Ax=b
John Creighton
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 Physics ********************
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:
F_{ox} r_{oy} - F_{oy} r_{ox} = F_{1x} r_{1y} - F_{1y} r_{1x}
F_{ox} r_{oz} - F_{oz} r_{ox} = F_{1x} r_{1z} - F_{1z} r_{1x}
F_{oy} r_{oz} - F_{oz} r_{oy} = F_{1y} r_{1z} - F_{1z} r_{1y}
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.
Thank you very much for your help.
Robert Israel
John Creighton <JohnCre...@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
Gregory C. Jones
>
> *********************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 Physics ********************
>
> 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.
g
John Creighton
"Gregory C. Jones" wrote:
> John Creighton 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.
>
> 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 Physics ********************
> >
> > 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.
John Creighton
Robert Israel wrote:
> In article <3B165D92...@hotmail.com>,
> John Creighton <JohnCre...@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
Gregory C. Jones
>
> Gregory C. Jones wrote:
> > Given |r_1| < |r_0|, a solution does not always exist either.
>
> This is distressing.
No -- this is an elementary fact. Let's say you have a force F_0
at vector r_0, and F_1 and r_1 in the same plane. The magnitudes of
the moments are
|r_0| |F_0| sin(theta_0) = |r_1| |F_1| sin(theta_1)
where theta_1 is the angle between r_1 and F_1,
theta_2 is the angle between r_2 and F_2.
Since |F_0|=|F_1|, this reduces to
|r_0| sin(theta_0)=|r_1| sin(theta_1)
and if the left hand side is bigger than |r_1|, no real angle theta_1
will work, since |sin(theta_1)| is bounded by 1.
g
Ed Green
>*********************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.
John Creighton
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.
John Creighton
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.
John Creighton
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
>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.
>> 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.
John Bailey
<JohnCre...@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"
<arthur...@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
John Creighton
John Bailey wrote:
This sounds very interesting. Unfortunately you link doesn't work at the
moment.
John Bailey
<JohnCre...@hotmail.com> wrote:
>
>
>John Bailey wrote:
>
>> On Sun, 03 Jun 2001 00:45:29 GMT, John Creighton
>> <JohnCre...@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"
>> <arthur...@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.
An abstract is at:
http://www.jssst.or.jp/jps/jpsj/1999/p684/p68411/p68411h/p68411h.html
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
John Creighton
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.
John Bailey
<JohnCre...@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.
John
John Creighton
John Bailey wrote:
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
Ed Green
>John Bailey wrote:
>> On Wed, 28 Feb 2001 05:11:40 -0500, "Art Marks"
>> <arthur...@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.
John Creighton
Ed Green wrote:
> From: John Creighton JohnCre...@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.
John Creighton
John Creighton wrote:
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 Creighton
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.
Ed Green
>Date: 6/5/2001 12:50 PM Eastern Daylight Time
>Message-id: <3B1D0EA8...@hotmail.com>
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.
Ed Green
>>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 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.
Ed Green
>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.
John Creighton
Ed Green wrote:
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.
John Creighton
Ed Green wrote:
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 unstrained state. Then dynamics will carry it to its strained shape.
P.S. is a facet the same as a mesh point?
John Bailey
>>>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 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 we are getting close! Somewhere back in time, I recall
knowing the formula P=T/R where P is the pressure at a given point on
the surface, T is the tension in the (assumed to be) inextensible bag
and R is the radius of curvature at that point. The obstacle part of
the problem arises when the math must account for the transition to a
suspension ring or some such attachment which then supports the
balloon load. If you only model a free flight ballon, a la an
aerosonde, I suppose you can avoid the obstacle problem.
Using good old P= T/R I can express the solution for the 2D version of
the problem, including accounting for pressure gradient with altitude
as a ODE. Extending this solution to the real life 3D case without
getting hopelessly lost in tensor geometry is the next challenge.
John
John Creighton
John Bailey wrote:
I could see how you would do this. The tension would be given by the angle
the bag is relative to the vertical and the amount of pressure force it must
support. The radius would give you the change in angle. Of course the
problem would be slightly annoying. Because the solution we may get may be
for a balloon that has more gas in it then the one we were looking for. I
also wonder if every solution we get would be valid. Might some solutions
have a non zero slope at the top of the balloon and consequently need to be
thrown out?
It is trickier though in three space. We have to deal with two radiuses of
curvature. BTW does any one know any good sources to learn about tensors on
the net. I think this is something I should learn sometime.
Ed Green
Date: 6/9/2001 11:24 AM Eastern Daylight Time
>> 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 unstrained state. Then dynamics will carry it to its strained shape.
Well, if I am correct you can probably neglect the elasticity
of the bag, the thickness, and even the weight; you will get
something that behaves in a very balloon like way if you just
take an inextensible ideally flexible membrane able to support
tension. But if you neglect tension you will not get something
acting very much like any kind of balloon, because it is bag
tension, coupled with the principal local radii of curvature,
which supports the pressure difference between inside and out.
>P.S. is a facet the same as a mesh point?
Sorry. I was thinking of your balloon as polygonal, defined
by a finite number of planar faces. Each face would be a facet,
just another name. Does this have anything to do with the
scheme of finite approximation you have adopted?
Ed Green
Yes. Was it you that posted the citation for the dripping faucet
paper? Well, on about page 3 of that paper the full form of
the formula is given: P = T(1/R1 + 1/R2).
R1 and R2 are the principle radii of curvature (it looks like a factor
of two is missing somewhere). This will be important if we want
to model anything other than a spherical balloon; particularly
a teardrop shape, where the radii in the tail have different
algebraic signs.
>The obstacle part of
>the problem arises when the math must account for the transition to a
>suspension ring or some such attachment which then supports the
>balloon load. If you only model a free flight ballon, a la an
>aerosonde, I suppose you can avoid the obstacle problem.
Or one where the gas bag is the only load bearing element;
I think high altitude weather balloons are like that.
>Using good old P= T/R I can express the solution for the 2D version of
>the problem, including accounting for pressure gradient with altitude
>as a ODE. Extending this solution to the real life 3D case without
>getting hopelessly lost in tensor geometry is the next challenge.
Hmm... How would one set this up in a smooth PDE form?
Suppose we just considered the gas bag for now. Should
we assume cylindrical symmetry, and try to set up an
equation for r(z)? Then we still get an ODE, though it
may take some work to get it! First we have to find R1,
R2 at each z.... well, never mind "first"... and I am
worried about the contraint "inextensible". We would
need some kind of unstressed shape for the bag, and
I think this constraint would be an integral constraint.
Finally, a real bag may develop reentrant folds, like
a curtain. It may be necessary to allow the bag to
"contract", though have a maximum surface area.
John Creighton
> >P.S. is a facet the same as a mesh point?
>
> Sorry. I was thinking of your balloon as polygonal, defined
> by a finite number of planar faces. Each face would be a facet,
> just another name. Does this have anything to do with the
> scheme of finite approximation you have adopted?
Yes. With my paramaterization, a mesh point would occur at the intersection of a
line of latitude and a line of longitude. I originally wasn't going to call them
mesh points but recently in an application section of a linear algebra book I saw
them refer to the discrete points they used to estimate the equilibrium
temperature distribution of a body as mesh points. The term mesh points comes from
the field of computer graphics. A mesh plot is a plot of a surface. It is usually
made by the intersection of two sets of parallel lines. I believe these lines are
usually perpendicular.
If I associate with each of these mesh points a mass an area and a spring constant
with its neighbors I can use the laws of dynamics to find the equilibrium state of
the balloon (provided I introduce a damping force). It would also be interesting
to look at this problem in terms of facets. The description might be more
complicated but it will probably also be more accurate. I do believe though that
both descriptions will converge on the same result. I think there is something
deeper hear, something that sort of strings its way through, measure theory,
topology, vector calculus, tensors, differential equations. I think there is some
pattern dying to come out. We have a geometry and there are several different ways
to analyze it. I think they all converge on the same result. Amongst the
computations and the current problem where is the pattern hiding. Has it been
already found or will we discover it?
Ed Green
>> >P.S. is a facet the same as a mesh point?
>>
>> Sorry. I was thinking of your balloon as polygonal, defined
>> by a finite number of planar faces. Each face would be a facet,
>> just another name. Does this have anything to do with the
>> scheme of finite approximation you have adopted?
<snip>
>If I associate with each of these mesh points a mass an area and a spring
>constant
>with its neighbors I can use the laws of dynamics to find the equilibrium
>state of
>the balloon (provided I introduce a damping force). It would also be
>interesting
>to look at this problem in terms of facets. The description might be more
>complicated but it will probably also be more accurate. I do believe though
>that
>both descriptions will converge on the same result.
Yes. Ok, I guess you can model the balloon as a net of
mass points linked by springs. The springs will develop the
tension, and as the masses become negligible and the
springs very stiff, you will approach the flexible inextensible
bag; not immediately clear to me how you handle pressure --
an additional outward normal force on masses, but you have
to take into account at least nearest neighbor masses to
find the "outward normal" and the effective area to be assigned
the mass. Maybe you have thought this out already.
> I think there is something
>deeper hear, something that sort of strings its way through, measure theory,
>topology, vector calculus, tensors, differential equations. I think there is
>some
>pattern dying to come out. We have a geometry and there are several different
>ways
>to analyze it. I think they all converge on the same result. Amongst the
>computations and the current problem where is the pattern hiding. Has it been
>already found or will we discover it?
Yes! Hey, you begin to sound like me. :) There are patterns
weaving their way through everything, tugging at the boarders
of our intuition, struggling to get out; we must work at helping
them reveal themselves. The patterns will continue to tug
until we allow our brains to ossify.
John Creighton
Ed Green wrote:
To find the pressure force I considered the force on an adjacent facet.
I then moved them to act through the mesh points. Some adjustments were
made to try to keep the net moment about the center of mass
in the system the same.
John Creighton
John Bailey wrote:
I am interested to see it solved with a partial differential equation
approach.
I would most certainly read it. Mostly because I want to see the connection
between different mathematical pictures of the same problem. However,
I am looking for a more general technique.
It's kind of like: Newton's method for finding roots is great but is nice to
have the good old by section method kicking around just in case. Similarly
partial differential equation, and tensor geometry are great tools to gain
insight into many problems but each problem has to be set up and tackled
independently.
If my surface approximations work, then I will be able to extend them to a
much
wider rang of problems and all that will be required is a change of the
initial
conditions (i.e. the unstrained surface) Some examples include:
A blimp
A rotating spherical space station
and possibly even an inflatable Truss. (Just because it may not be practical
that
is not going to stop me from analyzing it)
John Creighton
From you email I conclude that the optimum ustrained shape for
a balloon of uniform thickness and material would be the shape
of the tear drop given by the paper about a dripping
faucet. The reason this shape would be optimal is because every
point on the bag would be at the same stress. If as Ed Green
Suggested that the bag is unextentable then the unstrained shape
in this particular case would also be the same as the strained shape.
I do wonder though if this optimal shape is of an use for a real
balloon. There is a huge difference in scale between a water droplet
and a balloon. I am not sure if the conditions are comparable and
if it will be possible to design a real balloon where the tension
Jonathan Gardner
If you can find the solution in 2-space, 3-space should be trivial. I think
R can be expressed as a differential. I'll do the 3-space right now. For
instance, if you were to consider a balloon which was actually a piece of
rubber stretched flat on top of a sealed box, then I think the simple model
that follow will give you a good solution. You need to transform this to a
radial axis, and I don't have the equations to do that with me.
Consider each point on the surface. Any point experiences two forces, the
sum of which is the net force. If there is a net force, then the point will
move, and we are assuming that each point is massless, and so they cannot
have a velocity and so the net force must be 0 at all points on the surface.
The first force is the force caused by his neighbors and the distance away
from the neighbors, in the direction of his neighbors, all summed up. Is
this not an integral? His neighbors will be tugging on him with a spring
force, so k * displacement. Don't forget that there is a point away from
his neighbors that he should be at where there will be zero spring force.
The second force is the pressure. It is equal to the pressure of the gas
outside minus the pressure on the inside, times the area of that point
(which is dx * dy). Make sure you get the signs right.
You have to have these two forces equal and opposite. I will play with this
at home and bring you a solution when I encounter it, but this seems the
most logical way to approach it.
After you have solved the force for any position on the surface, you need
to specify the BC (boundary conditions). This is trivial for a box, but may
be more difficult for a balloon.
John Creighton
in retrospect should be obvious:
-The pressure in the balloon is approximately the same inside and out. (This is
useful to get a good estimation of the volume of a balloon)
-The addition of a load to a balloon will only change the shape considerably in
the region of the balloon where the load is applied. (If you don't believe me,
hold a balloon at the point where it is tied and pull it under water. If the
balloon is inflated enough you will notice very little change in the shape of the
balloon except in the region where the force is applied.
These insights gave me a hint on how to deal with folds in a real balloon.
The idea is to forget about the dynamics for a second. Think of the balloon as a
surface with a fixed area. Consider the balloon creasing along lines of longitude.
We know aproximatly the final volume of the balloon. We want to estimate the
radius of curvature of the sections between the creases and the amount of overlap
of the sections along each crease.
John Creighton
> outside minus the pressure on the inside, times the area of that point
> (which is dx * dy). Make sure you get the signs right.
You give me a hit. Perhaps now I can set up the partial differential equation.
Solving it will be another problem.
John Bailey
<JohnCre...@hotmail.com> wrote:
>
>From you email I conclude that the optimum ustrained shape for
>a balloon of uniform thickness and material would be the shape
>of the tear drop given by the paper about a dripping
>faucet.
I agree with your reasoning re optimality, however I had assumed that
the shape would be based on the same ODE's as the fluid droplet but
with different boundary conditions and possible different
normailizations. It appears the solution of the ODE's is fairly
straightforward, however not so straightforward enough that I have yet
debugged my attempt to replicate the droplet work. If and when I do,
it should be a simple matter to shift gears and address the balloon
problem.
>The reason this shape would be optimal is because every
>point on the bag would be at the same stress. If as Ed Green
>Suggested that the bag is unextentable then the unstrained shape
>in this particular case would also be the same as the strained shape.
>
>I do wonder though if this optimal shape is of an use for a real
>balloon. There is a huge difference in scale between a water droplet
>and a balloon. I am not sure if the conditions are comparable and
>if it will be possible to design a real balloon where the tension
>is the same on every point on the surface.
Until I realized the ODE's were the same form and the computations
could be direct and sequential, no trees, lattices or meshes, I tended
to agree with you. With a translation of the different physics coming
back to the same math, is there any other game in town? If the
balloon shape is tailored, its a different problem, if the balloon
relies on pleats to adjust its shape, its hopelessly complex. problem.
John
John Bailey
<JohnCre...@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.
This has been a great thread!
Unfortunately, I am about to leave for the weekend plus a week for
New England. I think the doable version of the balloon shape problem
is essentially complete. An Excel spreadsheet version of this
solution is at:
http://www.frontiernet.net/~jmb184/interests/sci.physics/balloon2.xls
Insert notes (an Excel tool) was used to put notes explaining many of
the entries.
The first five lines contain a viewable version of the formulae which
are in lines 7 through 10. To use the spreadsheet, replicate row 10,
columns A through -J downward for as far as you like. Adjust delta_S
to match your choice.
The formulae replicate the ODE for the surface of the balloon, using
the names outlined in the article on droplet shape.
http://buss96.phys.metro-u.ac.jp/~brk/drip/7976.pdf
A pdf print image of the Excel is also available at my site:
http://www.frontiernet.net/~jmb184/interests/sci.physics/balloon2.pdf
Remaining work is to determine appropriate normalization and initial
conditions for the balloon solution.
John
A few points of difference:
1) At the minimum, R_1 = R_2. , by symmetry. This way of determining
curvature avoids the indeterminate 0/0 that results from using
cos(pi/2)/r which is given in the paper.
2) For a gas balloon, a pressure with a gradient so the pressure drops
at a rate proportioanl to z should replace the water droplet
assumption of a pressure directly proportional to z.
3) For Balloon load, pick a radius r at which a ring will be used to
transfer the load from the supported structure to the skin of the
balloon. Using the Tension assumed and the angle theta, determine
the horizontal and vertical components of force, distributed around
the ring. The horizontal components will cancel themselves. The
resultant vertical component of force is the payload lift the balloon
can support.
Its solved, in principal--the details when I return. :-)
John
Jonathan Gardner
> If I associate with each of these mesh points a mass an area and a spring
> constant with its neighbors I can use the laws of dynamics to find the
> equilibrium state of the balloon (provided I introduce a damping force).
Don't waste time trying to find the dynamic solution unless you are well
versed with multi-dimensional wave mechanics. It gets messy very quickly,
and you will probably start rederiving some stuff in the atomic model for
quantum mechanics. (I still have a bad taste in my mouth from this stuff...
YECH!)
Instead, think about the forces that act at each point, and try to get the
net force at that point to be zero. This way, you will have the static
solution that won't have a time counterpart, and you won't get into wave
mechanics.
The only two forces acting at a "mesh point" is going to be the pressure
(which is related to the area) and the tension/spring force determined by
the topology at that point. Make those equal and opposite. (if you like,
you can throw in gravity and whatever, add BC that will represent a weight
at the bottom of the ballon that will distort the shape.) You will end up
with a perfect sphere in this case. In fact, start with a sphere, describe
the topology as a sphere described with only one variable - r. One of the
BC will be the radius at which the tension at all points is zero. (Or in a
different coordinate system, you can change the default shape to something
useful.)
The trick to this problem is to NOT solve it in x,y,z. You want to solve it
in spherical coordinates or in cylindrical coordinates.
PS. Do we need to consider the area as dependent on the topology as well,
or is it constant? I need to pull out a pen and a paper and pull out my
calculus books, see if it makes a difference.
--
Jonathan Gardner
"Infinity isn't all that large - except at the end." -- Uncle Al
Please don't take anything I say seriously, even if you are really gullible.
Jonathan Gardner
> I've made considerable progress with my surface approximations and
> realized what in retrospect should be obvious:
>
> -The pressure in the balloon is approximately the same inside and out.
> (This is useful to get a good estimation of the volume of a balloon)
This is not a good assumption. If the pressure is almost the same on the
inside and out, then you wouldn't have to tie it off when you blew it up,
would you? In fact, it would be very easy to blow it up in the first place
right? All you are doing is moving air around if the pressure is the same.
This is not so in real life. The pressure is very different on the outside
and the inside.
> -The addition of a load to a balloon will only change the shape
> considerably in the region of the balloon where the load is applied. (If
> you don't believe me, hold a balloon at the point where it is tied and
> pull it under water. If the balloon is inflated enough you will notice
> very little change in the shape of the balloon except in the region where
> the force is applied.
This is what physicists do in this case: If it won't make a difference, get
rid of it. The truth is, I would suspect that the entire shape of the
balloon will change if weight is applied. You will get a teardrop shape,
which is very different from a sphere on both ends.
The main reason why balloons are shaped the way they are is because of
their default shape, I think.
John Creighton
Jonathan Gardner wrote:
> John Creighton wrote:
>
> > I've made considerable progress with my surface approximations and
> > realized what in retrospect should be obvious:
> >
> > -The pressure in the balloon is approximately the same inside and out.
> > (This is useful to get a good estimation of the volume of a balloon)
>
> This is not a good assumption. If the pressure is almost the same on the
> inside and out, then you wouldn't have to tie it off when you blew it up,
> would you? In fact, it would be very easy to blow it up in the first place
> right? All you are doing is moving air around if the pressure is the same.
>
> This is not so in real life. The pressure is very different on the outside
> and the inside.
I mean relatively. There are 101 000 N per m^2 of pressure in the atmosphere.
I would be suppressed if the additional pressure change in a balloon is more
then one tenth of this.
John Creighton
John Bailey wrote:
> >The reason this shape would be optimal is because every
> >point on the bag would be at the same stress. If as Ed Green
> >Suggested that the bag is unextentable then the unstrained shape
> >in this particular case would also be the same as the strained shape.
> >
> >I do wonder though if this optimal shape is of an use for a real
> >balloon. There is a huge difference in scale between a water droplet
> >and a balloon. I am not sure if the conditions are comparable and
> >if it will be possible to design a real balloon where the tension
> >is the same on every point on the surface.
>
> Until I realized the ODE's were the same form and the computations
> could be direct and sequential, no trees, lattices or meshes, I tended
> to agree with you. With a translation of the different physics coming
> back to the same math, is there any other game in town?
Well perhaps for this problem ODE's are the best approach but I do
believe there is a more general technique hiding hear. Hmmm trees.
That gave me a hit. What all these data structures have in common are
nodes and branches (Of course if I studied any topology that would
probably be obvious). I foresee iterative solutions of surfaces converging
on there static equilibrium. Give me some time.
> If the
> balloon shape is tailored, its a different problem, if the balloon
> relies on pleats to adjust its shape, its hopelessly complex. problem.
>
> John
I look forward to you solution. I have done as much as I wish to do with
my original
technique. I look forward to you solution.
Determining The Shape of A balloon That is lifting a Mass M.
Introduction:
Some things that seem relatively simple can be relatively complicated to
analyze. In this paper we ask the simple question what is the shape of a
balloon? To deal with this problem we describe the surface with a mesh an
apply the principles of dynamics to find the equilibrium state of the
balloon. This technique is part of a more general topological concept. The
idea is that there are many finite approximation of a surface that are
composed of nodes and branches it is possible to construct a sequence of
these approximation where in the limiting case the approximation will
behave exactly as the original surface. This paper just deals with one of
these techniques and this simple technique could be extended to analyze
such problems as the stresses on a rotating space station or an inflatable
truss.
.....
http://www.geocities.com/s243a/physics/TheShapeOfABalloon/index.htm
.....
Conclusion:
The techniques described above produced a plot that qualitatively looked
as one would expect a balloon carrying a load to look. The algorithm took
over four seconds to produce the plot on a Pentium three processor. This
if far to slow for any useful dynamic application but sufficient for
static analysis. However even for static applications it makes it time
consuming to compare results over several ranges of values.
It is not clear why in the simulation it was unnecessary to introduce
damping forces. It is likely in a lot of applications this will not be the
case. Even for static analysis it would be desirable to able to watch how
the balloon dynamically reaches equilibrium. This might make it easier to
spot any fundamental modes of oscillation that may occur in a structure.
To make dynamic observations of the transition to equilibrium possible
techniques could be introduced to select the mesh points to apply the laws
of dynamics to at random. This would allow for a more gradual transition
that could be observed through animation.
Further approximation techniques could be applied to allow the algorithm
to iteratively converge on the correct solution. This could be done by
further partitioning the surface after an initial approximation in made.
To do this a new data structure will be required. This will be dealt with
in future papers.
Ed Green
>John Creighton wrote:
>
>> I've made considerable progress with my surface approximations and
>> realized what in retrospect should be obvious:
>>
>> -The pressure in the balloon is approximately the same inside and out.
>> (This is useful to get a good estimation of the volume of a balloon)
>
>This is not a good assumption. If the pressure is almost the same on the
>inside and out, then you wouldn't have to tie it off when you blew it up,
>would you?
Mr. Gardiner: In your busy investigations of dyslexia and
other functional dyschiralities you understandably have
not had the time to review this thread in sufficient detail to
determine what was being discussed. Naturally it would be
unfair to expect an individual of your manifold talents; able
to finish a chemstry exam in 10 short minutes and then --
oh, divine condescension -- doodle for 20 minutes so while
your less fortunate classmates writhed on the floor in weird
contortions, with their thumbs twisting at odd angles, should
not feel too inferior; it would be unfair to expect you to
read the context like us mere mortals: so I shall tell you!
Mr.Creighton was referring to a large scale aerostat such
as a hot air balloon, and not to the elastomeric child's toy.
Do stop by again however, Mr. Gardiner.
Ed Green
>Don't waste time trying to find the dynamic solution unless you are well
>versed with multi-dimensional wave mechanics. It gets messy very quickly,
>and you will probably start rederiving some stuff in the atomic model for
>quantum mechanics. (I still have a bad taste in my mouth from this stuff...
>YECH!)
<snip>
>Please don't take anything I say seriously, even if you are really gullible.
Caveat emptor. You have been warned.