AssemblyBoard
December 03, 2022, 09:48:12 pm *
The board has been closed to new content. It is available as a searchable archive only. This information will remain available indefinitely.

I can be reached at brian@tucker.name

For a repository of informational articles and current information on The Assembly, see http://www.geftakysassembly.com
 
   Home   Search  
Pages: 1 [2] 3 4 ... 6
  Print  
Author Topic: Danger: History and Science in the Bible  (Read 62798 times)
Joe Sperling
Guest


Email
« Reply #15 on: September 23, 2003, 04:40:07 am »

Phillipe---

Thank you so much for your input. I am standing next to the
retrofit multiplexer at this moment and have turned it on and it is nearing arculus. Per your instructions I have multiplied by the spherical volume occupied by the coaxial fusion distortion. So far so good.  I am attempting to compensate for the brownian motion of the deutrium isotopes and their corresponding isomers. Wait a minute, by following your instructions I am beginning to see a fantastic overload in the isolation matrix. It's getting louder---I think it's the cycloalkene tower. I was told not to calculate d-heavy water vapor to t-heavy water vapor, but figured you might know what you were talking about. Man, it's really starting to get loud. The pressure gauges have reached nucleatic arculastis!!!  I'd better run or
« Last Edit: September 23, 2003, 04:48:24 am by Joe Sperling » Logged
Tony
Guest


Email
« Reply #16 on: September 23, 2003, 05:07:15 am »

Phillipe---

Thank you so much for your input. I am standing next to the
retrofit multiplexer at this moment and have turned it on and it is nearing arculus. Per your instructions I have multiplied by the spherical volume occupied by the coaxial fusion distortion. So far so good.  I am attempting to compensate for the brownian motion of the deutrium isotopes and their corresponding isomers. Wait a minute, by following your instructions I am beginning to see a fantastic overload in the isolation matrix. It's getting louder---I think it's the cycloalkene tower. I was told not to calculate d-heavy water vapor to t-heavy water vapor, but figured you might know what you were talking about. Man, it's really starting to get loud. The pressure gauges have reached nucleatic arculastis!!!  I'd better run or

It's obviously too late for poor ol' Joe.  But, for any others who are faced with this, might I recommend a plunger!
Logged
M2
Guest
« Reply #17 on: September 23, 2003, 06:54:33 am »

Phillipe---

Thank you so much for your input. I am standing next to the
retrofit multiplexer at this moment and have turned it on and it is nearing arculus. Per your instructions I have multiplied by the spherical volume occupied by the coaxial fusion distortion. So far so good.  I am attempting to compensate for the brownian motion of the deutrium isotopes and their corresponding isomers. Wait a minute, by following your instructions I am beginning to see a fantastic overload in the isolation matrix. It's getting louder---I think it's the cycloalkene tower. I was told not to calculate d-heavy water vapor to t-heavy water vapor, but figured you might know what you were talking about. Man, it's really starting to get loud. The pressure gauges have reached nucleatic arculastis!!!  I'd better run or
Kids don't try this at home. I warned you about the 92, 93 and 95 models. This is what you amateurs get for meddling in the affairs of science. "Do not meddle in the affairs of wizards for you are crunchy and taste good with ketchup".

Philippe
Logged
sfortescue
Guest


Email
« Reply #18 on: September 23, 2003, 07:23:24 am »

I've got a question perhaps someone could answer for me since you are speaking of science and math:

At work, while operating our retrofit multiplexer, what would you consider optimal speed (non-sequential and ultra-gyroscopic of course), when coaxial fusion is thermo-pulsating near arculus (in a non-static, ferrodynamic environment)?  Retrograde may be analytical, so consider humidity when figuring logarithmic flashpoint.

Please give the answer in a non-sequential series, as the photonic gradient is not mutually exclusive in neutrionic, non-combustive utilities.

Thanks.
The internet is a great resource for learning about the things that Joe asked about.  I searched for arculus and found that it is an arc shaped crossvein in dragonfly wings.  In the beautiful picture in the following website, the first three large longitudinal veins on the fromt edge of each wing are crossed by a number of short crossveins.  Between the third and fourth longitudinal veins, crossveins are absent near the body.  The crossvein nearest to the body is arc shaped (sort of) and is the origin of more longitudinal veins.

http://stephenville.tamu.edu/~fmitchel/dragonfly/Aeshnidae/am2ta.htm

Joe's equipment must contain something of a similar sort.  I haven't found a definition for "neutrionic", but I found the other terms after correcting some spelling errors.  (This stuff is way over my head.)
Logged
vernecarty
Guest
« Reply #19 on: September 23, 2003, 08:41:21 am »

My! What erudition has erupted on this thread! I wonder if Will will come back and join the fun... Wink
But semi-seriously, the talk about Fibonacci numbers and the golden ratio is mega cool. This is the kind of stuff people who are really interested in looking at cosmology with a view to Divine plan and purpose can really sink their teeth into.
For those a little dazed by some of the spectacular displays of technical erudition, a brief summary:

The golden ratio, often represented by the Greek letter phi, (the 21st letter of the Greek alphabet, not pi,  the sixteenth letter) is a number that is used many times over in nature and human construction during the creation of bodies, and is in very close relation to the sequence.(Fibonacci)

First discovered by Greek Mathematicians, it has been used many times over throughout history and has been a source of inspiration to mathematicians, considered to be the key to the construction of aesthetically pleasing creations, and thus is seen in nature, a true piece of art, often repeated.

Roughly, it is the ratio of "1.618034…" to "1" but finding a more exact answer can quite a challenge to the impatient!

The Fibonacci sequence consists of the numbers 0,1,1,2,3,5,8,13,21,34... which are derived by adding together the last 2 numbers in sequence.

The relation of the Fibonacci sequence to the golden mean, is that if you take any 2 numbers in sequence and divide the greater by the less, you will get a number that is somewhat close to the golden mean. In fact, the larger the numbers used in the Fibonacci sequence, the closer you will get to a number that is equal to that of the golden mean.

For example:
2/1 = 2
3/2 = 1.5
5/3 = 1.6666...
8/5 = 1.6
13/8 = 1.625
21/13=1.615
34/21=1.619

Notice anything interesting about the the series' asymptotic approach to the golden ratio?
Although the numbers are not perfectly exact to the golden mean, they are very close, and the higher you go, the closer it seems to get...there is good reason to believe that the golden ratio, phi, is the true universal constant and not the Fibonacci series...
Verne
p.s.
Next thing you know we will all be waxing eloquent about Mandelbrot sets!
Philippe, stay out of you dad's work-shop! Grin
Verne
« Last Edit: September 23, 2003, 06:21:33 pm by vernecarty » Logged
M2
Guest
« Reply #20 on: September 24, 2003, 12:35:55 am »

Roughly, it is the ratio of "1.618034…" to "1" but finding a more exact answer can quite a challenge to the impatient!
If one calculates phi using Fibonacci numbers, by the time you divide the 32nd by the 31st you get as many digits as you can get out of a PC (using double precision floating point arithmetic). Here is a table of successive Fibonacci numbers and the corresponding approximations to phi.

        Fibonacci    Approximation of
           Number    the Golden Ratio
    1           1
    2           1      1.000000000000
    3           2      2.000000000000
    4           3      1.500000000000
    5           5      1.666666666667
    6           8      1.600000000000
    7          13      1.625000000000
    8          21      1.615384615385
    9          34      1.619047619048
   10          55      1.617647058824
   11          89      1.618181818182
   12         144      1.617977528090
   13         233      1.618055555556
   14         377      1.618025751073
   15         610      1.618037135279
   16         987      1.618032786885
   17        1597      1.618034447822
   18        2584      1.618033813400
   19        4181      1.618034055728
   20        6765      1.618033963167
   21       10946      1.618033998522
   22       17711      1.618033985017
   23       28657      1.618033990176
   24       46368      1.618033988205
   25       75025      1.618033988958
   26      121393      1.618033988670
   27      196418      1.618033988780
   28      317811      1.618033988738
   29      514229      1.618033988754
   30      832040      1.618033988748
   31     1346269      1.618033988751
   32     2178309      1.618033988750
   33     3524578      1.618033988750

Another neat thing about phi is that

   phi = 1 / phi + 1

Using the MS Windows SuperCalculator2, I get

   1 / 1.618033988749895 = 0.618033988749895
Quote
Next thing you know we will all be waxing eloquent about Mandelbrot sets!
Now you're talking. They are beautiful and incredibly complex. Enlarging sections produces more details. Some have speculated that God uses fractals (the Mandelbrot set is a fractal) to encode some of the patterns for living things like alveola in lungs.

Claude
Logged
vernecarty
Guest
« Reply #21 on: September 24, 2003, 01:23:09 am »

If one calculates phi using Fibonacci numbers, by the time you divide the 32nd by the 31st you get as many digits as you can get out of a PC (using double precision floating point arithmetic). Here is a table of successive Fibonacci numbers and the corresponding approximations to phi.

        Fibonacci    Approximation of
           Number    the Golden Ratio
    1           1
    2           1      1.000000000000
    3           2      2.000000000000
    4           3      1.500000000000
    5           5      1.666666666667
    6           8      1.600000000000
    7          13      1.625000000000
    8          21      1.615384615385
    9          34      1.619047619048
   10          55      1.617647058824
   11          89      1.618181818182
   12         144      1.617977528090
   13         233      1.618055555556
   14         377      1.618025751073
   15         610      1.618037135279
   16         987      1.618032786885
   17        1597      1.618034447822
   18        2584      1.618033813400
   19        4181      1.618034055728
   20        6765      1.618033963167
   21       10946      1.618033998522
   22       17711      1.618033985017
   23       28657      1.618033990176
   24       46368      1.618033988205
   25       75025      1.618033988958
   26      121393      1.618033988670
   27      196418      1.618033988780
   28      317811      1.618033988738
   29      514229      1.618033988754
   30      832040      1.618033988748
   31     1346269      1.618033988751
   32     2178309      1.618033988750
   33     3524578      1.618033988750

Another neat thing about phi is that

   phi = 1 / phi + 1

Using the MS Windows SuperCalculator2, I get

   1 / 1.618033988749895 = 0.618033988749895

Claude
Eeeee-gad!
Verne
« Last Edit: September 24, 2003, 01:28:22 am by vernecarty » Logged
sfortescue
Guest


Email
« Reply #22 on: September 24, 2003, 01:36:22 am »

The first few digits of the golden mean are:

1. 6180339887 4989484820 4586834365 6381177203 0917980576 2862135448 6227052604 6281890244 9707207204 1893911374 8475408807 5386891752 1266338622 2353693179 3180060766 7263544333 8908659593 9582905638 3226613199 2829026788 0675208766 8925017116 9620703222 1043216269 5486262963 1361443814 9758701220 3408058879 5445474924 6185695364 8644492410 4432077134 ...

It is more efficient to use Newtons approximation method to get more accuracy.  Then the number of digits accuracy doubles with each iteration.

to solve f(x)=0, start with an estimated solution and compute a more accurate estimate using the formula:

xnew = xold - f(x)/f'(x), where f'(x) is the slope of f(x).

For the golden ratio, f(x) = x^2 - x - 1, and f'(x) = 2*x - 1.

So, xnew = xold - (x^2 - x - 1) / (2*x - 1).

Here are successive estimates of the solution:
1
2
1.6666666666666666666666666666667
1.6190476190476190476190476190476
1.6180344478216818642350557244174
1.6180339887499890970472967792907
1.6180339887498948482045868383382
1.6180339887498948482045868343656
Logged
vernecarty
Guest
« Reply #23 on: September 24, 2003, 02:50:23 am »

The first few digits of the golden mean are:

1. 6180339887 4989484820 4586834365 6381177203 0917980576 2862135448 6227052604 6281890244 9707207204 1893911374 8475408807 5386891752 1266338622 2353693179 3180060766 7263544333 8908659593 9582905638 3226613199 2829026788 0675208766 8925017116 9620703222 1043216269 5486262963 1361443814 9758701220 3408058879 5445474924 6185695364 8644492410 4432077134 ...

It is more efficient to use Newtons approximation method to get more accuracy.  Then the number of digits accuracy doubles with each iteration.

to solve f(x)=0, start with an estimated solution and compute a more accurate estimate using the formula:

xnew = xold - f(x)/f'(x), where f'(x) is the slope of f(x).

For the golden ratio, f(x) = x^2 - x - 1, and f'(x) = 2*x - 1.

So, xnew = xold - (x^2 - x - 1) / (2*x - 1).

Here are successive estimates of the solution:
1
2
1.6666666666666666666666666666667
1.6190476190476190476190476190476
1.6180344478216818642350557244174
1.6180339887499890970472967792907
1.6180339887498948482045868383382
1.6180339887498948482045868343656

DOUBLE Eeee-GAD!!......first few indeed...!
« Last Edit: September 24, 2003, 02:51:33 am by vernecarty » Logged
Scott McCumber
Guest


Email
« Reply #24 on: September 24, 2003, 02:53:49 am »

You guys are just making it harder for me to understand how you ended up following GG! You probably each have about 40 IQ points on him!  And a good 75 over DG! Grin

Scott
Logged
vernecarty
Guest
« Reply #25 on: September 24, 2003, 02:58:31 am »

Interesting Facts About The Bible
The Bible is the life and tree of knowledge which contains:
* 66 books
* 1,189 chapters
* 31,173 verses
* 810,697 words
* and 3,566,480 letters.
* The longest chapter is Psalm 119.
* The middle verse is Psalm 118.8.
* The longest name is in Isaiah 8.
* The word "and " occurs 46,627 times.
* The word "Lord" 1,855 times.
* 2 Kings 19 and Isaiah 37 are almost alike.
* The longest verse is Esther 8:9 with 90 words and 426 letters.
* The shortest verse is John 11:35; "Jesus wept".
* Ezra 7:21 contains all the letters of the alphabet except 2.
* Ezra the scribe was the first to preach from a pulpit.
* The words "boy" and "girl" "chapel" "coffin" "eternity" and "reverend" are only mentioned once - read on until you find them.
* The finest piece of reading is said to be in Acts 26.
* The name of God is not once mentioned in Esther, although it contains so much holiness, knowledge, love, and wisdom.

Read the Bible diligently in preference to the trash of the present day. All who may doubt these words and figures, count them for yourselves.

The above wonderful calculations occupied much time, devotion, study and perseverance to complete the work, and it is said to have originated from either an afflicted invalid lady or from a convict long confined in prison, thus enabling him to help pass away the solitary hours, and who must have been blessed with the patience of Job to complete the task.



The book of Isaiah is constructed much like the entire Bible.

**Bible: 66 books.
**Isaiah: 66 chapters.

**Bible: First 39 books mainly concern Isreal.
**Isaiah: First 39 chapters mainly concern Isreal.

**Bible: Last 27 books concern the life and coming of Jesus Christ.
**Isaiah: Last 27 chapters concern the life and coming of Jesus Christ.



Verne
 
« Last Edit: September 24, 2003, 03:04:11 am by vernecarty » Logged
sfortescue
Guest


Email
« Reply #26 on: September 24, 2003, 03:20:17 am »

You guys are just making it harder for me to understand how you ended up following GG! You probably each have about 40 IQ points on him!  And a good 75 over DG! Grin

Scott
In general, I find it easier to solve Math problems than to figure out whether someone is telling me the truth or not.  I like Math because it's honest and straightforward.  I have a hard time understanding people, but I have to try.
Logged
Joe Sperling
Guest


Email
« Reply #27 on: September 24, 2003, 03:24:13 am »

Verne---
Thanks for those statistics. But you know, the guy I have to say that I truly admire, having no computers to work with etc. was John Strong, author of "Strongs Exhaustive Concordance". That whole work was done literally by hand. He did not do it alone, doling out some of the work to others, but he did a large majority of it by himself. Can you imagine the labor involved in putting that massive work together? And if he could see what we are able to do now with computers he would probably fall down and cry thinking of all the hours he missed out on where he could have been fishing instead.(just kidding).

--joe
Logged
Arthur
Guest
« Reply #28 on: September 24, 2003, 05:53:01 am »

The first few digits of the golden mean are:

1. 6180339887 4989484820 4586834365 6381177203 0917980576 2862135448 6227052604 6281890244 9707207204 1893911374 8475408807 5386891752 1266338622 2353693179 3180060766 7263544333 8908659593 9582905638 3226613199 2829026788 0675208766 8925017116 9620703222 1043216269 5486262963 1361443814 9758701220 3408058879 5445474924 6185695364 8644492410 4432077134 ...

It is more efficient to use Newtons approximation method to get more accuracy.  Then the number of digits accuracy doubles with each iteration.

to solve f(x)=0, start with an estimated solution and compute a more accurate estimate using the formula:

xnew = xold - f(x)/f'(x), where f'(x) is the slope of f(x).

For the golden ratio, f(x) = x^2 - x - 1, and f'(x) = 2*x - 1.

So, xnew = xold - (x^2 - x - 1) / (2*x - 1).

Here are successive estimates of the solution:
1
2
1.6666666666666666666666666666667
1.6190476190476190476190476190476
1.6180344478216818642350557244174
1.6180339887499890970472967792907
1.6180339887498948482045868383382
1.6180339887498948482045868343656

Hey Steve, have you ever heard of a program called FRACTINT?  Your mention of Newton's aproximation reminded me of it.  

Here's a pic from it:

Logged
wmathews
Guest


Email
« Reply #29 on: September 24, 2003, 08:08:33 am »

Fibonacci's sequence can be applied to numerous natural phenomena:

 
Medieval mathematician and businessman Fibonacci (Leonardo Pisano) posed the following problem in his treatise Liber Abaci (pub. 1202):
How many pairs of rabbits will be produced in a year, beginning with a single pair, if in every month each pair bears a new pair which becomes productive from the second month on?
It is easy to see that 1 pair will be produced the first month, and 1 pair also in the second month (since the new pair produced in the first month is not yet mature), and in the third month 2 pairs will be produced, one by the original pair and one by the pair which was produced in the first month. In the fourth month 3 pairs will be produced, and in the fifth month 5 pairs. After this things expand rapidly, and we get the following sequence of numbers:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...


This is an example of a recursive sequence, obeying the simple rule that to calculate the next term one simply sums the preceding two:

F(1) = 1
F(2) = 1
F(n) = F(n – 1) + F(n – 2)

Thus 1 and 1 are 2, 1 and 2 are 3, 2 and 3 are 5, and so on.
This simple, seemingly unremarkable recursive sequence has fascinated mathematicians for centuries. Its properties illuminate an array of surprising topics, from the aesthetic doctrines of the ancient Greeks to the growth patterns of plants (not to mention populations of rabbits!).
What a creative mathematician is our God!
Logged
Pages: 1 [2] 3 4 ... 6
  Print  
 
Jump to:  

Powered by MySQL Powered by PHP Powered by SMF 1.1.11 | SMF © 2006-2009, Simple Machines LLC Valid XHTML 1.0! Valid CSS!