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Author Topic: The Holographic Universe and Pi = 4 in Kinematics!  (Read 208806 times)

Offline MarkE

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #225 on: June 03, 2014, 06:48:23 PM »
You have yet to show how my response to TK's example used static geometry as you wrongly and falsely asserted.  It is your pants which are very full.

Gravock
LOL, no I don't have to play your games your way  just because you say so.  It's perfectly fine to stand back as you load up your pants and stay far back as you play with the contents.

Offline profitis

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #226 on: June 03, 2014, 07:07:51 PM »
@gravityblock..did mathis have anything to say about the standing of the second law thermodynamics?

Offline TinselKoala

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #227 on: June 03, 2014, 08:13:05 PM »
TK's example was based on movement and not static geometry, but he conveniently left out the time variable which is found in the real world, just as plane geometry conveniently leaves time out of the equation.  My reply to his example did not use static geometry.  Please show me otherwise!  Also, there are many high road low road videos showing the same result.  Also, I clearly stated that both cars had the same net fall in my original post, so you can not say you thought one track was level and the blue car would never reach its destination.  It is your pants that is full, and this is another psychological projection by you.

Gravock

Don't lie.

Note the frequent appearance of the quantity "SECONDS".... a TIME VARIABLE found in the real world.

Quote
The circumference of the orbit (assuming pi = 3.1416 and a circular orbit) is 2 x pi x 149.6 million km, or about 939.97 million km.

The tangential speed computed from the radius and the conventional value of pi is therefore 939.97 million km / 31,558,118 seconds or about 29785 meters/second.

The diameter of the orbit is about 299.2 million km. Traversing this distance at the tangential velocity of 29814 m/sec will therefore take about 10035553 seconds. Four times that is 40,142,212 seconds... but a year is only 31,558,118 seconds.  Curiously.... 10035553 x  3.1416 = about 31,527,693 seconds.... nearly exactly the number of seconds in a sidereal year.

Conclusion.....  The value of pi, for the real figure of the Earth's orbit, taking TIME and velocity into account, must be very close to 3.1416, and not close at all to 4.

Please feel free to show a working that demonstrates otherwise.

Or just continue to flail and flop about, gasping and protesting .... but your pants are developing a pretty smelly leak, when you have to start lying to try to support your silly points.

Offline gravityblock

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #228 on: June 04, 2014, 12:07:48 AM »
@gravityblock..did mathis have anything to say about the standing of the second law thermodynamics?

Mathis has an article on entropy and he defines heat as photon density.

Gravock

Offline sarkeizen

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #229 on: June 04, 2014, 12:12:45 AM »
He has an article on entropy.  Mathis defines heat as photon density.
So uh...when are you going to get back to me on my question.  It was clear, in plain English and somehow devoid of profanity. :)

In case you forgot, it's right here.
Explain what part of this diagram is expressly stated as non-euclidian.  http://www.milesmathis.com/vel5.jpg
I can give you an easier question if you can't answer this one.

Offline gravityblock

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #230 on: June 04, 2014, 12:20:38 AM »
sarkeizen,

Are you saying the taxicab geometry is euclidean?

Gravock

Offline sarkeizen

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #231 on: June 04, 2014, 12:28:51 AM »
Are you saying the taxicab geometry is euclidean?
What I'm saying is that this diagram (http://www.milesmathis.com/vel5.jpg) implies that pythagoras theorem is untrue.  You say I'm wrong because pythagorian theorem doesn't hold in non-euclidian geometry.

Do you understand that in order for your comment to be relevant something in that diagram must be in non-euclidian geometry?

If you do somehow understand that then please indicate what.

Offline gravityblock

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #232 on: June 04, 2014, 12:29:50 AM »

   Gravityblock, you're oh so very clever, like your argument add in time
   and maturity should come.
                                  John.

Are you familiar with the "Reciprocal System"?

Gravock

Offline gravityblock

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #233 on: June 04, 2014, 12:36:49 AM »
What I'm saying is that this diagram implies that pythagoras theorem is untrue.  You say I'm wrong because pythagorian theorem doesn't hold in non-euclidian geometry.

Do you understand that in order for your comment to be relevant something in that diagram must be in non-euclidian geometry?

If you do somehow understand that then please indicate what.

Compare your diagram to the image below, then you should know how to distinguish between euclidean and non-euclidean geometry.

Gravock

Offline sarkeizen

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #234 on: June 04, 2014, 12:44:12 AM »
Compare your diagram to the image below
So you're refusing to answer a clearly worded, plain English question again?

i) If I say that this diagram: http://www.milesmathis.com/vel5.jpg showing a bunch of "steps" implies that the pythagorean theorem is false.  Do you agree with me or not?
ii) If not, then is your basis for your objection that the pythagorean theorem is not applicable to non-euclidean geometry?.  Yes or no?
iii) If yes, then clearly that diagram has to represent something in non-euclidean geometry.  Agree or disagree?

If you disagree how can your objection to my proof about the diagram be unrelated to the diagram?  If you agree then please tell me what (perhaps everything) is non-euclidean.

Offline gravityblock

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #235 on: June 04, 2014, 01:24:11 AM »
I think it would be rather hilarious to take a walk in the city with gravock. When you come to that vacant lot and want to cut across the diagonal to get over to the next Starbuck's... he will be constrained to make little right-angled segments that are parallel to the streets, while you simply walk the diagonal and get your decaf nonfat Grande Latte halfway drunk by the time he walks in the door.

Obviously you, sarkeizen, and MarkE would jump into the green car and take the straight and shortest path, while I jump into the yellow car taking the curved and longest path and win the race!  In your example above, you once again conveniently left out the time element by not allowing me to have the same acceleration along the rectilinear path as one would have by travelling a curved path in the real world with a time variable.  You on the other hand would only have a velocity across the diameter.  You do not win TK, for there is no such thing as an orbital velocity.  It is an acceleration along the perimeter or circumference of a curved path and only a velocity across the diameter.

Gravock

Don't lie.

Note the frequent appearance of the quantity "SECONDS".... a TIME VARIABLE found in the real world.

Quote from: TinselKoala
The circumference of the orbit (assuming pi = 3.1416 and a circular orbit) is 2 x pi x 149.6 million km, or about 939.97 million km.

The tangential speed computed from the radius and the conventional value of pi is therefore 939.97 million km / 31,558,118 seconds or about 29785 meters/second.

The diameter of the orbit is about 299.2 million km. Traversing this distance at the tangential velocity of 29814 m/sec will therefore take about 10035553 seconds. Four times that is 40,142,212 seconds... but a year is only 31,558,118 seconds.  Curiously.... 10035553 x  3.1416 = about 31,527,693 seconds.... nearly exactly the number of seconds in a sidereal year.

Conclusion.....  The value of pi, for the real figure of the Earth's orbit, taking TIME and velocity into account, must be very close to 3.1416, and not close at all to 4.

Please feel free to show a working that demonstrates otherwise.

It is you who is trying to mislead the reader.  The Starbuck's example I am now referring to, which conveniently had no appearance of quantity of seconds or a time variable in order to give me an acceleration, has nothing to do with a previous post made by you which does have quantity "seconds" or a time variable.  There is no doubt the reader will see this deliberate misdirection made by you.

Gravock

Offline sarkeizen

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #236 on: June 04, 2014, 01:36:04 AM »
It is you who is trying to mislead the reader.

Is answering a few clear, plain English, yes/no questions so hard?  Here they are again, in case you forgot...

i) If I say that this diagram: http://www.milesmathis.com/vel5.jpg showing a bunch of "steps" implies that the pythagorean theorem is false.  Do you agree with me or not?
ii) If not, then is your basis for your objection that the pythagorean theorem is not applicable to non-euclidean geometry?.  Yes or no?
iii) If yes, then clearly that diagram has to represent something in non-euclidean geometry.  Agree or disagree?

If you disagree how can your objection to my proof about the diagram be unrelated to the diagram?  If you agree then please tell me what (perhaps everything) is non-euclidean.

Offline gravityblock

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #237 on: June 04, 2014, 01:37:19 AM »
So you're refusing to answer a clearly worded, plain English question again?

i) If I say that this diagram: http://www.milesmathis.com/vel5.jpg showing a bunch of "steps" implies that the pythagorean theorem is false.  Do you agree with me or not?
ii) If not, then is your basis for your objection that the pythagorean theorem is not applicable to non-euclidean geometry?.  Yes or no?
iii) If yes, then clearly that diagram has to represent something in non-euclidean geometry.  Agree or disagree?

If you disagree how can your objection to my proof about the diagram be unrelated to the diagram?  If you agree then please tell me what (perhaps everything) is non-euclidean.

If you are not able to distinguish between euclidean and non-euclidean geometry, then you are beyond help.  I can not see the relevant parts for you, it is something you must see for yourself.  In your diagram, you are trying to use euclidean geometry to prove a non-euclidean geometry to actually be euclidean.  The fact that the Pythagorean theorem is proven false, should be a good indicator for you that it is not euclidean geometry and is actually non-euclidean.  You are caught in a contradiction, and there is no way out for you, except to assert that you do not see the relevant parts of your diagram which expressly states it as being non-euclidean.

Gravock

Offline sarkeizen

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #238 on: June 04, 2014, 01:46:27 AM »
it is not euclidean geometry and is actually non-euclidean.
So in other words this diagram http://www.milesmathis.com/vel5.jpg is non-euclidean right?

Offline gravityblock

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Re: The Holographic Universe and Pi = 4 in Kinematics!
« Reply #239 on: June 04, 2014, 01:50:48 AM »
So in other words this diagram http://www.milesmathis.com/vel5.jpg is non-euclidean right?

It has both euclidean and non-euclidean geometry in it.  Your question wrongly implies it has one or the other, when in fact it has both.  This is a deliberate misdirection made by you.

Gravock