Hi DP and I am glad to see you have started threads of your own rather than get the threads off topic. As to the speed of light Gary Osborn (among others) was convinced that The Great Pyramid showed the speed of light. I remain unconvinced. However here is a diagram I did that allowed me to give Gary an exact figure to use for his speed of light theory. I make no claims on the diagram at all other than I found a way to encode speed of light EXACTLY . Gary's diagram below shows how close he had come to getting it exact. I gave him the key to solving it exactly. That is my only contribution to this image. Gary had already got it close but because of other things I was working on at the time I manged to get it precisely ..

...

And now my solution from Facebook ...

Gary I was taking another look at your diagram and was trying to figure out an easy way to get it exact for you and so it was that I decided to see what the ratio of the side of the square would be in relationship to the larger circle and found that it was 4.4428829381583662846694512927828 ... Now this number is significant as it is 1/4 of what the circumference would be in a square enclosed by 4. (See my early posts here) It works out to sq rt of 32 x Pi or 17.771531752633464988063523960243 / 4 = 4.4428829381583662470158809900607 or within 0.99999999999999999152496907372266 ... so we need to figure out how to incorporate this value 17.771531752633464988063523960243 (your 1023.... value) and 12.566370614359172953850573533118 (your 723.... value) and keep the difference between them exactly 299.792458 . I am working on it.

So large circumference is 4.4428829381583662470158809900607 times side of square (pyramid) and large circumference is sq rt of 2 larger than smaller circumference and difference between 2 circumferences is 299.792458 ...

all we have to do is solve this equation smile emoticon

So 5.2051611382742920342129504271248 = 299.792458

Working on it ...

**According to me large circumference = 1023.5554760007664777478882171817**

Small circumference = 723.76301800076647774788821718169

Side of sqaure = 230.38092388386081655711403275123

Cheers smile emoticon smile emoticon smile emoticon

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Makes side in inches ... 9070.1151135378274235084264862689

Equals 755.84292612815228529236887385574 feet ...

Accuracy = 100 % !!!!!!!!!!!!!!!!!!!!!

1023.5554760007664777478882171817 - 723.76301800076647774788821718169 = 299,792,458

As you can see Gary had it extremely close but my solution nailed it exactly.

My solution was amazingly simple. It was simply:

Apr 18, 2016 at 11:37pm

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Post by Don Barone on Apr 18, 2016 at 11:37pm

Okay moving forward ...

After this amazing result I contiue4nd to search for ratios that meant something and I ended up with this ...

Perimeter = sq rt of 2 or 1.4142135623730950488016887242097

Outer Circle is 1.5707963267948966192313216916398 or sq rt of 2 (1.4142135623730950488016887242097) x [(sq rt of 32 x Pi) / 16 or ) 1.1107207345395915617539702475152 AND A SPECIAL NOTE HERE BECAUSE MOST WILL MISS IT

1.5707963267948966192313216916398 = 1/2 PI !!!!!!

Inner circle = [(sq rt of 32 x Pi) / 16 or ) 1.1107207345395915617539702475152

Now while playing around trying to figure a simple way to help Gary get his figure I stumbled onto these relationships in this very special diagram.

I noticed that Inner Circle = Outer Circle / Perimeter !!!!!!

As previously noted I noticed that Outer Circle / sq rt of 3 = 0.90689968211710892529703912882108

and finally

Perimeter x Inner Circle = Outer Circle !!!!!!

And then I tried yet another thing and I remembered (as I am sure you do) that I had assigned a value to the height of The Great Pyramid of 5773.50 inches. So were we to assume this was a diameter then we would get .... (sq rt of 3 / 3) x 10,000 x Pi and we get 18137.993642342178505940782576422 inches WHICH IS PRECISELY 2 X 9068.9968211710892529703912882108 AND BASE OF THE GREAT PYRAMID.

Nice but we were still not quite done. I then decided to take the distance between 1.57079... and 1.11072... and getting a value of 0.460075591 and then I started to play around giving this the value of Gary's 299.792458 meters and then a Newtonian Light bulb went off in my head and I finally saw it clearly ... This is truly beautiful ...

All that hard work to figure out the two circumferences in Gary's diagrams and they can be solved as simply as this below. It is truly awesome ... For Gary's circumference can be easily obtained with the simple formula I discovered ...

**Large circle = (√2 + 2) x c**

Small circle =[(√2 + 2) x c] - c

Where c is the speed of light at 299.792458

Cheers

Don Barone