ColorWeight (double version)

The color weight without weight.

Description

This project is an introduction at the triple color temperature.
The range software is:
301^3 color combinations
100^2 spacetime combinations (meters and seconds)
1000 weight values (Kilograms)
It is this:
max image = 301^2 * 100^2
GB = 301^2
time = 100
space = 100
weight = 1000 --> R = 301 = GB diagonal
The software statements are:
point of half = clock verse dynamic solution without weight
perception = clock verse dynamic solution without weight
perception of the color = clock verse dynamic solution of the color without weight
perception of the color weight = clock verse dynamic solution
the color weight without weight = 0
the 301 static sectors instant weight without weight = 0
point of half of the color weight = clock verse dynamic solution
point of half of the 301 static sectors instant weight = clock verse dynamic solution
point of half of the 301 static sectors instant weight = point of half with weight in 301 static sectors instant
the color without weight = null
0 = null --> the color weight without weight = the color without weight --> weight = 1
0 = 1 --> null = weight
perception of the color = point of half in 301 static sectors instant
perception of the color with weight = clock verse dynamic solution
point of half of the 301 static sectors instant weight = perception of the color with weight
The perception of the color weight is the perceptive phenomena.
The incredible phenomena is the color weight like emotive dimension.

• MATLAB release: MATLAB 6.1 (R12.1)

Other Files

ColorWeight/ColorWeight.m,
ColorWeight/ColorWeight2.m,
ColorWeight/formulaT.txt,
ColorWeight/PhysicsColor.m,
ColorWeight/PhysicsMat.m,
ColorWeight/PhysicsMat10.m,
ColorWeight/PhysicsMat2.m,
ColorWeight/PhysicsMat3.m,
ColorWeight/PhysicsMat4.m,
ColorWeight/PhysicsMat5.m,
ColorWeight/PhysicsMat6.m,
ColorWeight/PhysicsMat7.m,
ColorWeight/PhysicsMat8.m,
ColorWeight/PhysicsMat9.m,
ColorWeight/RV.txt,
ColorWeight/RV2.txt,
ColorWeight/RV3.txt,
ColorWeight/RV4.txt,
license.txt

[us]

when we run your nice

     colorweight2

using lena.jpg as an input with parameters

to select a row of time in seconds (1-30100): 12
to select a column of space in meters (1-30100): 12
to select the red color (0-300): 12
to insert a positive number of weight [Kg] with range 1-1000: 12

we get this result

point of half: clock verse dynamic solution without weight
time = 0 % <- ok, same val
space = 0 % <- ok, same val
velocity = -11 % <- YES, we got the same value(!)...
acceleration = -11.917 (!?!) % <- SEE NOTE...
mass = 0 % <- ok, same val
weight force = -11 % <- ok, same val

however, when we use our own algorhythm, we get
acceleration = -11.921

which is grossly different and now jeorpadizes our next submission to NEUROLOGY...

please, clarify immediately as this is soon going to cause some real problems - if not havoc - for the scientific world...

us

[PXlab]

acceleration = -11.9167
acceleration = NN/N^2-W= (1/12)-12 = (0.0833)-12 = (12/144)-12
N = 12 --> time Z
NN = 12 --> space X
W = 12 --> XZ, the video plane
---
acceleration = -11.921
acceleration = (0.0790)-12 = (79/1000)-12
---
X --> 12 = 79 --> 67 of difference in the space
Z^2 --> 144 = 1000 --> 856 of difference in the time^2
X/Z^2 --> 67/856 = 0.0783 = number of acceleration difference in the two solutions
---
X/(12*Z) --> 1/12 in my result, this is a velocity/12
X/(12*Z) --> 79/(12*sqrt(1000)) in your result, this is a velocity/12
1/12 = 79/(12*sqrt(1000)) --> 79 = sqrt(1000) --> 47.3772 = number of velocity difference in the two solutions
---
my velocity = your velocity = -11
my time = your time = 0
---
47.3772 = 0 * 0.0783 --> vel diff / time = acc diff --> 0 = 605.0728 = time of union of the two equal solutions
---
-11.9167 = -11.921 --> 0.0043 = 0 in a time of 605.0728
---
0.0043 [m/s^2] * 605.0728 [s] = 2.6018 [m/s] = number of velocity for the same solution
---
the difference of our solutions is in the space: 2.6018 [m] in X with 1[s] of reference in Z
my zero space is equal to your zero space with a factor of 2.6018 meters
or
the difference of our solutions is in the frequency: 2.6018 [Hz=1/s] in 1/Z with 1[m] of reference in X
my zero time is equal to your zero time with a factor of 2.6018 Hertz
---
0.0783-0.0043 = 0.0740 --> 0.0783 = 0 in a time of 605.0728
---
The point is that with one decimal there is not difference, the results are differents in the second decimal (1=2) and in the third decimal (6=1) for a total of 16=21 and this is 5 in 1000. A value of 0.005 like difference is a not difference if 5 is 0 in the third decimal. I think that the difference could be the definition of the zero value and this could be a not difference of results. Is a proportional factor, only one. And a proportional time factor is always present like union of two systems. I don’t know your algorithm, I hope this is a valid solution.

[Matt Fig]

us,

You're not use correct metric for this time frame evaluation. Pi==NaN with the reference parameter of half, in this case. So it is not of concern to you for this solution, and scientific resolution is fine.
Color is motion in detection of point reference metric!! This is the valid point you miss in your algorithm.

[PXlab]

0 = space
0 = time
0 = 67 --> difference in the space
0^2 = 856 --> difference in the time^2
0/0^2 = 0.0783 = number of acceleration difference in the two solutions
0.0783 = 0 in a time of 605.0728 is 0/0^2 = 0 in X/Z^2
0 = 0 --> 0^2 = null --> 0 = sqrt(null) --> this is a zero time --> 856 = null
sqrt(856) = 29.2575
zero time --> 0 = 29.2575
---
0.0783 = 0 in a time of 605.0728 is 0 = 0 in X and this is the X = 0
0^2 in Z^2 is null in Z^2
0 in Z is sqrt(null) in Z --> the zero time: 0 in Z is 29.2575 in Z
---
X = 0 is 0
Z = 0 is 29.2575
the zero time is the XZ union: X=Z is 0=0 and it is 0=29.2575
---
605.0728 /29.2575 = 20.6809 = number of zero time ripetition to have a time of 605.0728
20.6809 ripetitions of 29.2575 are a time of 605.0728 --> 0 = 0 in X --> 0.0783 = 0
---
0.0783-0.0043 = 0.0740 --> 0.0783 = 0 in a time of 605.0728
0.0783-0.0043 = 0.0740 --> 0.0783 = 0 with 20.6809 ripetitions of 29.2575
0.0783-0.0043 = 0.0740 --> 0.0783 = 0 with 20.6809 ripetitions of Z = 0
---
0 = 0 in X is Z = Z in X --> 0.0783 = 0 --> a time of 605.0728
---
with 20.6809 circles of zero time we obtain 0 = 0 in X and this is the 0^2 = null
0.0783-0.0043 = 0.0740 --> 0/0^2 = 0 in X/Z^2
---
-11.9167 = -11.921 --> 0.0043 = 0 in a time of 605.0728
-11.9167 = -11.921 --> 0.0043 = 0 with 20.6809 ripetitions of 29.2575
-11.9167 = -11.921 --> 0.0043 = 0 with 20.6809 ripetitions of Z = 0
---
0.0783-0.0740 = 0.0043 --> 0/0^2 = 0 in X/Z^2
0.0043 = 0.0043 --> 20.6809 circles of zero time are 0 = 0 in X --> 0^2 = null
for the same X space like 0 = 0 are necessary 20.6809 circles of zero time
29.2575 = 360° --> the zero time
20.6809 circles of zero time are the equal solutions
20.6809 ripetitions of Z = 0 are the equal solutions
a time of 605.0728 is an union time of two different solutions and it is the time of separation of two equals solutions
---
605.0728 / 60 = 10.0845 = number of minutes to obtain a time of 605.0728 [s]
0.0845 min = 0.0845 * 60 sec = 5.07 sec = 5 sec + 7/100 of sec
10 min + 5 sec + 7/100 of sec --> 0 = 0 in X --> 0.0783 = 0
---
0.0043 [m/s^2] * 605.0728 [s] = 2.6018 [m/s] = number of velocity for the same solution
0.0043 [m/s^2] * ( 10 min + 5 sec + 7/100 of sec ) = 2.6018 [m/s]
2.6018 [m/s] like diagonal of the spacetime matrix 2D
I write this input in ColorWeight2:
to select a row of time in seconds (1-30100): 1
to select a column of space in meters (1-30100): 26018 --> the same solution * 10000
to select the red color (0-300): 0 --> black in the first color level, we are in the same solution
to insert a positive number of weight [Kg] with range 1-1000: 1 --> seconds = Kg
Result:
red profundity = 0 --> 2D of space
green profundity = 0 --> time forward
blue profundity = 260 --> time behind <-
PHYSICS POINT
red time [s] = 2
green time [s] = 2
blue time [s] = -0.7333
red space [m] = 18
green space [m] = 18
blue space [m] = 2.4000
red velocity [m/s] = 9
green velocity [m/s] = 9
blue velocity [m/s] = 1.2000 <- (*)
red acceleration [m/s2] = 18
green acceleration [m/s2] = 18
blue acceleration [m/s2] = 2.4000 <- (*) (X)
red mass [kg] = 0.0556
green mass [kg] = 0.0556
blue mass [kg] = 0.0074 <- (X)
red weight force [kg*m/s2] = 1
green weight force [kg*m/s2] = 1
blue weight force [kg*m/s2] = 0.1333 <- (X)
clock verse dynamic solution: point of half with weight in 301 static sectors instant
[s] = 1
[m] = 18
[m/s] = 18 <- (**)
[m/s2] = 18 <- (**)
[kg] = 0.0556 <- (****)
[kg*m/s2] = 1 <- (*****)
zero up = 18
zero down = 1
point of half: clock verse dynamic solution without weight
time = 0
space = 17
velocity = 17 <- (***)
acceleration = 17 <- (***)
mass = -0.9444 <- (****)
weight force = 0 <- (*****)
---
(*) --> velocity = acceleration/2 --> time = 0.5 in the physics point
(**) --> time = 1
(***) --> time = 1
clock verse dynamic solution = point of half --> time = time
(****) --> clock verse dynamic solution + point of half --> 2 * time --> 0.0556 - 0.9444 = 2 * mass = -0.8888
(*****) --> 0 = 1
2 * time = 2 * 0.5 = 1 = double time in the physics point --> 2 * mass = -0.8888 --> mass = -0.4444
[s=Kg] --> double time in the physics point = -0.4444 --> physics point time = -0.2222 = half physics point
(X) --> blue mass * blue acceleration = blue weight force --> 0.0178 = 0.1333 --> 0.1155 of difference
if 0.0044 of difference is zero then a double difference is 0.2222 and it is 50.5 times the zero
---
10000 times the same solution is 26018 [m] like horizontal of the space line 1D and it is 50.5 times the zero
the same solution is 2.6018 [m] like horizontal of the space line 1D and it is 50.5/10000 times the zero
0.005 times the zero=0.0044 is a not difference because the definition of the zero value is the not different solution
29.2575 = 360° --> the zero time --> the 0.0044 time
0.005 * 29.2575 = 0.005 * 360° --> 0.005 times the 0.0044 time --> 0.1463 = 1.8° --> time 2.2e-5
1463 = 18000° --> time 0.22 --> 1463 = 50 circles
---
our systems has: velocity = weight force = -11 = -50 * 0.22
What do you think of this solution?
Can I ask how many input has your system?

[PXlab]

for the same X space like 0 = 0 are necessary 20.6809 circles of zero time
zero time = 0.0044 time
for the same X space like 0 = 0 are necessary 20.6809 circles of 0.0044 time
20.6809 circles of 0.0044 time = 0.091 time
for the same X space like 0 = 0 is necessary a 0.091 time
0.22 / 0.091 = 2.4176
1463 = 18000° --> time 0.22 --> 1463 = 50 circles --> 2.4176 * time 0.091
our systems has: velocity = weight force = -11 = -50 * 2.4176 * 0.091
-11 = -50 * 2.4176 * 0.091 --> -11 = -11.0001 --> 1/10000 = element of a 100^2 matrix
0.0001*50=0.005 --> our difference
my conclusion is this: our difference is 50 elements of the 100x100 matrix
our difference --> 10x5 matrix --> space = 2 time or time = 2 space --> 0 = 2 * 0
each 100 values of X or Z there is an element of difference --> zero
each 200 values of Z or X there is an element of difference --> 2 * zero
5:1000 --> one value of difference each 200 values
1000 = 10^3 = a cube 10x10x10 = 5x5x5x2x2x2 = 5x10x20
5:1000 --> 5:5x10x20
10 --> zero/10
20 --> zero/5
Our difference is a confirmation of equal solutions, both ok.
Is this resolving?

[PXlab]

Matt say
Color is motion in detection of point reference metric.
Read this
Color is motion in the video profundity dimension and this is the XZ weight.
Color is motion in detection of point reference metric second.
Video profundity dimension is a metric second weight reference.
weight / ( time * altitude ) = 1 --> weight = time * altitude
Video profundity dimension is a weight^2 reference.
Color is motion in the weight^2 reference.
Color is motion in detection of point reference Kg square.
Union of the Matt words with this statements: [m=Kg^2]
Axial: T = Z^4/X^2 --> Kg in Z^2/X
T = future time
Z = video altitude
X = video width
---
Y = past time
T / Y = t = present instant
T = Y^2 --> t = Y
T = Z^3 --> X^2 = Z
Y = Z^3 --> T = Y --> t = 1
Y = Y^2 --> Z^3 = T --> 1 = Y
---
XT/Z^2 = Z^2/X --> m/Kg=Kg
1/Kg = Kg --> m=1
T = 1 --> X/Z^2=Z^2/X --> X^2=Z^4
In a spacetime matrix 100x100 this is:
X=10 --> X^2=100
Z=sqrt(10) --> Z^2 = 10
Z^2=10 --> Z^4 = 100
It is T=1.
100 = 100 --> X^2 = Z^4
T = 1 --> Z^3=1 --> Z^2 = 1/Z --> sqrt(10) = 1/10 --> 10 = 1/100 --> 1 = 1000
1 = 1000 --> 1 = Z^6 --> 1 = T * Y --> 1 = X^12
1 = a cube 10x10x10 --> 1 = Z^6 = TY = X^12
5:1000 --> one value of difference each 200 values
1*1*1 = 1000^3 --> 1 = Z^18 = (TY)^3 = X^36
5^3:1000^3=125:1000000000 --> one value of difference each 200^3=8000000 values
2D --> one value of difference each 200^2=40000 values
2D --> one value of difference each spacetime matrix 100x100 per 4 times
In the two systems there is one value of difference each 200 values: 40000/200=200
200 values of two systems difference * 200 times is one value of difference and it is 40000 values
50 elements each 10000 * 200 times are 40000 elements and this is the spacetime matrix 100x100 per 4 times
one value of difference is the spacetime matrix 100x100 per 4 times if each value is a matrix element
I think this: the two systems difference is the spacetime matrix 100x100 per 4 times each 200 values because one value of difference is the spacetime matrix 100x100 per 4 times. In formula I say this:
the difference value quarter = spacetime matrix 100x100
the difference value quarter each 50 values = spacetime matrix 100x100 each 50 values
In the previous comment I wrote: our difference is 50 elements of the 100x100 matrix.
I can write now: our difference is the different value quarter of the 100x100 matrix.
Our difference is the different value of a matrix with 2500 elements.
What is the meaning of this statement? In my program 2500 elements are a quarter of spacetime reference.
A different value of our spacetime quarter: 50 elements of difference in 10000 are 12.5 elements in 2500 and 12.5*200 is 2500. The number of two system union appear to be 200.
I write in the program the start solution: (12,12,12,12).
The weight inserted in ZERO UP and ZERO DOWN + 100 say this:
velocity / gravity [s] axes: Y 112
The weight inserted in ZERO UP and ZERO DOWN + 200 say this:
velocity / gravity [s] axes: Y 212
The utility is this: 2*(212-112)=200
The number of two system union is 2 times the velocity/gravity in 3° level less the velocity/gravity in 2° level with 2° and 3° level in one hundred union. The one hundred union is situation of the altitude dimension (T axes of future time). Two times this situation is the number of two system union in T axiality). The 100+100 is the systems union and 100 (the one hundred situation) is half union of the double system in T because 212 in Y less 112 in Y is a complete union in Y:
half double system union in T = 100 in Y --> 100 double system in T = 100 in Y
double system in the present instant = 1
Yes, I say; my system in the present instant is 0.5 and your system in the present instant is 0.5 and so the present instant is a double model unity.

[PXlab]

m=ms
m=Kg2
ms=Kg2
---
ms=Kg --> TY=Z2/X
---
Kg2=Kg --> Kg=1 --> Z2=X
Z=sqrt(10)
X=10
---
20.6809 circles of 0.0044 time = 0.091 time --> 360° = circle = matrix 100x100 --> 2500 = 90°
the number of two system union appear to be 200 --> 2500/12.5 = 200 --> 90°/12.5=7.2°
200 = 7.2° --> number of two system union
60 sec = 360° --> 1 min = circle --> 1 sec = 6°
60 min = 360° --> 1 hour = circle --> 1 min = 6°
60 sec = 1 min --> 360° = 6° --> factor = 60
60 min = 1 hour --> 360° = 6° --> factor = 60
factor^2 = 3600 --> number of sec in one hour
200 = factor^2/18 --> number of sec in 1/18 of hour
7.2° = factor^2/18 --> number of sec in 1/18 of hour
360°/7.2° = 50 --> 1 hour / 50 = 7.2° with 1 hour = 360°
360°/7.2° = 50 --> 1 min / 50 = 7.2° with 1 min = 360°
factor --> 1 min = 1 hour
factor^2 --> 1 min * 1 min = 1 hour * 1 hour
3600 --> 360° * 360° = 360° * 360°
factor^2 = 3600 --> 1 min square = 360° square --> number of sec in one hour --> 3600 sec = 129600°
factor^2 = 3600 --> 1 hour square = 360° square --> number of sec in one hour --> 3600 min = 129600°
129600/7.2=18000
18000° * number of two system union = 129600° --> way sec and way min
129600/200=648
648 * number of two system union = 129600° --> way sec and way min
18° = 0.648 --> 129.6° = 129.6° --> way sec / 1000 and way min / 1000
9° = 0.324 --> 64.8° = 64.8° --> way sec / 2000 and way min / 2000
---
212+112=324 --> velocity/gravity in 3° level more the velocity/gravity in 2° level with 2° and 3° level in one hundred union and define this value VG --> 9000° = VG --> 18000° = 2 * VG
2 * VG * number of two system union = 129600° --> way sec and way min
VG * number of two system union = 64800° --> way sec / (2*VG) and way min / (2*VG)
number of two system union = 64800° / VG --> way sec / (2*VG) and way min / (2*VG)
number of two system union = 7.2 --> way sec / 18000° and way min / 18000°
---
I write in the program the solution of identity: (1,26018,0,1).
Physics point
red space [m] = 18
green space [m] = 18
red acceleration [m/s2] = 18
green acceleration [m/s2] = 18
clock verse dynamic solution
[m] = 18
[m/s] = 18
[m/s2] = 18
static solution: instant sector n° 1
acceleration [m/s2] = 18
physics correlations
altitude / time [m/s] axes: T / Y 18
time / weight [s/Kg] axes: XY / Z^2 18
VFA = velocity * force * acceleration = altitude / ( time * weight^2 ) = 18
altitude / weight [m/Kg] axes: T / XZ 18
altitude * weight / time [m*Kg/s] axes: XZT / Y 18
altitude / ( time * weight ) [m/(s*Kg)] axes: T / XZY 18
gravity [m/s^2] axes: T / Y^2 18
altitude / time^2 [m/s^2] axes: T / Y^2 18
weight / mass [Kg*m/s^2] axes: XZT / Y^2 18
altitude / ( time * weight ) [m/(s*Kg)] axes: T / XZY 18
altitude / weight [m/Kg] axes: XT / Z^2 324
altitude * time / weight [m*s/Kg] axes: XTY / Z^2 324
---
18° = 2*VG/1000
---
m=° --> red space = green space = 2*VG/1000
m=° & s2=1 --> red acceleration = green acceleration = 2*VG/1000
m=° & s=1 --> altitude / time = 2*VG/1000
s=° & Kg=1 --> time / weight = 2*VG/1000
m=° & Kg=1 --> altitude / weight = 2*VG/1000
m=° & Kg=s --> altitude * weight / time = 2*VG/1000
m=° & Kg=1/s --> altitude / ( time * weight ) = 2*VG/1000
m=° & s^2=1 --> gravity = altitude / time^2 = 2*VG/1000
m=° & Kg=s^2 --> weight / mass = 2*VG/1000
---
18° = 0.648 --> 129.6° = 129.6° --> way sec / 1000 and way min / 1000
---
2*VG/1000 = 0.648 --> VG = 648/2 = 324
9000° = 324 --> number of two system union
9000/324=27.7778
200/7.2=27.7778
Two systems union is VG value and it is in Y dimension: 605.0728 = time of union of the two equal solutions.
9000° = 324 --> 605.0728
324*1.8675=605.0728
9000°*1.8675=16808°
605.0728 = 16808° --> time of union = time of union --> 1 = 27.7778°
---
90° : 15 sec=min = 27.7778° : X sec=min --> X = 4.6296 [sec=min] --> time of union
4.6296 sec --> speed up
4.6296 min --> speed down
speed up = speed down --> factor = 60
The double speed way is the present instant like diagonal of 60 elements in a matrix of 60 rows (speed up, sec) per 60 column (speed down, min).
60 = 60 --> factor^2 = 3600
The equal speed of the two ways is the speed^2 definition like number of seconds in one hour and this is 60 circles with circle equal minute.
With a current association I say this:
right point of current up = 60 = PDX/2
right point of current down = 60 = PDX/2
PDX^2/4=3600=PSX --> the perceptive plane 2D, the speed^2 definition like 60 circles with circle equal minute, the left point of current.
Current=PSX+PDX=3720
---
Are we the same opinion? Say to me, regards.

[sis MA]

 

[Tyler Jandreau]

This is absolute nonsense. Marco, pasting your inane fprintf('') statement output does not explain your "theory".

[James]