Imagine upon the equator of a perfectly spherical Earth of uniform density rests a railway track fully encircling the globe. Upon the track is a stationary train. Which is to say that the train is at rest relative to the track.
A particular sealed carriage of cubic proportions is empty of furnishings, passengers and all other material distractions, except for an atmosphere identical in all respects, including the barometric pressure gradient, to that which is external to the carriage, and except for an unrestrained helium filled balloon.
The balloon is inflated with that precise pressure that enables it to remain still, bouyant and at rest with respect to the carriage in a position exactly in the centre of the carriage.
The train then moves off maintaining an acceleration of 1g.
Ignoring relativistic effects for the time being, describe and explain the motion, if any, of the balloon with respect to the carriage.
The courageous (count me out) may also if they wish describe the behaviour of the balloon as relativistic effects become significant.:)

Wey man thats easy. The bloon gans neewhere cos it's like murshunless as the air inside the carrudge is canny still. As the train gans aroond the uckwaita, the bloon gans up and doon as the hurizun changes.
U culd aalways fill the bloon with watta and fling the fkuca oot the winda like. Nee mair bloon. Ne mair problim like. ???:ok:


Divvent knaa nowt aboot the rellytevisticness tho. Is that how far it'd stretch with watter in it?

Ok, in for a penny in for a pound.

At first the balloon will drift to the back of the carriage at an acceleration of 1g and seem to stick to the rear wall (obvious).

There it will remain until you answer the following question. :)

Does the train reach orbital velocity and thus start to attain a low but increasing orbit until it breaks clear of Earth's gravity or are we to assume something is keeping it on the tracks regardless of the speed the train is going ?


NOOOOO!!!

You bloody fool Blackace, Rainbow you cool dude you.

The balloon will move forwards, that's because the air is denser and will move to the back of the carriage pushing the balloon forwards. There it will stay until you answer the above question. :)

Imagine the air as if it were water, except much more compressible, with a pressurwe gradient. Also remember that the balloon is balanced at a point where the air density exactly counteracts the pull of the earth's downward G force. Remember that gravity and acceleration are felt as the same thing.

The carriage starts to accelerate upwards, as it does so it pushes against the air which starts to compress and sunk downwards, at which stage the balloon also sinks downwards.

If the acceleration was sufficiently severe, e.g. several hundred gravities, the air would compress to a solid against the rear wall and the balloon would stretch out flat until it reached it's limit of elasticity and burst, with the helium then itself compressing to a solid layer.

Assuming a lower acceleration, at some stage the air collapsing downwards would reach a density where it would rebound off the rear wall and pass back through the remaining descending air, flow around the balloon until the density pressing against the balloon at any time would overcome the balloon's downward acceleration and push it up again. The pressre would would then agin collapse and move down wards until the pressure built up again.

Effectively, the balloon would move down towards the rear wall and then oscillate up and down a few times until settling at a new, lower, level, when the new pressure gradient of the air stabilised against the new G force.

p.s. Relitavistic affects relate to speed, not acceleration, so the above affects will not change, unless the rate off acceleration changes.

If the rate of acceleration increases when approaching a significant proportion of the speed of light, in itself no mean trick with the mass of the carriage also having increased to an almost infinite weight, exactly the same thing will happen, except to an outside observer it will seem to happen.....very....very.....sloooooowwwwwwllllllyyyyy....

Is the track on a conveyor belt? Would that make the answer different?

If the balloon is anchored (i.e. string) , it will move forward.
Did this in a car once. It's counter-intuitive.

But what happens if a fly "bimbles into the front of the train at say, 5 m.p.h?"

kansasw - Thanks for a coffee/monitor moment!

Rainbow... you lost me at 'equator' :eek:

The intuitive answer is that the balloon remains suspended where it was as the 'air' mass inside the carriage is accelerated 'as a whole' with the acceleration of the carriage. However, how is the air mass inside the carriage accelerated? The only force acting on the internal air mass is the back wall of the carriage - and that stays at the back.

Can of worms really-

1) Is it 'air' as we know it? We assume so.

2) Is the density of air in the carriage uniform? i.e. it's not warmer at the top

3) Is the 'rear' elevation wall of the carriage a perfectly uniform and flat surface? (and perpendicular to the acceleration force)

4) Is the Balloon 'balloon shaped' or does it have a perfectly uniform flat surface exactly parallel to the theoretically uniform total (back of carriage shaped) force of air molecules (having been accelerated from the back wall of the carriage with 100% consistency) that will accelerate the balloon?

It's seductive to argue the balloon would stay in situ as the back wall of the carriage would accelerate the carriage air mass as a whole and as the mass of the combined balloon/helium material is equal to the equivalent mass of 'air' it it displaces it would do the same.

However, the mass of air molecules that encounter the rear facing surface of the balloon will be initially hitting a mass of balloon material (not helium) that is far denser (no helium no floating) and so will NOT accelerate the mass in situ with the less dense air around it it - regardless of the aerodynamic effect of the balloon shape. Consider a membrane of tissue paper in the middle of the carriage that exactly fitted the compartment.

My take is that the balloon would slowly drift to the back and stay there. Whether it gets as far as the back of the carriage before the whole system recovered equilibrium depends on the size/shape of the balloon, the size/shape of the carriage, the density of the balloon material itself and the overall density of the atmosphere in the carriage...

...and probably many other things too.

Hmmm,if one had a object travelling east at the speed of light and another at a distance of roughly 186,000 miles or one light second traveling west toward it, at the speed of light, what is their closing speed, and when will they collide?
:rolleyes:

According to Einstein they are still only closing at the speed of light. When they collide is meaningless beyond the term 'simultaneously'...

I stand by my original corrected supposition that as the train moves forward the balloon does also, it will stick to the front of the carriage (the direction the carriage is travelling in) and remain there until the rest of the question is explained in greater detail as per the following.

ie.

1. Are there any ropes or holes of a description infinite involved ?

2. Is there a horse anywhere ?

3. Is Tony Draper an admin and thus can I be banned for upsetting him ?

4. Is the train stuck to the track or can it take off when it reaches orbit velocity ?

I'm keeping the ball, its mine. What you do to me at playtime is your decision and could be the subject of a serious sexual assault case if the need arises.

The balloon is inflated with that precise pressure that enables it to remain still, bouyant and at rest with respect to the carriage in a position exactly in the centre of the carriage.


How's that again?

The bouyant part tells me it will (unless otherwise restrained) rest against the ceiling of the carriage.

For the balloon to float unrestrained means that it has been ballasted to neutral bouyancy, and thus Archimedes will have a Principle say in its dynamic motion. It will remain in the center of the carriage. But, it may wobble a bit since its center of mass is below the center of bouyancy.

If the balloon is anchored (i.e. string) , it will move forward.
Did this in a car once. It's counter-intuitive.
Not counter-intuitive to me, although it might be a surprise for a moment.
The air mass in the car tries to move aft because the acceleration vector of gravity is now rotated so it points down and AFT.
The bouyancy vector of the balloon is opposite the acceleration vector, i.e. up and FORWARD.
Great little thought problem! :ok:

what is their closing speed, and when will they collide?Closing velocity is given as c.

Time being a route dependent variable, there is no universal time constant and precisely when the event occurs will depend upon the relative position, velocity and direction of the observer.

For the two train drivers it will be mercifully quick and they will neither see it coming nor feel anything. ;)

The air mass in the car tries to move aft because the acceleration vector of gravity is now rotated so it points down and AFT.

IMHO this is getting near the mark.

The experiment has been set up so that the balloon stays put in the middle. For it to do this, the weight of the elastic material must have been taken into account. So, a softish-- but more or less spherical balloon.

The carriage soon settles into a 1g man made longitudinal acceleration which it then maintains, but the standard G is still there, therefore the air will now have a pressure gradient fore and aft, and also from top to bottom. The balloon will seek a position in this vectored pressure point which will be slightly to the rear. The vertical gradient will settle to its original state

Its position will be analogous to the seas on our planet. These are not simply pulled by the moon, but by gravitational sheer. The pressure gradient is akin to that sheer.

Wibble

The real issue here is the application of Newtonian physics. This is clearly too simplistic for such a complex issue and one needs a more relativistic version:
Newton’ Second Law: F=ma.
This relates F and a via the mass m of the object being accelerated which in this case is both the train and the balloon. Since m is a numerical invariant the force on the mass is always parallel to the resultant acceleration.
Once into special relativity, of course, things become slightly more complex. The Force turns out not always to be parallel to the Acceleration a! Using a matrix notation it becomes clear:
. Let m be the invariant mass, v be the velocity as a column vector (whose entries are expressed as fractions of c and whose magnitude is the speed v as a fraction of c), let vt be the velocity as a row vector, and let 1 be the 3 x 3 identity matrix. Also let the Greek "gamma" be γ = (1 - v2) -1/2. Then the actual result turns out to be
F = γ m (1 + γ2 v vt) a
and
a = (1 - v vt) F / (γ m)
It is therefore quite clear that the balloon will move forward and sideways with a greater sideways vector at speeds approaching c.
However, this also demonstrates that at c, and speeds in excess of c, the balloon will become a pink elephant and cease to be airborne.
All this, naturally makes the assumption that there is no leakage of helium from the balloon whch would not be, in fact, the case. Given the semi permeable nature of the separaing film between the helium atmosphere and the surrounding air (mostly N2) combined with the fact that the helium atmosphere will be under greater pressure, one could expect the balloon to lose buoyancy, the resultant increase in relative density turning it into a flat pink elephant at relativistic speeds.

But, if the ballon exited an open window, would it circle the globe before the train?

Boy,

You guys have WAY too much time on your hands! :rolleyes:

Love

LP

Now, if we change the acceleration inside the train, the density gradient changes slightly - so air moves slightly, and the balloon moves with it.

Correct, as the train in constantly accelerating at 1g, the air density gradient will be from the front of the train to the back, with the denser elements in the air tending to pool at the back. The air near the back wall will be denser than the air on the front wall.

As the balloon is filled with Helium, it is less dense than the atmosphere surrounding it and it will tend to float to the least dense area of the atmosphere.

so as the train accelerates the balloon will move forward in the direction of that acceleration.

I still want to know if the train is fixed to the track or can it get to its orbit velocity and begin to rise ?

If the train is fixed as its velocity increases there will be a centrifugal force exerted on the atmosphere in the train, that will force the dense air up to the ceiling and therefore the balloon will then roll down the rear wall to the floor. So it will be stuck on the rear wall at floor level.

Why not just tie it to the nearest seat rail? It will remain there at the end of its string until the ticket collector comes round and chastises the owner for being so stupid as to bring a great big balloon on the train..

so as the train accelerates the balloon will move forward in the direction of that acceleration.


No no no!!! With respect, you have fallen into the trap for which this question was devised.

The new longitudinal gradient will cause the equilibrium point to be moved rearward...so the balloon will move to that point.





Let m be the invariant mass, v be the velocity as a column vector (whose entries are expressed as fractions of c and whose magnitude is the speed v as a fraction of c), let vt be the velocity as a row vector, and let 1 be the 3 x 3 identity matrix. Also let the Greek "gamma" be ? = (1 - v2) -1/2. Then the actual result turns out to be
F = ? m (1 + ?2 v vt) a
and
a = (1 - v vt) F / (? m)
It is therefore quite clear that the balloon will move forward and sideways with a greater sideways vector at speeds approaching c.


I was coming to that!!! :*

The new longitudinal gradient will cause the equilibrium point to be moved rearward

Fiendish. Not many options but there is always hope. OK, OK, decision time - risky I know, but:

CHANCERY LANE

Let's idealise and simplify the experimental apparatus, and indeed the experiment itself, while attempting to maintain the integrity of the original question.
Magically, the balloon is ideal in that it is perfectly elastic, impermeable and massless. Sadly, the train derailed well before relativistic effects became measureable, and well before any forces inclining it into orbit became measurable. Centrifugal force, if such a notion exists, was not apparent here.
Happily, the experiment survived for more than sufficient time for remote observors to compare and record the behavior of the balloon while the carriage was at rest, and while the carriage was well maintaining a 1g acceleration from west to east along the track.
For the duration of the experiment, strangely, the Earth stood still relative to the cosmos. You get that...
Maybe this will help..I need all the help I can get. The air pressure gradient within the carriage at rest is detirmined by the Earth's natural gravitational constant force of acceleration '1g'. Which tends to cause the air pressure on the floor of the carriage to be greater than that near the roof. Our balloon found its 'at rest' point, or has been said quite well, its 'equilibrium point' midway between the two, with the carriage at rest.
Now let's begin and maintain an acceleration of our carriage also of 1g towards the east. Which you would think would tend to pile up air pressure along the west wall of the carriage.
So we have now two forces of acceleration acting on the atmospheric pressure gradient influencing our balloon within our carriage; each of equal magnitude, but pependicular to each other. As has been said, it rotates. But does it migrate taking our balloon with it?
Sum the two forces of acceleration...surprisingly the sum does not equal 2g.
Good grief, enough for now. Are we on the right track?:)

No no no!!! With respect, you have fallen into the trap for which this question was devised.

The new longitudinal gradient will cause the equilibrium point to be moved rearward...so the balloon will move to that point.

LOL this is a very old question and can be proved by experiment, just go out and buy a helium balloon and get on a train with it.

You will see exactly what this guy saw....

I was riding the subway yesterday on the way back from a concert, and I saw a girl holding a helium balloon and when the train started moving the balloon moved forward with the train, I thought that because of the acceleration and the inertia the balloon would move back, but it actually went forward. When the train stopped also that balloon moved back not forward like everything else in my section of the train. What was going on there???

Why does a helium balloon float in the first place?

The heavier air (atomic mass of nearly 60) displaces the lighter Helium (atomic mass of 4).

Same thing happens when the subway accelerates. All the gases, like everything else, are pushed to the back of the subway, but the heavier air displaces the Helium. The helium balloon is floating the same as it ever was - away from the accelerating force, whether it be gravity or the force from an accelerating train.

Thank god experimentation still proves something.

The cited experimental evidence (fakepilot saw the same thing in post #7) proves only that a TETHERED balloon behaves that way, not that a free neutrally-bouyant one will. :}

Le Pen

You guys have WAY too much time


Absolutely not!

You are clearly still wrapped in your belief that Euclydian and Newtonian theory still apply!

At sub light speed (let’s call this c-) time, although a variable remains a known factor in the equation. However at speeds close to c and c+ it can become an unknown variable in the equation. Einstein introduced an alternative form of the field equations to accommodate a static universe solution in his theory.

Of course, it's all relative really!:E

Of course, it's all relative really! :E


So are aunties!! :}

It was all going so well, no conveyor belts....well, not many, and plenty of hot air, just what's needed to get to the nitty-gritty. But then...........


>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>But does it migrate taking our balloon with it?
Sum the two forces of acceleration...surprisingly the sum does not equal 2g.
Good grief, enough for now. Are we on the right track?

I poo poo your argument...and even poo-poo, well in advance, any riposte that you may contemplating. The balloon's final vertical position will be unchanged; it is still subject to G only, no g-force, so summing of the vectors is a red-herring. It is only the longitudinal pressure gradient that will be affected and the equilibrium point will be pressed to the rear....or west if you like. :8

proves only that a TETHERED balloon behaves that way, not that a free neutrally-buoyant one will

I see your point, yes I missed that. Thinking, Be Right Back. :)

OK here goes.

I think Its the buoyancy point that's required.

Here was my train of thought....

If you had a rectangular box with the balloon floating exactly in the middle and you turned that box through 90 degrees so its on its edge, would the balloon move. Answer no.

If you then accelerated that rectangle upwards so the force pulling the air down is greater than 1g would the balloon move. well the density gradient is now greater so the air at the bottom is now more dense than it was before so yes it will, it will move slightly upwards because it floats. Its the same as a submarine that will move up in the ocean if you increase gravity and hence the density of the water surrounding it.

Returning to our problem then, the only two things I need to know are what is the value of the accelerating force and what is its vector.

In our accellerating train its 1 g acting down and 1 g acting backwards, the total force is therefore 1.4 g and its vector is 135 degrees.

That vector is towards the rear bottom corner of the train so the air gradient will be densest there and as the air is now more dense than it was before the balloons buoyancy point will be raised slightly away from that vector.

So initially the balloon with move at an angle of 315 degrees, reach its new buoyancy point and then stop. So it moves slightly up and forwards.

As the train gets faster the force of gravity is gradually overcome by centrifugal force so now the only force is the accelerating force which is like us turning the original rectangle on its side, as we have already seen the buoyancy point here is in the dead centre. So as centrifugal force overcomes gravitational force the balloon will gradually migrate back to its original starting position.

As centrifugal force increases we get the opposite effect instead of 1 g we can assume -1 g the opposite of what we had before. the vector of the combined forces is now 45 and not 135 degrees, so the balloon will again gradually move away from its centre position and move down and slightly forwards , however this time as centrifugal force increases the balloon is gradually forced lower and lower until its resting on the floor.

Have I missed anything this time ?

If I was any good at math I might suggest the balloon describes a spiral.

Hmmm, one sees a flaw here,in order to drive the carridge at the speed of light the wheel rims would have to be traveling faster than light, and thats a no no.
:rolleyes:

As the train gets faster the force of gravity is gradually overcome by centrifugal force so now the only force is the accelerating force which is like us turning the original rectangle on its side, as we have already seen the buoyancy point here is in the dead centre. So as centrifugal force overcomes gravitational force the balloon will gradually migrate back to its original starting position.What causes the balloon to move rearwards to its original starting position in the centre of the carriage during this phase if the constant acceleration is maintaining the increased air density at the rear of the carriage? The shift of the vertical component of the density gradient due to centrifugal forces increasing surely interacts with but cannot undo that which the constant longitudinal component initially caused and maintains.

What causes the balloon to move rearwards to its original starting position in the centre of the carriage

the starting force is 1 g with the train stationary, this rises to 1.4 g at 135 degrees when the longitudinal force is applied, but then it returns to 1 g when centrifugal force is equal to gravitational force albeit now 90 degrees offset.

So, what forces the balloon to move back is the reducing air density gradient which eventually equals the initial air density gradient but now 90 degrees out.

If you agree that turning the rectangle on its side would not cause the balloon to move, then it must follow that the balloon will end up back at the starting point when centrifugal force equals gravitational force, as this is essentially all we have done.

The balloon will always move in the opposite direction to the accelerating vector, as centrifugal force gets greater than gravitational force and the longitudinal force remains at 1 g, the resulting vector will increase exponentially as it goes from 90 degrees to 0 degrees, at 0 degrees the length of the vector would be infinity. As the balloon must move in the opposite direction to this vector it will end up on the floor.

'S funny that. As soon as I looked at your post, even before beginning to read it, I saw what I'd missed. The zero gravitational point, when the original horizontal eqilibrium is replicated in a vertical plane due to centrifugal forces, and the longitudinal 1g taking its place at 90 degrees.

So, if the density vector shifts in an arc from 180 degrees (gravity and zero acceleration), back thru 90 degrees (gravity equalled by centrifugal forces, 1g acceleration at 90 degrees) and on up from there as centrifugal forces increase, but the vector never quite reaching 0 degrees as the accelerative force will always form some fractional component acting into it, what is the track of the balloon that allows it to come back through the same position?

Does it describe first an upward and forward elliptical path, coming back to its original position when the vector is at 90 degrees. Then a downward and forward elliptical path as the vector goes beyond 90 degrees. Then, when its close to its start position a second time, but unable to return to that exact position as the vector can never reach 0 degrees once the acceleration has begun, is driven, by ever increasing centrifugal forces shifting the density gradient upwards, downwards at a fractioanlly forward angle to a point on the floor very slightly ahead of its original starting point?

from what I can see it should be something like this. With the direction of travel of the train going from right to left.

http://blackaceproduction.com/Images/path.JPG

A = Starting point.
B = Longitudinal force 1 g, gravitational force equals 1 g.
A = (second time) Centrifugal force equals gravitational force, longitudinal force is still 1 g.
C = Centrifugal force much greater than longitudinal force which is still 1 g.

moves from A to B in a straight line as the longitudinal acceleration begins instantly. Then from B to A in an arc, then from A to C where its then on the floor of the carriage.

Of course someone's going to come in and shoot this down in flames as usual. :)

As I’ve understood, the vertical acceleration of gravity, acting on air in a cube shaped box, results in a pressure gradient from top to bottom in that box. By this I mean that the pressure of air at the top of the box is lower than that at the bottom. The pressure gradient is most likely NOT linear.

Increase the vertical acceleration, to 2g say, and there is a new pressure gradient. This time the pressure is even higher at the bottom of the box and lower at the top. The change in pressure at the middle of the box is hard to predict, without complex mathematics.

Decrease the acceleration, to 0g, and there is no pressure gradient, easy.

In our case we have two acceleration forces acting, 1g downwards and 1g rearwards. Combined, this results in an acceleration equal to the square root of 2, approximately 1.14g towards the rear bottom edge of the box. This is easy to work out by Pythagoras theorem. So now we have a pressure gradient which results in changes in air pressure in relation to position measured diagonally from the rear bottom edge of the box instead of from the bottom of the box as before.

Just as the balloon found the equilibrium point while the box was stationary, with the simple vertically varying pressure gradient, the balloon will find the new equilibrium point in the new pressure gradient, which is oriented diagonally towards the rear bottom edge of the box.

Now make a simplifying assumption that the balloon’s skin is not very elastic at all, like the silver helium filled balloons we see. Then, if the balloon finds itself in lower pressure air the balloon not change shape and will move towards the higher pressure air, ie rearward and downward. Conversely, if the balloon finds itself in higher pressure air, it will migrate towards the lower pressure air, ie forward and upward.

A balloon with perfectly elastic skin would change shape to suit the surrounding air pressure, rather than moving.

Neither a perfectly inelastic nor a perfectly elastic balloon exist and so the balloon both flexes AND moves when it finds itself in a different pressure of air.

Now the really difficult thing. It is very hard to predict the exact profile of the pressure gradient. The experiments reported so far appear to imply that the pressure gradient changes such that the balloon finds itself in higher pressure air and moves forwards and upwards (if not for the tether) towards lower pressure air. But of course, we have no idea where the balloons were positioned relative to the vessel (train carriage or car) in these experiments. In our experiment, the balloon is positioned at the centre of the box.

This is a question of fluid dynamics, which I’m not too clever with, but the basic idea is as I described. There is most likely a relatively basic formula for describing the profile of a pressure gradient given an acceleration, direction, size of vessel and fluid type. I don’t think we’re going to get to the bottom of this any time soon though!!!

Now the really difficult thing. It is very hard to predict the exact profile of the pressure gradient.

No it isn't, the lowest pressure in the train will always be 1 atmosphere unless its hermetically sealed and the question never said it was.

It follows then that any increase in pressure gradient must mean the air pressure at the centre must increase.

Look here under acceleration.

Left 1 Atmos, Right 1.02 Atmos....Middle = 1.01 Atmos.

Increase the accelleration....

Left 1 Atmos, Right 1.04 Atmos....Middle = 1.02 Atmos.

Decrease the acceleration, to 0g, and there is no pressure gradient, easy.

At no time does the force in this experiment drop below 1 g

A balloon with perfectly elastic skin would change shape to suit the surrounding air pressure, rather than moving.

Isn't a bubble of gas in liquid similar to our balloon ?

For those confused by the use of the term "hermatically sealed", it simply means "air tight".

I took A particular sealed carriage in the original post to mean that air could neither enter nor leave the carriage, ie, that the carriage is "hermetically sealed" or for the rest of us, "airtight".

Admittedly, if the carriage were not airtight, you would be right, the lowest pressure would be 1 atmosphere and the balloon would move forward and upwards.

The mention of the pressure dropping below 1g was just for example and to ease understanding...

In the case where the carriage IS airtight, as implied, the answer to the question relies on the resulting pressure gradient from the 1g horizontal accelleration combined with the 1g vertical accelleration. This is not easy to work out, and it is not linear as you suggest...

http://www.auf.asn.au/metimages/atmosdensity.gif

Fair point about the bubble comparison. I know what you're getting at and you're right.

Ahh my apologies, I never saw that.

In that case the pressure gradient would always remain the same at the centre regardless of the force implied wouldn't it ?

So the balloon would never move ?

God I am such an idiot at times.

This is of course assuming that the compression of the pressure gradient is linear which you say it isn't.

right, at a constant speed, the balloon goes nowhere in a relative sense, i.e. goes along with the train, the box, the air inside of it. They are all subjected to the same laws. Accelerate the whole shebang at the same time and there will be differences in relative position between the objects (matter) directly proportional to their mass... now prove me wrong if you can.

Compare the train carriage to the planet earth. Compare the back wall with the ground. Compare a small helium balloon weighing say 10 grammes with an enormous helium filled balloon weighing several tonnes. By now I hope it is obvious that both balloons could be inflated such that they float side by side at say 10 metres altitude.

By your argument, due to the different masses of the balloons, they would be in different positions, which is now clearly not true.

now prove me wrong if you can

I'll do my best :)

In light of what "Hoping" says I can see that the more force you put on the carriage in any direction the more rigidly stuck in the middle of the carriage the balloon will be.

More force = bigger pressure gradient.

even if you did reduce the force to 0 g the balloon would still not move as the whole thing is now weightless.

I agree stuff moves because of its mass, but by far the most massive object in the train is the atmosphere itself. As the balloon is buoyant in that atmosphere the movement due to its mass will be almost negligible. Its all relative.

chornedsnorkack

that's where i f***ed up and Hoping set me straight, it does say in the original question that it is sealed, this assumes no air can enter or escape.

Hence I've changed my opinion to one of complicated patterns of movement to one of the balloon remaining stuck in the middle of the carriage regardless of the forces involved and their direction. In fact I'm now in agreement with the first guy who posted an answer to this question :)

Complicated patterns of movement initially followed by an equilibrium where the balloon has found its rightful place in the carriage...

(I would love to have added that that place would be "first class" since the balloon is full of hot air, but then again, its full of helium...)

Firstly, in idealising the balloon I wanted to advance a notion that it is, among other things, perfectly deformable, yet I indicated wrongly that it is perfectly elastic. To correct my error; the surface area of the balloon is a constant. Much of my input here is inspired by heroic efforts above. By the way, the X's seem to be the only way I can paragraph my text...
X
For my own sake I'm going to attempt thinking outside the square, or in this case, the cube. Somewhat pressed for time so please forgive any typos, poor expression and repetitive argument. Not that things improve at other times..Thanks:)
X
Consider that our carriage has a twin; identical in every respect, alone and isolated deep in intergalactic space unaffected by any influence of any field including any gravitational field.
X
Magically, the twin is maintaining an acceleration of 1g by a mysterious kinetic force beneath it so that the twin is accelerating upwards at 1g. ('Beneath' & 'upwards' in intergalactic space?? It surprised me too..)
X
Now before we commence any east/west accelerations let's compare let's compare the circumstances or the set of events within the twin in space with that within its counterpart stationary on the tracks on Earth.
X
A hypothetical observor visiting would have an identical and indistinguishable experience within each of the carriages. He would not (ok.. she...) she would not detect any difference in her weight while in each of the carriages. She would not detect any difference in the air pressure gradient between each of the carriages. The balloon in each would be positioned nicely in the centre of each carriage.
X
They seem identical experiences because they are identical experiences. The frame of reference within each carriage is the same. It's the principle of equivalence. Irregardless of events and perspectives external to the carriages, events within each carriage, including the behaviour of idealised helium filled balloons, will be identical given identical forces acting upon each of the carriages.
X
Now let's concentrate upon our Space Carriage (SpaceCar) accelerating 'up' at 1g (within which the set of events are identical to our stationary earth bound twin) and then impose and maintain upon the SpaceCar an additional acceleration of 1g vectored from west to east (in intergalactic space...hmmm).
X
At the same time back on Earth the EarthCar moves off on the track and maintains an acceleration of 1g vectored west to east.
X
Now what are the event differences experienced by the SpaceCar interior as compared to the EarthCar interior? From a perspective within each Car the experience seems identical because it is identical. The forces and accelerations acting upon each are the same. The behaviour of the balloon within each car after the addition of the east/west acceleration will be the same.
X
So why add the SpaceCar to the mix? Because in my mind I find it easier to sum two accelerations acting upon the SpaceCar than those acting upon the EarthCar, even though the relevant accelerations are identical.
X
Summing the accelerations forced upon the SpaceCar we find we have a 'down' to 'up' acceleration of 1g, and we find we have a 'west' to 'east' acceleration of 1g. A parallelogram of vectors or Pythagoras as mentioned above will give us a single resultant vector solution.
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Namely, the two accelerations sum to a single acceleration equal to (square root of 2)g or approx 1.41g vectored from the western edge of the floor of the SpaceCar diagonally across and up through the eastern edge of the roof of the SpaceCar passing through the centre of the SpaceCar. The solution for the EarthCar is, need I say, identical.
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How do these new circumstance affect events within each carriage? (Recall that the effects are identical? I thought so..:) )
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What happens to the air pressure gradient? What happens to the air pressure gradient orientation relative to the carriage? What happens to the 'point of equilibrium'? What happens to the balloon? (Remember the balloon?):)
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I think that in a day or two I'm going to have to stick my neck out!:uhoh:

Flippin' ek !! Occam's razor comes to mind.

I think that in a day or two I'm going to have to stick my neck out!

MAUDE !! Bring me my best chopping axe.

Well I prefer Barry Scott's first answer ... and he can clean the carriage with Cilit Bang as a bonus

What happens to the air pressure gradient? What happens to the air pressure gradient orientation relative to the carriage? What happens to the 'point of equilibrium'? What happens to the balloon?

I've already decided it makes no difference, whatever force you apply and whatever direction that force comes from the balloon will remain stationary.

I will stay on that and wait your answer, I hope after all the time I spent thinking about this problem when you do stick your neck out it wont be a bloody affair.

Thanks Maude, no not just yet, but sharpen the blade a bit will you.

Sticks neck out.
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Let's simplify: What we have simply done to the set of events within the carriage, using the principle of equivalence, is to increase the gravitational constant of acceleration from 1g to 1.4g
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Now remember that the atmosphere within the carriage is in all respects identical to that external to it, including the barometric pressure gradient. So what need for the carriage? Or the train or track?
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Abracadabra and all of the experimental apparatus disappears except for the Earth, its atmosphere, and the balloon there in the air, all of which is motionless and stable, with the local gravitational constant remaining at 1g.
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I think you can guess what we are going to do next...
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Suddenly every specification describing the Earth remains the same except for its mass which increases by a factor of 1.4
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What are the consequences of this somewhat significant mass increase? Well, for a start we know that the force of gravitational attraction of a mass is proportional to its mass, so the Earth's gravitational constant has immediately increased from 1g to 1.4g
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What affect has this upon the atmosphere? Well, the effect will be to attract a lot more air closer to the Earth's surface, increasing its density there. (Can we leave aside thermal effects and say that this event is temperature neutral?)
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What affect will a local atmospheric density increase have on the behaviour of our helium filled balloon? Well, surely the balloon will become more bouyant in the thicker air, will rise, ascend, gain altitude; in a direction opposite to that of the gravitational attractive force and toward the ascended 'point of equilibrium'.
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How far will it rise? I don't know.
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Now, let's transpose this logic, such as it is, (with Maude at the grindstone) to our balloon in the carriage and predict:
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The balloon in the carriage, as the train moves off and maintains an acceleration of 1g, will migrate in the opposite direction to that of the vector of the sum of the actual and artificial attractive gravitational forces.
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The balloon migrates away from the thicker air towards the thinner air seeking its 'point of equilibrium' (ascending while moving in the direction of train travel) from its central position towards the centre of the eastern intersection of the carriage's eastern wall and the carriages eastern end of its ceiling. (Can I put it plainer?:) )
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How far does our balloon migrate, or seek to migrate within the carriage? I don't know.
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So, there it is, but where is Maude? I'm waiting, as the remedy for all ills awaits, for a potentially fatal flaw to be exposed in my argument, and for the life of me, I must prepare for it.:uhoh:

There is a fatal flaw I think. I was arguing all along exactly what you say, but "Hoping" made me realise that was wrong.

You say the air is made denser, yes we agree on that, if the container was leaky. Think about this.

At 1g.

The top of the container would be 1 Atmos and the bottom would be lets say 1.3 Atmos, the middle would be 1.15 Atmos, for arguments sake lets say our buoyancy point is 1.15 Atmos. So our balloon is in the middle.

If we now increase gravity the top would be 1 Atmos, the bottom would be higher, maybe 1.6 Atmos so the middle would be 1.3 Atmos, so our balloon is no longer at buoyancy and floats up to where the 1.15 Atmos is.

That is fine for a leaky capsule, but our capsule isn't leaky, you said yourself in the question it was sealed.

So now try the same with a sealed capsule.

At 1g

The top of the container would be 1 Atmos and the bottom would be lets say 1.3 Atmos, the middle would be 1.15 Atmos, So our balloon is in the middle.

If we now increase gravity, the bottom would be higher, maybe 1.6 Atmos, what would the top be ?

Its sealed so if the pressure went up at the bottom the pressure at the top must come down. The buoyancy point would remain exactly in the middle and the balloon wouldn't move. The pressure gradient has increased yes, but its balanced because its a sealed container.

So I think the balloon wouldn't move.

Some good points there blackace..
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As others have indicated, this is a dashed more challenging matter than it appears on the surface. It seems that only some pretty deep physics/fluid dynamics/computer modelling and/or an actual experiment can answer the conjectures.
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You know, nonetheless it's somewhat amazing that there is a wide variety and complexity of opinion with regard to the simple behavior of balloon in a box when we kick the box!
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Perhaps it all comes down simply to the size of the box.
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Let's allow that you're right in suggesting that the balloon remains stationary relative to the carriage (call it the small box of everyday carraige dimension of the order of some metres per edge) upon acclerations.
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Now, let's allow that my 'whole earth' model where there is no carriage but an increase to 1.4g causes the balloon to move, to rise in a thickening atmosphere.
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Now allow several experiments with a carriage of variable dimensions. Let's say the largest carriage is a cube 1000km to a side (yes, a thousand kilometres..) and call it the big box (on train 1g acellerated).
Well, that system would surely behave consistantly with my whole earth model and the balloon would move within the carriage as described in another post above.
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So, if you allow my big box no train/whole earth 1.4g model (balloon moves rel to box) , and if you allow your small box on train model (balloon doesn't move rel to box), then somewhere between these two extremes of box dimensions there's another box with dimensions that are 'just right'...
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..the on train Goldilocks box, where the balloon is at a tipping point between moving or not... I'm getting a headache...time for something simpler..
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On board was I on a loaded bulk iron ore carrier off the coast of Brazil steaming true north at 10 knots through a flat sea and southerly breeze of 10 knots approaching the line with some steel ball bearings in my pocket and as I crossed the equator I rolled one ball bearing along her centre line fore and aft towards the stern at 10 knots to the flat steel deck....:)
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(Why can I paragraph preview posts but not submitted posts? I would be grateful anyone..)

This might solve it rainbow, it is from matelot in the Computer Forum.

User Contol Panel/ Edit Options/ Misc. Options, then in the dropdown list select Enhanced Interfaced Full WYSIWYG Editing.

Thanks Henry Crun, here we go (para 1)

I hope it works (para 2)

If it does, then many thanks to matelot too (para 3)

Henry, you & matelot are very very clever, many thanks, rainbow:)

(If matelot ever wants to crew out of Rio...;) )

Lying awake stricken with insomnia one has come to the conclusion that this experiment has shot its bolt and therefore one proposes another,it won't be expensive we shall use the same kit,and for duration of the experiment we shall ignore such things as friction and the tensile strength of steel railway wheels ect as is customary.
In order to carry out of this thought experiment we have caused to be built a railway line with no gradients circling the entire globe at the equator ignoring such trivialities as mountain ranges oceans ect.
Now you have your railway carriage with yourself seated on the floor as observer, you can even have your helium balloon, though for our purpose said balloon its totaly superfluous.
The only difference is this railway carriage is firmly clamped to our shiny new equatorial perminent way,it is free to move forward but not sideways or up,now we apply your provobial 1g constant acceleration, now pretty soon your going lickety split, the clickety clack of the wheels has dopplered up through a buzz to a scream to way beyond human hearing, in a supprisingly short time at said 1g constant acceleration you achieve a forward velocity of 17,500 MPH,(this thought experiment is being carried in good honest Imperial measure).
This is of course escape velocity from our good earth, and indeed were the clamps to be released suddenly the carriage along with you and your balloon would shoot off at a tangent and achieve orbit, albeit just one orbit, as the laws of orbital mechanics indicate at perigee it would return to the railway track with a loud bang.
Anyway because your transport is firmly clamped to the rails orbit does not occur,and at the point when the railway carriage reaches escape velocity the acceleration suddenly ceases, now this is the question,if you reach down with the palms of your hands and give yourself a wee push in a upward direction,would you float off the floor in the manner of those orange boiler suited chaps in the vomit Comet?
hmmmm?
:rolleyes:

sticks neck out Mk11

Yes, for reasons that Mr Draper shall explain later:)

If we take a container, whether a train or a test tube and subject it to a constant acceleration for sufficient time, the particles with the greatest mass will settle at the end farthest from the acceleration force and the lighter particles will be displaced progressively towards the other end - as occurs in a centrifuge. But the train is not a centrifuge, as the acceleration is of limited duration; meanwhile, the postulation of giving the balloon neutral buoyancy means that the density of the balloon is necessarily the same as the surrounding air - regardless of its content. Though the helium displaces a lower mass than an equal volume of air, the postulated neutral buoyancy may only be achieved by constructing the balloon of material equal in weight to the difference in mass between the two gases. The balloon will therefore accelerate at the same rate as the surrounding air.

Sitting in the train, passengers' posteriors will acclerate at the same rate as the train. Their heads and upper bodies will be displaced by a rotating couple - those facing forwards will be pressed into their seats, those facing to the rear will find their upper bodies move away from the seat. Instinctive muscular reaction will restrain this movement. This effect is more easily seen when standing in a tube train and that is why they are equipped with roof straps.

Without the straps, the humans will tend to move towards the rear of the train - or if you like towards the bottom of a very large test tube in a centrifuge - because their mass is greater than that of the surrounding gas. If the train accelerated for long enough and the humans were unconscious, they would all move to the rear of the train and displace the air that was previously there towards the front.

Is this why everyone calls me an airhead?

Thanks for giving me a headache, Rainbow.

OK, Newton's first says the balloon won't move until a force is exerted on it.
So the instant the box starts to move, the balloon stays still, until the resulting compression is transmitted to it. So, initially, the balloon will move rearwards relative to the box.
Soon after, the horizontal pressure gradient will become equal to the vertical gradient due to g as previously described. (And is non-linear).

As you have constrained the surface area of the balloon, and thus the density, it will end up in the position of equal density in the new pressure distribution. This will be to the rear of the original center position.

It will undergo damped oscillations towards that equilibrium position.

OK. Maybe one little wobble until settling in the spot.

Paris, I thought I said that in post #4... ;)

Dear Paris Hilton

I regret that you are suffering a headache, but really, there's no need to thank me alone for it; consider it a consequence of this forum and all of its inhabitants, such as we are.

I would hope that your headache is nothing compared to the extreme migraine these few pages, over these few days, have caused me, and others it seems.

Should it be of any comfort to you, I have found that a cup of tea, an aspirin and a good lie down is at times a sufficient remedy for the outrages we collectively suffer here.

Yours faithfully
rainbow
;)

Thanks for the advice.

You could have just asked the standard question about the tethered helium balloon going the 'wrong' way in a car.

Now if the balloon were free to expand or contract, it would be a different scenario.

Oh no! Why did I say that?

Off to see my makeup artist.

http://i108.photobucket.com/albums/n28/cpt_jackson/Paris.jpg

I would hope that your headache is nothing compared to the extreme migraine these few pages, over these few days, have caused me, and others it seems.

Dear Mr Rainbow.

As the psychiatrist in charge of Mr Blackaces recovery let me first say he has at least stopped showing his private parts in public to the visitors.

Currently he is being held in a secure unit and he keeps mentioning your name. I would appreciate a visit from your good self as it would improve his chances of rehabilitation. He keeps saying he would love to see you, but he tends to get into a wild uncontrollable fit at the very mention of your name.

With Respect.

Dr Moonshine.

So, even Paris Hilton gets headaches...:hmm:

I knew she was too good to be true. ;)

But what happens if a fly "bimbles into the front of the train at say, 5 m.p.h?"

Simple, the fly, thinking it will make a smooth landing, will lower the undercart to settle onto the face of the train.

However because its eyesight is cr*p it will only notice its made a mistake and the train is really going at 65mph when the said fly's ar$e has gone through its eyeballs. :ok:

This is an attempt to answer Mr D’s postulation.

As the train accelerates towards near light speed (c-) on its improbable track, it will increase in mass if we apply Relativity Theory to the mass of the train which in this case is an unknown. And, what is more a variable unknown.

However, Newtonian Theory (one refuses to call it a Law) implies that this forward motion must be transferred to the planet Earth which at this point has a known mass.

As the train accelerates at a constant 1 g it’s mass will increase near to infinity – or certainly to a point where it greatly exceeds that of the earth.

The earth starts turning at an increased velocity. There will be a point at which the Earth and the train have the same velocity albeit in different directions.

Eventuially howver the mass of the train will be so great that the Earth will spin a near light speeds while the train remains stationary!

Now, if there were to be a fly that then hit the train window……….

:rolleyes:

I think -




occasionaly.

:\

Paris, why did you say that?

I have constrained the surface area of the balloon, but not 'thus' its density. The balloon is deformable, indeed perfectly so, but its surface area is finite.

Hence, while the balloon is not free to expand beyond its surface area, it is free to contract. These circumstances are not unlike balloons, released from ground level for scientific or meteorlogical purposes, which appear limp and flacid at first but magisterially expand when the altitude suits them.

I hope this clarification of initial conditions is of more help to you than a cup of tea, an aspirin and a good lie down, while considering Mr Draper's set of events, was of comfort to me.

I understand blackace is improving. Perhaps Dr Moonshine might be interested in a locum placement hereabouts...one wearies of aspirin.:)

OK. Enough. We will resolve this matter over a game of yahoo pool.

Rainbow. You're fired.

Took me long enough but I have finally come up with irrefutable proof that the balloon will not move.

Take the sealed capsule to the moon and drop it. according to physics (Galileo) all objects fall at the same rate regardless of the mass of the object.

The capsule, the air inside it, and the balloon, will all fall at the same rate together (the air inside is falling and now has a neutral pressure gradient).

If the balloon does not move (relative to the capsule) during a transition from +g to 0g, then it can not move when the transition is the other way from 0g to +g. Or 0g to 5g, or 1g to 5g etc. It makes no difference what the acceleration is, or what direction it is.

Why oh why did I not think of this sooner.

:ugh: