Solved Circumference of Oval and Ellipse, both simultaneously

[Archimedes Plutonium]

Newsgroups: sci.math
Date: Mon, 15 Jan 2018 11:02:17 -0800 (PST)

Subject: Solved the problem of Circumference for Ellipse, Oval. See no 2.71...
involved, but see an easy trig function solution
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Mon, 15 Jan 2018 19:02:18 +0000


Solved the problem of Circumference for Ellipse, Oval. See no 2.71... involved, but see an easy trig function solution

- show quoted text -
Good news, excellent news, for I solved both Ellipse Oval circumference, bad news, though, see no 2.71... in it.

Another good news, the formula involves A = BCD not addition of A = B+C.

Another good news, it is all very simple, simple indeed, for it is no more than turning a plane sheet of paper into a cylinder or turning it into a cone.

So, now get a plane sheet of paper to do the circle.

__________


__________



__________


Now I did not draw the end lines. But if we bend that paper around to be a cylinder that middle line becomes a circle with exactly that as circumference.

Now we do the ellipse and we need two sheets of plane paper



___
   /
  /
 /
/__

plus a second one identical to that and we bend both into a semi-cylinder forming a whole cylinder.

The ellipse circumference is 2 times one of those straight line diagonals.

Now the oval.

Same as above ellipse, only the bending is not a cylinder but a cone.

We define all Ellipse that exist or ever existed comes from a cylinder diagonal cut.

We define all Ovals, the very definition of oval itself, as all diagonal cuts of a cone.

In this process, I cannot see a 2.71... emerging

But rather instead, I see a 3.14.... emerging and that a trig function will be the formula in

A= BCD

AP

On Monday, January 15, 2018 at 19:13:52 +0000, Archimedes Plutonium wrote:
Solved Circumference of Oval and Ellipse, both simultaneously. See no 2.71...., but rather a trig function

Newsgroups: sci.math
Date: Mon, 15 Jan 2018 11:52:02 -0800 (PST)

Subject: only need an angle Re: Solved Circumference of Oval and Ellipse, both
simultaneously. See no 2.71...., but rather a trig function
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Mon, 15 Jan 2018 19:52:02 +0000

only need an angle Re: Solved Circumference of Oval and Ellipse, both simultaneously. See no 2.71...., but rather a trig function

On Monday, January 15, 2018 at 1:13:59 PM UTC-6, Archimedes Plutonium wrote:
> Newsgroups: sci.math
> Date: Mon, 15 Jan 2018 11:02:17 -0800 (PST)
>
> Subject: Solved the problem of Circumference for Ellipse, Oval. See no 2.71...
>  involved, but see an easy trig function solution
> From: Archimedes Plutonium <plutonium....@gmail.com>
> Injection-Date: Mon, 15 Jan 2018 19:02:18 +0000
>
>
> Solved the problem of Circumference for Ellipse, Oval. See no 2.71... involved, but see an easy trig function solution
>

Alright, all I need here is an angle. All 0 degree angles will be circles for 0 degrees is parallel to base of cone or cylinder.

Any angle off of 0 degrees is a diagonal to the cylinder or cone.

All diagonals of cylinder will end up being a ellipse, all diagonals of a cone will end up being a oval.


> Good news, excellent news, for I solved both Ellipse Oval circumference, bad news, though, see no 2.71... in it.
>
> Another good news, the formula involves A = BCD not addition of A = B+C.
>

The formula for circle is A=BCD as that of Circumference = 3.14..x 2 x radius

The formula for circle is Circumference = parallel straightline  (curling up one sheet of paper)

The formula for ellipse is Circumference = diagonal straight line x 2  (curling 2 semi-cylinders)

The formula for oval is Circumference = diagonal straight line x 2  (curling up 2 semi cones)


> Another good news, it is all very simple, simple indeed, for it is no more than turning a plane sheet of paper into a cylinder or turning it into a cone.
>
> So, now get a plane sheet of paper to do the circle.
>
> __________
>
>
> __________
>
>
>
> __________
>

As of yet, I do not know what trig functions are involved. As of yet, for circle, the apparent trig function is where 0 degrees is 1 and that would be cosine. But each ellipse is composed of two semicylinders glued together. Each Oval is two semicone glued together.

Remember, we define Oval as all those figures that come from a cone diagonal cut.



>
> Now I did not draw the end lines. But if we bend that paper around to be a cylinder that middle line becomes a circle with exactly that as circumference.
>
> Now we do the ellipse and we need two sheets of plane paper
>
>
>
> ___
>    /
>   /
>  /
> /__
>
> plus a second one identical to that and we bend both into a semi-cylinder forming a whole cylinder.
>
> The ellipse circumference is 2 times one of those straight line diagonals.
>
> Now the oval.
>
> Same as above ellipse, only the bending is not a cylinder but a cone.
>

Now you may ask, why semi-cylinder, and semi-cone? If it was not obvious to you before. That if you take a sheet of paper, a plane and draw a straightline, and with only that one piece of paper, you cannot join the ends in one curling. You need to separate sheets of paper to join the endpoints.

> We define all Ellipse that exist or ever existed comes from a cylinder diagonal cut.
>
> We define all Ovals, the very definition of oval itself, as all diagonal cuts of a cone.
>
> In this process, I cannot see a 2.71... emerging
>
> But rather instead, I see a 3.14.... emerging and that a trig function will be the formula in
>
> A= BCD
>

Now, let us talk about the mechanics of the cut, as the cut enters a cylinder and enters a cone. That cut is a specific angle and it is this angle that is going to be used in the formula of circumference a unique angle. As for the circumference, it is going to be an exact circumference, because, well we simply measure the distance of the straightline before we curl into a cylinder or cone.

AP

[Michael Moroney]

Archimedes Plutonium <plutonium....@gmail.com> writes:

>Solved the problem of Circumference for Ellipse, Oval. See no 2.71...
>involved, but see an easy trig function solution

Solved, the problem of whether the ellipse is a conic section.

Some preliminaries:

Top view of the conic section and depiction of the coordinate system used
in the proof:

^ x
|
-+- <= x=h
.' | `.
. | .
| | |
' | '
`. | .'
y <----------+ <= x=0

Cone (side view):
.
/|\
/ | \
/b | \
/---+---' <= x = h
/ |' \
/ ' | \
/ ' | \
x = 0 => '-------+-------\
/ a | \

Proof:

r(x) = a - ((a-b)/h)x and d(x) = a - ((a+b)/h)x, hence

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

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

qed

[Archimedes Plutonium]

Excellent, fine fantastic, a bit of luck for luck is rare. A piggyback ride on the back of the ellipse corcumference.

Notice that each ellipse perimeter has a unique attendant oval. Now this attendent oval is probably going to be related to the perimeter, vis a vis, by a constant, similiar to the fact volume cone is 1/3 cylinder.

So there is a good chance that once i know the perimeter of a ellipse and angle of cut that the attendant oval is a constant of that perimeter.

So what i thought was a difficult calculation may well turn out the easiest.

AP

[Serg io]

Bonus Question:

who solved these first ?
(we have their solution in writings passed down through history)



Bonus Problem:

solve using spherical coordinates

[benj]

Can't have solutions, Sergio, because a few posts back Archiepoo
"Proved" that ellipses were NOT conic sections because only "Ovals" were
and ellipses weren't "ovals".

I suggest that you wait a few years and do lots of cooking in aluminum
pans and soon you will also come to understand this genius insight.

[Archimedes Plutonium]

So benj is smarter than sergio

Have to keep that in mind

What is wrong with sergio?

Too much Hillary on the mind

Too much coal fired polluted air from Beijing