You've got a point there.Kepler certainly worked on floating points.NaM? Not a mortal?Doesn't multiplication by zero kill an immortal?

- Cuchulainn
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You've got a point there.Kepler certainly worked on floating points.NaM? Not a mortal?Doesn't multiplication by zero kill an immortal?

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

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Tips 6: The Cross itself is mortal. How would Kepler gather the immortals?Kepler certainly worked on floating points.NaM? Not a mortal?Doesn't multiplication by zero kill an immortal?

- Traden4Alpha
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Kepler worked with two points but three points were ...You've got a point there.Kepler certainly worked on floating points.NaM? Not a mortal?

- Traden4Alpha
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Did he encircle them? (Or did he enellipse them?)Tips 6: The Cross itself is mortal. How would Kepler gather the immortals?Kepler certainly worked on floating points.NaM? Not a mortal?

Both so to say! This is why the cross is irrational kind off.Did he encircle them? (Or did he enellipse them?)Tips 6: The Cross itself is mortal. How would Kepler gather the immortals?Kepler certainly worked on floating points.

Last edited by Collector on June 3rd, 2017, 1:25 pm, edited 1 time in total.

- Traden4Alpha
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So then this thread is in the conic section of the forum.Both so to say! This is why the cross is irrational kind off.Did he encircle them? (Or did he enellipse them?)Tips 6: The Cross itself is mortal. How would Kepler gather the immortals?

okay then, Tips 7: the Music of the spheres indicated spheres and certain harmonic formations of spheres.

Only indivisible spheres are immortal (the atoms of Democritus and Leuppikus).

Kepler would have gathered the Divine 7 by sphere packing (Kepler conjecture).

And closed the gates to the cross with 2 Gate keepers, the Guardians of the Cross.

And yes the average of the cross arms is irrational! And is a very important ratio in terms of Magic Numbers hidden in Nature.

Only indivisible spheres are immortal (the atoms of Democritus and Leuppikus).

Kepler would have gathered the Divine 7 by sphere packing (Kepler conjecture).

And closed the gates to the cross with 2 Gate keepers, the Guardians of the Cross.

And yes the average of the cross arms is irrational! And is a very important ratio in terms of Magic Numbers hidden in Nature.

- Traden4Alpha
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The average of the cross arms can be rational!

The average of the cross arms is r*(2 + √12). But if r is the product of any rational number and the inverse of that irrational factor, the average will be rational.

That said, at least one of the arms must have irrational length.

The average of the cross arms is r*(2 + √12). But if r is the product of any rational number and the inverse of that irrational factor, the average will be rational.

That said, at least one of the arms must have irrational length.

good point! √12, yes the 12 disciples had to show up somewhere.The average of the cross arms can be rational!

The average of the cross arms is r*(2 + √12). But if r is the product of any rational number and the inverse of that irrational factor, the average will be rational.

That said, at least one of the arms must have irrational length.

Last edited by Collector on June 4th, 2017, 1:01 pm, edited 2 times in total.

For the Glory and the Queen try to Solve Part 2 of The Big Quest

Tips 8

The Queen is Holding the Secret to the Whole World in Her Hand!

Kepler will be useful! also Mother Nature!

- Traden4Alpha
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In warped space-time, do circles exist?

Do irrationals even exist in the "real" world???? If space is quantized (dare I say atomized), then perhaps all distances are integers.

Do irrationals even exist in the "real" world???? If space is quantized (dare I say atomized), then perhaps all distances are integers.

Only the tiny immortal spheres exist (they are invariant), other circles (spheres) are not perfect circles (spheres). And this plays a significant role in this quest.In warped space-time, do circles exist?

Do irrationals even exist in the "real" world???? If space is quantized (dare I say atomized), then perhaps all distances are integers.

The Great Tips 9:

Kepler-Cannonbals (large not divine spheres, but half divine and hard to break) and the Battle of the Alamo.

Calculate the number of days (include "hours", that is also fraction of a day)` of the siege (by using the Cross of Kepler) and you have exactly half of what you need to know the Magic Number that Controls this world!

In the end don't forget the Royal Orb: The Cross is Above, The Orb is Below!

Okay, the Battle of the Alamo started Feb 23 the year 1836 (1836+(31+23)/365=1836.15). This was naturally a hint of the proton to electron mass ratio which is \(\frac{m_P}{m_e}\approx 1836.15\)

With Kepler Cannon balls I tried to get us involved in Kepler sphere packing.

Assume for a moment a proton is 1836.15 sphere packed electrons and that the electrons are spheres with radius r. Now let us sphere pack 1836.15 spheres using the Kepler Conjuncture, this gives a volume of approximately

\(\frac{1836.15\times\frac{4}{3}\pi r^3}{\frac{\pi}{3\sqrt{2}}}=10386.84732\times r^3 \)

This means the radius of the large sphere can be (approximated) by

VolumeLargeSphere=\(\frac{4}{3}\pi R^3\)

\(R=\left(VolumeLargeSphere\times \frac{3}{4\pi}\right)^{1/3}=\left(10386.84732\times\frac{3}{4\pi}\right)^{1/3}\approx 13.54r

\)

This proton sphere however has a jagged surface because it is sphere packed like the Kepler cross, so we use the Kepler cross to find the average radius of the proton sphere. This gives an average proton radius of 13.4r, where r is the radius of the electron.

Now divide this by the electron radius. This gives 13.4 (also the Sige of Almo lasted exactly 13.4 days, I should know)

My calculator went out of battery so could someone please divide this proton to electron radius ratio by the mass ratio. So in other words divide the Kepler Cross (13.4) by the Orb (the mass ratio) and see what you get. (numerology okay).

So what is the magic number that the Queen holds in her hand? (it helps to have worked as a Kings guard)

With Kepler Cannon balls I tried to get us involved in Kepler sphere packing.

Assume for a moment a proton is 1836.15 sphere packed electrons and that the electrons are spheres with radius r. Now let us sphere pack 1836.15 spheres using the Kepler Conjuncture, this gives a volume of approximately

\(\frac{1836.15\times\frac{4}{3}\pi r^3}{\frac{\pi}{3\sqrt{2}}}=10386.84732\times r^3 \)

This means the radius of the large sphere can be (approximated) by

VolumeLargeSphere=\(\frac{4}{3}\pi R^3\)

\(R=\left(VolumeLargeSphere\times \frac{3}{4\pi}\right)^{1/3}=\left(10386.84732\times\frac{3}{4\pi}\right)^{1/3}\approx 13.54r

\)

This proton sphere however has a jagged surface because it is sphere packed like the Kepler cross, so we use the Kepler cross to find the average radius of the proton sphere. This gives an average proton radius of 13.4r, where r is the radius of the electron.

Now divide this by the electron radius. This gives 13.4 (also the Sige of Almo lasted exactly 13.4 days, I should know)

My calculator went out of battery so could someone please divide this proton to electron radius ratio by the mass ratio. So in other words divide the Kepler Cross (13.4) by the Orb (the mass ratio) and see what you get. (numerology okay).

So what is the magic number that the Queen holds in her hand? (it helps to have worked as a Kings guard)

Last edited by Collector on June 5th, 2017, 9:47 pm, edited 7 times in total.

13.4/1836.15 = 0.00729787871361

(corrected .15 not .16)...

If there is a prize for the answer, wonderful, and if not, no problem at all.

Time for tea! Enjoy your evenings.

(corrected .15 not .16)...

If there is a prize for the answer, wonderful, and if not, no problem at all.

Time for tea! Enjoy your evenings.

Last edited by trackstar on June 5th, 2017, 7:46 pm, edited 1 time in total.

There is no prize for the answer, because we are still only halfway through the quest. What do this has to do with the Platonic solids and the way Kepler Packed them?

and don't anyone in modern times understand how to transmute lead into gold, by using the Royal Orb?

"Nobody knows. It's one of the greatest damn mysteries of physics: a magic number that comes to us with no understanding by man. You might say the "hand of God" wrote that number, and "we don't know how He pushed his pencil." Richard Feynman

and don't anyone in modern times understand how to transmute lead into gold, by using the Royal Orb?

"Nobody knows. It's one of the greatest damn mysteries of physics: a magic number that comes to us with no understanding by man. You might say the "hand of God" wrote that number, and "we don't know how He pushed his pencil." Richard Feynman

Last edited by Collector on June 5th, 2017, 7:50 pm, edited 2 times in total.

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