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Topic: 4)are all German mathematicians like Siegfried Bosch
, Albrecht Böttcher, Dietrich Braess- as dumb as Franz, tea
ching a Conic section is ellipse, when in truth it is an ova
l?

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plutonium.archimedes@gmail.com

Posts: 17,473
Registered: 3/31/08
4)are all German mathematicians like Siegfried Bosch
, Albrecht Böttcher, Dietrich Braess- as dumb as Franz, tea
ching a Conic section is ellipse, when in truth it is an ova
l?

Posted: Oct 1, 2017 7:24 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

3)are all German mathematicians like Hans Georg Bock,
Hans-Jürgen Borchers, Walter Borho -- as dumb as Franz, teaching a Conic section is ellipse, when in truth it is an oval?


Synthetic Geometry & Analytical Geometry Proofs that Conic section = Oval, never an ellipse-- World's first proofs thereof

_Synthetic Geometry proofs that Cylinder section= Ellipse// Conic section= Oval

First Synthetic Geometry proofs, later the Analytic Geometry proofs.

Alright I need to get this prepared for the MATH ARRAY of proofs, that the Ellipse is a Cylinder section, and that the Conic section is an oval, never an ellipse

PROOF that Cylinder Section is an Ellipse, never a Oval::
I would have proven it by Symmetry. Where I indulge the reader to place a circle inside the cylinder and have it mounted on a swivel, a tiny rod fastened to the circle so that you can pivot and rotate the circle. Then my proof argument would be to say--when the circle plate is parallel with base, it is a circle but rotate it slightly in the cylinder and determine what figure is produced. When rotated at the diameter, the extra area added to the upper portion equals the extra area added to bottom portion in cylinder, symmetrical area added, hence a ellipse. QED

Now for proof that the Conic section cannot be an ellipse but an oval, I again would apply the same proof argument by symmetry.

Proof:: Take a cone in general, and build a circle that rotates on a axis. Rotate the circle just a tiny bit for it is bound to get stuck or impeded by the upward slanted walls of the cone. Rotate as far as you possibly can. Now filling in the area upwards is far smaller than filling in the area downwards. Hence, only 1 axis of symmetry, not 2 axes of symmetry. Define Oval as having 1 axis of symmetry. Thus a oval, never an ellipse. QED

The above two proofs are Synthetic Geometry proofs, which means they need no numbers, just some concepts and axioms to make the proof work. A Synthetic geometry proof is where you need no numbers, no coordinate points, no arithmetic, but just using concepts and axioms. A Analytic Geometry proof is where numbers are involved, if only just coordinate points.

Array:: Analytic Geometry proof that Cylinder section= Ellipse//Conic section = Oval, never ellipse

Now I did 3 Experiments and 3 models of the problem, but it turns out that one model is superior over all the other models. One model is the best of all.

That model is where you construct a cone and a cylinder and then implant a circle inside the cone and cylinder attached to a handle so that you can rotate the circle inside. Mine uses a long nail that I poked holes into the side of a cylinder and another one inside a cone made from heavy wax paper of magazine covers. And I used a Mason or Kerr used lid and I attached them to the nail by drilling two holes into each lid and running a wire as fastener. All of this done so I can rotate or pivot the circle inside the cylinder and cone. You need a long nail, for if you make the models too small or too skinny, you lose clarity.

ARRAY, Analytic Geometry Proof, Cylinder Section is a Ellipse::


E
__
.-' `-.
.' `.
/ \
; ;
| G c | H
; ;
\ /
`. .'
`-. _____ .-'
F

The above is a view of a ellipse with center c and is produced by the Sectioning of a Cylinder as long as the cut is not perpendicular to the base, and as long as the cut involves two points not larger than the height of the cylinder walls. What we want to prove is that the cut is always a ellipse, which is a plane figure of two axes of symmetry with a Major Axis and Minor Axis and center at c.

Side view of Cylinder EGFH above with entry point cut at E and exit point cut at F and where c denotes the central axis of the cylinder and where x denotes a circle at c parallel with the base-circle of cylinder

| |
| | E
| |
| |
|x c |x
| |
| |
| |
|F |
| |
| |
| |


So, what is the proof that figure EGFH is always an ellipse in the cylinder section? The line segment GH is the diameter of the circle base of cylinder and the cylinder axis cuts this diameter in half such that Gc = cH. Now we only need to show that Fc = cE. This is done from the right triangles cxF and cxE, for we note that by Angle-Side-Angle these two right triangles are congruent and hence Fc = cE, our second axis of symmetry and thus figure EGFH is always an ellipse. QED



Array proof:: Analytic Geometry proof that Conic section= Oval// never ellipse

ARRAY, Analytic Geometry Proof, Conic Section is a Oval, never an ellipse::


A
,'" "`.
/ \
C | c | D
\ /
` . ___ .'
B

The above is a view of a figure formed from the cut of a conic with center c as the axis of the cone and is produced by the Sectioning of a Cone as long as the cut is not perpendicular to the base, and as long as the cut is not a hyperbola, parabola or circle (nor line). What we want to prove is that this cut is always a oval, never an ellipse. An oval is defined as a plane figure of just one axis of symmetry and possessing a center, c, with a Major Diameter as the axis of symmetry and a Minor Diameter. In our diagram above, the major diameter is AB and minor diameter is CD.

Alright, almost the same as with Cylinder section where we proved the center was half way between Major Axis and Minor Axis of cylinder, only in the case of the Conic, we find that the center is half way between CD the Minor Diameter, but the center is not halfway in between the Major Diameter, and all of that because of the reason the slanted walls of the cone cause the distance cA to be far smaller than the distance cB. In the diagram below we have the circle of x centered at c and parallel to base. The angle at cx is not 90 degrees as in cylinder. The angle of cAx is not the same as the angle cBx, as in the case of the cylinder, because the walls of the cone-for line segments- are slanted versus parallel in the cylinder. Triangles cAx and cBx are not congruent, and thus, the distance of cA is not equal to cB, leaving only one axis of symmetry AB, not CD.

/ \A
x/ c \x
B/ \

Hence, every cut in the Cone, not a hyperbola, not a parabola, not a circle (not a line) is a Oval, never an ellipse.

QED

AP





On Sunday, October 1, 2017 at 2:32:13 PM UTC-5, Me(FRANZ) wrote:
> On Sunday, October 1, 2017 at 9:25:13 PM UTC+2, Me(FRANZ) wrote:

> Cone (side view):
>
> .
> /|\
> / | \
> /b | \
> /---+---´ <= x = h
> / |´ \
> / ´ | \
> / ´ | \
> x = 0 => ´-------+-------\
> / a | \
>
> r(x) = a - ((a-b)/h)x
> d(x) = a - ((a+b)/h)x
>
> y(x)^2 = r(x)^2 - d(x)^2 = ab - ab(2x/h - 1)^2 = ab(1 - 4(x - h)^2/h^2
>
> => (1/ab)y(x)^2 + (4/h^2)(x - h)^2 = 1 ...equation of an ellipse
>
> Considerations:
>
> => y(h/2 + x')^2 = sqrt(ab - ab(2(h/2 + x')/h - 1)^2) = ab - ab(2x'/h)^2
>
> => y(h/2 + x') = sqrt(ab) * (sqrt(1 - (2x'/h)^2) ...symmetric relative to h/2
>
> => y(h/2) = sqrt(ab)
>
> ======================================================
>
> @Archie: Yes, this proves that a cone section "as depicted in my diagram" is an ellipse.


Below is a German, Franz, who "flys off the handle"


From: Me <franz.fri...@gmail.com>

> ... if the ellipse is a conic, then what the hell is a cylinder section
> -- Jan and Franz would likely say a rectangle.


No, we (I guess) would both say: an ellipse too.

(Yes, ok, I admit that his is slightly counterintuitive, but alas it's just a fact.)


"21st-century German mathematicians"

A
Karim Adiprasito
B
Hans Werner Ballmann
Victor Bangert

Martin Barner
Kai Behrend

Rudolf Berghammer
Jörg Bewersdorff

zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
Hans Georg Bock
Hans-Jürgen Borchers
Walter Borho
Siegfried Bosch
Albrecht Böttcher
Dietrich Braess
Simon Brendle
Egbert Brieskorn
Kathrin Bringmann
Franz Thomas Bruss
C
Joachim Cuntz
D
Wolfgang Dahmen
Albrecht Dold
E
Arthur Engel (mathematician)
F
Gerd Faltings
Klaus Fischer (mathematician)
Hans Föllmer
Helmar Frank
Jens Franke
Eberhard Freitag
Peter Friz
G
Helmuth Gericke
Siegfried Gottwald
Hans Grauert
H
Heiko Harborth
Günter Harder
Gisbert Hasenjaeger
Günter Heimbeck
Erhard Heinz
Dagmar R. Henney
Werner Hildenbrand
Diederich Hinrichsen
Helmut Hofer
Gerhard Huisken
Bertram Huppert
Daniel Huybrechts
J
Willi Jäger
Jens Carsten Jantzen
Hartmut Jürgens
K
Friedrich Kambartel
Manuel Kauers
Reinhardt Kiehl
Wilhelm Klingenberg
Horst Knörrer
Ulrich Kohlenbach
Bernhard Korte
Christoph Koutschan
Matthias Kreck
Martin Kreuzer
Erwin Kreyszig
Daniela Kühn
L
Joachim Lambek
Walter Ledermann
Felix Leinen
M
Curt Meyer
N
Hans-Joachim Nastold
R
Wolfgang Rautenberg
Gerhard Ringel
Peter Roquette
S
Gunther Schmidt
Karl-Otto Stöhr
W
Andreas Winter
Z
Günter M. Ziegler

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Me
6:15 PM (6 minutes ago)


Let's consider the Sectioning of a Cylinder and a Cone.

^ x
E|
-+-
.' | `.
/ | \
. | .
G | +c | H
. | .
\ | /
`. | ´
y <----------+ ´
F

> The above is a view of a ellipse with center c and is produced by the
> Sectioning of a Cylinder as long as the cut is not perpendicular to the base,
> and as long as the cut involves two points not larger than the height of the
> cylinder walls. What we want to prove is that the cut is always a ellipse,
> which is a [certain] plane figure of two axes of symmetry with a Major Axis
> and Minor Axis and center at c.
>
> So, what is the proof that figure EGFH is always an ellipse in the cylinder
> section [as well as in the cone section]?


Here's is an easy prove for it:

Cylinder (side view):

| | |
|-------+-------+ <= x = h
| | ´|
| | ´ |
| |´ |
| ´ | |
| ´ | |
x = 0 => ´-------|-------|
| r | |

d(x) = r - (2r/h)x

y^2 = r^2 - d(x)^2 = r^2 - r^2(2x/h - 1)^2 = r^2(1 - 4(x - h)^2/h^2

=> (1/r^2)y^2 + (4/h^2)(x - h)^2 = 1 ...equation of an ellipse

Considerations:

=> y(h/2 + x')^2 = sqrt(r^2 - r^2(2(h/2 + x')/h - 1)^2) = r^2 - r^2(2x'/h)^2

=> y(h/2 + x') = r * (sqrt(1 - (2x'/h)^2) ...symmetric relative to h/2

=> y(h/2) = r (= Gc = cH)



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