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The Holographic Universe and Pi = 4 in Kinematics!

Started by gravityblock, May 06, 2014, 07:16:02 PM

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gravityblock

MarkE,

Take a square with a perimeter of 4, such as in step 2 of the illustration we have been discussing.  When you rotate this square by a 1/4, do you agree a "circular path" will be inscribed within?

Gravock
Insanity is doing the same thing over and over again, and expecting a different result.

God will confuse the wise with the simplest things of this world.  He will catch the wise in their own craftiness.

MarkE

Quote from: gravityblock on May 20, 2014, 01:18:29 AM
MarkE,

Take a square with a perimeter of 4, such as in step 2 of the illustration we have been discussing.  When you rotate this square by a 1/4, do you agree a "circular path" will be inscribed within?

Gravock
Take any object with one or more points that are at a fixed distance from the center, and rotate that object about its center.  The points remaining a fixed distance from the center by definition follow the same circular path.   The path between those points around the original shape has nothing to do with the rotation of that object.

Pop quiz:  Take any solid object that is a close approximation to a circle:  soda can, pill bottle, whatever you like.  Place it on a piece of paper and carefully draw a square that the circular end of the object inscribes.  Next wrap a string around circle, and cut that string where it meets the start.  Now lay the string out and see how much of the perimeter of your square you can cover.  Are we done with this mindless game yet?

verpies

Quote from: MarkE on May 20, 2014, 07:45:33 AM
Next wrap a string around circle, and cut that string where it meets the start.  Now lay the string out and see how much of the perimeter of your square you can cover.  Are we done with this mindless game yet?
No, I've been pointing out that a circumference of a geometric circle measured by a string yields different length than circumference of a circular path measured inertially.

Quote from: MarkE on May 17, 2014, 08:31:41 PM
Quote from: verpies on May 17, 2014, 04:17:15 PM
I agree with that definition, but not all distance has to be measured by a flexible string
Straw man:  No one has stated that that "all distance has to be measured by a flexible string.

@Gravityblock
MarkE is correct in his diagram about the measurements of a geometric timeless circle (in his diagram the string comes up short) but he is incorrect about a circular path traversed by a massive object. 
In the latter the time to complete one revolution is 4 times longer than the time to traverse the diameter by velocity whose magnitude is equal to tangential velocity.
If you let him conflate the static situation with the dynamic one, he will be successful in refuting your postulate.

verpies

Quote from: MarkE on May 19, 2014, 12:09:13 PM
Plane geometry does not have anything to do with time.
But a circular path laying on a plane does.

Quote from: MarkE on May 19, 2014, 12:09:13 PM
LOL, sure you lack evidence but it is the whole world of science that has been wrong for centuries. 
There is not much theoretical work done on the subject of proportions in circular path of motion because people like you summarily dismiss the difference between circular motion and an abstract geometric circle devoid of the time aspect, before even looking into the issue.
I don't need to have a thousand publications to convince me - one is enough.  A mind like you can also convince me with logical arguments either way.
The ultimate evidence would come from a well constructed experiment measuring a circular motion, as well as good logical analysis without prejudice.

Quote from: MarkE on May 19, 2014, 12:09:13 PM
Now your problem is to show how a circular path doesn't have the identical geometry of the "abstract circles" you keep claiming that it differs from while still being circular.
I don't see it as a big problem.
The points belonging to abstract geometric circles do not involve time, while the circular paths of motion - do.  ( by the very definition of motion ).
The difference is very clear and nonproblematic.

Quote from: MarkE on May 19, 2014, 12:09:13 PM
Then you are utterly and completely wrong.Then once again you are completely wrong.
Don't repeat I'm wrong without an accompanying proof regarding this issue in circular paths of motion.

Quote from: MarkE on May 19, 2014, 12:09:13 PM
Yes, valid vector math might provide useful evidence.  If you do it right, you should see your errors.
My simple question to you "how can a force that is always perpendicular to the tangent velocity can change a parallel component of that velocity" is a beginning of such vector analysis.
Why don't you answer it?

Quote from: MarkE on May 19, 2014, 12:09:13 PM
Mathis has been disproven.
Where can I read a rigorous critique of his paper ?

MarkE

Quote from: verpies on May 20, 2014, 01:59:27 PM
No, I've been pointing out that a circumference of a geometric circle measured by a string yields different length than circumference of a circular path measured inertially.
There is only one kind of circle.  Despite all gentle efforts to guide you and GravityBlock you insist on flunking fourth grade math.  If it's a game of yours and GravityBlocks, it is a silly and tedious one.
Quote
Straw man:  No one has stated that that "all distance has to be measured by a flexible string.
Straw man:  no such statement was made.  If you object to the idea that a flexible string can be used to measure either the circumference of a circle or the perimeter of a square, then you are free to offer such an objection no matter how inane it might be.
Quote


@Gravityblock
MarkE is correct in his diagram about the measurements of a geometric timeless circle (in his diagram the string comes up short) but he is incorrect about a circular path traversed by a massive object. 
A path is circular if it maps out to a circle.  If it did the circle that the path maps out, like all circles would have a C/D ratio of a circle, which is far from 4.0.
Quote
In the latter the time to complete one revolution is 4 times longer than the time to traverse the diameter by velocity whose magnitude is equal to tangential velocity.
This is just sad.
Quote
If you let him conflate the static situation with the dynamic one, he will be successful in refuting your postulate.