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From: Archimedes Plutonium on 6 Feb 2010 02:54


Archimedes Plutonium wrote:
> Alright, I have to retrench somewhat here. Looking at Mathworld's
> pseudosphere and funnel
> equations, that the contact nested inside is a line contact and not
> area.
>
> So I have to retrench the model and operation. What I do is cut the
> sphere into half
> and then cut a hole into what used to be the poles and slide the two
> hemispheres
> down the two spines of the pseudosphere and as I slide it down I keep
> an eye out
> as to when I reach a maximum surface area contact between the sphere
> and
> pseudosphere.
>
> This revised model operation reminds me of bicycle axles and the race
> cup and the
> bearings in the race cups. So that the bearings freely move around
> inside the race
> and that race is like a pseudosphere. But in that bicycle analogy, the
> bearings want
> as little of surface contact as possible. Here I want the opposite of
> the maximum
> surface area contact.
>
> If my hunch is correct it will have a maximum surface area contact of
> 18 degrees
> arc on both hemispheres adding up to a total of 36 degrees arc.
>
> I think I am getting closer and closer to the perfect model that
> easily explains and proves
> the final answer.
>
> Another way of finding the final answer is to manipulate the equations
> and see where intersections of the sphere with pseudosphere occur. I
> am not expert on those equations
> and leave it to those who are very much everday familar with those
> equations.
>
> So is the final answer of a Pseudosphere of radius 1 unit with Sphere
> of radius 1 unit,
> of the maximum surface area contact that of 10% of the sphere surface?
>

I suppose a better way of stating the above, without cutting and spine
lowering.

We are given a sphere and pseudosphere of 1 unit radius.

Now we ask what is the maximum surface area of contact when the sphere
is
rolled around once on the northern-pseudosphere and then rolled around
once on the southern-pseudosphere.

As in all mathematics, once the problem is clear and the definitions
well-defined,
only then can a proper math take place. Just like my precision
defining of finite-number,
I have not yet reached a precision outline of the problem of surface
area contact.
Hopefully this is the precise problem.

So if we have these two objects and allowed to roll the sphere around
the northern
pseudosphere once and then the southern portion once, what is the
maximum contact
area between the two objects? Is it 10% of the surface area of the
sphere?

Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
From: gudi on 6 Feb 2010 10:47
On Feb 6, 12:54 pm, Archimedes Plutonium
<plutonium.archime...(a)gmail.com> wrote:
> Archimedes Plutonium wrote:
> > Alright, I have to retrench somewhat here. Looking at Mathworld's
> > pseudosphere and funnel
> > equations, that the contact nested inside is a line contact and not
> > area.
>
> > So I have to retrench the model and operation. What I do is cut the
> > sphere into half
> > and then cut a hole into what used to be the poles and slide the two
> > hemispheres
> > down the two spines of the pseudosphere and as I slide it down I keep
> > an eye out
> > as to when I reach a maximum surface area contact between the sphere
> > and
> > pseudosphere.
>
> > This revised model operation reminds me of bicycle axles and the race
> > cup and the
> > bearings in the race cups. So that the bearings freely move around
> > inside the race
> > and that race is like a pseudosphere. But in that bicycle analogy, the
> > bearings want
> > as little of surface contact as possible. Here I want the opposite of
> > the maximum
> > surface area contact.
>
> > If my hunch is correct it will have a maximum surface area contact of
> > 18 degrees
> > arc on both hemispheres adding up to a total of 36 degrees arc.
>
> > I think I am getting closer and closer to the perfect model that
> > easily explains and proves
> > the final answer.
>
> > Another way of finding the final answer is to manipulate the equations
> > and see where intersections of the sphere with pseudosphere occur. I
> > am not expert on those equations
> > and leave it to those who are very much everday familar with those
> > equations.
>
> > So is the final answer of a Pseudosphere of radius 1 unit with Sphere
> > of radius 1 unit,
> > of the maximum surface area contact that of 10% of the sphere surface?
>
> I suppose a better way of stating the above, without cutting and spine
> lowering.
>
> We are given a sphere and pseudosphere of 1 unit radius.
>
> Now we ask what is the maximum surface area of contact when the sphere
> is
> rolled around once on the northern-pseudosphere and then rolled around
> once on the southern-pseudosphere.
>
> As in all mathematics, once the problem is clear and the definitions
> well-defined,
> only then can a proper math take place. Just like my precision
> defining of finite-number,
> I have not yet reached a precision outline of the problem of surface
> area contact.
> Hopefully this is the precise problem.
>
> So if we have these two objects and allowed to roll the sphere around
> the northern
> pseudosphere once and then the southern portion once, what is the
> maximum contact
> area between the two objects? Is it 10% of the surface area of the
> sphere?

I don't understand what you understand by "Rolling". There is no
curvature
and such, the gross results for area and volumes results have been so
similar
and so the name "pseudo".For differential areas I here make an outline
but it does
hold necessarily for volumes at a differential element level.

Without looking into any other details in your postings but just to
give you
an idea of which segments of surface have equal area in either case
correspond,
I have put in the following sketch.

http://i50.tinypic.com/2bpmbk.jpg

Consider the following:

The area enclosed between two axially separated planes i.e., axial
difference delZ sliced apart in case of sphere (commonly known as
a spherical segment)

AND

the annular area between radially cut/separated cylindrical shells
separated by delR along radius in case of the pseudosphere somewhat
looking like a frustum of cone if meridian curvature is neglected,
are equal, when delR = delZ.

It is not difficult to derive these as the areas are part of surfaces
of revolution.

In case of sphere/pseudosphere of radius / pseudoradius a,
fully rotated Areas are = 2 pi a delZ and 2 pi a delR respectively.

If further the polar angle difference is same and equal to delTheta,
Area/a = delTheta * delZ = delTheta * delR is the condition for equal
area mapping.

Hope this helps you to determine what or how much to "Roll" in each
case for each of the equal area differential shell surface elements.

Narasimham

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