constant negative curvature surface-

On Aug 25, 11:32*am, gudi wrote:
On Aug 25, 3:00*am, JEMebius wrote:



Robert Israel wrote:
gudi writes:

Can anyone please suggest a parametrization of the surface ? (x,y,z) =
f( u,v ),
for Gauss curvature = -1 . The edge may be cuspidal line, with larger
rotations
from center.

http://i25.tinypic.com/65q042.jpg

It looks to me like the surface in the picture is pretty much flat near the
origin. *I don't see how it could have Gaussian curvature -1 there.

I guess the picture shows part of the surface that is generated by a cubic parabola z =
A.x^3 rotating around the Z axis, meanwhile changing A in a sinusoidal manner in three
periods per revolution.

In cylinder coordinates (r, phi, z):

z = Amax . sin(3.phi) . r^3

The corresponding Taylor series in x and y starts with 3rd-degree terms, therefore the
curvature at the origin is indeed zero.

For constant negative curvature please seehttp://en.wikipedia.org/wiki/Pseudosphere.
More inhttp://en.wikipedia.org/wiki/Tractrixandhttp://en.wikipedia.org/wiki/....

Happy studies: Johan E. Mebius

Thanks.Please see my reply to Robert Israel also.(May be you mean z =
Amax sin(6 phi) r^3).

Narasimham

http://www.flickr.com/photos/24869933@N07/3569707781/

Here is an LCD display attempting to render the image:

http://www.flickr.com/photos/24869933@N07/3569733893/

http://www.flickr.com/photos/2486993...th/3619373444/

this is edge of the same image

http://www.flickr.com/photos/2486993...n/photostream/

Musatov

However, as would be a possibility to circulate along that matter, for
instance, when the E would be along the G(E)...

Therefore, then the E, would be an inverse as a continued matter, when
especially would be along an application, which is x -- x coef. -1, all
along...

However, that a definitely would be continued along the G(E), as it would be
a possible to get the E when G(E) would be an open matter along the E,
simply as that, a definitely as a matter a fact...

--
Ahmed Ouahi, Architect
Best Regards!


"M.M.M." kirjoitti
...
On Aug 25, 11:32 am, gudi wrote:
On Aug 25, 3:00 am, JEMebius wrote:



Robert Israel wrote:
gudi writes:

Can anyone please suggest a parametrization of the surface ? (x,y,z)
=
f( u,v ),
for Gauss curvature = -1 . The edge may be cuspidal line, with larger
rotations
from center.

http://i25.tinypic.com/65q042.jpg

It looks to me like the surface in the picture is pretty much flat
near the
origin. I don't see how it could have Gaussian curvature -1 there.

I guess the picture shows part of the surface that is generated by a
cubic parabola z =
A.x^3 rotating around the Z axis, meanwhile changing A in a sinusoidal
manner in three
periods per revolution.

In cylinder coordinates (r, phi, z):

z = Amax . sin(3.phi) . r^3

The corresponding Taylor series in x and y starts with 3rd-degree terms,
therefore the
curvature at the origin is indeed zero.

For constant negative curvature please
seehttp://en.wikipedia.org/wiki/Pseudosphere.
More
inhttp://en.wikipedia.org/wiki/Tractrixandhttp://en.wikipedia.org/wiki/....

Happy studies: Johan E. Mebius

Thanks.Please see my reply to Robert Israel also.(May be you mean z =
Amax sin(6 phi) r^3).

Narasimham

http://www.flickr.com/photos/24869933@N07/3569707781/

Here is an LCD display attempting to render the image:

http://www.flickr.com/photos/24869933@N07/3569733893/

http://www.flickr.com/photos/2486993...th/3619373444/

this is edge of the same image

http://www.flickr.com/photos/2486993...n/photostream/

Musatov

On Aug 25, 4:37*pm, "M.M.M." wrote:
On Aug 25, 11:32*am, gudi wrote:





On Aug 25, 3:00*am, JEMebius wrote:

Robert Israel wrote:
gudi writes:

Can anyone please suggest a parametrization of the surface ? (x,y,z) =
f( u,v ),
for Gauss curvature = -1 . The edge may be cuspidal line, with larger
rotations
from center.

http://i25.tinypic.com/65q042.jpg

It looks to me like the surface in the picture is pretty much flat near the
origin. *I don't see how it could have Gaussian curvature -1 there.

I guess the picture shows part of the surface that is generated by a cubic parabola z =
A.x^3 rotating around the Z axis, meanwhile changing A in a sinusoidal manner in three
periods per revolution.

In cylinder coordinates (r, phi, z):

z = Amax . sin(3.phi) . r^3

The corresponding Taylor series in x and y starts with 3rd-degree terms, therefore the
curvature at the origin is indeed zero.

For constant negative curvature please seehttp://en.wikipedia.org/wiki/Pseudosphere.
More inhttp://en.wikipedia.org/wiki/Tractrixandhttp://en.wikipedia.org/wiki/....

Happy studies: Johan E. Mebius

Thanks.Please see my reply to Robert Israel also.(May be you mean z =
Amax sin(6 phi) r^3).

Narasimham

http://www.flickr.com/photos/24869933@N07/3569707781/

Here is an LCD display attempting to render the image:

http://www.flickr.com/photos/24869933@N07/3569733893/

http://www.flickr.com/photos/2486993...th/3619373444/

this is edge of the same image

http://www.flickr.com/photos/2486993...n/photostream/

Musatov- Hide quoted text -

- Show quoted text -

xxein: I'm not here to view comic books.

"Ahmed Ouahi, Architect" wrote:
[snip]

pseudosphere

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2

M.M.M. wrote:
xxein wrote:
On Aug 25, 4:37*pm, "M.M.M." wrote:
On Aug 25, 11:32*am, gudi wrote:





On Aug 25, 3:00*am, JEMebius wrote:

Robert Israel wrote:
gudi writes:

Can anyone please suggest a parametrization of the surface ? (x,y,z) =
f( u,v ),
for Gauss curvature = -1 . The edge may be cuspidal line, with larger
rotations
from center.

http://i25.tinypic.com/65q042.jpg

It looks to me like the surface in the picture is pretty much flat near the
origin. *I don't see how it could have Gaussian curvature -1 there.

I guess the picture shows part of the surface that is generated by a cubic parabola z =
A.x^3 rotating around the Z axis, meanwhile changing A in a sinusoidal manner in three
periods per revolution.

In cylinder coordinates (r, phi, z):

z = Amax . sin(3.phi) . r^3

The corresponding Taylor series in x and y starts with 3rd-degree terms, therefore the
curvature at the origin is indeed zero.

For constant negative curvature please seehttp://en.wikipedia.org/wiki/Pseudosphere.
More inhttp://en.wikipedia.org/wiki/Tractrixandhttp://en.wikipedia.org/wiki/....

Happy studies: Johan E. Mebius

Thanks.Please see my reply to Robert Israel also.(May be you mean z =
Amax sin(6 phi) r^3).

Narasimham

http://www.flickr.com/photos/24869933@N07/3569707781/

Here is an LCD display attempting to render the image:

http://www.flickr.com/photos/24869933@N07/3569733893/

http://www.flickr.com/photos/2486993...th/3619373444/

this is edge of the same image

http://www.flickr.com/photos/2486993...n/photostream/

Musatov- Hide quoted text -

- Show quoted text -

xxein: I'm not here to view comic books.
How unfortunate for you to dismiss truth because of your own
prejudices.

"xxein" wrote in message
...
On Aug 25, 4:37 pm, "M.M.M." wrote:
On Aug 25, 11:32 am, gudi wrote:





On Aug 25, 3:00 am, JEMebius wrote:

Robert Israel wrote:
gudi writes:

Can anyone please suggest a parametrization of the surface ?
(x,y,z) =
f( u,v ),
for Gauss curvature = -1 . The edge may be cuspidal line, with
larger
rotations
from center.

http://i25.tinypic.com/65q042.jpg

It looks to me like the surface in the picture is pretty much flat
near the
origin. I don't see how it could have Gaussian curvature -1 there.

I guess the picture shows part of the surface that is generated by a
cubic parabola z =
A.x^3 rotating around the Z axis, meanwhile changing A in a sinusoidal
manner in three
periods per revolution.

In cylinder coordinates (r, phi, z):

z = Amax . sin(3.phi) . r^3

The corresponding Taylor series in x and y starts with 3rd-degree
terms, therefore the
curvature at the origin is indeed zero.

For constant negative curvature please
seehttp://en.wikipedia.org/wiki/Pseudosphere.
More
inhttp://en.wikipedia.org/wiki/Tractrixandhttp://en.wikipedia.org/wiki/....

Happy studies: Johan E. Mebius

Thanks.Please see my reply to Robert Israel also.(May be you mean z =
Amax sin(6 phi) r^3).

Narasimham

http://www.flickr.com/photos/24869933@N07/3569707781/

Here is an LCD display attempting to render the image:

http://www.flickr.com/photos/24869933@N07/3569733893/

http://www.flickr.com/photos/2486993...th/3619373444/

this is edge of the same image

http://www.flickr.com/photos/2486993...n/photostream/

Musatov- Hide quoted text -

- Show quoted text -

xxein: I'm not here to view comic books.
--------------------------------------------------

You are here to view an artefactual/superficially imposed yin-yang of sorts.

On Aug 25, 10:18*pm, M-Theory wrote:
M.M.M. wrote:
xxein wrote:
On Aug 25, 4:37*pm, "M.M.M." wrote:
On Aug 25, 11:32*am, gudi wrote:

On Aug 25, 3:00*am, JEMebius wrote:

Robert Israel wrote:
gudi writes:

Can anyone please suggest a parametrization of the surface ? (x,y,z) =
f( u,v ),
for Gauss curvature = -1 . The edge may be cuspidal line, with larger
rotations
from center.

http://i25.tinypic.com/65q042.jpg

It looks to me like the surface in the picture is pretty much flat near the
origin. *I don't see how it could have Gaussian curvature -1 there.

I guess the picture shows part of the surface that is generated by a cubic parabola z =
A.x^3 rotating around the Z axis, meanwhile changing A in a sinusoidal manner in three
periods per revolution.

In cylinder coordinates (r, phi, z):

z = Amax . sin(3.phi) . r^3

The corresponding Taylor series in x and y starts with 3rd-degree terms, therefore the
curvature at the origin is indeed zero.

For constant negative curvature please seehttp://en.wikipedia.org/wiki/Pseudosphere.
More inhttp://en.wikipedia.org/wiki/Tractrixandhttp://en.wikipedia.org/wiki/....

Happy studies: Johan E. Mebius

Thanks.Please see my reply to Robert Israel also.(May be you mean z =
Amax sin(6 phi) r^3).

Narasimham

http://www.flickr.com/photos/24869933@N07/3569707781/

Here is an LCD display attempting to render the image:

http://www.flickr.com/photos/24869933@N07/3569733893/

http://www.flickr.com/photos/2486993...th/3619373444/

this is edge of the same image

http://www.flickr.com/photos/2486993...n/photostream/

Musatov- Hide quoted text -

- Show quoted text -

xxein: *I'm not here to view comic books.

How unfortunate for you to dismiss truth because of your own
prejudices.- Hide quoted text -

- Show quoted text -

What "truth"?

I see nothing but a very rough draft of a drawing.

Your draftsmanship needs a lot of work.

Try CAD software.

Tom Davidson
Richmond, VA

Unfortunately or a fortunately, it is a simply an algebra matter...

However, when the E would be along a locality as a definitely a convex
multiplicand...

Therefore, it would be an under the Pi which would be an under multiplicand
as Pi (xy)_ Pi(x) Pi(y) ( x, y eps. E ), simply as that...

Whether, I do mention a simply or a simple, does not means anything is a
simple along that matter, a definitely as a matter a fact...

--
Ahmed Ouahi, Architect
Best Regards!


"Uncle Al" kirjoitti
...
"Ahmed Ouahi, Architect" wrote:
[snip]

pseudosphere

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2

Uncle Al wrote:
"Ahmed Ouahi, Architect" wrote:
[snip]

pseudosphere

Or looking from the inside out?

/BAH

However, mister /BAH, or a just to swim, first of all, along the following
equation, all along...

y² - x² = z
--- --- ---
b² a² c²

a, b 0, c =/= 0

Therefore, as otherwise, to follow the space more, when a definitely the E
would be along a K as along its own stricturation which it would be
(x,y) -- xy of the E x E along the E....

Whatsoever, along that matter would be found the E along the following
metamorphosis as it would be a definitely the

E , as which E would be deduced along the E...
1 1

However, along an infinite space, which would produce what would follows...

(alpha, a).(beta, b) = (alpha beta, alpha b + beta a + ab )

Therefore, as would be deduced from the E along an adjunctive matter of an
unity, as along an operation of the

E -- E , as a simply as that, a definitely as a matter a fact...
1


--
Ahmed Ouahi, Architect
Best Regards!


"jmfbahciv" jmfbahciv@aol kirjoitti
...
Uncle Al wrote:
"Ahmed Ouahi, Architect" wrote:
[snip]

pseudosphere

Or looking from the inside out?

/BAH