*********************The Math********************** The math problem is straight forward. I am looking for the solution space of a system of the form Ax=b. I know I can through algebra get the matrix in upper triangle form. However if one of my coefficients is zero then I know it is not a legitimate row operation to multiply a row by zero. Moreover, I am not sure if the final result I arrive at will be correct. I suppose a case method could be done but I was wondering if there was a more elegant way to approach the problem.
Perhaps there is a technique to find just one solution. Then I can use the fact that all solutions can be expressed as the sum of a solution to Ax=b and the solutions to Ax=0. The non trivial solutions to Ax=0 clearly occur when the determinate of A is equal to zero.
The idea is to move a force keep while keeping the magnitude and moment of the force the same. The moment about P(x,y,z) is equal to r x F where r is the vector going from P(x,y,z) to the point where the force (F) is acting.
Trying to keep the moment the same we get the equation: r_0 x F_0 = r_1 x F_1 This gives the system of linear equations:
I am trying to find the solution space of this system. I then hope to find a vector F_1 in this solution space with the same magnitude as the original vector F_0
************************************************************ For all curios the reason I am trying to solve this problem is I wish to model the shape a balloon will take. This is purely for my own interest. My field of study is electrical engineering not mechanical and I have a physics under grad.
In article <3B165D92.6CFDA...@hotmail.com>, John Creighton <JohnCreight...@hotmail.com> wrote:
>The idea is to move a force keep while keeping the magnitude and moment >of the force the same. The moment about P(x,y,z) is equal to r x F where >r is the vector going from P(x,y,z) to the point where the force (F) is >acting. >Trying to keep the moment the same we get the equation: >r_0 x F_0 = r_1 x F_1
OK, it's easier to do this geometrically rather than going to coordinates. Call the left side b. Since the cross product of two vectors is always orthogonal to those vectors, there is no solution unless b is orthogonal to r_1. If it is orthogonal to r_1 (and r_1 is not 0), then use the identity A x (B x C) = (A.C)B - (A.B)C: the general solution is
F_1 = |r_1|^(-2) (b x r_1) + c r_1
where c is an arbitrary constant.
Robert Israel isr...@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2
> *********************The Math********************** > The math problem is straight forward. I am looking for the solution > space of a system of the form Ax=b. I know I can through algebra get the > matrix in upper triangle form. However if one of my coefficients is zero > then I know it is not a legitimate row operation to multiply a row by > zero.
It is never legitimate to multiply a row by zero. This destroys the associated equation as a constraint to the solution.
> Perhaps there is a technique to find just one solution. Then I can use > the fact that all solutions can be expressed as the sum of a solution to > Ax=b and the solutions to Ax=0. The non trivial solutions to Ax=0 > clearly occur when the determinate of A is equal to zero.
Is r_1 known?
|F_1|=|F_0| is a nonlinear equation; linear algebra isn't going to help really.
> The idea is to move a force keep while keeping the magnitude and moment > of the force the same. The moment about P(x,y,z) is equal to r x F where > r is the vector going from P(x,y,z) to the point where the force (F) is > acting.
> Trying to keep the moment the same we get the equation: > r_0 x F_0 = r_1 x F_1
Let r_0, r_1, and F_0 be given. They must be in the same plane.
Rotate r_0 and F_0 as vectors in their plane until the line which extends parallel to F_0 and through the point r_0 passes through the point r_1. Then slide F_0 along this line until it acts at point r_1. The moment has not changed.
The solution is not unique; we may "flip" F_0 so that it points radially inward or outward, but ignoring this the solution ought to be unique.
Given |r_1| < |r_0|, a solution does not always exist either.
> > *********************The Math********************** > > The math problem is straight forward. I am looking for the solution > > space of a system of the form Ax=b. I know I can through algebra get the > > matrix in upper triangle form. However if one of my coefficients is zero > > then I know it is not a legitimate row operation to multiply a row by > > zero.
> It is never legitimate to multiply a row by zero. This destroys > the associated equation as a constraint to the solution.
> > Perhaps there is a technique to find just one solution. Then I can use > > the fact that all solutions can be expressed as the sum of a solution to > > Ax=b and the solutions to Ax=0. The non trivial solutions to Ax=0 > > clearly occur when the determinate of A is equal to zero.
> Is r_1 known?
> |F_1|=|F_0| is a nonlinear equation; linear algebra isn't going to help > really.
True, however once we solve r_0 x F_0 = r_1 x F_1 then we are left with a quadratic equation.
> > The idea is to move a force keep while keeping the magnitude and moment > > of the force the same. The moment about P(x,y,z) is equal to r x F where > > r is the vector going from P(x,y,z) to the point where the force (F) is > > acting.
> > Trying to keep the moment the same we get the equation: > > r_0 x F_0 = r_1 x F_1
> Let r_0, r_1, and F_0 be given. They must be in the same plane.
> Rotate r_0 and F_0 as vectors in their plane until the line which > extends parallel to F_0 and through the point r_0 passes through the > point r_1. Then slide F_0 along this line until it acts at point r_1. > The moment has not changed.
> The solution is not unique; we may "flip" F_0 so that it points radially > inward or outward, but ignoring this the solution ought to be unique.
> Given |r_1| < |r_0|, a solution does not always exist either.
This is distressing.
> g
I have to think both Robert and Gregory. Both of their responses where vary helpful and gave me much insight into the problem.
If you curios as to why I wanted to keep the magnitude of the force the same but change the moment read the following:
I had divided a sphere into a bunch of mesh points along lines of latitude and lines of longitude. I was able to know approximately the area of the sections that the mesh points formed the corners of by taking cross products. I wanted to consider the mesh points as particles which had forces acting on them based on pressure and the modules of elasticity of the material that formed the surface of the sphere. The force due to pressure is equal to the pressure change times the area. However the sections I took the area of did not have the mesh points at their center. I new I could consider an equivalent force going through the mesh points and its magnitude would be similar. However I did not want to introduce a net moment about the center of mass when I moved the forces.
Robert Israel wrote: > In article <3B165D92.6CFDA...@hotmail.com>, > John Creighton <JohnCreight...@hotmail.com> wrote:
> >The idea is to move a force keep while keeping the magnitude and moment > >of the force the same. The moment about P(x,y,z) is equal to r x F where > >r is the vector going from P(x,y,z) to the point where the force (F) is > >acting.
> >Trying to keep the moment the same we get the equation: > >r_0 x F_0 = r_1 x F_1
> OK, it's easier to do this geometrically rather than going to coordinates. > Call the left side b. Since the cross product of two vectors is always > orthogonal to those vectors, there is no solution unless b is orthogonal > to r_1. If it is orthogonal to r_1 (and r_1 is not 0), then use the > identity A x (B x C) = (A.C)B - (A.B)C: the general solution is
> F_1 = |r_1|^(-2) (b x r_1) + c r_1
> where c is an arbitrary constant.
Of courese once we put in the costraint |F_1|=|F_0| we find c= |F_o|*cos(apha)/|r1| Thus we arive at the equation |r_1|^(-2) (b x r_1) + (|F_o|*cos(apha)/|r1|) r_1=|Fo| Once apha is found we can then find F_1
> Robert Israel isr...@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2
JohnCreight...@hotmail.com wrote: >*********************The Math********************** >The math problem is straight forward. I am looking for the solution >space of a system of the form Ax=b. I know I can through algebra get the >matrix in upper triangle form. However if one of my coefficients is zero >then I know it is not a legitimate row operation to multiply a row by >zero. Moreover, I am not sure if the final result I arrive at will be >correct. I suppose a case method could be done but I was wondering if >there was a more elegant way to approach the problem.
>Perhaps there is a technique to find just one solution. Then I can use >the fact that all solutions can be expressed as the sum of a solution to >Ax=b and the solutions to Ax=0. The non trivial solutions to Ax=0 >clearly occur when the determinate of A is equal to zero.
Yes, that is true.
In general det|A| = 0 implies Ax is mapped into a subspace of the original vector space, so in general either there will be many solutions to Ax = b or none; a bunch of stuff is collapsed onto b, or nothing.
Ah, linear algebra.
>************************************************************ >For all curios the reason I am trying to solve this problem is I wish to >model the shape a balloon will take. This is purely for my own interest. >My field of study is electrical engineering not mechanical and I have a >physics under grad.
Hey, that is an interesting problem.
There was a kind of conundrum floating around here about one a while ago which caused me to have a brief insight event; but I'm better now.
Let's see if I can remember it: You have a helium filled gas bag in air, say. The zeroth order analysis is "air supports a weight equal to the weight of air displaced by the balloon".
That's buoyancy for you.
Then we might get curious about a more detailed analysis of the shape and forces on the envelope, as you are.
So how could we befuddle ourselves about this... hm. We could say "the external pressure varies insignificantly over the height of the balloon. Therefore the net force on any shape balloon is (requires a bit of thought) zero.
Oops! Threw buoyancy out with the bath water.
So variation of pressure with height is evidently crucial, even for a child's balloon, or we get no buoyancy.
Here is the oolie: We might guess the pressure differential across the fabric is zero. But this can't be true identically over the height of the balloon, because the gasses inside and outside have different densities.
Here is your thought question, if I may be so bold:
Identify the factors we are ignoring and use them to suggest an explanation for the teardrop shape characteristic of some weather balloons.
Another version of the oolie is to focus on the bag and assert that if the pressure is equal across the fabric at all points, there is no net force on the bag! However, even forgetting the weight of the bag itself, which must be supported, it is the bag which transmits buoyant force to the load.
So how can an object with zero net buoyant force on it generate net lift?
I think it was "Old Man" who called my attention to this. I hope he is well.
Ed Green wrote: > Here is your thought question, if I may be so bold:
> Identify the factors we are ignoring and use them to suggest > an explanation for the teardrop shape characteristic of some > weather balloons.
I think you answered this bellow but the answer is simple. The weather balloon must support its own weight and the weight of whatever it carries.
> Another version of the oolie is to focus on the bag and assert > that if the pressure is equal across the fabric at all points, > there is no net force on the bag!
Clearly you know this isn't true. However, you make me wonder if when modeling the shape of the balloon I should consider the pressure gradient on the inside as well as on the outside.
> However, even forgetting the > weight of the bag itself, which must be supported, it is the bag > which transmits buoyant force to the load.
> So how can an object with zero net buoyant force on it > generate net lift?
Ed Green wrote: > Here is your thought question, if I may be so bold:
> Identify the factors we are ignoring and use them to suggest > an explanation for the teardrop shape characteristic of some > weather balloons.
I think you answered this bellow but the answer is simple. The weather balloon must support its own weight and the weight of whatever it carries.
> Another version of the oolie is to focus on the bag and assert > that if the pressure is equal across the fabric at all points, > there is no net force on the bag!
Clearly you know this isn't true. However, you make me wonder if when modeling the shape of the balloon I should consider the pressure gradient on the inside as well as on the outside.
> However, even forgetting the > weight of the bag itself, which must be supported, it is the bag > which transmits buoyant force to the load.
> So how can an object with zero net buoyant force on it > generate net lift?
And you also know the answer to this is, it can't.
Ed Green wrote: > Here is your thought question, if I may be so bold:
> Identify the factors we are ignoring and use them to suggest > an explanation for the teardrop shape characteristic of some > weather balloons.
I think you answered this bellow but the answer is simple. The weather balloon must support its own weight and the weight of whatever it carries.
> Another version of the oolie is to focus on the bag and assert > that if the pressure is equal across the fabric at all points, > there is no net force on the bag!
Clearly you know this isn't true. However, you make me wonder if when modeling the shape of the balloon I should consider the pressure gradient on the inside as well as on the outside.
> However, even forgetting the > weight of the bag itself, which must be supported, it is the bag > which transmits buoyant force to the load.
> So how can an object with zero net buoyant force on it > generate net lift?
And you also know the answer to this is, it can't.
>> Another version of the oolie is to focus on the bag and assert >> that if the pressure is equal across the fabric at all points, >> there is no net force on the bag!
>Clearly you know this isn't true. However, you make me wonder >if when modeling the shape of the balloon I should consider the pressure >gradient on the inside as well as on the outside.
Yes. As I said, it was the poster known as Old Man who got me to see this. The pressure gradient inside is irrelevant in determining lift if we take the shape of the bag as given; but it is certainly relevant in determining that shape.
>> However, even forgetting the >> weight of the bag itself, which must be supported, it is the bag >> which transmits buoyant force to the load.
>> So how can an object with zero net buoyant force on it >> generate net lift?
>And you also know the answer to this is it can't.
Well, yes; but the point is there must be varying differential pressures across the bag to develop a net buoyant force.
The variation in the differential pressure will be a function of the shape of the bag and also a function of tension in the bag. A solution will simultaneously locally satisfy a condition relating these factors and an overall force balance including weight, as you must realize. You may not have realized the first condition.
>> Here is your thought question, if I may be so bold:
>> Identify the factors we are ignoring and use them to suggest >> an explanation for the teardrop shape characteristic of some >> weather balloons.
from an earlier thread on sci.physics: The Volume of a Balloon
On Wed, 28 Feb 2001 05:11:40 -0500, "Art Marks"
<arthurruh...@netscape.net> wrote: >Can anyone tell me how I can get the approximate volume of a balloon?
Probably this is killing a fly with a sledgehammer, but I believe much of the problem of a balloon's shape (thinking lighter than air craft, not toys) maps into one part of the droplet formation problem. This latter problem of course, is of interest in ink jet nozzle design and other high tech small object challenges. If there is enough interest, it would be worthwhile to review the following article and reapply their approach to droplet size to the reoriented and rescaled situation of a balloon. Simulation of a Dripping Faucet Nobuko Fuchikami, Shunya Ishioka, and Ken Kiyono http://buss96.phys.metro-u.ac.jp/~brk/drip/7976.pdf (quoting) The formation of drops is an intriguing phenomenon widely observed in everyday life. Although scientific researches on this subject date back to the seventeenth century, 1) great progress has been achieved only recently, mainly in detailed studies on the behavior of drops near the breakup point. Breakup of a drop is a critical phenomenon corresponding to a singularity of a nonlinear partial differential equation obeyed by the uid with free surface. (end quote) It would be nice to know that a hot-air balloon's designed margin of safety was calculated using accurate models of its break-up point.
John Bailey wrote: > On Sun, 03 Jun 2001 00:45:29 GMT, John Creighton > <JohnCreight...@hotmail.com> wrote:
> >Ed Green wrote:
> >> Here is your thought question, if I may be so bold:
> >> Identify the factors we are ignoring and use them to suggest > >> an explanation for the teardrop shape characteristic of some > >> weather balloons.
> from an earlier thread on sci.physics: The Volume of a Balloon
> On Wed, 28 Feb 2001 05:11:40 -0500, "Art Marks" > <arthurruh...@netscape.net> wrote:
> >Can anyone tell me how I can get the approximate volume of a balloon? > Probably this is killing a fly with a sledgehammer, but I believe much > of the problem of a balloon's shape (thinking lighter than air craft, > not toys) maps into one part of the droplet formation problem. This > latter problem of course, is of interest in ink jet nozzle design and > other high tech small object challenges. If there is enough interest, > it would be worthwhile to review the following article and reapply > their approach to droplet size to the reoriented and rescaled > situation of a balloon. > Simulation of a Dripping Faucet > Nobuko Fuchikami, Shunya Ishioka, and Ken Kiyono > http://buss96.phys.metro-u.ac.jp/~brk/drip/7976.pdf > (quoting) > The formation of drops is an intriguing phenomenon widely observed in > everyday life. Although scientific researches on this subject date > back to the seventeenth century, 1) great progress has > been achieved only recently, mainly in detailed studies on the > behavior of drops near the breakup point. Breakup of a drop is a > critical phenomenon corresponding to a singularity of a nonlinear > partial differential equation obeyed by the uid with free surface. > (end quote) > It would be nice to know that a hot-air balloon's designed margin of > safety was calculated using accurate models of its break-up point.
> John
This sounds very interesting. Unfortunately you link doesn't work at the moment.
>> On Sun, 03 Jun 2001 00:45:29 GMT, John Creighton >> <JohnCreight...@hotmail.com> wrote:
>> >Ed Green wrote:
>> >> Here is your thought question, if I may be so bold:
>> >> Identify the factors we are ignoring and use them to suggest >> >> an explanation for the teardrop shape characteristic of some >> >> weather balloons.
>> from an earlier thread on sci.physics: The Volume of a Balloon
>> On Wed, 28 Feb 2001 05:11:40 -0500, "Art Marks" >> <arthurruh...@netscape.net> wrote:
>> >Can anyone tell me how I can get the approximate volume of a balloon? >> Simulation of a Dripping Faucet >> Nobuko Fuchikami, Shunya Ishioka, and Ken Kiyono >> http://buss96.phys.metro-u.ac.jp/~brk/drip/7976.pdf
>This sounds very interesting. Unfortunately you link doesn't work at the >moment.
I am certain I downloaded their pdf file but I cannot find it at the moment.
For a taste of the article, here are some extracts: 1) In this paper, we present a new algorithm for simulating a dripping faucet based on Lagrangian description instead of Eulerian one like Navier-Stokes equations. We decompose a drop into many parts (at most 300 ~400 disks or less) and describe the dynamics in terms of time evolution equations obeyed by each part under the in uence of gravity, surface tension and viscosity. The authors promises a simplified model which is outlined at the end of their paper. 2) (quoting) We (will) construct an improved mass-spring model based on a detailed analysis of our numerical simulation. The model reveals the basic mechanism of the complex behavior of the dripping faucet system. A key of the new model is that the mass dependence of the spring-constant is taken into account. (end quote)
John Bailey wrote: > >Can anyone tell me how I can get the approximate volume of a balloon? > Probably this is killing a fly with a sledgehammer, but I believe much > of the problem of a balloon's shape (thinking lighter than air craft, > not toys) maps into one part of the droplet formation problem. This > latter problem of course, is of interest in ink jet nozzle design and > other high tech small object challenges. If there is enough interest, > it would be worthwhile to review the following article and reapply > their approach to droplet size to the reoriented and rescaled > situation of a balloon.
I assume a helium bubble would be a rescaled and reorientation of a water droplet but I am not sure the same results would hold for a balloon. For a water droplet or a soap bubble the surface tension is the same on every point on the surface. This would not be true for a balloon.
> Simulation of a Dripping Faucet > Nobuko Fuchikami, Shunya Ishioka, and Ken Kiyono > http://buss96.phys.metro-u.ac.jp/~brk/drip/7976.pdf > (quoting) > The formation of drops is an intriguing phenomenon widely observed in > everyday life. Although scientific researches on this subject date > back to the seventeenth century, 1) great progress has > been achieved only recently, mainly in detailed studies on the > behavior of drops near the breakup point. Breakup of a drop is a > critical phenomenon corresponding to a singularity of a nonlinear > partial differential equation obeyed by the uid with free surface. > (end quote) > It would be nice to know that a hot-air balloon's designed margin of > safety was calculated using accurate models of its break-up point.
> John
P.S. I'll try to read the paper A.S.A.P. It sounds very interesting.
>> >Can anyone tell me how I can get the approximate volume of a balloon? >> Probably this is killing a fly with a sledgehammer, but I believe much >> of the problem of a balloon's shape (thinking lighter than air craft, >> not toys) maps into one part of the droplet formation problem. This >> latter problem of course, is of interest in ink jet nozzle design and >> other high tech small object challenges. If there is enough interest, >> it would be worthwhile to review the following article and reapply >> their approach to droplet size to the reoriented and rescaled >> situation of a balloon.
>I assume a helium bubble would be a rescaled and reorientation of a water >droplet but I am not sure the same results would hold for a balloon. For a >water >droplet or a soap bubble the surface tension is the same on every point >on the surface. This would not be true for a balloon.
I think you are right. An high altitude balloon covers a ball of air with its fixed (slightly elastic) skin. Its a version of the obstacle problem. (quoting) At its simplest, an obstacle problem arises when an elastic string is held fixed at two ends, A and B, and passes over a smooth object which protrudes between the two ends. (end quote) The solution conforms to one given shape for a portion of its path and then free of that constraint and under another. I ran into this problem in converting Black-Scholes option pricing from European Options to American Options. It is messy. Formulating the droplet solutions into ones which reflect the obstacle problem is more than I would care to attack.
John Bailey wrote: > On Sun, 03 Jun 2001 19:51:31 GMT, John Creighton > <JohnCreight...@hotmail.com> wrote:
> >John Bailey wrote:
> >> >Can anyone tell me how I can get the approximate volume of a balloon? > >> Probably this is killing a fly with a sledgehammer, but I believe much > >> of the problem of a balloon's shape (thinking lighter than air craft, > >> not toys) maps into one part of the droplet formation problem. This > >> latter problem of course, is of interest in ink jet nozzle design and > >> other high tech small object challenges. If there is enough interest, > >> it would be worthwhile to review the following article and reapply > >> their approach to droplet size to the reoriented and rescaled > >> situation of a balloon.
> >I assume a helium bubble would be a rescaled and reorientation of a water > >droplet but I am not sure the same results would hold for a balloon. For a > >water > >droplet or a soap bubble the surface tension is the same on every point > >on the surface. This would not be true for a balloon.
> I think you are right. An high altitude balloon covers a ball of air > with its fixed (slightly elastic) skin. Its a version of the obstacle > problem. (quoting) At its simplest, an obstacle problem arises when > an elastic string is held fixed at two ends, A and B, and passes over > a smooth object which protrudes between the two ends. (end quote) > The solution conforms to one given shape for a portion of its path and > then free of that constraint and under another. > I ran into this problem in converting Black-Scholes option pricing > from European Options to American Options. It is messy. Formulating > the droplet solutions into ones which reflect the obstacle problem is > more than I would care to attack.
Well, I've got the easy stuff done:
Paramaterized the surface Calculated the Volume Assigned area vectors to each approximately square section.
Now comes the trickier stuff. I hope to solve the problem by starting the balloon off in an unstained shape then letting the dynamics carry it to its equilibrium shape. I plan to introduce a damping force to help the balloon settle to its equilibrium state.
>John Bailey wrote: >> On Wed, 28 Feb 2001 05:11:40 -0500, "Art Marks" >> <arthurruh...@netscape.net> wrote:
>> Simulation of a Dripping Faucet >> Nobuko Fuchikami, Shunya Ishioka, and Ken Kiyono >> http://buss96.phys.metro-u.ac.jp/~brk/drip/7976.pdf >> (quoting) >> The formation of drops is an intriguing phenomenon widely observed in >> everyday life. Although scientific researches on this subject date >> back to the seventeenth century, 1) great progress has >> been achieved only recently, mainly in detailed studies on the >> behavior of drops near the breakup point. Breakup of a drop is a >> critical phenomenon corresponding to a singularity of a nonlinear >> partial differential equation obeyed by the uid with free surface. >> (end quote)
>This sounds very interesting. Unfortunately you link doesn't work at the >moment.
I got it to work, but not on first try. I had to back out to the parent web-site and click back through to the .pdf file.
(If only my sainted grandfather could hear me talking like this ;).
It is a very nice preprint.
>> It would be nice to know that a hot-air balloon's designed margin of >> safety was calculated using accurate models of its break-up point.
>> John
Of course this gives me the opportunity for more pedantry, by pointing out the differences between buoyant gas bags and faucet drips:
The surface of the gas bag is essentially inextensible and can support varying amounts of tension up to failure by tearing.
The surface of the water drop is essentially infinitely extensible and always has the same tension; and the mode of "failure" is generation of new closed surfaces, like an amoeba.
Skimming this paper did call my attention to one mistake in my thinking: in the formula P = T/R for differential pressure across a spherical interface, the generalized form is P = (1/2)T(1/R1 + 1/R2), R1 and R2 the principle radii of curvature (I may have misplaced a factor of 2).
I knew this but in thinking about the tear drop shape of a weather balloon I was thinking this implied a pressure inversion on the concave part of the balloon. I was forgetting that the "tail" of a teardrop had a second radius of curvature which was positive and diminishing.
I'm not sure how this applies to folds of wrinkled fabric.
Ed Green wrote: > From: John Creighton JohnCreight...@hotmail.com
> >Ed Green wrote:
> <snip>
> >> Another version of the oolie is to focus on the bag and assert > >> that if the pressure is equal across the fabric at all points, > >> there is no net force on the bag!
> >Clearly you know this isn't true. However, you make me wonder > >if when modeling the shape of the balloon I should consider the pressure > >gradient on the inside as well as on the outside.
> Yes. As I said, it was the poster known as Old Man who > got me to see this. The pressure gradient inside is irrelevant > in determining lift if we take the shape of the bag as given; but > it is certainly relevant in determining that shape.
I wonder how relevant thorough? The gas on the inside should be much lighter and therefore have much less of a pressure gradient. I wonder if the results would be reasonably close if the pressure gradient on the inside was treated as constant. It is worth noting that the the pressure gradient on the inside should counteract the upside down tear drop shape.
> > From: John Creighton JohnCreight...@hotmail.com
> > >Ed Green wrote:
> > <snip>
> > >> Another version of the oolie is to focus on the bag and assert > > >> that if the pressure is equal across the fabric at all points, > > >> there is no net force on the bag!
> > >Clearly you know this isn't true. However, you make me wonder > > >if when modeling the shape of the balloon I should consider the pressure > > >gradient on the inside as well as on the outside.
> > Yes. As I said, it was the poster known as Old Man who > > got me to see this. The pressure gradient inside is irrelevant > > in determining lift if we take the shape of the bag as given; but > > it is certainly relevant in determining that shape.
> I wonder how relevant thorough? The gas on the inside should be much > lighter and therefore have much less of a pressure gradient. I wonder > if the results would be reasonably close if the pressure gradient on the > inside was treated as constant. It is worth noting that the the pressure > gradient on the inside should counteract the upside down tear drop > shape.
I found another reason to consider the preassure gradiant on the inside. If there was no preassure gradiant then the force from the air in the balloon would be the same in all directions. Infact it would be weightless.
John Bailey wrote: > I think you are right. An high altitude balloon covers a ball of air > with its fixed (slightly elastic) skin. Its a version of the obstacle > problem. (quoting) At its simplest, an obstacle problem arises when > an elastic string is held fixed at two ends, A and B, and passes over > a smooth object which protrudes between the two ends. (end quote) > The solution conforms to one given shape for a portion of its path and > then free of that constraint and under another. > I ran into this problem in converting Black-Scholes option pricing > from European Options to American Options. It is messy. Formulating > the droplet solutions into ones which reflect the obstacle problem is > more than I would care to attack.
> John
Since there seemed to be some interest in this problem I decided to post on the web what I got done so far. http://www.geocities.com/s243a/physics/TheShapeOfABalloon/index.htm I have: -Paramaterized the surface -Found the pressure forces on the sections formed by the latitude longitude grid -Found an equivalent force through the mesh points.
note* The pressure forces take into account the variation in pressure with altitude both inside and outside the balloon
*Also note You should be able to understand the general procedures with out reading the code I have provided. However, if anyone is interested, I am more then willing to explain the matlab code.
>From: John Creighton JohnCreight...@hotmail.com >Date: 6/5/2001 12:50 PM Eastern Daylight Time >Message-id: <3B1D0EA8.1C7BD...@hotmail.com>
>John Creighton wrote:
>> Ed Green wrote:
>> > From: John Creighton JohnCreight...@hotmail.com
>> > >Ed Green wrote:
>> > <snip>
>> > >> Another version of the oolie is to focus on the bag and assert >> > >> that if the pressure is equal across the fabric at all points, >> > >> there is no net force on the bag!
>> > >Clearly you know this isn't true. However, you make me wonder >> > >if when modeling the shape of the balloon I should consider the pressure >> > >gradient on the inside as well as on the outside.
>> > Yes. As I said, it was the poster known as Old Man who >> > got me to see this. The pressure gradient inside is irrelevant >> > in determining lift if we take the shape of the bag as given; but >> > it is certainly relevant in determining that shape.
>> I wonder how relevant thorough? The gas on the inside should be much >> lighter and therefore have much less of a pressure gradient. I wonder >> if the results would be reasonably close if the pressure gradient on the >> inside was treated as constant. It is worth noting that the the pressure >> gradient on the inside should counteract the upside down tear drop >> shape.
>I found another reason to consider the preassure gradiant on the inside. >If there was no preassure gradiant then the force from the air in the >balloon would be the same in all directions. Infact it would be weightless.
Hmm... good points.
I wonder about the upside down teardrop shape. I don't have a good explanation for it other than thinking it must depend on an external tension applied to the bag at the low point.
I see your point: if we filled a gas bag with helium in vacuum it would slump over under its own weight, and in fact if we suported the bag at a single point, we would wind up with a pendant "right side up" tear drop shape.
>>> >Can anyone tell me how I can get the approximate volume of a balloon? >>> Probably this is killing a fly with a sledgehammer, but I believe much >>> of the problem of a balloon's shape (thinking lighter than air craft, >>> not toys) maps into one part of the droplet formation problem. This >>> latter problem of course, is of interest in ink jet nozzle design and >>> other high tech small object challenges. If there is enough interest, >>> it would be worthwhile to review the following article and reapply >>> their approach to droplet size to the reoriented and rescaled >>> situation of a balloon.
>>I assume a helium bubble would be a rescaled and reorientation of a water >>droplet but I am not sure the same results would hold for a balloon. For a >>water >>droplet or a soap bubble the surface tension is the same on every point >>on the surface. This would not be true for a balloon.
>I think you are right. An high altitude balloon covers a ball of air >with its fixed (slightly elastic) skin. Its a version of the obstacle >problem. (quoting) At its simplest, an obstacle problem arises when >an elastic string is held fixed at two ends, A and B, and passes over >a smooth object which protrudes between the two ends. (end quote) >The solution conforms to one given shape for a portion of its path and >then free of that constraint and under another.
I had to read this a few times before I understood what you were saying.
But you are treating an approximate case, so far as balloons are concerned; unless you are thinking of the geometry of a net thrown over the gas bag: that would conform to your definition of an "obstacle problem". But the gas bag itself is filled with an expansive medium (I wrote that to avoid immediately reiterating "gas". Clever, eh. ;), not hung over a ball of gas.
The shape of the gas bag is determined by local differential pressure, given the two principal radii of curvature at each point, and the tension, and by the requirement to support its own weight and the weight of any load pendant from the bag, which also partially controls the tension.
I think in this analysis the weight of the trapped gas comes out in the wash... it is automatically supported when its pressure is opposed at each point by the external pressure + differential bag pressure.
>From: John Creighton JohnCreight...@hotmail.com >John Bailey wrote:
>> I think you are right. An high altitude balloon covers a ball of air >> with its fixed (slightly elastic) skin. Its a version of the obstacle >> problem. (quoting) At its simplest, an obstacle problem arises when >> an elastic string is held fixed at two ends, A and B, and passes over >> a smooth object which protrudes between the two ends. (end quote) >> The solution conforms to one given shape for a portion of its path and >> then free of that constraint and under another. >Since there seemed to be some interest in this problem I decided to >post on the web what I got done so far. >http://www.geocities.com/s243a/physics/TheShapeOfABalloon/index.htm >I have: >-Paramaterized the surface >-Found the pressure forces on the sections formed by the latitude longitude >grid >-Found an equivalent force through the mesh points.
>note* >The pressure forces take into account the variation in pressure with altitude >both inside and outside the balloon
<snip>
Have you explicitly considered bag tension? This will be vitally important in determing the shape.
If you have a "polygonal balloon", I think you could incorporate tension by assigning a constant value to each facet, and then calculating the force on neighboring facets considering the angle formed between them and the length of intersection, or the edge. You would then need some way of adjusting the tensions on the fly to reach an equilibrium of forces.
Ed Green wrote: > >From: John Creighton JohnCreight...@hotmail.com
> >John Bailey wrote:
> >> I think you are right. An high altitude balloon covers a ball of air > >> with its fixed (slightly elastic) skin. Its a version of the obstacle > >> problem. (quoting) At its simplest, an obstacle problem arises when > >> an elastic string is held fixed at two ends, A and B, and passes over > >> a smooth object which protrudes between the two ends. (end quote) > >> The solution conforms to one given shape for a portion of its path and > >> then free of that constraint and under another.
> >Since there seemed to be some interest in this problem I decided to > >post on the web what I got done so far. > >http://www.geocities.com/s243a/physics/TheShapeOfABalloon/index.htm > >I have: > >-Paramaterized the surface > >-Found the pressure forces on the sections formed by the latitude longitude > >grid > >-Found an equivalent force through the mesh points.
> >note* > >The pressure forces take into account the variation in pressure with altitude > >both inside and outside the balloon
> <snip>
> Have you explicitly considered bag tension? > This will be vitally important in determing the shape.
> If you have a "polygonal balloon", I think you could incorporate > tension by assigning a constant value to each facet, and then > calculating the force on neighboring facets considering the > angle formed between them and the length of intersection, > or the edge. You would then need some way of adjusting > the tensions on the fly to reach an equilibrium of forces.
I'm getting to that. I will take into account the modules of elasticity, the thickness of the bag and the strain. The initial shape of the balloon will be in an unstained state. Then dynamics will carry it to its strained shape.
