Geometric Perception Patterns 3.2

8/14/2019 Geometric Perception Patterns 3.2

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c = 3.0 * 10^8 m ê sC = 2 pr

This is the circumference of our initial circle of radius r

C2 = 2 pr1

This is the circumference of our second circle, the base of the cone, of radius r1

r^2 = r1^2 + h^2

This is the initial radius squared expressed as the slant

of the cone in terms of the height of the cone, h, and the radius

of the base of the cone, r1

r =,Hr1^2 + h^2L

t = qr

t ê q = r

The arc length taken out of a circle at a given time is =

t = C - C2 = 2 pr - 2 pr1 = qr Ø Equation 7

r1^2 ã r^2 - h^2

r1 =,Hr^2 - h^2L

h § r

t = time

1 second = 6 degrees

Printed by Mathematica for Students



Solve@r1 ^ 2 + h ^ 2 ã r^2, hD::h Ø - r2

- r1

2 >, :h Ø r2- r

1

2 >>Ht ê qL = r

t = C - C2 = 2 p r - 2 p r1 = 2 p Ht ê qL - 2 p HHt ê qL^ 2 - h ^ 2Lt == 2 p Ht ê qL - 2 p HHt ê qL^ 2 - h ^ 2L

Add 2 p HHrL^2 - h^2L to both sides

t + 2 p HHt ê qL^2 - h^2L = 2 p Ht ê qL

Subtract t from both sides and remember that (t/q ) = r

2 p HHt ê qL^2 - Hh^2LL = 2 p Ht ê qL - t = 2 pr - t

Divide by 2 p on both sides.

Ht ê qL - t ê H2 pL = HrL - t ê H2 pL = HHt ê qL^2 - Hh^2LL = HHrL^2 - Hh^2LL = r1

Square both sides. Substitute : (t/q ) = r

HHt ê qL - Ht ê H2 pLLL^2 = HHrL - Ht ê H2 pLLL^2 = HHt ê qL^2 - Hh^2LL = Hr1^2LHHrL - Ht ê H2 pLLL^2 = HHrL^2 - Hh^2LL

Add h^2 to both sides.

((r) - (t/(2 p)))^2 + (h^2) = (r)^2

Substitute : q r=t

((t/q) - (t/(2 p)))^2 = ((r) - (q*r/(2 p)))^2

HHrL - Hq * r ê H2 p LLL^ 2

r -

r q

2 p

2

2 Geometric Perception Patterns 3.2.nb




ExpandB r -

r q

2 p

2Fr2

-

r2q

p

+

r2q

2

4 p2

r2- r

2

q

p

+ r2

q

2

4 p 2

+ h ^ 2 = HHrL - Ht ê 2 p LL^ 2 + Hh ^ 2L = r ^ 2

Hr ^ 2L - r2-

r2q

p

+

r2q

2

4 p 2

= h ^ 2

SimplifyBHr ^ 2L - r2-

r2q

p

+

r2q

2

4 p 2

Fr2 H4 p - qL q

4 p 2

= h ^ 2

r2

H4 p - q

Lq

4 p2

= h

SimplifyB r2 H4 p - qL q

4 p 2

Fr2 H4 p - qL q

2 p

For perception of a point of light, we will take this formula and substitute the speed of light times time in for the height of

the cone. Then, we will say that the time is in terms of angular transformation, just like the angle related to the base of the cone.

h := c t

Such that t seconds = 6 q

Thus, we may write :

r2 H4 p-qL q

2 p= h = c t = c 6 q = H3.0 * 10^8L H6L HqL

r2 H4 p - qL q

2 p

= h = c t = H18.0 * 10^8L q

Solve

B:r2 H4 p - qL q

2p

==

H18.0 * 10^8

L* q

>, q

F:8q Ø 0.<, :q Ø

2.32352 µ 106

r2

2.36506 µ 1025

+ 184900. r2

>>

Geometric Perception Patterns 3.2.nb 3




PlotB 2.3235219265950113`*^6 r2

2.3650572504748037`*^25+ 184900.` r2

, 8r, - 100000, 100000<F

-100000 -50 000 50 000 100 000

2.µ10-10

4.µ10-10

6.µ10-10

8.µ10-10

1.µ10-9

Now, we would like to see what the plot of h is purely in terms of r over the course of theta, because theta is clearly purely in

terms of r when travelling at the speed of light.

r2 J4 p - J 2.32352µ106

r2

2.36506µ1025

+184900. r2NN J 2.32352µ10

6r2

2.36506µ1025

+184900. r2N

2 p

=

r2 H4 p - qL q

2 p

= h = H18.0 * 10^8L * q =

SolveB r2 J4 p - J 2.3235219265950113`*^6 r2

2.3650572504748037`*^25+184900.` r2NN J 2.3235219265950113`*^6 r

2

2.3650572504748037`*^25+184900.` r2N

2 p

ã

r2 J 12.566370614359172` r

2

1.279100730381181`*^20+r2 N

p

-

r2 J 12.566370614359172` r

2

1.279100730381181`*^20+r2 N

2

4 p 2

, rF99r Ø - 5.46414 µ 10

9=, 8r Ø 0.<, 9r Ø 5.46414 µ 109==

SimplifyB r J 12.566370614359172` r2

1.279100730381181`*^20+r2N

p

-

r2 J 12.566370614359172` r2

1.279100730381181`*^20+r2N2

4 p 2

-

r2 J4 p - J 2.3235219265950113`*^6 r2

2.3650572504748037`*^25+184900.` r2NN J 2.3235219265950113`*^6 r

2

2.3650572504748037`*^25+184900.` r2N

2 p

F

2.26195 µ 1010

r4

I1.2791 µ 1020

+ r2M2

- 242.602

r4 I2.97202 µ 1026

- 4.65661 µ 10-10

r2MI2.36506 µ 1025

+ 184900. r2M2





PlotB r2 J4 p - J 2.3235219265950113`*^6 r2

2.3650572504748037`*^25+184900.` r2NN J 2.3235219265950113`*^6 r

2

2.3650572504748037`*^25+184900.` r2N

2 p

, 8r, - 100, 100<F

-100 -50 50 100

5.µ10-7

1.µ10-6

1.5µ

10-6

PlotB: r2

J12.566370614359172` r

2

1.279100730381181`*^20+r2

Np

-r

2

J12.566370614359172` r

2

1.279100730381181`*^20+r2

N2

4 p 2 >, 8r, - 100, 100<F

-100 -50 50 100

5.µ10-7

1.µ10-6

1.5µ10-6

However, for the proper forms of the equation in order to graph the spiral, whose variable functions deliver information

of the environmental contour to the perceiver, we must solve for r and make a series substitution. We will see that the transforma-

tion is illustrated by different sections of the personal experience.

SolveB r2 H4 p - qL q

2 p

== h, rF::r Ø -

2 h p

4 p q - q2

>, :r Ø

2 h p

4 p q - q2

>>

For the time being and the purposes of this paper, we will ignore the negative solutions. Granted, they exist reasonably

and for height, probably spatially. However, we are only concerned with the distance that light approaching the perceiver travels.

Also, it does not make much sense to have a negative radius.





PlotB 2 H3.0 * 10^8L H6L HqL p

4 p q - q2

, 8q, - 10, 10<F

-10 -5 5 10

5.0µ109

1.0µ1010

1.5µ1010

2.0µ1010

r ^ 2 =

2 h p

4 p q - q2

^ 2

4 p2h

2

4 p q - q2

= r 2

We will write the inductively provable statement that:

r 2

= H1 + 3 + 5 + ... + H2 n - 1LLAlthough the square of r is truly continuous, for the purpose of discussing the theory, it is useful to be able to quantize r' s

value. A gauge of scale could be used to make any necessary alterations to the mathematical truths for making actual measure-

ments if necessary.

Thus, for n = 1, 4 p2

h2

4 p q-q2

= 1

SolveB 4 p 2h

2

4 p q - q2

== 1, hF

::h Ø -

4 p q - q2

2 p

>, :h Ø

4 p q - q2

2 p

>>





PlotB 4 p q - q2

2 p

, 8q, - 10, 10<F

-10 -5 5 10

0.4

0.5

0.6

0.7

0.8

0.9

1.0

We see easily that it goes right up to 1 and comes straight back down over the course of 4 Pi radians. Hence, if we do all of them

together, we see that it increases from 1 to 2 to 3, respectively, but comes back down each time at the same spot on the x axis.

PlotB::-

4 p q - q2

2 p

>, : 4 p q - q2

2 p

>>, 8q, - 15, 15<F

-15 -10 -5 5 10 15

-1.0

-

0.5

0.5

1.0

+

SolveB 4 p 2h

2

4 p q - q2

== 1, qF::q Ø 2 1 - 1 - h

2p>, :q Ø 2 1 + 1 - h

2p>>





PlotB2 1 - 1 - h2p , 8h, - 10, 10<F

-10 -5 5 10

1

2

3

4

5

6

PlotB::2 1 - 1 - h2p >, :2 1 + 1 - h2

p >>, 8h, - 10, 10<F

-10 -5 5 10

2

4

6

8

10

12

Earlier, we stated that the change in size implied a change in distance. Let' s say that this change of distance is the

derivative of the function for height of the transformed circle through a cone.

DB H4 p - qL q

2 p

, qF4 p - 2 q

4 p H4 p - qL q

· H4 p - qL q

2 p

„ q

4 p - q q3ê2

- 2 p H4 p - qL q + 8 p2 ArcSinB q

2 p F4 p

Considering that the contour of the world is a sinusoidal function of size and shape, we may conclude that :





Clear@x, y, t, a, b, tmini, tmaxiDa =

H4 p - qL q

2 p

;

b =

4 p - 2 q

4 p H4 p - qL q

tmini = - 10 p ;

tmaxi = 10 p ;8x@t_ D, y@t_ D< =8Ha - bL * Cos@qD + b * Cos@Ha - bL ê b * qD, Ha - bL * Sin@qD - b * Sin@Ha - bL ê b * qD<;

hypocycloid =

ParametricPlot@8x@qD, y@qD<, 8q, tmini, tmaxi<,

PlotStyle -> 88Blue, [email protected]<<,

AspectRatio -> Automatic, AxesLabel -> 8"x", "y"<D4 p - 2 q

4 p

H4 p - q

Lq

4

6

y





-1.0 -0.5 0.5 1.0 1.5x

-2

2





-4

There is also an interesting special case of what happens to the integrated change of circumference when the height is equal to

the radius (i.e. we no longer have a cone, just a distance). Firstly, however, we will look at trying to integrate the change in

circumference with respect to theta by substituting in for h = c6 q

DC = 2 p r - 2 p r1

2 p r - 2 p r1 = C - C2

2 h p

4 p q-q2

= r

,Hr^2 - h^2L = r1

Our equation for r thus turns into :

2 I1.8 µ 109

qM p

4 p q - q

2

= r

h = c6q = (3.0*10^8) (6) (q )

H3.0 * 10^8L H6L HqL1.8 µ 10

9q

From that, we can say that DC is equal to :

2 p

2 H1.8`*^9 qL p

4 p q - q2

- 2 p I,Hr ^ 2 - H1.8`*^9 qL^ 2LMRemember from earlier that, with the same model, we were able to show that :

SolveBq ==

12.566370614359172` r2

1.279100730381181`*^20+ r2

, rF::r Ø -

I0. + 6.47236 µ 1013

ÂM q

- 4.11558 µ 108

+ 3.27507 µ 107

q

>, :r Ø

I0. + 6.47236 µ 1013

ÂM q

- 4.11558 µ 108

+ 3.27507 µ 107

q

>>

And we want to be able to plug in for r in terms of theta.





2 p

2 H1.8`*^9 qL p

4 p q - q2

- 2 p . H0.` + 6.472364372540152`*^13 ÂL q

- 4.11557987`*^8 + 3.2750744`*^7 q

^ 2 - H1.8`*^9 qL^ 2

7.10612 µ 1010

q

4 p q - q2

- 2 p - 3.24 µ 1018

q2

-

I4.18915 µ 1027

+ 0. ÂM q

- 4.11558 µ 108

+ 3.27507 µ 107

q

‡ 0

2 p 7.106115168784338`*^10 q

4 p q - q2

- 2 p - 3.24`*^18 q2

-

H4.189150057092708`*^27+ 0.` ÂL q

- 4.11557987`*^8 + 3.2750744`*^7 q

„ q

‡ 0

2 p 7.10612 µ 1010

q

4 p q - q2

- 2 p - 3.24 µ 1018

q2

-

I4.18915 µ 1027

+ 0. ÂM q

- 4.11558 µ 108

+ 3.27507 µ 107

q

„ q

· 7.106115168784338`*^10 q

4p q - q

2

- 2 p - 3.24`*^18 q2

-

H4.189150057092708`*^27+ 0.` ÂL q

- 4.11557987`*^8 + 3.2750744`*^7 q

„ q

· 7.10612 µ 1010

q

4 p q - q2

- 2 p - 3.24 µ 1018

q2

-

I4.18915 µ 1027

+ 0. ÂM q

- 4.11558 µ 108

+ 3.27507 µ 107

q

„ q

This conclusion shows us that there is not any possible way to integrate

a change in our perception. The solutions to the change that we perceive

in radius Hi.e. r Ø r1L are too imaginary to integrate. In many ways,

this shows us that our perception of the phenomenon is somewhat indescribable. Yet,

if h = r, we may find something plausibly integratable. Let' s find out.

2 h p

4 p q-q2

= r

,Hr^2 - h^2L = r1

r > r1

r := h

Thus,

SolveB 2 p

4 p q - q2

== 1, qF88q Ø 2 p<<

Thus, this shows us that our model is valid as well as the math' s being correct; that when we set r = h, the total angular transfor-

mation is 2 p.





SolveB 2 h p

4 p q - q2

ã h, hF88h Ø 0<<

· 7.106115168784338`*^10q

4 p q - q2

- 2 p - 3.24`*^18 q2 - H4.189150057092708`*^27+

0.`Â

Lq

- 4.11557987`*^8 + 3.2750744`*^7 q

„ q

· 7.106115168784338`*^10 q

4 p q - q2

„ q

7.10612 µ 1010 Kq H- 4 p + qL + 4 p H4 p - qL q ArcSinB q

2 p

FOH4 p - qL q

‡ 2 p I- 3.24`*^18 q2M

H2 p L H4.189150057092708`*^27 + 0.` H- 1L^H1 ê 2L L q

- 4.11557987`*^8 + 3.2750744`*^7 q

^H1 ê 2L „ q

It looks as though it is far too hard to integrate it in terms of theta, for it is too imaginary. Thus, we must put it back in terms of r,

and integrate something more tangible, like length.

DC == 2 p

2 H1.8`*^9 qL p

4 p q - q2

- 2 p I,Hr ^ 2 - H1.8`*^9 qL^ 2LM

q ==

12.566370614359172` r2

1.279100730381181`*^20+ r2





In[1]:= 2 p

2 JH1.8`*^9L J 12.566370614359172` r2

1.279100730381181`*^20+r2NN p

4 p J 12.566370614359172` r2

1.279100730381181`*^20+r2N - J 12.566370614359172` r

2

1.279100730381181`*^20+r2N2

-

2 p

.r ^ 2 -

H1.8`*^9

L12.566370614359172` r2

1.279100730381181`*^20+ r2

^ 2

Out[1]= - 2 p r2

-

5.1164 µ 1020

r4

I1.2791 µ 1020

+ r2M2

+

8.92981 µ 1011

r2

I1.2791 µ 1020

+ r2M -157.914 r

4

I1.2791µ1020+r

2M2+

157.914 r2

1.2791µ1020+r

2

In[2]:= · - 2 p r2-

5.116402921524724`*^20 r4

I1.279100730381181`*^20 + r2M2

+

I8.929807683926348`*^11 r2

M ì I1.279100730381181`*^20 + r2

M-

157.91367041742973` r4

I1.279100730381181`*^20 + r2M2

+

157.91367041742973` r2

1.279100730381181`*^20+ r2

„ r

3.141592653589793` rr2

I1.279100730381181`*^20 + r2M2

I1.279100730381181`*^20 + r2M -

2.`2.0092067288834284`*^20

r+ 1.5707963267948966` r

r2 I1.6360986784616704`*^40- 2.558201460762362`*^20 r2 + r4MI1.279100730381181`*^20+ r2M2

+

0.` I1.279100730381181`*^20 + r2M

r2 I1.6360986784616704`*^40- 2.558201460762362`*^20 r2+ r4M

I1.279100730381181`*^20 + r2M2

ì





r 1.6360986784616704`*^40 - 2.558201460762362`*^20 r2+ r4

+

1

r

0.`r2

I1.279100730381181`*^20 + r2M2

I1.279100730381181`*^20 + r2MLogA1.279100730381181`*^20 + r2E -

0.5` I1.279100730381181`*^20+ r2M I1.279100730381181`*^20 + 1.` r2M

r2 I1.6360986784616704`*^40- 2.558201460762362`*^20 r2+ r4M

I1.279100730381181`*^20 + r2M2

I1.6360986784616704`*^40- 2.558201460762362`*^20 r2+ 1.` r4M3ê2

- 6.283185307179587` r 1.6360986784616704`*^40 - 2.558201460762362`*^20 r2+ 1.` r4 ì

I- 1.279100730381181`*^20- 1.` r2M + 6.283185307179587` r - 1.279100730381182`*^20

1.6360986784616704`*^40- 2.558201460762362`*^20 r2+ 1.` r4

+

1.` r2 1.6360986784616704`*^40- 2.558201460762362`*^20 r2+ 1.` r4 ì

I- 1.6360986784616704`*^40+ 2.558201460762362`*^20 r2- 1.` r4M

0.` I- 1.279100739983322`*^20+ r2M3ê2 I- 1.2791007207790399`*^20 + r2M3ê2

II- 1.279100739983322`*^20+ r2M I- 1.2791007207790399`*^20+ r2MM3ê2

+

1.` - 1.279100739983322`*^20+ r2- 1.2791007207790399`*^20+ r2

I- 1.934965393265402`*^21 + 15.127544989934918` r2M ì I- 1.279100739983322`*^20+ r2M I- 1.2791007207790399`*^20 + r2M

- 1.279100739983322`*^20 + 1.` r2- 1.2791007207790399`*^20+ 1.` r2

-

J1.0000000000000002`I- 1.279100739983322`*^20+ r2M3ê2 I- 1.2791007207790399`*^20+ r2M3ê2

LogA5.47827358680159`*^101+ 4.2829102170624404`*^81 r2EN í II- 1.279100739983322`*^20 + r2M I- 1.2791007207790399`*^20+ r2MM3ê2

+





1.0000000000000004`I- 1.279100739983322`*^20+ r2M3ê2 I- 1.2791007207790399`*^20+ r2M3ê2

LogB- 4.28291021706244`*^81+ 3.348376023353612`*^61 r2- 3.3483760233536126`*^61

- 1.279100739983322`*^20 + 1.` r2- 1.2791007207790399`*^20+ 1.` r2 F ì

II- 1.279100739983322`*^20 + r2

M I- 1.2791007207790399`*^20+ r2

MM3ê2

+

1.` I- 1.279100739983322`*^20+ r2M3ê2 I- 1.2791007207790399`*^20+ r2M3ê2

LogB- 2.558201460762362`*^20 + 2.` r2+

2.` - 1.279100739983322`*^20 + 1.` r2- 1.2791007207790399`*^20+ 1.` r2 F ì

II- 1.279100739983322`*^20 + r2M I- 1.2791007207790399`*^20+ r2MM3ê2 ì r2 1.6360986784616704`*^40- 2.558201460762362`*^20 r2

+ r4

I-

2.0927350145960093`*^60+

4.908296035385014`*^40 r

2-

3.837302191143545`*^20 r4+ 1.` r6M == C



Documents

Geometric Perception Patterns 3.2