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Author Topic: The Holographic Universe and Pi = 4 in Kinematics!  (Read 214676 times)

verpies

• Hero Member
• Posts: 3480
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #60 on: May 19, 2014, 08:19:59 AM »
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.
That was Gravityblock's whole point.  The path length does not change with finer subdivision - only area does.
You were supposed to be a good opponent and refute his observation that real physical circles have the same circumference as physical squares - not agree with Gravityblock.

BTW:  A real physical circle must be defined by some real physical process, not an abstract one.

You still have not replied directly which diagram (Diag.3 or Diag.4 or Diag.5) correctly depicts reality in this physical process, in your opinion.

verpies

• Hero Member
• Posts: 3480
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #61 on: May 19, 2014, 08:42:42 AM »
Wrong and wrong.  This is basic calculus and physics.Invented math and physics yield nonsense answers.
Refute my statements rigorously.  Show me the error in logic or math.
An argument by assertion is not the way to do it.

First of all:  How can V(t1) be smaller than the tangential velocity VT(t0) if the force acting on it was always perpendicular between t0 and t1 ?

MarkE

• Hero Member
• Posts: 6830
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #62 on: May 19, 2014, 08:44:13 AM »
Quote
Quote
Wrong and wrong.  This is basic calculus and physics.Invented math and physics yield nonsense answers.

Refute my statements rigorously.  Show me the error in logic or math.
An argument by assertion is not the way to do it.

First of all:  How can V║(t1) be smaller than the tangential velocity VT(t0) if the force acting on it was always perpendicular between t0 and t1 ?

http://theory.uwinnipeg.ca/physics/circ/node6.html

verpies

• Hero Member
• Posts: 3480
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #63 on: May 19, 2014, 09:02:35 AM »
Refute my statements rigorously.  Show me the error in logic or math.
An argument by assertion is not the way to do it.

First of all:  How can V║(t1) be smaller than the tangential velocity VT(t0) if the force acting on it was always perpendicular between t0 and t1 ?
http://theory.uwinnipeg.ca/physics/circ/node6.html
That theory page does not show rigorously that when "the direction of the centripetal acceleration is inwards along the radius vector"  then circular motion is produced - it just asserts it, like you.

The question how can V(t1) be smaller than the tangential velocity VT(t0) if the force acting on it was always perpendicular between t0 and t1, still stands unanswered.

MarkE

• Hero Member
• Posts: 6830
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #64 on: May 19, 2014, 10:11:29 AM »
That was Gravityblock's whole point.  The path length does not change with finer subdivision - only area does.
You were supposed to be a good opponent and refute his observation that real physical circles have the same circumference as physical squares - not agree with Gravityblock.

BTW:  A real physical circle must be defined by some real physical process, not an abstract one.
LOL, you can enjoy yourself misstating what I have said if that pleases you.

Gravityblock's method does not reproduce the path of travel along the circumference.  The path that his method generates constantly approaches and then turns away from the circumference.  With enough and small enough square elements, his method can accurately approximate the circle's area and outline using the vertices that touch or approach the circumference.  As his method constantly turns away from the circumference it never improves its approximation of the circumference's path.  It doesn't matter how small or how many elements he uses.

The shortest path between two points is only orthogonal line segments when one of those segments is zero length.  The distance following a Manhattan route between vertices on the circumference is therefore always greater than the straight line distance between the same vertices.

MarkE

• Hero Member
• Posts: 6830
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #65 on: May 19, 2014, 10:19:35 AM »
http://theory.uwinnipeg.ca/physics/circ/node6.html

That theory page does not show rigorously that when "the direction of the centripetal acceleration is inwards along the radius vector"  then circular motion is produced - it just asserts it, like you.

The question how can V(t1) be smaller than the tangential velocity VT(t0) if the force acting on it was always perpendicular between t0 and t1, still stands unanswered.
http://dev.physicslab.org/Document.aspx?doctype=3&filename=CircularMotion_CentripetalAcceleration.xml

verpies

• Hero Member
• Posts: 3480
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #66 on: May 19, 2014, 10:55:31 AM »
LOL, you can enjoy yourself misstating what I have said if that pleases you.
It does not please me to misstate you, but drawing conclusions from your statements and synthesizing a larger statement from several of your statements is not a misstatement. It is an acceptable way of conducting conversation.  If I make an error along the way I expect you to point it out.

Gravityblock's method does not reproduce the path of travel along the circumference.  ...

The shortest path between two points is only orthogonal line segments when one of those segments is zero length.  The distance following a Manhattan route between vertices on the circumference is therefore always greater than the straight line distance between the same vertices.
This is the same as stating that at the limit the chord approaches the curve/arc.
BTW: Note that this is a condensation and interpretation of your words and your animation - a valid debating technique, not a misstatement.

And you would be correct if all of the points on the curve/circle had the same time coordinates - like in an abstract geometric circle.
But neither Gravityblock not I are analyzing abstract circles.  We are analyzing real circles made by real physical processes where the points on the circle do not have the same temporal coordinates.
In such real circles, the chord does not approach the curve and this very issue is the subject of this paper.

You are welcome to prove your assertion that at the limit the chord approaches the curve/arcs created by real physical processes.

MarkE

• Hero Member
• Posts: 6830
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #67 on: May 19, 2014, 11:11:44 AM »
It does not please me to misstate you, but drawing conclusions from your statements and synthesizing a larger statement from several of your statements is not a misstatement. It is an acceptable way of conducting conversation.  If I make an error along the way I expect you to point it out.
LOL, no that's called making things up.  If you can't win the point, just build a man of straw to slay.
Quote

This is the same as stating that at the limit the chord approaches the curve/arc.
BTW: Note that this is a condensation and interpretation of your words and your animation - a valid debating technique, not a misstatement.

And you would be correct if all of the points on the curve/circle had the same time coordinates - like in an abstract geometric circle.
Plane geometry does not involve time.
Quote

But neither Gravityblock not I are analyzing abstract circles.  We are analyzing real circles made by real physical processes where the points on the circle do not have the same temporal coordinates.
Here we go:  A special pleading to "circles" that are not "circles" except when you two want them to be circles.  Yet they fail tests for basic properties of circles.
Quote
In such real circles, the chord does not approach the curve and this very issue is the subject of this paper.
Neither you nor Gravityblock have established that your special "circles" are in fact circles.
Quote

You are welcome to prove your assertion that at the limit the chord approaches the curve/arcs created by real physical processes.
LOL, now you don't believe first semester calculus.  You are free at any time to as you say use actual facts to argue your specious and silly case.

verpies

• Hero Member
• Posts: 3480
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #68 on: May 19, 2014, 12:58:20 PM »
Plane geometry does not involve time.
But we are not discussing abstract timeless plane geometry.  You are assertions about Pi are correct in abstract time geometry.
We are not discussing abstract geometry, we are discussing real physical problems from the start.

http://dev.physicslab.org/Document.aspx?doctype=3&filename=CircularMotion_CentripetalAcceleration.xml
This link you posted refers to an article describing a physical circle, created by real forces acting on a real mass.  There is no avoiding the time variables in this one.

That article states:
"Note that in both cases, Δv points to the center of the circle reflecting that the acceleration is also directed towards the center of the circle"
..but it is just an empty assertion.

That article does not prove that the acceleration vector and force that causes the circle lays on a line that passes through the center of the circle.
That article correctly subtracts two tangent velocity vectors.  On my diagram that is VT(t0) - VT(t1) but it fails to prove that the result of this subtraction lays on a line that passes through the center of the circle.
What proof did you or that article give that the acceleration/force vector lays on such line?  What proof did you or that article give that the acceleration/force vectors do not lay on the dashed lines depicted on the diagram below that does not pass through the center ?

gravityblock

• Hero Member
• Posts: 3286
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #69 on: May 19, 2014, 01:09:30 PM »
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.

MarkE,

The plot of a convergent sequence {an} is shown in blue in the illustration below. Visually we can see the sequence is converging to the limit 0 as n increases.  Similarly, we can visually see the exponentially larger quantity of smaller squares in each successive squaring method is converging while the path length does not change.

Gravock

MarkE

• Hero Member
• Posts: 6830
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #70 on: May 19, 2014, 01:14:59 PM »
But we are not discussing abstract timeless plane geometry.  You are assertions about Pi are correct in abstract time geometry.
Circles are basic constructs of plane geometry.  [/quote]
We are not discussing abstract geometry, we are discussing real physical problems from the start.  [/quote]Citing the rantings of internet cranks as reference "papers" hardly seems like a discussion of anything real.
Quote

This link you posted refers to an article describing a physical circle, created by real forces acting on a real mass.  There is no avoiding the time variables in this one.
Here we go back to your special pleadings of "abstract" and "physical" circles.  Until such time as you can actually delineate what it is that makes a circle:  "physical", distinct from textbook circles, and still qualifies them as circles, you might as well say "brominsmores".
Quote

That article states:
"Note that in both cases, Δv points to the center of the circle reflecting that the acceleration is also directed towards the center of the circle"
..but it is just an empty assertion.
That article does not prove that the acceleration vector and force that causes the circle lays on a line that passes through the center of the circle.
So you assert.  Read it again.
Quote
That article correctly subtracts two tangent velocity vectors.  On my diagram that is VT(t0) - VT(t1) but it fails to prove that the result of this subtraction lays on a line that passes through the center of the circle.
What proof did you or that article give that the acceleration/force vector lays on such line?  What proof did you or that article give that the acceleration/force vectors do not lay on the dashed lines depicted on the diagram below that does not pass through the center ?
Let's see your vector math that can actually hold an object on a circular path while the accelerating force does not point radially through the center of that circle.  Be sure that whatever "circle" you use satisfies the requirements: a closed path where all points on the circumference are equidistant from the center.

verpies

• Hero Member
• Posts: 3480
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #71 on: May 19, 2014, 01:20:21 PM »
Here we go:  A special pleading to "circles" that are not "circles" except when you two want them to be circles.
...
Neither you nor Gravityblock have established that your special "circles" are in fact circles
I was clear from the first post about the difference between abstract circles and physical circles.
Are you claiming that the circle in that article you quoted is not a circle?

Yet they fail tests for basic properties of circles.
They don't.
Your argument that Pi=3.14 defines a circle instead of the circle defining the Pi is putting the cart before the horse and it is a fallacy.  But I thank you for sensitizing me to this line of argument reversal.  I will be ready for it with other opponents.

Until such time as you can actually delineate what it is that makes a circle:  "physical", distinct from textbook circles, and still qualifies them as circles, you might as well say "brominsmores".
A circle is a set of equidistant points on a spatial plane from the center of the circle.  The difference between an abstract and physical circle is whether these points have time coordinates or not.  Physical circles do and those coordinates are not the same.  Time is hard to diagram and most likely that's why you are confused about the distinction.

Citing the rantings of internet cranks as reference "papers" hardly seems like a discussion of anything real.
That paper is relevant because it discusses the approach of the chord to the arc in real physical circles at the limit.
Attack author's arguments not the author.  We are beyond burning Brunos and the likes of him.

LOL, now you don't believe first semester calculus.
Actually I don't thing that's applicable in case of physical circles.

You are free at any time to as you say use actual facts to argue your specious and silly case.
Mathis proves that the chord does not approach the arc at the limit in kinematic circles quite exhaustively with rigorous arguments.  I shouldn't have to repost his paper here - a link should be sufficient.
If you are resorting to ridiculing his rigorous analysis instead of refuting his arguments, that means that you have run out of ammunition.

MarkE

• Hero Member
• Posts: 6830
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #72 on: May 19, 2014, 01:31:42 PM »
MarkE,

The plot of a convergent sequence {an} is shown in blue in the illustration below. Visually we can see the sequence is converging to the limit 0 as n increases.  Similarly, we can see the exponentially larger quantity of smaller squares in each successive squaring method is converging while the path length does not change.

Gravock
Which does absolutely nothing for getting the path traveled following orthogonal segments to better approximate the path length of the circumference.

gravityblock

• Hero Member
• Posts: 3286
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #73 on: May 19, 2014, 01:43:00 PM »
Which does absolutely nothing for getting the path traveled following orthogonal segments to better approximate the path length of the circumference.

The path length isn't changing as it converges, and we can visually see this, so this is the exact path length of the circumference.  The "better" approximation you speak of simply doesn't exist as you wrongly assert.

Gravock

MarkE

• Hero Member
• Posts: 6830
Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #74 on: May 19, 2014, 01:44:19 PM »
I was clear from the first post about the difference between abstract circles and physical circles.
Really?  What is that difference?  What qualifies a "physical circle" to be a circle and what properties may it have that are different than an "abstract circle"?
Quote

Are you claiming that the circle in that article you quoted is not a circle?
You are off in the bushes again.
Quote

Your argument that Pi=3.14 defines a circle instead of the circle defining the Pi is putting the cart before the horse and it is a fallacy.  But I thank you for sensitizing me to this line of argument reversal.  I will be ready for it with other opponents.
Slay those men of straw.  I have stated clearly that Pi is defined as the ratio of a circle's circumference to its diameter, and the mutual claim you make with GravityLock that the ratio is numerically equal to four is patently false.
Quote

A circle is a set of equidistant points on a plane from the center of the circle.  The difference between an abstract and physical circle is whether these points have time coordinates or not.  Physical circles do and those coordinates are not the same.
"Those coordinates" are not the same as what?  A plane has two axes.  A circle is a construct of plane geometry.  Be sure that your "physical circle" conforms to those requirements.
Quote
Time is hard to diagram and most likely that's why you are confused about the distinction.
Again, a circle is a construct of plane geometry.  There are only two dimensions.  If you cannot draw it on a piece of paper then it isn't plane geometry.
Quote
The paper is relevant because it discusses the approach of the chord to the arc in real physical circles at the limit.
The paper's premise is utter and total BS.  Mathis introduces the line RBD which never appears in Lemma VI.  Lemma VI declares that as B approaches A that the angle subtended between B-A-D approaches zero.  This is visibly obvious.  As B approaches A, B rises to A and the line between B and A comes closer and closer to being parallel with the line between A and D.  Ergo in the limit the slope of the line between A and B becomes tangent to the circle, parallel to the line between A and D and the subtended angle:  B-A-D goes to zero.  Ergo the cited article is in error.
Quote
Attack author's arguments not the author.  We are beyond burning Brunos and the likes of him.