Add onto my 150th book// Proof Research into Cylinder, Cone, Sphere, Ellipsoid, Ovoid, Torus; fixing the mistakes of Apollonius and Euclid // Math Research series, book 2 by Archimedes Plutonium

So I need to establish if the pattern of 78.5% for circle inside square and ellipse inside rectangle holds for oval. My latest newly revised definition of oval is this.

Oval-- is the adjoining of 1/2 of an ellipse with a different 1/2 of a different ellipse. And where this oval fits snug inside a isosceles trapezoid and intersects the isosceles trapezoid in 4 and only 4 points of the oval.

So all I have to do to call it a 100% success story is see if 78.5% of the isosceles trapezoid is the area of the enclosed oval.

The pattern of 78.5% held for circle inside square, and held for ellipse inside rectangle, does it hold for oval inside of isosceles trapezoid. If it does, it nails down permanently the definition of oval.

--- various quotes from my 150th book ---
Proof Research into Cylinder, Cone, Sphere, Ellipsoid, Ovoid, Torus; fixing the mistakes of Apollonius and Euclid // Math Research series, book 2
by Archimedes Plutonium

Our radius of the circle is 1 so its area 1/2 of 6.28.. *radius^2 is 3.1415...*(1^2), hence area is 3.1415
and our square area was 4. So here we have a percentage of 3.1415 / 4 = 78.5%

Ellipsoid volume is easy because we see first if ellipse is a 78.5% pattern inside a rectangle, and if so, then we just switch out the radius of a sphere with the semi axes of the ellipsoid.

78.5% of a rectangle area equals the area of inscribed ellipse inside that rectangle
78.5% of rectangle perimeter approximates the perimeter (circumference) of inscribed ellipse

We can say the area of a inscribed circle inside a square is always 78.5% of the area of the square.

78.5% of a square area equals the area of inscribed circle inside that square
78.5% of square perimeter equals the perimeter (circumference) of inscribed circle

We examine any given cone and we inscribe it inside a associated square-pyramid. Remember, the volume of square-pyramid was found to be V = (1/3) side^2 (height).

Here we cross-reference each square of the square-pyramid with the height to each circle of the cone with height.

Each of the squares as we go higher in the square-pyramid is related to a circle in the cone by that of 78.5%.

Now we cross reference the volume of the inscribed cylinder with a rectangular box that encloses the cylinder. Cylinder volume = V = ((1/2)pi*r^2)(height).

Here again the percentage 78.5% comes forth.

78.5% of a rectangular box surface area equals the inscribed cylinder surface area
78.5% of rectangular box volume equals the inscribed cylinder volume.

Since the cone with associated cylinder is directly related to square-pyramid with associated rectangular box by the same percentage 78.5% we can then replace the (side^2) in square-pyramid with the ((1/2)pi*r^2) in cylinder, leaving us with a proof that the volume of the cone is V = (1/3) ((1/2)pi*r^2)(height).

--- end quoting ---

So, all I need do is find out if the inscribed oval inside a isosceles trapezoid with only 4 points of intersection is 78.5% area that of the isosceles trapezoid.

At the moment I do not have any eggs in the house to make preliminary practical measurement.

But I suspect I can find out the answer from simply using the knowledge of ellipse inscribed in rectangle is 78.5% area.

And I hypothesis the oval is two different ellipses, each 1/2 and stuck together at what can be called the oval midpoint-line. Now this hypothesis would say that any given ellipse can only have a unique other smaller ellipse and a larger ellipse connected to the midpoint-line and produce two ovals. One oval for the smaller ellipse and one oval for the larger ellipse stuck to the given ellipse. I have no evidence that this is true-- for it is all coming from my mind's-eye but that is easily susceptible to error.

And the reason I see a unique oval from a half ellipse with a larger ellipse and a smaller ellipse attached at midpoint line, the reason I see these 2 as unique because of what is called a smooth curve in geometry. A smooth curve for me is one that has no vertices. So if I were to attach a 1/2 ellipse to another 1/2 ellipse, a vertex arises in most cases except for 2 different ellipses, one larger and one smaller.

And it is at this midpoint-line of oval that will be two of the 4 points of intersection of the oval with isosceles trapezoid.

And so I have the feeling that I can arrive at a proof of 78.5% simply from the facts and data of ellipses with rectangles.

Where the smallest ellipse of a oval is the smaller of the parallel sides and so less area, while the larger ellipse requires the length of rectangle to become larger to make the slanted sides of isosceles trapezoid.

Does the decrease in area for smaller ellipse become compensated by increase in area for larger ellipse of the Oval?? It must as the rectangle ellipse is 78.5%, and if 78.5% for oval inside of isosceles trapezoid is true.

So I have the sneaky suspicion that I can prove this without even test sampling on a egg oval.

AP, King of Science, especially Physics

Alright I did manage to find some pictures on the Internet of oval inside a isosceles trapezoid-- Research Gate "egg inscribed in an isosceles trapezoid"

And below that is a oval inscribed inside a quadrilateral figure shaped like this.

/\
\/ where the 4 points of intersection are on the sides of the oval, not at poles of oval.
Alamy "hand geometrical drawing..."

I bring this up because it appears the oval is 66% of area in isosceles trapezoid while in the two triangles it is the desired 78.5%.

I have to see if this figure has a name.

AP

On Monday, December 19, 2022 at 12:40:27 AM UTC-6, Archimedes Plutonium wrote:
> Alright I did manage to find some pictures on the Internet of oval inside a isosceles trapezoid-- Research Gate "egg inscribed in an isosceles trapezoid"
>
> And below that is a oval inscribed inside a quadrilateral figure shaped like this.
>
> /\
> \/ where the 4 points of intersection are on the sides of the oval, not at poles of oval.
> Alamy "hand geometrical drawing..."
>
> I bring this up because it appears the oval is 66% of area in isosceles trapezoid while in the two triangles it is the desired 78.5%.
>
> I have to see if this figure has a name.
>
The name of the figure is the Kite.

I was measuring the areas in the Alamy picture on internet and it does come up to about 78.5%

Measurements done in mm.
Entire kite area is .5(34x51) + .5(34x22) = 867 + 374 = 1241
Area of 4 triangles is .5(16x12) + .5(12x6) + 2*.5(27x5)= 96 + 36 + 135 = 267
So the area of egg oval inside of kite is 1241 - 267 = 974

Is 974 that of 78.5% of 1241???

Well multiply 1241 by .785 gives 974

Or in Sigma Error, no error.

So here, I have found the enclosing quadrilateral-- a kite, and I had first though it to be the isosceles trapezoid.

But I must ask the question, a egg oval shape is common in biology, and the kite, I would imagine is common in biology of birds wings and flying. Am I correct on that assumption that the kite is the figure of aerodynamic importance?

So we have two figures of biology of common occurence-- oval egg and kite for flying.

Now my above is not a proof, but well on its way of being a proof.

So much more I need to prove, such as is the oval in that Alamy picture having 2 foci and are the foci related to the intersection line and is the two diagonals of the kite, the center of the oval?

Much progress but much more needs to be done. Especially the proofs of uniqueness. Draw any kite, and only a unique oval fits inside, and that oval is a composition of 2 different ellipses. Where exactly does one ellipse end and the other second smaller ellipse begin????

AP

Alright, this is getting better all the time. It is the Kite that encloses the Oval, not the isosceles-trapezoid.

So my next important question since a given ellipse produces a unique rectangle, does a kite produce a unique rectangle and thus produces a unique Oval.

Yes, looking at the kite we see it is assembled from a square inside a rectangle of the kites 4 triangles, 2 of the triangles are formed from a square, the last 2 triangles form the tail of the kite.

This uniqueness provides a roadmap of what the oval is. A unique ellipse from a rectangle and a unique oval from rectangle turned into a kite.

On Monday, December 19, 2022 at 2:38:25 AM UTC-6, Archimedes Plutonium wrote:
> On Monday, December 19, 2022 at 12:40:27 AM UTC-6, Archimedes Plutonium wrote:
> > Alright I did manage to find some pictures on the Internet of oval inside a isosceles trapezoid-- Research Gate "egg inscribed in an isosceles trapezoid"
> >
> > And below that is a oval inscribed inside a quadrilateral figure shaped like this.
> >
> > /\
> > \/ where the 4 points of intersection are on the sides of the oval, not at poles of oval.
> > Alamy "hand geometrical drawing..."
> >
> > I bring this up because it appears the oval is 66% of area in isosceles trapezoid while in the two triangles it is the desired 78.5%.
> >
> > I have to see if this figure has a name.
> >
> The name of the figure is the Kite.
>
> I was measuring the areas in the Alamy picture on internet and it does come up to about 78.5%
>
> Measurements done in mm.
> Entire kite area is .5(34x51) + .5(34x22) = 867 + 374 = 1241
> Area of 4 triangles is .5(16x12) + .5(12x6) + 2*.5(27x5)= 96 + 36 + 135 = 267
> So the area of egg oval inside of kite is 1241 - 267 = 974
>
> Is 974 that of 78.5% of 1241???
>
> Well multiply 1241 by .785 gives 974
>
> Or in Sigma Error, no error.
>
> So here, I have found the enclosing quadrilateral-- a kite, and I had first though it to be the isosceles trapezoid.
>
> But I must ask the question, a egg oval shape is common in biology, and the kite, I would imagine is common in biology of birds wings and flying. Am I correct on that assumption that the kite is the figure of aerodynamic importance?
>
> So we have two figures of biology of common occurence-- oval egg and kite for flying.
>
> Now my above is not a proof, but well on its way of being a proof.
>
> So much more I need to prove, such as is the oval in that Alamy picture having 2 foci and are the foci related to the intersection line and is the two diagonals of the kite, the center of the oval?
>
> Much progress but much more needs to be done. Especially the proofs of uniqueness. Draw any kite, and only a unique oval fits inside, and that oval is a composition of 2 different ellipses. Where exactly does one ellipse end and the other second smaller ellipse begin????
>
> AP

Alright I can see from the Alamy egg enclosed inside of Kite, where we can take any and every ellipse, where the adjoining of a different smaller ellipse takes place and thus creates the oval.
It is the line segment of the broadest line of the kite, formed from the two triangles that make a square in the half rectangle of a ellipse (and this half rectangle needs shortening. It is at that point of the curve that rises then falls off once crossing that point. It is this point where you can continue with the ellipse or add on a smaller ellipse to form a oval. A unique smaller ellipse that forms the oval.

So the rectangle and kite dictate where the oval is formed.

> Alamy "hand geometrical drawing..."
>
> I bring this up because it appears the oval is 66% of area in isosceles trapezoid while in the two triangles it is the desired 78.5%.
>
> I have to see if this figure has a name.
>
The name of the figure is the Kite.

I was measuring the areas in the Alamy picture on internet and it does come up to about 78.5%

Measurements done in mm.
Entire kite area is .5(34x51) + .5(34x22) = 867 + 374 = 1241
Area of 4 triangles is .5(16x12) + .5(12x6) + 2*.5(27x5)= 96 + 36 + 135 = 267
So the area of egg oval inside of kite is 1241 - 267 = 974

Is 974 that of 78.5% of 1241???

Well multiply 1241 by .785 gives 974

Or in Sigma Error, no error.

So here, I have found the enclosing quadrilateral-- a kite, and I had first though it to be the isosceles trapezoid.

But I must ask the question, a egg oval shape is common in biology, and the kite, I would imagine is common in biology of birds wings and flying. Am I correct on that assumption that the kite is the figure of aerodynamic importance?

So we have two figures of biology of common occurence-- oval egg and kite for flying.

Now my above is not a proof, but well on its way of being a proof.

So much more I need to prove, such as is the oval in that Alamy picture having 2 foci and are the foci related to the intersection line and is the two diagonals of the kite, the center of the oval?

Much progress but much more needs to be done. Especially the proofs of uniqueness. Draw any kite, and only a unique oval fits inside, and that oval is a composition of 2 different ellipses. Where exactly does one ellipse end and the other second smaller ellipse begin????

AP

And this leads to a mystery question, instead of attaching a smaller 1/2 ellipse or a larger 1/2 ellipse at
this juncture point defined by the kites broadest line segment. Can we attach a semicircle at that juncture point? If so, we then have two different classes of Ovals. One class is the joining of 2 different 1/2 ellipses and the other class is the adjoining of a semicircle to a 1/2 ellipse.

Looking at the egg (bought some today) it looks very much semicircle. And logically it makes sense that ovals should be of two primary sources of ellipse and circle, because rectangle and squares are two primary sources in quadrilaterals. Hard not to imagine that rectangle and squares relate and that ellipse and circle must then relate.

So, well, hard to prove. Let me call that inflection point in the kite broadest line segment the maximum point where the curve then declines to the other pole. So all I need to do is prove that a second semiellipse can fit, smaller than the first semiellipse and that a semicircle can fit.

On Monday, December 19, 2022 at 11:37:50 PM UTC-6, Archimedes Plutonium wrote:
> Alright I can see from the Alamy egg enclosed inside of Kite, where we can take any and every ellipse, where the adjoining of a different smaller ellipse takes place and thus creates the oval.
> It is the line segment of the broadest line of the kite, formed from the two triangles that make a square in the half rectangle of a ellipse (and this half rectangle needs shortening. It is at that point of the curve that rises then falls off once crossing that point. It is this point where you can continue with the ellipse or add on a smaller ellipse to form a oval. A unique smaller ellipse that forms the oval.
>
> So the rectangle and kite dictate where the oval is formed.
> > Alamy "hand geometrical drawing..."
> >
> > I bring this up because it appears the oval is 66% of area in isosceles trapezoid while in the two triangles it is the desired 78.5%.
> >
> > I have to see if this figure has a name.
> >
> The name of the figure is the Kite.
>
> I was measuring the areas in the Alamy picture on internet and it does come up to about 78.5%
>
> Measurements done in mm.
> Entire kite area is .5(34x51) + .5(34x22) = 867 + 374 = 1241
> Area of 4 triangles is .5(16x12) + .5(12x6) + 2*.5(27x5)= 96 + 36 + 135 = 267
> So the area of egg oval inside of kite is 1241 - 267 = 974
>
> Is 974 that of 78.5% of 1241???
>
> Well multiply 1241 by .785 gives 974
>
> Or in Sigma Error, no error.
>
> So here, I have found the enclosing quadrilateral-- a kite, and I had first though it to be the isosceles trapezoid.
>
> But I must ask the question, a egg oval shape is common in biology, and the kite, I would imagine is common in biology of birds wings and flying. Am I correct on that assumption that the kite is the figure of aerodynamic importance?
>
> So we have two figures of biology of common occurence-- oval egg and kite for flying.
>
> Now my above is not a proof, but well on its way of being a proof.
>
> So much more I need to prove, such as is the oval in that Alamy picture having 2 foci and are the foci related to the intersection line and is the two diagonals of the kite, the center of the oval?
>
> Much progress but much more needs to be done. Especially the proofs of uniqueness. Draw any kite, and only a unique oval fits inside, and that oval is a composition of 2 different ellipses. Where exactly does one ellipse end and the other second smaller ellipse begin????
>
> AP

Alright, I need to include the Stadium figure. And see if it has that magical number of 78.5% that the enclosed circle with square, ellipse with rectangle and oval with kite.

I need to make several definitions.

Oval = the appending of 2 different semiellipse or semicircle+ semiellipse.
All ellipse and oval have 4 special points-- these points are where the number to right or left is rising or declining, one or the other but not both. Call them poles. On a ellipse and oval, we call the portion of them the leg of the ellipse or leg of the oval. On the stadium, the leg of the stadium is the two parallel straight line segments.

Now with those definitions, I seek to find out if a ellipse or oval, enclosed inside a stadium yields that magical number of 78.5% of area. Obviously a unique ellipse fits inside of any given stadium with 4 points of intersection. But with the oval, a semicircle oval +semiellipse is up for question. I need to prove also, if a oval exists that is semicircle + semiellipse.

AP

I am glad I wrote my 150th book for I find myself continually using it for reference. Proof Research into Cylinder, Cone, Sphere, Ellipsoid, Ovoid, Torus; fixing the mistakes of Apollonius and Euclid // Math Research series, book 2

by Archimedes Plutonium

150th published book

78.5% of a rectangular box surface area equals the inscribed cylinder surface area

78.5% of rectangular box volume equals the inscribed cylinder volume.

Now I cannot help but notice that phi the golden ratio is 1.618.... while 0..785^2 is 0.616 for a Sigma Error of just 618/616 = 0.3%, and in physics we say within 0.5% or less they are equal. The interpretation here is the quantum mechanics psi squared that pi digits and golden ratio are the same thing.

Some philosophical geometry. I call it that because the answers are not clear enough.
1) Is pi involved in the parabola? I would say no because those are polynonomials.
2) Is pi involved in hyperbola? Not sure for it has the features of a circle equation. And instead of conics being >< apex to apex, if we had then <> base to base, then the hyperbola cut would be indead a ellipse, perhaps a circle.

3) Can we say the only smooth curves in geometry are circle, ellipse, oval? Anything else that is closed looped and a curve is not smooth but has a vertex or more.

4) The proof that the area of circle is pi*r^2 involves slicing the circle up into fine rectangles that form a rectangle of r height and base of pi*r. But can the proof of area begot from the fact of 1/4 of circle enclosed in square of radius^2? Can we argue that multiply this square of r^2 times pi is area circle?

AP

So to have a smooth curve, no vertices, requires 4 points as maximums where to the left and right of that point is a rise or fall but not both.

What about a stadium? we see the ends as 2 maximum but not the 4, instead we see a constant, no rise, no fall as the parallel line segments take off.

Quadrilaterals have 4 vertices as they close loop around.

Triangles have 3 vertices and closed looped.

A parabola has 1 maximum. A hyperbola has 2 maximums.

Curiously and caught my attention fast, a polynomial of 5th degree has 4 maximums.

AP

🐁 of Math and 🐀 of Physics Archimedes "tard" Plutonium
<plutonium.archimedes@gmail.com> fails at math and science:
> Now I cannot help but notice that phi the golden ratio is 1.618.... while 0.785^2 is 0.616 for a Sigma Error of just 618/616 = 0.3%, and in physics we say within 0.5% or less they are equal.

"We" do? "We" who? Any physicist or mathematician who thinks two
significant figures being the same is "equal" needs to be fired. Who
would want to drive across a bridge that spans a gap of 618 meters but
was only 616 meters long??? According to StupidPlutonium, the bridge is
equal to the gap! Hahaha!!!

> The interpretation here is the quantum mechanics psi squared that pi digits and golden ratio are the same thing.

Only the interpretation of the insane...

Especially since it's all numerology, not math or science.

Moscow electric Blackouts██۞██
Kibo Parry M. is he using bola or javelin short circuit???
On Tuesday, December 20, 2022 at 10:26:03 PM UTC-6, Michael Moroney wrote:

> Moscow electric Blackouts██۞██

James Parry dunce in sci.math & sci,physics
On Tuesday, December 20, 2022 at 11:49:40 AM UTC-6, Michael Moroney wrote:
> Pluto, so you finally tracked down James Parry in Moscow?
>
> But once again, that sentence no verb.
>
> Kibo WAS short circuiting Moscow electricity?
> Kibo WILL BE short circuiting Moscow electricity?
> Kibo PLANS ON short circuiting Moscow electricity?
> Kibo THINKS ABOUT short circuiting Moscow electricity?
>
> ..?
>
> Now that you found Kibo, will you stop pretending that I am Kibo?
> Hopefully he won't be using your aluminum pole idea, otherwise there
> will be an ex-Kibo. As in:
>
> Kibo WAS VAPORIZED WHILE short circuiting Moscow electricity.

Kibo, is Xi aware of the fact that Outer Manchuria is 28 times bigger than Taiwan????
>
>
>
> > And as the Baby Xi grew up from the rice paddies and reeds of Outer Manchuria, stolen by the Naxi and Zani Dictator Putin in Moscow, Xi learned in school in chemical engineering that Taiwan was 1/28 the size of Outer Manchuria, as Putin bombs Ukraine. And the nascent Xi orders 1,000 divisions to the Outer Manchuria border to regain back the stolen Old China.
> >
> > > On Thursday, December 15, 2022 at 3:01:41 AM UTC-6, simon ross wrote:
> > > > On Thursday, December 15, 2022 at 3:42:11 AM UTC-5, Chris M. Thomasson wrote:
> > > > > On 12/15/2022 12:29 AM, Archimedes Plutonium wrote:
> > > > > > Chris in a javelin throw to knock out Moscow high voltage lines
> > > > > >> ........///// /////////////////////////////////////////////////////////////////////////|
> > > > > >> ......///// /////////////////////////////////////////////////////////////////////////..|
> > > > > >> ....///// /////////////////////////////////////////////////////////////////////////....|
> > > > > >> ..///// /////////////////////////////////////////////////////////////////////////......|
> > > > > >> ///// /////////////////////////////////////////////////////////////////////////........|
> > > > > >> |__| |__|__|__| |__|__|__|__|__| |__|__|__|__|__| |__|__|__|__|__|......../
> > > > > >> |__| |__|__|__| |__|__|__|__|__| |__|__|__|__|__| |__|__|__|__|__|....../
> > > > > >> |__| |__|__|__| |__|__|__|__|__| |__|__|__|__|__| |__|__|__|__|__|..../
> > > > > >> |__| |__|__|__| |__|__|__|__|__| |__|__|__|__|__| |__|__|__|__|__|../
> > > > > >> |__| |__|__|__| |__|__|__|__|__| |__|__|__|__|__| |__|__|__|__|__|/
> > > > > >
> > > > > >
> Vitaly a freedom fighter in Russia
> > > > > >
> > > > > >> ||
> > > > > >> ||
> > > > > >> //////// ||
> > > > > >> |:---[.] ||
> > > > > >> |( _J ||
> > > > > >> | ^ ( _| ||
> > > > > >> / \_____) ||
> > > > > >> / _____ \ || Long Aluminum javelin to throw to short circuit
> > > > > >> | / \ | || .
> > > > > >> | | | | ||
> > > > > >> \ /\ /\__ | ||
> > > > > >> | | \/---
> > > > > >> \ \ )
> > > > > >> | >_____/_____)
> > > > > >> \__________/ ||
> > > > > >> / \ ||
> > > > > >> | | ~~
> > > > > >> \ \\
> > > > > >> \ |\
> > > > > >> \ | \
> > > > > >> \ | |
> > > > > >> | | |
> > > > > >> | | |
> > > > > >> | / |
> > > > > >> |________/____|
> > > > > >> (_________)____)
> > > > > >> ==============================
> > > > > >> freedom fighter from the tiresome dictator
> > > > > >
> > > > > > Yea, the dictator has no power and no hot water, no heat, no electricity for his house, offices.