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### Author Topic: The Holographic Universe and Pi = 4 in Kinematics!  (Read 236035 times)

#### gravityblock

• Hero Member
• Posts: 3287
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #45 on: May 18, 2014, 08:27:08 PM »
Let's try it with a "time variable" then. Draw a big circle and a square around it. You walk around the square and I'll walk around the circle, at the same speed. Who will walk completely around, first?

(And I note that you did not provide a single credible reference or support for your position. Nor did you provide an example of problem-solving using your value.)

You conveniently left out the first four steps.  We'll start walking at the same speed in the fifth tile or the fifth step in the illustration below, and we'll finish at the same time.

In addition to this, references and support for my position has been provided.  Your disagreement with those references doesn't make them not credible!  You have only asserted those references aren't credible without providing one scientific argument against any of those references posted in this thread.

Gravock

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #46 on: May 18, 2014, 08:35:26 PM »
If there's an increase in X, then there will be a proportional decrease in Y which maintains the same distances and perimeter.  Also, you're trying to derive 3.14 as Pi in this example by using radians where a full circle equals 2 * 3.14 or Tau.  This is no different than me saying 2 * 4  = 8 to represent the eight points on the circumference that lie on arcs that are multiples of 4/4 or 1.

There's a reason why the taxicab geometry correctly represents the true value of Pi being four in a real circle with a time variable.

Gravock
For any finite path approximation by turning the corners instead of traveling from vertex to vertex that are closest to the circumference you artificially increase your travel distance.  For every approach towards the circumference you make a matching turn away from it.  If you make enough and small enough squares subtracting the area they consume from the inset square will give you an approximation of the circle's area.  The more and smaller squares you use, the better the area approximation.  Because you keep turning away from the circumference your estimate of the perimeter length never improves.  If the method took the diagonal paths then the path approximation would improve with more and smaller squares and eventually approach Pi*D.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #47 on: May 18, 2014, 08:38:01 PM »
You conveniently left out the first four steps.  We'll start walking at the same speed in the fifth tile or the fifth step in the illustration below, and we'll finish at the same time.

In addition to this, I have provided references and support for my position.  Your disagreement with those references doesn't make them not credible!

Gravock
TinselKoala wins every time following the circular path.  You keep turning away from the perimeter and have to go back over and over again increasing your travel distance to 4/Pi Tinsel Koala's.

#### gravityblock

• Hero Member
• Posts: 3287
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #48 on: May 18, 2014, 08:46:27 PM »
For any finite path approximation by turning the corners instead of traveling from vertex to vertex that are closest to the circumference you artificially increase your travel distance.  For every approach towards the circumference you make a matching turn away from it.  If you make enough and small enough squares subtracting the area they consume from the inset square will give you an approximation of the circle's area.  The more and smaller squares you use, the better the area approximation.  Because you keep turning away from the circumference your estimate of the perimeter length never improves.  If the method took the diagonal paths then the path approximation would improve with more and smaller squares and eventually approach Pi*D.

TinselKoala wins every time following the circular path.  You keep turning away from the perimeter and have to go back over and over again increasing your travel distance to 4/Pi Tinsel Koala's.

TK does not win, for the square will be as uniform as the circle itself at the planck length.

Gravock

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #49 on: May 18, 2014, 09:25:13 PM »
TK does not win, for the square will be as uniform as the circle itself at the planck length.

Gravock
No, making the diversionary steps forced by the squares method you have specified smaller increases the number of diversionary steps.  By your own assertion, the length of such a path remains stuck at 4*D.  The path length of the circumference is Pi*D which has been approximated to eight digits as 3.1415953.

#### gravityblock

• Hero Member
• Posts: 3287
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #50 on: May 18, 2014, 09:41:27 PM »
No, making the diversionary steps forced by the squares method you have specified smaller increases the number of diversionary steps.  By your own assertion, the length of such a path remains stuck at 4*D.  The path length of the circumference is Pi*D which has been approximated to eight digits as 3.1415953.

No, because a real circle with a time variable will have a path length which is also stuck at 4*D at the planck scale as it traverses through space-time in a zig-zag or rectilinear motion.  The path length of the circumference is 4*D with no approximation.

Gravock

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #51 on: May 18, 2014, 09:53:58 PM »
No, because a real circle with a time variable will have a path length which is also stuck at 4*D at the planck scale due to traversing through space-time in a zig-zag or rectilinear motion.  The path length of the circumference is 4*D with no approximation.

Gravock
If you want to keep ignoring reality, you are free to do so.

#### gravityblock

• Hero Member
• Posts: 3287
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #52 on: May 18, 2014, 09:59:05 PM »
If you want to keep ignoring reality, you are free to do so.

Again, your post is not a scientific argument.

Gravock

#### Farmhand

• Hero Member
• Posts: 1583
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #53 on: May 18, 2014, 10:08:43 PM »
You conveniently left out the first four steps.  We'll start walking at the same speed in the fifth tile or the fifth step in the illustration below, and we'll finish at the same time.

In addition to this, references and support for my position has been provided.  Your disagreement with those references doesn't make them not credible!  You have only asserted those references aren't credible without providing one scientific argument against any of those references posted in this thread.

Gravock

Tile 5 or step 5 is impossible to achieve.

.

#### gravityblock

• Hero Member
• Posts: 3287
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #54 on: May 18, 2014, 10:14:35 PM »
Tile 5 or step 5 is impossible to achieve.

.

This is possible to achieve.  Step 5 only needs to be repeated down to the planck length and not repeated to infinity as the illustration says.  Step 5 should say, "repeat to the planck length".

Gravock.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #55 on: May 18, 2014, 10:15:54 PM »
Again, your post is not a scientific argument.

Gravock
LOL, I've covered why the squares method taken to its limit converges on the area but not the circumference.  And your scientific counterargument was?

#### gravityblock

• Hero Member
• Posts: 3287
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #56 on: May 18, 2014, 10:38:23 PM »
LOL, I've covered why the squares method taken to its limit converges on the area but not the circumference.  And your scientific counterargument was?

Look up Planck's constant - matter/energy is quantized. A circle is theoretical, there's no perfect circle in nature anywhere.  A real circle with a time variable is quantized at the planck scale with a zig-zag or rectilinear circumference, just as you find with the square in the squares method.  This is how there is a convergence on the rectilinear circumference at the planck scale.

Edit:  Also, in step two of the squaring method, we can see there are four points of the square which converge on the rectilinear circumference of the circle.  In each successive step of the squaring method, more and more points converge exponentially.  At the planck scale, all points will have converged on the rectilinear circumference of the circle.

Gravock

#### verpies

• Hero Member
• Posts: 3473
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #57 on: May 18, 2014, 11:52:03 PM »
In order to establish a circular path, a continuous acceleration must be centripetal:  orthogonal to the instant velocity which directs it to the center of a circle.
I was away today and regrettably I could not participate in the ongoing discussion about circular motion.

I just saw the statement above made by MarkE and I disagree with him that a centripetal force directed at the center of the circle depicted in Diag.4 will result in circular path without violating Newton's 1st law.
This is because an orthogonal force & acceleration cannot change any velocity component that is perpendicular to it.

This problem is germane to the discussion about Pi in kinematic circles and I invite everyone to discuss it before we return to the Pi issue.

I remind everyone that Newton 1st law pertains to the innate vector of motion as well to any components of that motion.
Please study the diagram attached below *

And let's keep the discussion civil and scientific.  It is OK to call attention to a Straw Man when one sees it but please do not immediately assume that it is constructed maliciously.  It can be a result of misunderstanding.
Since not everyone reading this discussion might be familiar with the names of these debating fallacies please link them to their definitions at RationalWiki.org

P.S.
@TinselKoala
You might find this paper about Taxicab geometry enjoyable.

@TinselKoala, @MileHigh, @Farmhand
I guarantee that the problem of circular motion illustrated here will bring you many minutes (or hours) of intellectual enjoyment even if you are a practical guy that prefers knobs and molten solder.  I welcome you to disagree with me since you make a good opponent.

Please begin by stating whether in your opinion Diag.3 or Diag.4 or Diag.5 correctly depicts reality.

------------------------------------------------------------------------------------------------------------------------------------------------------

Legend to the diagram below ( its hi-res version is here ) :
VT(t0): Tangential velocity at time t0.
V(t1) : Velocity at the time t1 that is parallel to the tangential velocity at the previous time (t0).
V(t1) : Velocity that is perpendicular to the velocity V(t1) at the time t1.

Example statements:
│VT(t0)│= │VT(t1)│= │VT(t2)│= etc... : A statement meaning that the magnitude of all tangential velocities is equal in all times t0, t1, t2, etc...
V(t1) ║ VT(t0): A statement meaning that velocity V(t1) is parallel to velocity VT(t0).
V(t1) < VT(t0): A statement meaning that the velocity V(t1) is smaller than the velocity VT(t0).
« Last Edit: May 19, 2014, 08:22:52 AM by verpies »

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #58 on: May 19, 2014, 05:25:54 AM »
Look up Planck's constant - matter/energy is quantized. A circle is theoretical, there's no perfect circle in nature anywhere.  A real circle with a time variable is quantized at the planck scale with a zig-zag or rectilinear circumference, just as you find with the square in the squares method.  This is how there is a convergence on the rectilinear circumference at the planck scale.

Edit:  Also, in step two of the squaring method, we can see there are four points of the square which converge on the rectilinear circumference of the circle.  In each successive step of the squaring method, more and more points converge exponentially.  At the planck scale, all points will have converged on the rectilinear circumference of the circle.

Gravock
Look up limits.  Each one of your jaunts along two edges of the approximating squares travels along one segment towards the perimeter and one away from it.    Dividing into a larger quantity of smaller squares does not change the path length.  It does not make the path a better approximation of the circumference. Y * X/X is still Y even for very large and very small values of X.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #59 on: May 19, 2014, 05:28:14 AM »
I was away today and regrettably I could not participate in the ongoing discussion about circular motion.

I just saw the statement above made by MarkE and I disagree with him that a centripetal force directed at the center of the circle depicted in Diag.4 will result in circular path without violating Newton's 1st law.
This is because an orthogonal force & acceleration cannot change any velocity component that is perpendicular to it.
Wrong and wrong.  This is basic calculus and physics.
Quote

This problem is germane to the discussion about Pi in kinematic circles and I invite everyone to discuss it before we return to the Pi issue.

I remind everyone that Newton 1st law pertains to the innate vector of motion as well to any components of that motion.
Please study the diagram attached below *

And let's keep the discussion civil and scientific.  It is OK to call attention to a Straw Man when one sees it but please do not immediately assume that it is constructed maliciously.  It can be a result of misunderstanding.
Since not everyone reading this discussion might be familiar with the names of these debating fallacies please link them to their definitions at RationalWiki.org

P.S.
@TinselKoala
You might find this paper about Taxicab geometry enjoyable.

@TinselKoala, @MileHigh, @Farmhand
I guarantee that the problem of circular motion illustrated here will bring you many minutes (or hours) of intellectual enjoyment even if you are a practical guy that prefers knobs and molten solder.  I welcome you to disagree with me since you make a good opponent.

Please begin by stating whether in your opinion Diag.3 or Diag.4 or Diag.5 depict reality.

------------------------------------------------------------------------------------------------------------------------------------------------------

Legend to the diagram below ( its hi-res version is here ) :
VT(t0): Tangential velocity at time interval t0.
V(t1) : Velocity at the time interval t1 that is parallel to the tangential velocity at the previous time interval (t0).
V(t1) : Velocity that is perpendicular to the velocity V(t1) at the time interval t1.

Example statements:
│VT(t0)│= │VT(t1)│= │VT(t2)│= etc... : A statement meaning that the magnitude of all tangential velocities is equal in all time intervals t0, t1, t2, etc...
V(t1) ║ VT(t0): A statement meaning that velocity V(t1) is parallel to velocity VT(t0).
V(t1) < VT(t0): A statement meaning that the velocity V(t1) is smaller than the velocity VT(t0).
Invented math and physics yield nonsense answers.