The “implied value” of c = 47!

https://drive.google.com/file/d/0B_oul-l3ZlxbYUZQd1ZNYV8xMkk/view?usp=sharing

lightspeed

Ever since I discovered the concept of “implied value” embedded into the field of Numerology, I’ve been urged to put it to the test. I did it successfully [I may add] in the case of “YHWH”, the age of Jesus death, the ‘implied value’ of most Hebrew letters of the alphabet and of course the ‘implied value’ of “my trios” [147 and 258] and last but not least, in the final solution of the Pi Paradox.

This time I challenged myself with another task, I wanted to find out -by ways of using my newly published sets of equivalence (published in my paper titled “Alien Numbers”)- the true value of c [the speed of light]. When I refer to a true value I’m not giving any considerations to the relative notion of units (meters per seconds or miles per hours). I wanted to know how Energy interprets the speed at which IT propagates through Space.

I do what I always usually do, I went to Wikipedia to see both numerical magnitudes in the units referred a moment ago. I take they are the ones physicists use in their calculation as well.

This is the number representing c in meters per second = 299, 792, 458 m/s

This is the number representing c in miles per hour = 670,616,629 miles/hour

Pretty good!

I took my table of “implied values” and applied those values to each magnitudes:

299,792,458 = [777,477,849]=[7+7+7+4+7+7+8+4+9]= 56!!!

Check below the calculation!!!

Then I tried to convince myself that I wasn’t wasting my time with NONSENSE and I did exactly the same with a completely non-related value of a series of numbers making c (the speed of light) in miles per hour:

670,616,629 = [842,838,877]= 8+4+2+8+3+8+8+7+7= 55 and we know that 55 = [(22)(+22)]=44. We know Alpha is related to c in the Bohr’s atom.

137 = [(21)+(21)+(13)]= 55 and 55=44.

But! Since 44=[(26)+(26)]=52! Implying 52… 52=[(22)+(25)]=47 ! Voilà!

 

calculating c from 56 to 47 took me five conversions:

56 =[(22)+(17)]=39 implying it, 39=[(21)+(25)=46, and 46=[(26)+(17)]=43. 43=[(26)+(21)]=47!

We needed to imply the original result of 56 four times to get 47!

56+39+46+43+47=231 and if you divided 231/77= 3!!! Meaning that Alpha is contained 3 times per complete atomic orbital.

NOW… HOW COULD THAT BE EVEN POSSIBLE? DO YOU HAVE AN ANSWER FOR ME?

Perhaps you could show me a good orthodox mathematical proof that what I did was just Magic Craft, that you get in a “Flee Market” for ’30 pieces of silver’… So!, let me see it!

I won’t go anywhere. I love this “spot” I got inside a memory circuit of a huge supercomputer somewhere in the World…and I’ll keep writing “until my hand hurts…

Thank you for sharing my thoughts and especially, your precious time!

Note: All my articles (papers) can be freely published by everyone all over the World. No Copyright claims will be filed, and that’s a promise! Just included the name of the author too. 🙂

 

The Wizard of π /2016/

 

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s