the geometric pattern of perception

8/14/2019 The Geometric Pattern of Perception

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The Geometric Patterns of Perception

by Parker EmmersonThe idea of this paper is to explain how, when a wedge of arc length, t, istaken out of a circle and the circle is then put back together, keeping the initialradius as the slant of the newly formed cone, the height increases and the inital radius changes angle to the circular base, whichhas a new radius. The difference between the radii leads to a change in circumference for the circle of our system. That changecan be considered a D @C Ddq . The change is also then placed in terms of time. So, we have relationary time aspects fusing withaxis at an infinite angle.

C = 2 p rThis is the circumference of our initial circle of radius rC2 = 2 p r 1

This is the circumference of our second circle, the base of the cone, of radius r 1

r^2 = r 1 ^2 + h^2This is the initial radius squared expressed as the slant

of the cone in terms of the height of the cone, h, and the radiusof the base of the cone, r 1

r = , Hr 1 ^2 + h^2 Lt = q rt êq = rt = C - C2 = 2 p r - 2 p r 1 = q r Ø Equation 7r 1 ^2 ã r^2 - h^2

r 1 = , Hr^2 - h^2 Lh < r

Solve @r 1 ^2 + h^2 ã r^ 2, h D:: h Ø - r 2

- r 12 >, : h Ø r 2

- r 12 >>

t = C - C2 = 2 p r - 2 p r 1

Put

r 1 = Hr^2 - h^2 Lin for the equation of the base of the cone of Equation 7.

t = C - C2 =

2 p r - 2 p r 1 = 2 p Ht êq L- 2 p HHt êq L^2 - h^2 L= t

Ht êq L= r

J2 p Hr L- 2 p HHr L̂2 - h^2 L N= t

Add 2 p HHr L̂2 - h^2 Lto both sides

t + 2 p HHt êq L̂2 - Hh^2 LL= 2 p Ht êq L= 2 p r

Printed by Mathematica for Students



Subtract t from both sides

2 p HHt êq L̂2 - Hh^2 LL= 2 p Ht êq L- t = 2 p r - t

divide by 2 p on both sides

Ht êq L- t ê2 p = HHt êq L^2 - Hh^2 LL= r 1 = HHr L^2 - Hh^2 LLSquare both sides

HHt êq L- Ht ê2 p LL̂2 = HHt êq L̂2 - Hh^2 LL= Hr 1 ^2 Ladd Hh^2 Lto both sides

Hh^2 L+ HHt êq L- Ht ê2 p LL HHt êq L- Ht ê2 p LL= Ht êq L̂2 = r^2

HHt êq L- Ht ê2 p LL̂2

HHr L- Hq * r ê2 p LL̂2

r -p

rq

2

2

-Hp * r * q L

2+Hr * q L

q

2

r -p r q

2

2

Expand Br -p * r * q

2

2

Fr 2

- p r 2q +

1

4p

2 r 2q

2

Subtract -p r q

2+ r

2

from both sides

Ht êq L̂2 - 8Ht êq L- t ê2 p < 8Ht êq L- t ê2 p <= Hh^2 LHr^2 êq ^2 L- 8Hr L- Hq * r Lê2 p < 8HHq * r Lêq L- Hq * r Lê2 p <Take the square root on both sides to find h

h = Sqrt @Ht êq L̂2 - 8Ht êq L- t ê2 p < 8Ht êq L- t ê2 p <D=

Hr L̂2 - 8Hr L- Hq * r Lê2 p < 8Hr L- Hq * r Lê2 p <

Hr L^2 - 8Hr L- Hq * r Lê2 p < 8Hr L- Hq * r Lê2 p <: r 2

- r -p r q

2

2

>

2 new ideas.nb




r 2- r 2

- p r 2q +

1

4p

2 r 2q

2

p r 2q -

1

4p

2 r 2q

2

p r 2q -

1

4p

2 r 2q

2= h

Square both sides

r 2 q -1

4p 2 r 2 q 2 = Hh^2 L

Divide by r 2 on both sides

p r 2q -

1

4

p 2 r 2

q2 ì Hr^2 L= IHh^2 Lë r 2 M

p q -1

4p

2q

2= IHh^2 Lë r 2 M

Hh^2 Lì p q -1

4p

2q

2= r 2

h = c3 q

HHH3.0 * 10^8 L* 3 * q L̂2 Lì p q -1

4p

2q

2= r 2

r 2

IHh^2 Lë r 2 M= x

Multiply by r^2; Divide by 1 - Hp * q L+1

4 Ip 2 * q 2 Mon both sides

EQUATION Ò X .1

Hh^2 Lì p * q -1

4 Ip2

* q2 M= r 2

set r 2= H1 + 3 + 5 ... H2 n - 1LL

H1 + 3 + 5 ... H2 n - 1LL= Hh^2 Lì p * q -

1

4 Ip 2

* q2

MFor n = 1

1 = Hh^2 Lì p * q -1

4Ip

2* q

2 M,

so

4 new ideas.nb




p * q -1

4Ip

2* q

2 M= Hh^2 L

p * q -1

4 Ip 2 * q 2 M= h

Plot B p * q -1

4 Ip 2 * q 2 M, 8q , - 10, 10 <F

- 10 - 5 5 10

0.2

0.4

0.6

0.8

1.0

Solve Bp * q -1

4 Ip 2

* q2 Mã 0, 8q <F

:8q Ø 0 <, : q Ø4

p>>

For n = 2;

H1 + H2 n - 1LL= 4 = Hh^2 Lì p * q -1

4Ip

2* q

2 MSo, multiplying on both sides,

4 p * q -1

4Ip

2* q

2 M= Hh^2 L

4 p * q -1

4Ip

2* q

2 M= h

new ideas.nb 5




Plot B 4 p * q -1

4Ip

2* q

2 M, 8q , - 10, 10 <F

- 10 - 5 5 10

0.5

1.0

1.5

2.0

For n = 3

Plot B 9 p * q -1

4Ip

2* q

2 M, 8q , - 10, 10 <F

- 10 - 5 5 10

0.5

1.0

1.5

2.0

2.5

3.0

For n = 4; 16 p * q -1

4Ip

2* q

2 M= h

For n = 5; 25 p * q -1

4Ip

2* q

2 M= h

For n = 6; 36 p * q -1

4 Ip 2

* q2 M= h

For n = 7; 49 p * q -1

4 Ip 2 * q 2 M= h

6 new ideas.nb




Plot B: p * q -1

4Ip

2* q

2 M, 4 p * q -1

4Ip

2* q

2 M,

9 p * q -

1

4 Ip 2

* q2

M, 16 p * q -

1

4 Ip 2

* q2

M,

25 p * q -1

4 Ip 2

* q2 M, 36 p * q -

1

4 Ip 2

* q2 M, 49 p * q -

1

4 Ip 2

* q2 M >, 8q , - 2, 2 <F

- 2 - 1 1 2

1

2

3

4

5

6

7

We have plotted what will happen to the height as thetachanges. Next we want to know r 1 and its change with theta if r is the

series r ^2 = H1 + 3 + ... H2 n - 1LLSome have noted the phalas and some have noted that the mind is sometimes thought of as the head.

Remember, r 1 = HHt êq L̂2 - Hh^2 LL= r 1 = Hr^2 - h^2 LFor n = 1;

r 1 = Hr^2 - h^2 L = 1 - p * q -1

4Ip

2* q

2 M

new ideas.nb 7




Plot B 1 - p * q -1

4Ip

2* q

2 M , 8q , - 5, 5 <F

- 4 - 2 2 4

0.2

0.4

0.6

0.8

1.0

For n = 2

Plot B 4 - 4 p * q -1

4Ip

2* q

2 M , 8q , - 5, 5 <F

- 4 - 2 2 4

1.6

1.7

1.8

1.9

2.0

For n = 3;

8 new ideas.nb




Plot B 9 - 9 p * q -1

4Ip

2* q

2 M , 8q , - 5, 5 <F

- 4 - 2 2 4

2.6

2.7

2.8

2.9

3.0

n = 4; r 1 = 16 - 16 p * q - 14 Ip 2 * q 2 M

n = 5; r 1 = 25 - 25 p * q -1

4Ip

2* q

2 M

n = 6; r 1 = 36 - 36 p * q -1

4 Ip 2 * q 2 M

n = 7; r 1 = 49 - 49 p * q -1

4 Ip 2

* q2 M

new ideas.nb 9




Plot B: 1 - p * q -1

4 Ip 2 * q 2 M , 4 - 4 p * q -1

4 Ip 2 * q 2 M ,

9 - 9 p * q -1

4 Ip 2 * q 2 M , 16 - 16 p * q -1

4 Ip 2 * q 2 M ,

25 - 25 p * q -1

4Ip

2* q

2 M , 36 - 36 p * q -1

4Ip

2* q

2 M ,

49 - 49 p * q -1

4 Ip 2 * q 2 M >, 8q , - 10, 10 <F

C = 2 p rThis is the circumference of our initial circle of radius rC2 = 2 p r 1

This is the circumference of our second circle, the base of the cone, of radius r 1

So, we integrate the circumference, to get a certain area,due to a representationality in the virtual world.

In our last discovery, we found out that the height of the cone is a function of the angle and the radius. Here, we will try toeliminate one of the variables, and we can do this, because it is the same variable as time when given a constant velocity, say of light going toward your eye. We will try to see about the concept of negative time and philosophically discuss its objectivemeaning. We begin by consulting Figure x.

from equations of a cone, we know that

Hh^2 LëIp * q -1

4 Ip 2 * q 2 MM= r 2

Thus, we can also say that for r^2=1, because n=1; that,

p * q -1

4Ip

2* q

2 M= Hh^2 Lë r 2= X^2

new ideas.nb 11




p * q -1

4Ip

2* q

2 M= Hh^2 Lë r 2= X^2;

r 2p * f @q D-

1

4Ip

2* f @q D2 M= Hh^2 L= c^2 * t^2

30 sec = p "time"so, for our time equality to work with the angle change, we are going to have to relate 360 degrees = 60 seconds = 2 p rad. Thatmeans to go from seconds to angles, we multiply a theta given in radians by 180/ p in order to go to degrees, and then convert tothose degrees to seconds by multiplying by (1/6 deg.), because 60 seconds=360 degrees.

So, this means that we can say that time is equal to the amount of the circle' s angularity/arc length that we remove from theinitial circle has a unit of time.

from time----->angle measure, we multiply by

From the equationct = c *

Hf

@q

DLS.T.

q is in degrees, and to go to time, we would convert

q deg. Hp rad. ê180 deg. L, and from radians, we say that thirty seconds go by every pi radians.,so to cancel the radians, we say every 30 seconds,

we cover pi radians of distance. Thus, the final conversion fom degrees to minutes runs :q deg. * Hp rad. ê180 deg. L H60 sec êp rad. L= t

Our multiplier then is just 1 sec. ê3 deg. to go to seconds.and we recently found that :

@Hh^2 LêHp r^2 LD= Hq - 1 ê4 p Hq ^2 LLThus, also we may say that it takes 3 deg. êsec. if we are given a time

c^2 * t^2 = r 2p * q -

1

4Ip

2* q

2 M= c^2 * H1 ê3Lq ^2

c^2 * H3 q L̂2 ì p * q -1

4Ip

2* q

2 M = r^2

r = c^2 * H3 q L̂2 ì p * q -1

4Ip

2* q

2 M

60 ê180

1

3

EQUATION Ò X .1

Hh^2 Lì p * q -1

4 Ip2

* q2 M= r 2

new ideas.nb 13




So, if we are given an angle, the generalized forms would look like this :

Ic^2 * HHH1 ê3Lq L̂2 LëIp * q -1

4 Ip 2 * q 2 MMM= r^2

What happens when we factor the equation

H3.0 * 10^8

L̂2 * HHt L̂2 Lì p * q -1

4Ip

2* q

2 M

Factor B H3.0 * 10^8

L̂2 * HHH1 ê3Lq L̂2 Lì p * q -1

4Ip

2* q

2 M

F

1.12838 µ 10 8-

q

- 4 + p q

H3.0 * 10^8

L̂2 * HHt L̂2 Lì p * Hq ê3L-1

4 Ip 2 * Hq ê3L2 M

3. µ 10 8t 2

p q

3-

p 2 q 2

36

An imaginary solution.

Then we' ll plot our time in terms of theta when solved for r.

14 new ideas.nb




Plot B H3.0 * 10^8

L̂2 * HHH1 ê3Lq L̂2 Lì p * q -1

4Ip

2* q

2 M

, 8q , - 10, 10 <F

- 10 - 5 5 10

5.0 µ 107

1.0 µ 108

1.5 µ 108

2.0 µ 108

2.5 µ 108

3.0 µ 108

Plot B H3.0 * 10^8

L̂2 * HHt L̂2 Lì p * 3 t -1

4Ip

2* 3 t 2 M

, 8t, - 10, 10 <F

- 10 - 5 5 10

- 3.0 µ 1016

- 2.5 µ 1016

- 2.0 µ 1016

- 1.5 µ 1016

- 1.0 µ 1016

- 5.0 µ 1015

5.0 µ 1015

new ideas.nb 15




Plot B H3.0 * 10^8

L̂2 * HHt L̂2 Lì p * 3 t -1

4Ip

2* H3 t L2 M

, 8t, - 10, 10 <F

- 10 - 5 5 10

5.0 µ 107

1.0 µ 108

1.5 µ 108

2.0 µ 108

2.5 µ 108

where c is the speed of light and 3 * t = the angle we have taken out of the circle, when keeping the initial radius of the circle as the slant of the new cone,

and our new radius of the base of the cone is determined

Hc^2 L= p * q -1

4 Ip 2 * q 2 Mr^2 êHHH1 ê3Lq L̂2 L

Hc^2 Lêr^2 = p * q -1

4 Ip 2 * q 2 M ì HHH1 ê3Lq L̂2 L

Solve B H3.0 * 10^8 L^2 * H3 q L^2 ì p * q -

1

4 Ip 2

* q2

M ã r, q F:: q Ø

12.5664 r

3.24 µ 10 18+ 9.8696 r

>>

::q Ø12.566370614359172` r

3.24`*^18 + 9.869604401089358` r >>

16 new ideas.nb




Plot B HH3.0 * 10^8 L̂2 Lì p * q -1

4Ip

2* q

2 M ì HHH1 ê3Lq L̂2 L , 8q , - 10, 10 <F

- 10 - 5 5 10

5.0 µ 107

1.0 µ 108

1.5 µ 108

2.0 µ 108

2.5 µ 108

3.0 µ 108

Plot BHH3.0 * 10^8 L̂2 Lì p * q -1

4Ip

2* q

2 M ì HHH1 ê3Lq L̂2 L, 8q , - 100, 100 <F

- 100 - 50 50 100

- 4.4 µ 1015

- 4.2 µ 1015

- 4.0 µ 1015

- 3.8 µ 1015

- 3.6 µ 1015

But to now place q in terms of r, we see a new plot

3p

p

4+

9. µ 10 12

r 2

= t

Factor B3p

p

4+

9.`*^12

r 2

F37.6991 r 2

3.6 µ 10 13+ 3.14159 r 2

new ideas.nb 17




Simplify B3p

p

4+

9.`*^12

r 2

F9.42478 r 2

9. µ 10 12+ 0.785398 r 2

Plot B3p

p

4+

9.`*^12

r 2

, 8r, - 1000, 1000 <F

- 1000 - 500 500 1000

2. µ 10 - 7

4. µ 10 - 7

6. µ 10 - 7

8. µ 10 - 7

1. µ 10 - 6

p

p

4+

9.`*^12

r 2

= q = dt = change in time

qp

4+

9.`*^12

r 2= p

q9.`*^12

r2

= p - qp

4r 2

= q H9.` * ^12 Lêp - qp

4

q H9.` * ^12 Lêp - qp

4= r

18 new ideas.nb




For n = 1;

q * H9.`*^12 Lì p - q *p

4= r = 1;

Thus, p - q *p

4= q * H9.`*^12 L;

and,

ì H9.`*^12 L+p

4= q o = o.a k p = q

p

4+ 9.`*^12

r 2;

= q *p

4+ q *

9.`*^12

r 2;

- q *p

4= q *

9.`*^12

r 2;

r 2p - q *

p

4= q * H9.`*^12 L;

q * H9.`*^12 Lì p - q *p

4= r 2 ;

:q *

H9.`*^12

Lì p - q *

p

4 >= r \

Solve Bc^2 * HHH1 ê3Lq L̂2 Lì p * q -1

4 Ip 2

* q2 M == r^2, q F

:: q Ø36 p r 2

4 c 2 + 9 p 2 r 2 >>

NBp ì H9.`*^12 L+p

4 F3.49066 µ 10 - 13

Solve B :q * H9.`*^12 Lì p - q *p

4 > == r, q F

Solve::eqf : :3. µ 10 6q

p + Times @á 3 à D>ã r is not a well - formed equation. à



new ideas.nb 19




Solve B:3.`*^6q

p -p q

4

>ã r, q F





Solve B:3. µ 10 6q

p -p q

4

>ã r, q F

Plot B 12.566370614359172` r

3.24`*^18 + 9.869604401089358` r, 8r, - 5, 5 <F

- 4 - 2 2 4

- 2. µ 10 - 17

- 1. µ 10 - 17

1. µ 10 - 17

2. µ 10 - 17

Plot B 36 p r 2

4 H3.0 * 10^8 L2+ 9 p 2 r 2

, 8r, - 5, 5 <F

- 4 - 2 2 4

2. µ 10 - 15

4. µ 10 - 15

6. µ 10 - 15

8. µ 10 - 15

20 new ideas.nb




Plot Bp

p

4+

9.`*^12

r 2

, 8r, - 50000000 000, 50 000 000 000 <F

- 4 µ 1010- 2 µ 1010 2 µ 1010 4 µ 1010

4

4

4

4

4

4

4

This plot maps what forever looks like.

Plot Bp

p

4+

1

r 2

, 8r, - 5, 5 <F

- 4 - 2 2 4

0.5

1.0

1.5

2.0

2.5

3.0

3.5

new ideas.nb 21




However, if we want to make sure that our units are correct, we will create a new change of variables by converting units.

p

p

4 +

9.`*^12

r 2

= q = dt = change in time

Converting theta into a change in time requires some thought, and also the multiplier Hp ê3L.

Speculation : Dark matter is present due to the incorrect integrations of the circle when compared with the luminosity of a change in height of the cone or the distance of an event to the subject.

CONCLUSION

Plot Bp

p

4+

9.`*^12

HHH3.0 * 10^8 L* 3 *q L^2 Lí Jp q-1

4p 2 q 2 N

2

, 8q , - . = ,.0005} ]

Plot Bp

p

4+

9.`*^12

HHH3.0 * 10^8 L* 3 *q L̂2 Lí Jp q-1

4p 2 q 2 N

2

, 8q , - .0005, .0005 < F

- 0.0004 - 0.0002 0.0002 0.0004

2

4

6

8

new ideas.nb 23




Plot Bp

p

4+

9.`*^12

HHH3.0 * 10^8 L* 3 *q L̂2 Lí Jp q-1

4p 2 q 2 N

2

, 8q , - .00005, .00005 < F

- 0.00004 - 0.00002 0.00002 0.00004

- 15

- 10

- 5

5

10

Plot B p

p

4+

9.`*^12

JJ3.` 10 8 N3 q N2

p q-p 2 q 2

4

2

, 8q , - 500, 500 < F

HHH3.0 * 10^8 L* 3 * q L̂2 Lì p q -1

4p

2q

2= r

Solve Bp

p

4+

9.`*^12

JJ3.` 10 8 N3 q N2

p q-p 2 q 2

4

2

ã q , q F

88q Ø 0. <, 8q Ø 4.0001 <<

HHH3.0 * 10^8 L3 q L̂2 Lì p * q -1

4Ip

2* q

2 M: = r

24 new ideas.nb




Plot B HHH3.0 * 10^8 L3 q L̂2 Lì p * q -1

4Ip

2* q

2 M, 8q , - 5, 5 <F

- 4 - 2 2 4

5.0 µ 108

1.0 µ 109

1.5 µ 109

2.0 µ 109

2.5 µ 109

Drawing Hypocycloids with these equations, we find that :

These are the Specific plots, which I invented by folding of a cone into circle, which is Divine Proportion.

new ideas.nb 25




Clear @x, y, t, a, b, tmini, tmaxi Da =

3 I1 + p 2 M2 p - 2 p + t + p 2 t

;

b = 9 1 - t - tp 2

+ 2p

;

tmini = - 10 * Pi;tmaxi = 10 * Pi;

8x@t_ D, y @t_ D<=8Ha - bL* Cos @t D+ b * Cos @Ha - bLêb * t D, Ha - bL* Sin @t D- b * Sin @Ha - bLêb * t D<;hypocycloid =

ParametricPlot @8x@t D, y @t D<, 8t, tmini, tmaxi <,PlotStyle -> 88Blue, Thickness @0.015 D<<,

AspectRatio -> Automatic, AxesLabel -> 8"x", "y" <D

- 5 5 10x

- 5

5

y

26 new ideas.nb




Clear @x, y, t, a, b, tmini, tmaxi Da = GoldenRatio *

3 I1 + p 2 M2 p - 2 p + t + p 2 t

;

b = 9 1 - t - tp 2

+ 2p ì GoldenRatio;





- 10 10 20 30x

- 30

- 20

- 10

10

20

30

y

new ideas.nb 27




Clear @x, y, t, a, b, tmini, tmaxi Da =

3 I1 + p 2 MGoldenRatio * 2 p - 2 p + t + p 2 t

;

b = GoldenRatio 9 1 - t - tp 2

+ 2p ì GoldenRatio;





- 4 - 2 2 4 6x

- 4

- 2

2

4

y

28 new ideas.nb




Clear @x, y, t, a, b, tmini, tmaxi Da = 10;b = 4;tmini = - 50 * Pi;tmaxi = 50 * Pi;

8x@t_ D, y @t_ D<= :Ha - bL* Cos B3

I1 + p 2

M2 p - 2 p + t + p 2 t F+ b * Cos BHa - bLêb *

3

I1 + p 2

M2 p - 2 p + t + p 2 t F,

Ha - bL* Sin B 9 1 - t -t

p 2+

2

p F- b * Sin BHa - bLêb * 9 1 - t -

t

p 2+

2

p F>;

hypocycloid =


AspectRatio -> Automatic, AxesLabel -> 8"x", "y" <Dy

new ideas.nb 29




9.0 9.5 10.0 10.5 11.0x0

5

30 new ideas.nb




- 5

new ideas.nb 31




CHAOS!

Clear @x, y, t, a, b, tmini, tmaxi Da = p ;b = GoldenRatio;tmini = - 10 p ;tmaxi = 10 p ;

8x

@t_

D, y

@t_

D<=

8Ha - bL* Cos @t D+ b * Cos @Ha - bLêb * t D, Ha - bL* Sin @t D- b * Sin @Ha - bLêb * t D<;hypocycloid =



- 3 - 2 - 1 1 2 3x

- 2

- 1

1

2

y

32 new ideas.nb

the geometric pattern of perception

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