air it can find, which is the top and forward section of the cab.

In this example the mass of the balloon and its inertia are overcome by the pressure gradient created by the car's forward acceleration and the balloon moves forward in the cab.

Does an incompressible lump of wood float in an incompressible liquid?

It's not about compressibility, but pressure gradients with liquids and relative density with a buoyant object.

We're agreed on the relative density: what's wrong with a simple (albeit compound) balance model to work out what goes up and what goes down?

Explain, please.

Heavy things go down, light things go up. In the accelerating car the apparent up and down have changed slightly (up towards the front of the car).

A full evaluation of the forces acting on a floating object will reveal, as you say, that it's the differential pressure between the upper and lower parts of the object which cause it to float.

But you seemed to take exception to MoronicBumble's treating the air as incompressible, which seemed like a reasonable (if unnecessary) simplification to me. The outcome would certainly be the same if the air was incompressible.

But the air isn't incompressible and it's relative density is what defines the pressure gradient.

His statement that the air pressure or density is the same at 1 foot altitude versus 7 feet is incorrect. There is a difference. The pressure gradient is a logarithmic scale as follows:

The must be a gradient for a buoyant object to move. No gradient, no movement or at the least any movement would be randomly driven.

No: I'm (we're) in agreement that the pressure gradient is significant. What I'm (we're) saying is insignificant is the density gradient.

MoronicBumble and I seem to have slightly parted in opinion here so I should have deleted the (we're)s.

To get back to you're diagram: yes the line is curved, but, to do most meaningful calculations in a small portion of the atmosphere you would only be concerned with the slope of the curve at that point, not it's curvature. The same way as we normally treat g as being constant when doing cricket (base) ball type ballistics.

Just to repeat myself: for the purposes of this challenge the pressure differentials in the car are significant, but, the density differentials are insignificant.

But we don't really need to analyse why things float we just need to understand that they do.

But we don't really need to analyse why things float we just need to understand that they do.

Well put! That's what I was trying to say (less concisely) in post #55.

In this example the mass of the balloon and its inertia are overcome by the pressure gradient created by the car's forward acceleration and the balloon moves forward in the cab.

The magnitude of the forces is completely haywire for an explanation based on pressure gradient. Calculate the difference in pressure at 1G acceleration. That difference is extremely small ( circa 1 : .999 ), The buoyancy effect, however, is very large (10 : 1 difference in density.) For the purposes of this experiment (the results of which have been seen by many parents on the way to birthday parties) the car can be considered a closed vessel in which pressure is everywhere the same. (As a closed car accelerates, you do not feel wind on your face as the "pressure gradient" causes the air molecules to move rearward. A fly flying around in your car is unaffected by this imaginary mass air flow.)

With a simple experiment, you can prove to yourself that the large pressure gradient in water with depth does not effect the buoyant force on a submerged, buoyant object. Push a block to the bottom of a pool with a long handled device like a pool skimmer. Measure the buoyant force at several levels.

This (car) experiment would operate in exactly the same fashion if the fluid were non-compressible in the conventional sense: Fill the interior with water and put a steel ball on the floor and a wood one on the ceiling. Accelerate the car. The wood ball will roll forward, the steel ball will roll aft.

My opinion is that you are mistaken, but we will just have to agree to disagree on this one until the answer is revealed.

We agree that the balloon will move forward.

I think we agree that if this car were accelerating at 1G, and if the balloon were tethered so that is does not touch the ceiling, that during this 1G acceleration the string would be at 45 degrees to the vertical. (I think we also agree that, for the sake of this discussion, the car's ventilation system is turned off and the windows are closed.)

I suspect, but am not sure, that we agree that if we substituted a fluid generally termed "non-compressible," (such as water) for air and another floating object such as a wooden ball for the balloon, that the observed effect would be the same: as the car accelerates, the ball rolls forward.

I think that our only disagreement may have to do with your statement: acceleration causes the air density to compress in the rear of the car's cabin and rarefy slightly in the front of the car's cabin.

(I think you mean that acceleration causes the air density to increase or the air to compress rather than "the air density to compress".) But our essential difference seems to be re compression and rarefaction. I think neither of these is required for buoyancy. Although compression and rarefaction occur (to a very small degree) in things we call "non-compressible" it is not compression and rarefaction that cause the pressure gradient that causes buoyancy. Compression and rarefaction are symptoms. The cause of the pressure gradient (and buoyancy) is the cumulative weight of the fluid.

Another disagreement may have to do with how far to go in the direction of basic physics. I prefer to say that a boat floats according to Archimedes buoyancy principals. You seem to be explaining why buoyancy works as it does. Which is perfectly OK... it just a different point of view. It is true that coal is just another form of solar energy.

Okay, it just looks like it may only be semantics is at play. Other than that we basically agree.

However, I am not sure that I understand your hypothesis for the mechanism that moves the balloon forward in the car. I wanted you to explain the forces that are at play and how they act on the balloon.

However, I am not sure that I understand your hypothesis for the mechanism that moves the balloon forward in the car.

My hypothesis is that the balloon floats in the air in the car, in accordance with Archimedes principals. If the car is not accelerating, the balloon will float straight upward with respect to the earth's gravity. If the car is accelerating, then the air in the car is no longer pulled straight down, but is pulled in a direction and magnitude that is the vector sum of the accelerations due to gravity and the car's motion. Thus, if the balloon were tethered (and able to swing fore and aft without colliding with the ceiling), it would work like a simple pendulum accelerometer (but this "pendulum" would be upside down). There is no mechanism other than ordinary buoyancy that moves the balloon forward and upward.

I agree completely with tkot's post 74 (except for this: It is easy to generalize the shape of the object to whatever, if one considers that the water pressure causes forces which are directed always vertically to each point of an object's surface.*) as an explanation of buoyancy effects. However, the question appears to ask not "Why does the balloon float?" nor "How does buoyancy work?" but instead "Why does the buoyant force have a forward component?"

In other words, the question has to do (mainly, I think) with the same effect as seen in the various centrifuge buoyancy demonstrations in which the steel ball goes out and the cork ball goes inward.

(Many of these questions are essentially didactic: it is not as if we can't simply drive a car and see what happens, and I am sure many of us have already seen this phenomena. So the discussion can become more about how you explain what happens than what actually happens.)

* The part I don't agree with is the words "always vertically" unless by that he means always in opposition to the vector sum of gravity and other accelerations. (In a centrifuge, light objects float mainly horizontally.)

However, the question appears to ask not "Why does the balloon float?" nor "How does buoyancy work?" but instead "Why does the buoyant force have a forward component?"

I cannot understand what exactly is not clear. I proposed the example of the object into the sea, to make it easier. In this case, it is clear that the water pressure at the bottom of a submerged object is higher than that on the top. (See #92 for an explanation of why pressure is applied at all.) So the fact that the buoyancy force is towards the sea surface is straight-forward. The same applies to the problem at hand, where acceleration is horizontal. (OK, one might add the gravity, but this is irrelevant in our case, as there is NO movement on the vertical axis, and moreover nobody really cares whether the balloon is spherical or not due to contact with the ceiling). In our case, the buoyancy force will be towards the front of the car.

The part I don't agree with is the words "always vertically" unless by that he means always in opposition to the vector sum of gravity and other accelerations. (In a centrifuge, light objects float mainly horizontally.)

When I said "vertically" I mentioned only the force due to the fluid pressure. Indeed the molecules of a fluid, press each particular point of an object orthogonally to its surface. There is no reason in the world why there should be any force along the local surface: It's not my intention to analyze possible fluid currents flowing around the object! Anyway, even if you forget that, just slice an object of arbitrary shape to infinitecimally thin slices and sum up to see that eventually any shape will follow the same buoyancy formula, which in turn derives from the pressure gradient (as I said in #74), which in turn derives from the compression of the fluid (as I said in #92).

would the friction (at acceleration) at the point that the balloon is touching the ceiling, cause it to rotate clockwise till there was no acceleration?

Do the math.

Park your car on a hill with a helium balloon inside. Measure the buoyant force. The force will be far greater than the "pressure gradient" which is so slight as to be unmeasurable.

See JRW's post.

See the simple and very common experiments with light and heavy balls in fluid containers on a centrifuge. The magnitude of buoyant forces are not explained by pressure gradients.

Just give me an explanation for why a helium balloon goes upward when released every time?

Tell me the exact physics behind why it goes up an no other direction.

Not sure which of your posts to respond to.

There is a gradient, but it has no bearing on this problem. Consider the balloon tethered at half the height of the cabin, the air at the level of the bottom of the balloon has a column of air above it, pressing down. The air under the balloon has the same pressure minus the difference between the displaced air and the lighter balloon and contents. The air, being gaseous, pushes the balloon up until it hits the roof when released.

If the the moonroof was open, the balloon would rise until it reached an altitude where the weight of the balloon and contents equaled the weight of the displaced air. It is the density gradient that matters, not the pressure.

It's true that the density gradient would regulate the point at which the balloon stops going up. But over small distances it's the pressure gradient that matters not the density gradient. If you treated the air as incompressible the answer you got to this challenge would be only infinitesimally different to the answer you get when you take the density gradient into account.

Just for the hell of it let's do some rough calculations. Forget about the car: lets assume that

1.) Air has a mass of 1 kg per m³ (I think it's about 1.2 but this is real rough)
2.) The helium balloon is a 1 m cube, and,
3.) has a total mass of ½ a Kg
4.) Atmospheric pressure at the top of the balloon is 10,000 Kg per m² (it's not much more at sea level: 1 bar is 10197.16 and 1 atmosphere is 10332.27; I wonder why they're different)

Now clearly the atmospheric pressure at the bottom of the balloon will be 10,001 Kg per m²

The balloon will have a total force acting on it of 1-½ Kg
It has a mass of ½ Kg so it will start to rise with an initial acceleration of g

The difference in density between the air at the top and bottom of the balloon will be 1 part in 10,000 and will be insignificant.

tkot has convinced me by referring to wikipedia on buoyancy that the initial acceleration will depend on the mass of the object and the mass of the volume of fluid it is displacing.

So in the above case the initial acceleration would actually be (¹/₃)g.

Waitasec. This is wrong (or at least not quite right).

The whole issue(s) of "pressure gradient", compressibility, etc. of the air isn't actually wrong, but it's not the significant cause or explanation. I.e., if air were incompressible, you'd still get the same effect (balloon moves forward), and virtually the same magnitude.

Put a fish tank of water, with a solid wooden ball floating in it, in the van, and accelerate. The ball will float FORWARD (relative to the tank walls). Or - as we'd more likely perceive it if the tank isn't full and sealed -- the water would "slosh" backwards, more rapidly than the ball -- displacing and thus leaving the ball 'behind' (at the front of the tank).

Water is effectively incompressible in a normal regime. Yes, as you dive down, the pressure does increase (1 atm * 33ft^-1) -- but you conflated that with actual compression. (Your home water supply is under pressure, but not compressed.) Further: a buoyant object does not, as you claim, float due to a pressure gradient and supposedly increasing density at the bottom. It floats because it's less dense than water, and the water flows under it, displacing it upwards - period.

Similarly, under inertial acceleration, a pressure gradient and/or compression of the fluid medium aren't at all necessary to describe or explain the effect (that the object moves forward relative to the medium). It's all (at least to a lot of decimal places) about the relative densities of the medium and the object.

As long as there's acceleration, there will be 'pseudo-gravity', with respect to the back of the van. Therefore an object that's buoyant wrt its medium, will be buoyant away from actual gravity (which under relativity we know to be effectively indistinguishable from acceleration), and away from this pseudo-gravity - thus, forwards.

You could draw a vector sum diagram, with one vector straight down at 9.8 m*sec^-2 and the other straight back at whatever the van's acceleration is, and calculate the angle and magnitude with trig and Pythagoras. The net force (due to acting on the balloon and the surrounding air will point diagonally down and back, which is the direction the (denser) air tries to take, displacing the (less dense) balloon in the opposite direction. (At least in Cartesian, Newtonian, Euclidean space-time. ;-)

The issue is you have an object that moves. Something has to cause the object to move. You are implying it moves opposite to gravity - antigravity, if you will.

A balloon does not possess antigravity properties. The balloon is not floating on the surface of air, so it is trapped in a tank of air.

Let's get back to the balloon moving forward. What force pushes the balloon forward? Forget antigravity - I think you will agree on that. So, what forces the balloon forward?

The answer is a difference of pressure of air behind the balloon is greater than the pressure in front of it. As the balloon travels forward the difference in pressure behind the balloon drops, but the pressure in front of the balloon also drops as the balloon continues along until it meets the front of the car's cab.

Let's ignore the balloon. Under acceleration, the air pressure at the back of the cab is higher than that of the air at the front. What is the relative pressure at the horizontal center of the cab? I will bet you will say it is something between the pressures at the horizontal extremes of the cab - and you would be right!

There is a horizontal pressure gradient from front to back of the car's cab that is a product of the car's acceleration.

When you reintroduce that balloon it must travel along that gradient toward the front of the cab (where the air pressure is least).

Additionally, what would happen if there was no gradient in the cab, yet we had acceleration. Let's pretend that it is insignificant or better yet zero as proposed by some posters. What would the balloon do? It would move backwards due to the effects of inertia because the air horizontal pressure would be equal on both horizontal sides of the balloon.

Let's take one more scenario for fun. Imagine the pressure was digital. That is, exactly half of the cab (back half) had a higher pressure than the front half under acceleration. Put the balloon in the exact horizontal middle of the cab and what happens? The balloon will move exactly 1/2 the balloons horizontal width forward and stop, pinned at the interface between pressure extremes by inertia.

The bottom line is that you can say that buoyancy causes an object to move away from the vector of gravity, but it does not describe the mechanism by which that happens. You can also say that an buoyant object floats on a medium that is more dense than the buoyant object. However, release an object at some point in the middle of that medium how does it know which way to go inside that medium? Are you saying it uses gravity as a cue? What sensing mechanism is used by the buoyant object to go away from gravities vector? What is the physics that makes that work?

Comments inline in bold caps:

The issue is you have an object that moves. NOT REALLY, NO. Something has to cause the object to move. THAT WOULD BE A FORCE. You are implying it moves opposite to gravity - antigravity, if you will. I'M NOT IMPLYING IT - HELIUM BALLOONS DO, EMPIRICALLY, RISE. ('ANTIGRAVITY' - NO, I WON'T, AND I DIDN'T. GRAVITY PULLS HELIUM BALLOONS DOWN, TOO.)

A balloon does not possess antigravity properties. The balloon is not floating on the surface of air, so it is trapped in a tank of air.

Let's get back to the balloon moving forward. What force pushes the balloon forward? Forget antigravity - I think you will agree on that. So, what forces the balloon forward?

The answer is a difference of pressure of air behind the balloon is greater than the pressure in front of it. NO. YOU'VE OMITTED THE DELTA BETWEEN THE RELATIVE DENSITIES, AND THUS INERTIAE, OF HELIUM AND AIR. BY YOUR REASONING, A BOWLING BALL WOULD BE AFFECTED SIMILARLY. As the balloon travels forward the difference in pressure behind the balloon drops, but the pressure in front of the balloon also drops as the balloon continues along until it meets the front of the car's cab. AGAIN - SUBSTITUTE BOWLING BALL FOR BALLOON IN THIS EXPLANATION - ALL YOUR FORCES ARE STILL IDENTICALLY AT WORK.

Let's ignore the balloon (YOU ALREADY HAVE. >;-) . Under acceleration, the air pressure at the back of the cab is higher than that of the air at the front. What is the relative pressure at the horizontal center of the cab? I will bet you will say it is something between the pressures at the horizontal extremes of the cab - and you would be right!

There is a horizontal pressure gradient from front to back of the car's cab that is a product of the car's acceleration.

When you reintroduce that balloon BOWLING BALL it must travel along that gradient toward the front of the cab (where the air pressure is least).

Additionally, what would happen if there was no gradient in the cab, yet we had acceleration. Let's pretend that it is insignificant or better yet zero as proposed by some posters. What would the balloon do? It would move backwards due to the effects of inertia because the air horizontal pressure would be equal on both horizontal sides of the balloon. SO: THE BALLOON AND BOWLING BALL ARE NOW ACTING IDENTICALLY - MOVING BACKWARDS - PURELY BECAUSE OF THE (HYPOTHETICALLY) EQUAL AIR PRESSURE ON FRONT AND BACK.

Let's take one more scenario for fun. Imagine the pressure was digital. QUANTUM / STEP-FUNCTION BALLOONS - NOW THIS IS FUN. ;-) That is, exactly half of the cab (back half) had a higher pressure than the front half under acceleration. Put the balloon in the exact horizontal middle of the cab and what happens? The balloon will move exactly 1/2 the balloons horizontal width forward and stop, pinned at the interface between pressure extremes by inertia.

The bottom line is that you can say that buoyancy causes an object to move away from the vector of gravity, but it does not describe the mechanism by which that happens. You can also say that an buoyant object floats on a medium that is more dense than the buoyant object. However, release an object at some point in the middle of that medium how does it know which way to go inside that medium? Are you saying it uses gravity as a cue? What sensing mechanism is used by the buoyant object to go away from gravities vector? What is the physics that makes that work?

LET ME TRY TO MAKE THIS SIMPLE - AS IT APPEARS YOU'VE NOT ACTUALLY READ, OR AT LEAST THOUGHT THROUGH WHAT I POSTED BEFORE (OR, SORRY, I'VE BEEN LESS THAN CLEAR - IF SO, MEA CULPA).

THROW A BOWLING BALL INTO A LAKE. IT WILL SINK (OR SO I'M TOLD).

NOW, RE-READ YOUR LAST PARAGRAPH W.R.T. THAT PHENOMENON. (IT WILL ALSO SINK. >;-)

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

(THEN, AB INITIO, RE-THINK YOUR WHOLE ARGUMENT ABOUT PRESSURE GRADIENTS IN THE FLUID MEDIUM W.R.T. BUOYANCY....)

So, you can't explain the effect and just throw a bowling ball into the argument. ;-)

Whatever.

So, you can't explain the effect and just throw a bowling ball into the argument. ;-)

Whatever.

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10 points for amusing. 0 points for responsiveness. And -10 points for being wrong - I can, and did, explain the effect.

(And, my 2 little boys, at the ages of 6 and 3.5, know that responding with "whatever" -- is absolutely offsides, because: (a) it's rude, and (b) -- far more seriously, it's intellectually lazy and sloppy. If they're going to be rude and insulting, they'd effing-well better be clever about it.)

Okay, the bowling ball is a red herring.

Obviously, the forces of acceleration on the ball overwhelm any other forces at play.

Back to the original question, which I either missed something in your answer or it was incomplete in its explanation. What exactly happens to the balloon when the car accelerates and why?

REPLY TO #109

Great responce about your two boys, loved it!

NO. YOU'VE OMITTED THE DELTA BETWEEN THE RELATIVE DENSITIES, AND THUS INERTIAE, OF HELIUM AND AIR. BY YOUR REASONING, A BOWLING BALL WOULD BE AFFECTED SIMILARLY.

I sense that there is a misunderstanding about what buoyancy is. To set things clear, lets agree (if I'm wrong correct me) that the buoyancy force is meant to be the force applied to an object inside a fluid BY THE FLUID and ONLY BY THE FLUID. The force caused by earth's gravity or any other force for that matter is something extra that has nothing to do with buoyancy itself. As psoderman already said, whether you put a bowling ball or a helium balloon (of the same volume) in the car, they will experience the same buoyancy force. Even the driver's head must experience about the same force.

The reason why a bowling ball and a balloon (or a head) behave differently, is because it's NOT ONLY the buoyancy forces that affect these objects. Remember that the objects' frame is accelerating, so there is an apparent force applied to them which is equal to m.a, where m is the mass and a is the acceleration. If m is small (as in the case of small densities, see helium balloon) then this force is small. It happens (as I already explained before in another post) that if the submerged object's density is smaller than that of the surrounding fluid then this inertial force cannot counterbalance the buoyancy force, so the latter wins. Otherwise the net force vector points to the other side.

If accelerating frames of reference pose a difficulty, just refer to a previous post where I mention the equivalence of this problem with that of a ball inside the sea. Here, gravity is our acceleration. Again, the buoyancy force is one thing, and the weight is another. Both combined produce a net force which either points upwards or downwards according to the difference in densities.

I cannot see how clearer it needs to get. Maybe somebody can rephrase these points in a better way, but what the heck, it is school physics.

Yes, I like this. We have to think about coordinate systems. Fixing the coordinate system to the car and letting the car accelerate with rate a is equivalent to keeping the coordinate system stationary and letting the balloon (mass m) accelerate. The apparent force F = ma is in the opposite direction of the buoyancy force. If m is small, buoyancy wins and the balloon moves to the front of the car. If m is large, buoyancy loses and the object moves to the back of the car. If m equals the mass of air displaced by the balloon, the buoyancy force and apparent force cancel out and the balloon does not move relative to the car. Thank you tkot.

I think the gradient looks more like this: stationary on the left, accelerating on the right:

Assuming the car is going right to left and the darker color is denser.

Not exactly. A submerged body in water will experience greater pressures on the lower side than the upper. If you integrate all the pressure (forces) around the body you will get an upward resultant force (buoyancy). The density of the body does not enter - a canon ball has the same buoyancy force as a submerged cork with the same volume. The canon ball sinks because it is heavier than the buoyancy force; i.e., heavier than the displaced volume of water. Therefore, the statement that something floats because the water flows under it does not make sense, it floats because it is lighter than the displaced water or buoyancy force.

A pressure gradient in water is the cause and buoyancy is the effect. We can assume water density is constant, but without a pressure gradient there is no upward force. Fortunately, we don't have to integrate the forces around the body because we know that the resultant buoyancy force equals the weight of the displaced water.

Bravo! Well stated. Simply take you kids on a long ride with a latex helium-filled balloon, having them hold the sting steady while you drive. It's a fun way to teach them and your non-engineering spouse about Pressure Gradients. Try it, you'll have a blast! (Well, hopefully not until your lesson is over. Haha)

what are you proposing is the mechanism for the ballon to move forward in the cab?

Buoyancy.

Perhaps you are thinking that the reason the balloon floats in the stationary car is the difference in density of the air at the cabin floor and ceiling? That is not a significant effect.

The buoyant force is very large as compared to the density differences top to bottom in a small volume of air. Perhaps you've seen a cork in a jar of water in a centrifuge?

At 10 feet up, air density is about .9994 that at sea level. This difference is not great enough to influence the balloon in any meaningful sense. The 10:1 difference in density between air and helium, however, will have a noticeable effect.

I get what you are saying and you are correct, but only with part of the story. The pressure gradient is what determines where in the cabin the buoyant balloon sits or gravitates to.

My point was that in order for a buoyant object to move in a direction, there must be a gradient, however small, or the balloon would have an equal chance of moving down as it does up or side to side.

Maybe a better experiment is looking at what happens to a helium balloon onboard the space station. You can place the balloon anywhere inside the station with the ventilation system off (it would only move with air currents). Sure, the helium is lighter than air, but there is no pressure gradient inside the space station and no mechanism to drive the balloon one way or the other. So, it remains where it was placed.

Pop the balloon and the helium would simply homogeneously disperse within the ship's atmosphere.

. . .The pressure gradient is what determines where in the cabin the buoyant balloon sits or gravitates to.

My point was that in order for a buoyant object to move in a direction, there must be a gradient, however small, or the balloon would have an equal chance of moving down as it does up or side to side. . . .

Anonymous --

You definitely need to read this overview of buoyancy.

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

Your arguments imply that Archimedes "missed the boat" in his buoyancy calculations.

Near the end, the Wiki article covers your pressure gradiant arguments also, and treats them as the relatively weak forces they actually are (in this, and most other applications)

When we read the "Official Answer" at the end of this month, it will be a "Reader's Digest" version of the Wiki article, guarantee!

Larry F

Did I read your link correctly where it stated that Archimedes principle only applies to an object not accelerating?

Did I read your link correctly where it stated that Archimedes principle only applies to an object not accelerating?

Yes, apparently you read something incorrectly. The article does state that buoyancy occurs in systems under acceleration, whether the acceleration of gravity, a "speeding up" type acceleration, or a change of direction acceleration (centripetal /centrifugal). That is the opposite of what you "read".

So, you still think Archimedes got it all wrong, huh? And you'r the first guy to figure out where he went wrong?

No, the principle is not wrong. However, the link was talking about the buoyant object itself undergoing acceleration (i.e., moving with changing velocity).

I understand what your point is and thank you for the link and it makes sense now.

No, the principle is not wrong. However, the link was talking about the buoyant object itself undergoing acceleration (i.e., moving with changing velocity).

What do you mean, "however"? The whole point of the original question is what happens to the balloon! (the "buoyant object itself" as you put it) That is what all the rest of us are talking about, also.

I understand what your point is and thank you for the link and it makes sense now.

It is not my point you need to "understand". If you truly understand what Archimedes said, you will recant your previous arguments as being irrelevant, and say that Archimedes' analysis / calculations of buoyancy is all that is needed to predict what will happen to the balloon as the car accelerates.

Sorry for not being clear. Perhaps I can reword it...

1. Archimedes' Principle is well established, I do not dispute it. The link you provided made a statement that Archimedes' Principle applies to an object that is not undergoing acceleration. From the link: "Archimedes' principle is a fluid statics concept. In its simple form, it applies when the object is not accelerating relative to the fluid." It was just an observation.

2. Thank you for providing the link. The link was informative and helped me better understand the principles related to the original problem, which I misstated earlier.

Air is not compressible under these conditions* in any meaningful sense.

If there is no pressure gradient, in which case different forces will be applied to the front and rear of the balloon, then why it should move at all? Especially towards the front, which is unintuitive, given the wish of the balloon's particles themselves to be inert?

I think that inertia causes the air particles move towards the rear of the car (in fact trying to adhere to their kinetic status up to that point, i.e. stay still). If it weren't for the back window pane, all particles would happily stay motionless, i.e. the car will leave without them. But in our case, the window will cause the air particles to accumulate, increasing the pressure and thus applying more and more difficulty along the way to other particles from accumulating. The particles right next to the window will feel exactly the pressure required to accelerate exactly as much as the car does. The particles further ahead will take it more easy. I think it is clear that a pressure gradient will form.

Moreover, I don't think that our problem has much to do on how compressible are the fluids in our experiment. The same would apply even if the car was filled with water and the balloon with alcohol. To make it a little bit more intuitive, imagine for a moment that we are not talking about an horizontal acceleration experiment, but we study earth gravity itself (acceleration = g). Isn't it true that deep in the sea the pressure is higher than that on the surface? We still have a pressure gradient there, although water is "not compressible". Buoyancy itself is the symptom of this pressure gradient!

Doesn't an "ideal" lump of wood float in an "ideal" liquid.

This is a simple buoyancy demonstration, not unlike this one.

No pressure gradient is required to explain the effect, in which a large buoyant force moves the baloon toward the front of the car. (At 1 G acceleration, the upward and forward forces would be equal.)

Consider the ordinary case when the car is stationary. The reason the balloon floats it not due to the infinitesimal difference in pressure and density at 2 feet ASL vs 5 feet ASL. It is due to buoyancy, fueled by the very large 10:1 difference in gas densities.

At a full G acceleration, the pressure gradient would mimic the vertical pressure gradient seen in still, free, air. This difference (1 : .999) is far too small to explain buoyancy effects.

Also consider that the pressure inside a helium balloon is greater than the pressure outside. So on the basis of pressure the balloon would sink... but of course it does not.

If the gradient is insignificant, why does the balloon not drop to the floor, or the side of the cab?

No pressure gradient is required to explain the effect

Well, I (and others along with me) am still anticipating an alternative explanation of the buoyancy effect... but, for some strange reason, you consider buoyancy as a causative and self-explained phenomenon, although in fact it is nothing more than a symptom of the difference between the forces on the two sides of the balloon.

Let me explain in few words what buoyancy really is:

First of all, to simplify the analysis, consider (as I already mentioned in #19) that we have a vertical (instead of horizontal) acceleration in the case of an object inside - for instance - the sea (i.e. vertical acceleration = 1g). To simplify it further, consider that this object is a box shaped object with area = A, thickness = δ and density = ρ, which is positioned so that area A is horizontal. The question now is: what's the difference between the force caused by the water that push the box's bottom towards the surface as compared with the force pushing the box's top towards the depths of the sea? The water which is by δ units below suffers an extra pressure due to the extra water of thickness δ and density ρ_w = 1 that lies above, which is translated as δρ_wg in force units per area units, i.e. our object will experience a force caused by the water's pressure gradient equal to Aδρ_wg = Vρ_wg (V being the volume) pushing it towards the surface. Now, if ρ = ρ_w then the object itself will weight the same amount so the net force will be zero and consequently it will not have to move. If however ρ < ρ_w then the object's weight won't suffice to counterbalance the pressure difference effect, so it will experience a net force equal to (ρ_w- ρ) Vg. This is buoyancy. And yes, those "infinitesimal" (as you call them) pressure differences, are the sole cause of the buoyancy effect.

It is easy to generalize the shape of the object to whatever, if one considers that the water pressure causes forces which are directed always vertically to each point of an object's surface. Then one uses some school trigonometry. Alternatively, one can slice an object horizontally to tiny slices and sum up (integrate) to get the total effect. Whatever the approach, one can see that the total buoyant effect is the same regardless of the shape.

Well, I (and others along with me) am still anticipating an alternative explanation of the buoyancy effect...

I agree with your explanation re the cause of buoyancy. I do not agree with AH's compression of air at the rear of the cabin theory, however, as a good explanation for the effect. The effect would be the same with a non-compressible fluid. The air in the cabin is better thought to slosh, rather than compress, in response to acceleration.

See my post number 87 for my only reservation re your explanation.

No pressure gradient is required to explain the effect

My statement here is unclear. "The effect" I was referring to is not buoyancy, per se. The pressure gradient AH referred to was one arising from the compressibility of air. Compressibility is not a requirement for this effect (the possibly counterintuitive motion of the balloon forward in the car) any more than it is a requirement for buoyancy. A pressure gradient does not require compressibility.

A pressure gradient does not require compressibility.

But of course pressure relates with some (slight might be) compression. In fact there is no such a thing as a totally non-compressible material! Pressure is nothing more and nothing less than force per unit area. So it is proportional to the statistical average of the force that each molecule of an object applies to another object. When the object is a gas, then this force is only explained by the thermal movement of the molecules. In other cases of "non-compressible" materials, there are also electric bonds among the molecules that prevent them from freely moving around. Nevertheless, the molecules can be displaced up to some extend when a force is applied to them (i.e. when they get in contact with another object or when they experience acceleration), in which the bonds will try to return them back to place. It's like a spring effect. The bonds can be quite strong, so even a tiny displacement (of a so called "non-compressible" material) can produce a sizable force.

Otherwise, there is no physical explanation why you feel higher pressure the deeper you dive into the sea. The only thing that applies force to you is the water that virtually touches you, NOT all the rest of the water that lies above or anywhere else. It's just that the water next to you is more compressed due to the cumulative effect of the weight above that makes exactly this water in its turn apply more force per unit area upon you. If there weren't any compression, then what do you think is the physical factor that tells deep water press harder?

the latin words for "squeeze" are . .

premo pressi pressum

both contained within the words "Pressure", and "Compression",
the former being a word that describes the content that in the second word is the resultant state.

so when someone writes . .
"A pressure gradient does not require compressibility"
huh???

To make a long story short: Pressure is one thing and compressibility is another. Specifically, compressibility is the measure of how much a given material's density reacts to changes in the pressure it is subjected to.

In what concerns the problem we've been posed:

It's true that there is no such thing as a fully "incompressible" material, but engineering relies on the knowledge of the simplifications you can introduce on a given problem's analysis without compromising the nature of the conclusions. So if you want to conclude wether a helium filled ballon floats to the front or to the rear of an accelerating vehicle you can treat both air and helium as being incompressible without damaging the quality of the conclusions you come to.

In fact there is no such a thing as a totally non-compressible material!

This is so obvious that it hardly requires and exclamation point! There are commonly understood engineering definitions for compressible and non compressible materials (and phases of materials). As I mentioned elsewhere, a bag filled with a liter of air at sea level will have lost more than 99% of its volume if taken down to the Gulf oil leak site. A similar bag filled with water will have lost less than 1%.

My body possesses no reliable sensors that can detect the .001% compression of water at normal swimming depths, but I can easily sense the pressure change. As far as my body can tell, the water might just as well be completely non-compressible.

A pressure gradient does not require compressibility.

My point here is that fluids considered by convention to be non-compressible can and do experience pressure gradients (an obvious one being the pressure vs depth gradient you mentioned.)

OK, so we agree that inertia sends the air particles toward the back of the car. It is also the inertia of the balloon that sends it toward the back of the car - instintaneously. It will take some measure of time for the particles to hit the back window and create this (seemingly very strong) pressure gradient. Even if that is the case, why wouldn't the air particles rushing toward the rear cause a pressure gradient that would carry the balloon along with it? The balloon moves backward relative to the car.

Face it folks, sometimes what's counter-intuitive is just plain wrong!

Bingo! There's a very simple experiment you can safely perform at home with a higher difference in densities: take one pingpong ball and a transparent recipient with a flat lid full of water (to the brim). close the lid with the pingpong ball inside the recipient and gently apply thrust in any direction. Watch the pingpong ball float "against" the intuitive "natural" motion.

...If the car were accelerating at a rate of one G, the balloon would move forward at roughly the same rate ...

I just noticed that. Sorry, for me being so picky, but the froward acceleration of the balloon won't be even close to the car's acceleration.

The force that applies to the balloon is the buoyant force F_b=m_a.a minus the inertial force F_i=(m_b+m_h).a, where m_a is the mass of the displaced air, m_b is the mass of the balloon and m_h is the mass of helium. I consider at this point the moment the balloon starts to move, otherwise additional force will appear due to the resistance of the air.

The mass that gets to move is the total mass of the balloon + helium and the mass of the displaced air (which needs to cover the empty space the balloon leaves behind its track), i.e. m_a+m_b+m_h therefore the acceleration of the balloon will be:

a_balloon = - ( m_a-m_b-m_h / m_a+m_b+m_h ).a_car or, if ρ represents density, and V volume and we set we A=ρ_a/ρ_h and B=m_b/ρ_hV then we get:

a_balloon = - (A-B-1 / A+B+1) . a_car

the acceleration of the balloon will tend to be equal to the car's acceleration, only when B is negligible and A is big. After a quick google referencing, A=ρ_a/ρ_h for air and helium is about 7, so the maximum acceleration we may have (i.e. when B=0) is about the 3/4 of the car's acceleration.

As I mentioned before, after the balloon starts moving considerably, this acceleration will be further reduced due to the resistance of the air.

I don't understand this. Why can't you treat the balloon and helium as one object, with a density of say, half that of air?

Why can't you treat the balloon and helium as one object...

Because that would be too logical, simple and straightforward, and would not provide for the use of unnecessary math.

I don't understand this. Why can't you treat the balloon and helium as one object, with a density of say, half that of air?

&

Because that would be too logical, simple and straightforward, and would not provide for the use of unnecessary math.

Fist of all, what I don't understand, is what it is that you don't understand... I don't think that I used any awesome maths, it's just school stuff. If you get allergic to even that is another thing, but you can at least try to explain to me what exactly I did wrong.

Secondly, I tried to answer to a quiz question, giving as much insight to this as I could. I think everybody here tries to do that either successfully, either not, but I think that nobody so far made a rough estimation of the acceleration that the balloon experiences. If you think that I missed something, I'm eager to read your opinions.

Finally, the reason I treated the balloon and the air as separate objects is that two masses can vary independently from each other and can take any values, with the constraint that the balloon is stuck at the ceiling of course. Moreover, the two masses are treated differently, the m_b is taken as it is, while the m_h is function of the balloon's volume. So what you need to know (given my "advanced math"), is the ratio of densities between air and helium (or whatever other fluids) the mass of the container (which can be weighed) and the volume of the balloon (which is easier to find, instead of having to weigh helium or any other fluid at hand).

What I don't understand is why I should take the combined helium+balloon object as one having the density half of that of air. I cannot see how this makes any logic. I'm not a chemist, but I tried google to check the densities of air and helium. Maybe you found some different values yourself? And anyway, if you need to get the averaged density of the balloon+helium object, don't you need to know the balloon and helium masses in the first place? Otherwise, you need a balance inside a vacuum to get the total weight of the balloon; if you get one such device, then go along...

Let's try to cut to the nub of this. Suppose you have a metal ball with a density exactly twice that of water suspended in a pool. You let it go; what's its initial instantaneous acceleration.

I would have guessed at ½g. It's easy to see that it very quickly reduces to considerably less than ½g, but, I'm not sure if the need to displace the water has an immediate effect.

I think I can see arguments both ways.

Look at all the forces acting on the ball, and look at it's mass. Easy.

But, for the ball to move you have to shove some water out of the way. Not so easy.

I strongly suspect that because the ball is stationary at the start, it can effectively start to accelerate without actually moving. But in reality the other forces would start to act so quickly that the period of maximum acceleration would be imperceptible.

But, for the ball to move you have to shove some water out of the way. Not so easy.

In #192 I mentioned the "Atwoods machine analogy", which shows what inertial mass to consider in order to calculate motion. Of course, this model assumes slow movement, so that no much turbulence is created from the air trying to fill the gap behind the balloon. And of course, one should additionally account for the drag. But the point is that the acceleration will start at time zero at the maximum (no drag due to air resistance or turbulent flow) and fall as the balloon moves. In any case, the maximum acceleration (the one at time zero) will be about 3/4 of the car's acceleration for a helium balloon, provided that I didn't miss anything else.

OK: thanks you've convinced me.

What I don't understand is why I should take the combined helium+balloon object as one having the density half of that of air.

Randall wrote "as one object, with a density of say, half that of air?" His word "say" means "for example." It does not mean that you should take the density of mass of this particular balloon as being 1/2 that of the air. But using even a balloon and contents density of 1/2 half that of air produces the correct answer to the challenge question. It is simple enough without being overly simple.

To "weigh" the combination of balloon rubber and its helium, a simple measurement of the buoyant force with a spring scale would suffice. A balance inside a vacuum is unnecessarily complicated.

His point is, I think, that in ordinary calculations of buoyant effect only the mass of the "buoyant object" (in this case, the combination of balloon rubber and helium) and its volume need be considered. A ship is made of a great many components with different densities, but to determine the location of the waterline, you need only know the mass of the entire ship and its contents: one need not individually weigh the air, the wood, the steel, the plastics, etc in the ship.

I cannot see how this makes any logic.

It is a very logical and reasonable simplification that is used almost universally in measuring buoyant force. For the purposes of this challenge problem, which is a very simple demonstration of buoyancy (for which there is a near perfect-perfect analogy seen in sliding a bubble balance along a flat surface) one can answer the question correctly (as MB did in post 5, as usbport did in post 9, as velisj did in post 24 [by analogy], and as JRW did in post 25) any assumption that puts the density of the balloon as an object (consisting of the skin and its filling) at a significantly lower value that that for air will yield the correct answer, namely that the balloon tends to move forward as the vehicle accelerates. Such assumptions are both reasonable and logical.

His word "say" means "for example

I know, I was just curious why we should take the density "as half that of air" or anything else for that matter. The whole point is how you measure the averaged density of the balloon, without knowing the mass of the balloon. But anyway, it doesn't matter, it's just an engineering problem having to do with how you get your data: You may be able to measure the volume, or the helium mass (in case you fill the balloon yourself) or use a balance in vacuum or whatever. One can twist the good-old buoyancy formulas to his liking and get results. Twisting the formulas is school stuff and there is no point to argue on who has the best one: my only effort lately was to give an estimation of the maximum acceleration that the balloon will experience. That's all.

To "weigh" the combination of balloon rubber and its helium, a simple measurement of the buoyant force with a spring scale would suffice

As for that, the scale will only measure the difference between the buoyant force and the balloon's total weight. The first is dependent on the mass of the displaced air, i.e. you need to know the volume of the balloon! It's only when the balloon's weight is insignificant that you can cancel the volume factor and make your life easier.

To wrap it up, in time zero it holds:

a_balloon = m_a-m_total / m_a+m_total . a_car ('a' stands for displaced air, 'total' stands for balloon+helium, happy?)

Now, IF the balloon's material is insignificant this equals to 3/4 a_car It's so simple indeed. If on the other hand one wants to consider the weight of the balloon, things will get a little bit more complicated. I won't repeat myself here, otherwise I will get hammered for once more!

I don't think you can add the displaced mass of air to the balloon mass to account for the movement through the air. That is best handled by introducing a drag term which is proportional to velocity squared. So, even a high initial acceleration and velocity would be quickly dampened by drag. The time history of displacement, velocity and acceleration would be done in a stepwise manner as they are all related and changing. Fortunately, the buoyancy force is constant as it only depends on pressure gradient which only depends on car acceleration.

I don't think you can add the displaced mass of air to the balloon mass...

But I think this is the case! As a quick test, try to eliminate this factor from the formula, and you will see some absurd results. I know because that's what I firstly thought myself until I got to a dead-end. To help you see it in a more engineering approach, consider the case of a massless balloon "filled" with a vacuum. This problem is equivalent with a balloon filled with air moving to the other direction, isn't it? So the displaced air is factor that we need to take into account.

Then, the final confirmation comes from wiki: Check in particular the "Atwoods machine analogy".

Of course I didn't take into account the drag, but I wanted to find the maximum acceleration, and that was right at the onset of motion. After the balloon starts to move it indeed experiences a drag (I mentioned that) analogous to the square of the velocity, as you say. Then, you will need some calculus to solve a differential equation. Not so difficult, but I'm a bit too rusty to do it. After all, one can also consider rotation of the balloon, friction due to contact with ceiling, change in it's volume due to pressure and the story can get long. What one knows for sure (unless I made a serious mistake in my maths) is that the acceleration - for the particular case of helium within air - won't exceed about 3/4 of that of the car's acceleration.

By removing the displaced mass of air from the equation, as I think you should, and using your density ratio of 7 to 1, I get an initial balloon acceleration at time 0 of a = 6g. At first blush this sounds ridiculous, but if you calculate the velocity at a small time step and the resulting drag (assuming a typical balloon size and drag coefficient) you find that the drag very quickly becomes large enough to the slow the balloon down dramatically. The buoyancy force is weak so the drag force can quickly counter it. It's like trying to throw a balloon in the air. It has low mass so it accelerates quickly, but very quickly slows down. It just doesn't want to go far very fast because of the drag and low inertia unlike a baseball. Without going through the time varying analysis it is difficult to predict the acceleration time history but I suspect is starts high and quickly approaches zero so the balloon just floats forward. I hesitate to show all the equations for this group, but they are really not complicated.

I hesitate to show all the equations for this group, but they are really not complicated.

In fact, they are very simple if presented in a spreadsheet: just a few formulas copied over and over for increments as small as you want to see. Observations of helium balloons rising supports your contention that the terminal speed, at which drag balances the net lifting force, is reached very soon after release.

It is easy to imagine an envelope for a helium balloon of one cubic meter volume that would have a mass of less than 100 grams: parachute cloth, thin plastic bag, etc. The mass of the helium would be about .18 kg. So the buoyant object mass would be .28 kg, whereas the mass displaced would be 1.2 kg. Using 10 m/s^2 as the acceleration due to gravity, the net buoyant force, in this example, would be roughly 12 N - 2.8 N = 9.2 N. Given F=MA, then 9.2 = .28 x 33. So the acceleration at time zero would be about 3.3 G. Therefore, your 6G looks about right, given that the 7:1 density ratio does not include the envelope.

By removing the displaced mass of air from the equation, ...

At first I did that myself. But then I checked the case where the gas in the balloon has a very small density, or even zero. The acceleration approaches (or equals) infinity! Of course the air drag will limit it somehow, but still it is an absurd result, because - as I already mentioned - one would expect this case to be equivalent with a balloon filled with air moving to the opposite direction, i.e. will have the same acceleration like that of the car. It's like holes traveling in a semiconductor: They behave exactly like the electrons, only that they move to the other direction. For one hole to go west, there needs to be one electron going east, consequently the inertial mass of the hole cannot be taken to be zero, but instead, it is equal to that of the electron!

That's the intuitive approach. Then I mentioned the equivalent machine depicted in the wiki page, which I think seals the case.

You can't have zero density or the balloon would collapse. The balloon pressure matches the local air pressure, so for a given temperature there is a density the balloon must have to support it's shape. A finite density or mass results in a finite acceleration, which is high at time 0 but very quickly decays because of drag. Neglecting drag of a balloon is neglecting a very obvious force. And it is easy to compute - why neglect it ?

if you forget the tension of the balloon than that's right - in any other case the pressure inside the balloon is higher than the air-pressure

You can't have zero density or the balloon would collapse.

Why the balloon needs to be liable to collapse"? Can't we device a balloon stiff enough to contain a perfect vacuum? It just needs to be strong enough to resist pressure equal to 1 atm. No big deal, except from making it light enough. Pressure in the balloon is not a factor in buoyancy, we only care about its volume along with the densities of the fluids.

Neglecting drag of a balloon is neglecting a very obvious force

When did I neglect the air drag? I repeat that what I needed to know is the MAXIMUM acceleration that happens at the onset of movement, where there is NO air drag. After the balloon moves, there is certainly an additional inhibiting force, but this just reduces the acceleration from the value I have calculated for time zero. Therefore at time zero the acceleration is 3/4 of that of the car, and this is the MAXIMUM as it then get lesser and lesser. If you insist to know the exact behavior of the balloon - and want to brush up your calculus skills - take acceleration as d²x/dt² the speed as dx/dt, set air drag to something proportional to the speed squared and solve the nice differential equation that turns up. But this is another issue whatsoever.

I also added somewhere else that if the balloon was filled with nothing but... nothing, i.e. a vacuum, then it would behave like a balloon filled with air being within empty space, that is, it would accelerate at car's acceleration (in the cars' framework) or in other words, stay still in the road's framework. No wonder, but why is that equivalence valid? If you add up (superimpose) the two cases i.e. a bubble moving in air and air moving in vacuum in opposite direction, you get the experiment at hand: At every time frame, you have a bubble moving leftwards to distance x and at the same time you have air moving rightwards to distance x to fill the gap behind. Of course, there is air drag to take into account, and this is due to the fact that the air will have to circumvent the balloon in order to find its place behind (and this creates some turbulence), but again, I disregard this drag for speeds close to zero. Alternatively, one can think of an extra thin balloon positioned in horizontal sense experiencing minimal air resistance. (So, forget about air resistance necessarily limiting enough an otherwise infinite acceleration).

Similarly, in case the balloon contains something else than vacuum, e.g. helium, then the equivalent problem is that of a balloon filled with an imaginary gas with density equal to the difference between air and helium moving inside a vacuum to the opposite direction.

By the way, what do you comment on the mechanical analogy depicted and explained in wiki?

I am unimpressed by the Atwood's machine analogy for a buoyant body under acceleration. I assume the equations came from T. Smid (ref. 5 in your link) who assumes that a buoyant body released from rest will have an initial acceleration of

a = F/(m+md) where F = buoyant force, m = mass of body, md = mass of displaced fluid

and not

a = F/m as most of us would assume.

Smid makes this assumption based on the idea that the buoyant force is not just acting on the body but it must act on the displaced fluid which resists acceleration because of inertia. If this assumption were valid it should hold true in any direction for any body moving through a fluid such as a tennis ball in flight, a canon ball, a torpedo or an airplane being launched from an aircraft carrier. No longer could we assume from Newton's law that the initial acceleration is the force on the body divided by its mass like we learned in school, but we must add a correction factor for the fluid that will soon be displaced by the body. Going one step further, if we released a body with weight W in air it's initial acceleration would not be g = W/m, but g = (W+w)/m. We wouldn't measure 9.807 m/s^2 at sea level but some other number dependent on the volume of the body dropped. This is absurd. The acceleration of gravity has been measured many times and tracks nicely with W/m for small drag.

So how do we deal with the displaced fluid that does indeed resist acceleration. As the moving body accelerates the displaced fluid flows around the body and reduces the net acceleration via drag. And how the fluid flows around is important. A cube would have flow separation, turbulence and high drag; an aerodynamic shaped body would have low drag. In small time steps we would compute the velocity, drag, resultant force, and new acceleration. Just like we have done for all kinematic problems. No need to add a phantom mass. (Smid's complaint that a buoyant body would have infinite acceleration if the mass went to zero is invalid because the buoyancy force goes to zero at the same rate. His energy analogy can also be resolved.)

To summarize, a buoyant body can have an initial acceleration much higher than g if it has small mass and decent volume. Of course it has to have enough mass to hold its shape in the fluid so there is a limit to possible maximum acceleration values at time zero. You can't have zero mass in a balloon. Even a rigid balloon supporting a vacuum would have mass because of the skin. Anyway, its motion is immediately dampened by drag. High acceleration at time t = 0, low acceleration at t = 0+. The complete time history of motion can be computed one step at a time (or via calculus).

Sorry, I meant to say Smid's assumption would lead to g = W / (m+md) for a free-falling object instead of the more common g = W / m.

No longer could we assume from Newton's law that the initial acceleration is the force on the body divided by its mass like we learned in school, but we must add a correction factor for the fluid that will soon be displaced by the body.

There is nothing that violates Newton's law: We account for all masses that move, whether this is the balloon or the displaced air. Both entities have mass, so Newton's law applies to both. There is no phantom mass. Now, how good is the Atwood's machine analogy in speed and acceleration anything else than almost zero is another point, but in any case one must take account of the displaced air that has to move; isn't this the way to truly apply Newton's law?

Newton's law applies even for significant speeds, only that one as to take account of an extra force: the air drag. You mention that this will eventually limit the acceleration of the balloon, that otherwise - according to your approach - would have an infinite acceleration. But check for a moment the case I mentioned in the previous post, that the balloon has a sausage (or better teardrop) shape which is very very thin and moves along its length. The air drag is thus minimal. Wouldn't such a balloon shoot through the front window and get into the outer-space in no time? Maybe I exaggerate, but you see my point.

Anyhow, in case the speed is almost zero, then we can assume that the Atwoods analogy is quite valid: the balloon advances for a distance during a given time, and at the same time a part of air moves magically from the front to the back of the balloon. This has nothing to do with drag. Two entities with mass take part in the process.

I stress that the air drag is due to the fact that the displaced air, doesn't just get it's place behind the balloon happily, but it has to do a tour around it. Of course it won't be the same atoms that will leave the front to get to the back, but one will push the other, causing some turbulence along their way, which will eventually distort the good old pressure map we were accustomed to when the balloon was stationary. This turbulence depends on an object's shape, that's why we have different drags for a tennis ball and an airplane, etc. Now, the way we generally formulate the air drag is as a force that is proportional to the object's speed squared (it's just a good approximation in most cases, nothing more), but the constant we need to use depends on the shape of the object and of course the characteristics of air - or of any other fluid - such as viscosity and density. In the end, inside this constant we already incorporate the effect of the motion of the air itself. After all, the drag is quite more significant than the effect of the displaced mass of air. That's why we don't consider this mass in the calculations when studying motion. In this case I agree with you that the mass of displaced air is nowhere to find in the formula (in fact it is either negligible or hidden inside the constant of the drag!) What we disagree is that you take this modeling of the air drag as the paradigm to analyze the motion at almost zero speeds. I think it is actually the other way round: At zero speeds everything seems to be analytic and can be set on paper to the last detail by using Newton laws, where in considerable speeds, we resort to using approximations due to our incapability to handle turbulence very well.

You can't have zero mass in a balloon. Even a rigid balloon supporting a vacuum would have mass because of the skin

You can think of a balloon big enough, so that the weight of the container is negligible as compared to that of the displaced air. (Volume increases faster than the surface!)

The acceleration of gravity has been measured many times and tracks nicely with W/m for small drag

Indeed acceleration of gravity has been measured, by either dropping an item in vacuum, or (much simpler) weighing a stationary object!

Well, you missed my key point that challenges the Atwood's machine analogy for an accelerating buoyant body. If the initial acceleration of a buoyant body must take into account the mass of the body plus the mass of the displaced fluid according to Atwood, then the same must be said for a body dropped in air. That is, the gravitational acceleration must be:

g = W/(m + md) according to Atwood's analogy

and not g = W/m as I was taught.

(W = weight, m = mass of body, md = mass of displaced air)

How could we determine the correct equation. We can measure two of the parameters and compute the third. For example, if I place an object on a spring scale I will measure its weight W, say 9.807 N. I can then drop the object and with high speed photography measure the acceleration rate g = 9.807 m/s^2. To calculate the mass I get:

m = W/g = 1 kg according to my equation, or

m = W/g - md = 1 - md kg according to Atwood's analogy

So, according to Atwood's analogy, the mass of a body depends on the mass of the displaced fluid, which depends on the body's volume. That doesn't make sense. We have to conclude that the Atwood's analogy for describing a body acceleration is invalid. The initial acceleration of a body in a fluid is just a = Ftotal /m and the initial acceleration of a buoyant body in water subject to buoyant force Fb is just a = (Fb - W) / m (which cannot become infinite because as m goes to zero, F and W go to zero at the same rate).

...the gravitational acceleration must be: g = W/(m + md) according to Atwood's analogy...and not g = W/m...

You also missed my point! The formula g = W/m indeed holds when there are no other parameters that affect a free fall. In the case we have an object dropping inside a fluid, then we don't have free fall. So g=W/m is not valid anymore. You have to take the generic Newton's law, i.e. substitute g (constant) for a (unknown variable), W for total F and m for total mass. You get to almost this formula you mention only when buoyancy is insignificant as compared to the object's weight and object is not allowed to run too fast so that air drag comes to play a role. Anyway, this is a conceptually wrong way to measure gravity.

the mass of a body depends on the mass of the displaced fluid

I don't claim, that the mass of the body gets altered, or that it depends on something else. What I say is that we need to take into account two masses, because two objects are in motion. We might argue whether Atwoods analogy (which seems to assume that displaced air magically moves through the balloon to get to the other side) is a good approximation (actually, it must be quite OK being in wiki), but nevertheless, we cannot do without taking account the displaced air in some way. Please check the analogy I proposed in another post with the positive holes (i.e. lack of electrons) moving in a semiconductor. Although holes are massless, they behave as if they had a mass equal to the electron they displace. They have an inertia that this mass dictates. Certainly, they cannot have infinite acceleration!

By the way, the only reason I used the Atwood's analogy was to show what would be the maximum acceleration, at time zero (i.e. at speed zero) where this analogy works. No air drag there, because there is no turbulence and consequent change of the pressure landscape. I have no wish to tell what happens after the balloon starts to move. This is difficult fluids mechanics, for which I have no clue. I'm not even sure about the drag being proportional to the square of the velocity. What I know is that the maximum acceleration is about 3/4 of that of the car and it will soon drop after the balloon starts to move.

There in one more problem with the Atwood's machine analogy for an accelerating buoyant body, which I believe came from the ideas of T. Smid. Smid assumes that when the resultant force is applied to the body in a fluid it acts on the mass of the body, m, plus the mass of the displaced fluid in contact with and just above the body, md. He adds those two masses (m+md) to find the acceleration using Newton's law (a = Ftotal / (m + md). But the displaced fluid mass above the body is in contact with a similar volume of fluid just above it, which in turn has another volume just above it, and so on all the way up to the surface. So why not say that the buoyant body is resisted by N volumes of displaced fluid ? The choice of one volume is quite arbitrary and not physically logical. Snid also pretends that his model is valid at time t = 0, but is no longer valid at t = 0+. But at time t = 0 the body is not moving so there can be no acceleration or resistance to it. The body has to move and push into the fluid to get resistance, even if the movement is only infinitesimal. You can put a force on the body, but only when you release it from rest and allow it to move can you get acceleration.

It is much more logical to consider the fluid mechanics. When the buoyant body is released from rest it pushes into the fluid with some initial acceleration a = Ftotal/m. At t = 0+ it immediately creates a positive pressure in the fluid above it and a negative pressure in the fluid it is leaving. The pressure difference forces the fluid to move around the body as it starts to rise. The pressure difference and fluid flow creates drag that reduces the body acceleration. Drag is configuration dependent and not easily defined analytically, but there is a huge database of drag coefficients as a function of body shape and Reynolds number. Thus we only have to estimate the drag and compute the acceleration change as the body moves. And yes, drag does vary with velocity squared.

Incidentally, air is a fluid so free-fall in air is analogous to free-fall (or free-rise) in water, only the accelerations are much different. Buoyancy forces can be important or unimportant in either case.

So why not say that the buoyant body is resisted by N volumes of displaced fluid ?

I'm afraid I didn't get that very well. I think that the volume (i.e mass) we need to account for, is that volume pushed off - and at the same time - left behind by an object as it moves through the fluid. It is physically sane to say so. Why we need to add up more volumes to this? In any case, the object will sense a force by the molecules that directly touch it, not all molecules all the way along. Of course there will be an accumulative effect which translates into different pressure at various depths, but this has already been accounted for when we calculated buoyancy, didn't we?

Anyway, I think the hole-electron analogy gives clearly the mechanics of the motion of something inside something else: The displaced something has to be taken into account... The only difference with the situation at hand, i.e. the case when the moving object is not a point mass but has a volume, is that the fluid particles don't just move through the object to get the the other side, but have to make the tour around he object. But after I gave a second though to that, I don't find this problematic: The mass of the displaced fluid - which goes to find its way to the back of the moving object - moves along the axis of motion as if it passed through the object. On the axis perpendicular to that of motion it also does move up and down (or down and up) but this motion at infinitecimally small speeds shouldn't concern us so much! That's because at small speeds it is expected that no turbulence or significant pressure anomalies will appear, so no extra forces will appear in the scene along the axis of motion - which is the only of any interest to us. At higher velocities, indeed the pressure landscape changes, but as I already said, I don't give a damn, because I already know that it will eventually inhibit the acceleration I calculated at time zero which I claimed is the maximum.

As for the proportionality of the air drag to the velocity squared, this hold only at considerable speeds and it is anyway an approximation. That's why I said that I'm not sure about the validity of this model at all circumstances. Certainly it's not valid at speeds close to zero, where the Atwood's analogy - not denying that it is an approximation itself - fits better.

Finally, at time zero, we can indeed have acceleration. If you have trouble to visualize this, take for instance a time period dt which you can make smaller and smaller until you get something that is zero enough for you. Or otherwise, see it this way: At the very moment you release a stone from your hand, it will have zero velocity but the acceleration of gravity is already there.