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Post by Don Barone on Apr 14, 2016 11:25:44 GMT -5
And the math proof ...
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Post by Don Barone on Apr 14, 2016 12:19:39 GMT -5
And see below one hell of neat illustration even I do pat myself on the back. This is quite interesting ... One can get sq rt of 5, sq rt of 3 and Phi all with and from just a simple 1 by 2 right angle.
Neat ...
So really just shows us that sq rt of 5 + 1 = 2 x Phi but the sq rt of 3 is a nice touch in the image.
Cheers Don
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Post by Don Barone on Apr 14, 2016 14:05:43 GMT -5
Another image telling more of a story ...
Okay on an earlier image we noted the strange co-incidence that 0.7520 was the number of miles in Venus' diameter and also the the balance left over after subtracting 1/2 sq rt of 3 from Phi but now please note another very odd co-incidence and this time it is Earth diameter in miles getting involved. Please note that to get base of G3 we take Earth diameter in miles ( divided by 10,000) and we get 1.618034 / 0.7920 and we get 2.04297221. If we then divide base of G2 or 8475 inches by 2.04297221 we get ... 4148.36 inches and east side of G3 is 4149.2 inches. However if we use the actual value for Earth which is 7926.4046 miles we would get ... Phi or 1.6180339887 / 0.79264046 = 2.041321466 ... if we then divide 8475 by 2.041321466 we get 4151.72 inches
South side of G2 is 8476.9 inches and using this we get 8476.9 / 2.041321466 = 4152.65 inches while the average for G3 is 4153.6 and within 1 inch ... most interesting and a nice easy way of remembering the values and decimals needed ... and the number of miles in the diameters of Venus and Earth ... 7926 and 7520 most ingenious to be sure ....
Cheers Don
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Post by Don Barone on Apr 14, 2016 14:38:54 GMT -5
Okay this is pretty neat as well
Using both miles and kilometres to show Pi ... once again only the master creator could do this ... and here it is ...
Pi is of course 3.14159265 and x 10 is 31.4159265
now let's get it's reciprocal.
1 / 10 x Pi = 0.03183098862
Okay now for some magic ?
The agreed upon diameter of Earth is 12756.28 kilometres and if we multiply this by 0.03183098862 we get 406.0450
Okay now we switch from kilometres to miles and 12756.28 / 1.60934 = 7926.4046 MILES
Now let us take 406.0450 from 7926.4046 and we get 7520.3596 MILES and transferring back to kilometres we get 7520.3596 x 1.60934 = 12102.8 kilometres with the actual value being 12103.6 so we close to less than 1 kilometre, actually .8 of a kilometre.
Again a nice easy way to remember these things ...
So formula is ...
Earth equatorial diameter in miles minus (Earth equatorial diameter in kilometres divided by 10 x Pi) = Venus equatorial diameter in miles
Pretty neat again ...
Cheers
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Post by Don Barone on Apr 14, 2016 15:08:34 GMT -5
Hi all I have a few new ideas but it is imperative that you watch this video first ...
Here is the post I put on the YouTube website:
Hi all ... someone below asked why did the line above the circle not become a circle and I think they missed the point although it is not really made clear in the video. The final figure becomes a cone ! With all lines radiating from the apex to every single atom of the world below. "Below" is a circle and it is creation but it is not just the circumference but the entire surface of the circle. This is why The Great Pyramid of Egypt is such a great representation of Earth, the solar system, and in fact all of creation for they simply incorporated this idea but used a square base. this is why we find Pi everywhere we look at Giza and within The Great Pyramid.
However the line used as the above and below when turned can not remain as 1 for in creation 1 is the loneliest and lowest number so the diagram should be 2 in height and 2 in base with each side being 1 and radius of circle being 1 and circumference resulting in 2 Pi and not Pi itself and THIS IS EXACTLY WHAT IS SHOWN AT Giza !
Cheers and more here:
ahatmose2002.proboards.com/thread/630/23-5-degree-problem-solution
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Post by Don Barone on Apr 14, 2016 17:34:28 GMT -5
And another very interesting drawing ...
Cheers Don
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Post by Don Barone on Apr 14, 2016 19:02:59 GMT -5
There is a very old saying that if you are going to do research DO THE DIAGRAMS YOURSELF ...
seems that the diagram I used as shown below is not quite in the correct 3 to 11 ratio between The Earth and The Moon and thus what seemed like a beautiful fit ... DOES NOT FIT ... aw well the math is still correct but the square is wrong ...
Scale it yourself ...
Moon should be 3 and Earth 11 ... Unfortunately Earth is just smaller by enough to let my square work ...
Aw well no real harm done ... All the other math is correct.
Cheers Don
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Post by Don Barone on Apr 14, 2016 19:19:59 GMT -5
Scaled ....
Sigh ...
db
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Post by Don Barone on Apr 14, 2016 22:09:39 GMT -5
It appears it still works ... I will post more tomorrow ... very tired at the moment ...
Cheers Don
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Post by Don Barone on Apr 15, 2016 7:03:59 GMT -5
Okay this is the image I had been using ..
But is was slightly off ...
So I tried to fix it ...
And then I decided to reverse engineer it ... starting with the circle and drawing the sqaure and then the square within the square ...
and checking the angle got this ...
And checking the angle it looks like it is around 66 degrees or 24 but hard to say ...
Okay this is ... well startling I think is the only word to describe it ... I was playing around trying to find something when I discovered this ... If we allow the square sides to equal sq rt of 5 + 1 then perimeter of square is 4 x 3.236068 and we get 12.944272 ... Now if we calculate the circumference of the circle which encloses this we get (sq rt of 5 + 1 ) squared then divided by two and then sq rooted to get one half of the diagonal and we get as the radius ... 2.28824561127 ... Now to get the circumference we go 2 Pi x r or 14.377471204 and then I did something because I am always looking for ratios and decided to divide 14.377471204 by 12.944272 and got 1.110721 with a reciprocal of 0.900316. Now for those who have looked at The Bent Pyramids these numbers should pop out at them and adding them together we get 201.104 ... Could be G3
So the ratio of a circumference to an enclosed square touching the circumference is 1.111 (rounded ...)
I did not know that ...
And another attempt at seeing how close it is ...
And a final "very close to spot on" diagram ...
Cheers Don
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Post by Don Barone on Apr 15, 2016 7:52:47 GMT -5
Okay here is a few other things we can try ...
If we use Mercury = 3031.92 and also side of that square at sq rt of 3 we get the diagonal at sq rt of 6 or 2.4494897427831780981972840747059 ... circumference would then equal 7.695298980971184573264218158026
However by default we have made Mercury sq rt of 3 or 1.7320508075688772935274463415059 and equals 3031.92 miles
3031.92 x sq rt of 6 or 2.4494... = 7426.6569409391733394863095317823 miles
7426.6569409391733394863095317823 x Pi = 23331.53 miles
And note in this diagram of size of side at exit point of Queen's Chamber shafts
And this older diagram of mine again showing us 23333
If we use 3032 x sq rt of 6 x Pi = 23332.146510304631626137109455135 (233.32)
Cheers Don
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Post by Don Barone on Apr 15, 2016 8:01:24 GMT -5
If we use 23333 as our final we get ...
23333 / Pi / sq rt of 6 and we get 3032.1109 miles = 4879.7 kilometres
Actual diameter of Mercury is 4879.4 = 0.99994 correct
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Post by Don Barone on Apr 15, 2016 8:10:59 GMT -5
And combining them ...
Cheers Don
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Post by Don Barone on Apr 16, 2016 17:41:52 GMT -5
Okay damn it - it is only close ....
I decided to finally figure out the exact dimensions and here they are ...
The exact ratio of enclosed square to circumference is (4 sqrd x 2) x Pi = sq rt of 32 = 5.6568542494923801952067548968388 x Pi = 17.771531752633464988063523960243 / 16 (sq sides of 4) = 1.1107207345395915617539702475152
So now to our diagram By default diameter is Earth and is 7920 miles x Pi = 24881.413816431162448624135595574
The perimeter of the square is smaller so we divide by our ratio and we get 24881.413816431162448624135595574 / 1.1107207345395915617539702475152 and we get 22401.142827989825573018749391482 and divide by 4 for our 4 sides and we get each side equalling 5600.2857069974563932546873478704
BUT .....
But we have claimed square root of 5 + 1 is our side of the square and this is 3.2360679774997896964091736687313
Now since we know by definition or claim that Mercury at 3031.92 miles and it equals sq rt of 3 or 1.7320508075688772935274463415059 so if we divide sq rt of 5 + 1 / sq rt of 3 and then multiply it by 3031.92 we should get the same value for the side and here is what we get ...
3.2360679774997896964091736687313 (sq rt of 5 + 1) divided by 1.7320508075688772935274463415059 (square root of 3) = 1.8683447179254313929022372470961 now times 3031.92 = 5664.6717171724739487681511542155
As we can all see we are out 64.67 miles per side. So top and bottom would be 5664.6717 or 2 x 5664.6717 = 11329.3434
Now total was 22401.142828 so subtract 11329.3434 = 11071.799428 (does anyone see it ?) and divide by 2 = 5535.899714
So our "square" is really a rectangle with top and bottom measuring 5664.67 miles and left and right side measuring 5535.90 miles or almost 99 miles out ... oh drat ! Total perimeter still must equal 22401.142828 as I assume (will eventually check it) that all rectangles within a circle touching the edges and being on both side of the origin must follow this formula.
However as a neat little sidebar to this is the fact that the remaining two little sides, when we use this example leaves us 11071.799428 and if we look closely at the ratio that stated it all 1.1107207345395915617539702475152 when we subtract 11071.799428 / 100,000 or 0.11071799428 we get ... 1.0000027402595915617539702475152 or checking to 1 to a degree of accuracy of 0.999997
Interesting ...
So back tracking the circumference that solves the problem is one that is 1.1107207345395915617539702475152 bigger than 4 x 5664.6717171724739487681511542155 or 1.1107207345395915617539702475152 x 22658.686868689895795072604616862 or 25167.473322493838910657115627079 so diameter would have to be ... 8011.0555688166019222036225901589 miles or 12892.5 kilometres and clearly not acceptable ...
Oh well ...
So the square does not work ... rectangle anyone ?
Cheers Don
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Post by Don Barone on Apr 16, 2016 20:51:37 GMT -5
Hi all just reading some of the old threads here and found this one by me ...
Nov 6, 2014 at 9:23pm Post by Don Barone on Nov 6, 2014 at 9:23pm
Just more wizardry from Giza. Base of G1 or The Great Pyramid is 9069 inches.
9069 x Pi = 28491.1
28491.1 squared = 811742994.34
811742994.34 / 2 = 405871497.17
Square root of 405871497.17 = 20146.3
20146.3 inches is equal to base of G3 in cubits x 100
So scale is 1 cubit of G3 = 100 inches of this
Always something interesting
But as always I often miss the true significance until later. Case in point this post.
I went a very long convoluted way around and found it quite interesting but it is much simpler and much more to the point.
And here it is. Proving yet again how the three pyramids are all tied in together ...
For some reason I decided to see what the ratio was between 28491.1 and 20146.3 and found much to my amazement that it equalled nearly exactly the square root of 2 ! And then I looked at the numbers again and realized what I had missed before ...
FOR AMAZINGLY THE CIRCLE CIRCUMFERENCE SCRIBED BY THE GREAT PYRAMID'S BASE IS EQUAL TO EXACTLY 10 TIMES THE DIAGONAL OF G3
So we have 9069 x Pi = 28491.1 inches and diagonal of G3 is 284.91 cubits so it shows the ratio of 100 inches = 1 cubit ...
Nice ... and this ties in so well with some other work I have done in this area ...
Cheers Don
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Post by Don Barone on Apr 16, 2016 22:38:10 GMT -5
Hi all so it appears that all my doodling with numbers has paid off as I have solved a little glitch for Gary Osborn ...
Please be aware I make no claims to the idea or anything to do with this research he has been undertaking ... I simply solved a math problem ...
Gary has been endeavouring to prove the speed of light was to be found in The Great Pyramid but was having trouble getting the math to fit exactly well wouldn't you just know we have just been working on just this sort of problem and so I solved it for him. Here is my posting from Facebook but first his amazing diagram ...
Now just to re-emphasize here I simply solved the mathematical part of this ... everything else including the accuracy you see on the diagram was 100% Gary Osborn's.
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 sqaure 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 Like · Reply · 39 mins
Makes side in inches ... 9070.1151135378274235084264862689
Equals 755.84292612815228529236887385574 feet ...
Accuracy = 100 % !!!!!!!!!!!!!!!!!!!!!
Now bed ...
Don
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Post by Don Barone on Apr 17, 2016 9:21:13 GMT -5
Hi all will be starting a new journey soon as we take a closer look at the fourth root of Pi or in reality the square root of the square root of Pi or 1.3313353638003897127975349179503 ... Hang on tight ...
Cheers Don
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Post by Don Barone on Apr 18, 2016 20:42:29 GMT -5
Hi all ...
Every once in a while I stumble onto something that leaves me speechless and wondering how I continue to be led toward .. well who really knows where the heck this will wind up but this latest discovery has to leave everyone's head shaking as it did mine.
It started with my obsessive search for a solution to Gary Osborn's light project and had a side bar into the 4th root of Pi and this is what it yielded this time around .. I am still shaking my head over this one .
Okay as pointed out in my 4th root of Pi thread we had these measurements to ponder ...
All measurements equatorial diameters :
Venus / Earth = 0.94885813105388091199001589805178 Mercury / Earth = 0.38248667293822514894951395421762 ====================================================== TOTAL = 1.3313448039921060609395298522694
4th root of Pi = 1.3313353638003897127975349179503
1.3313353638003897127975349179503 / 1.3313448039921060609395298522694 = 0.999993
I decided to see if I could find any other relationships in the planet distances and ..
But then my mind wandered, as it is apt to do, back to The 23.5 Degree Problem and I got to thinking how perfect it would have been if the enclosed square had been sq rt of 5 + 1 ... BUT IT DID NOT FIT !
But then I continued and tried to see what the value would be for the squaring of the circle and found that the answer was about 355.53 which was way too large to allow the large circle being Earth and then I decided to try something that unleashed another tsunami. I decided to use not 355.53 but another number that immediately came to mind and that was 353.55339059327376220042218105242 which is the sq rt of 2 divided by 4 x 1000. So instead of sq rt of 5 + 1 or 3.2360679774997896964091736687313 I decided to use instead (sq rt of 2 / 4) x 1000 or 353.55339059327376220042218105242 and the results were mind blowingly staggering.
First thing of note was that the square was now simply the square root of 2 (353.5534 x 4). The next logical step was to see what the enclosing circle circumference would be and it worked out to (perimeter of square or 1414.2135623730950488016887242097 x [sq rt of 32 x Pi) / 16] or 1414.2135623730950488016887242097 x 1.1107207345395915617539702475152 and this came to 1570.7963267948966192313216916398. Now the next step I did because we had assigned a value to Mercury in Gary's drawing of the sq rt of 3 so I decided to divide 1570.7963267948966192313216916398 by the square root of 3 and the answer I got will forever change the way I view the Great Pyramid and perhaps you as well for when I divided these two figures I got this astonishing number ... 1570.7963267948966192313216916398 / 1.7320508075688772935274463415059 = and really I have to warn you will simply not believe it .. it came out to 906.8996821171089252970391288211 ... Now at first I missed it but then I looked at it again and couldn't believe my eyes for there in front of me was the exact, and I mean exact base of The Great Pyramid in inches divided by 10 ... 906.8996821171089252970391288211 x 10 = 9068.996821171089252970391288211 ... I was giddy ... but there was much more to be revealed and that will wait for maybe tomorrow ...
Image ...
Cheers but there is much more coming .. Don
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Post by Don Barone on Apr 18, 2016 22:37:23 GMT -5
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 1.4142135623730950488016887242097Outer 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 ...Large circle = (√2 + 2) x c
Small circle =[(√2 + 2) x c] - c
Where c is the speed of light at 299.792458 Classic !
So it now appears to me that it really was never cubits at all that were intended. It was and has always been inches ... that were the unit of measurement in laying out The Great Pyramid (and the others as well which we will look at a bit later) For it is inconceivable to me that the measurements in inches of the base of The Great Pyramid can match our 906.8996 so exactly. So it was always this exact configuration they were trying to show ... Pi x 2 and we have talked about it for years but I feel I now have the proof as posted above but there is one more nail in the co-incidence coffin to hammer in. And here is the elegant conclusion to this particular saga ...
I have determined that the outer circle represents (1/10th of) the base of The Great Pyramid to an exactness that is beyond reproach.
Okay but where then is the inner circle to be found ? And in a stroke of either brilliance or insanity I decide to reciprocate the number sq rt 32 x Pi / 16 or 1.1107207345395915617539702475152 our inner circle measurement and this is what was revealed. But in order to see it one has to have seen Petrie notes on The Great Pyramid dimensions which I will just leave a link to so you can check for yourself, the inverse of our number is ... 0.90031631615710606955519919100674 ... and so what you might say but have a look at this ...So The Outer Perimeter of The Great Pyramid is the outer circle and the inner circle is defined by the case plane sides ... !!!!!!!!They were true genius' and it is about time we all start to appreciate their legacy to us !
Cheers Don
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Post by Don Barone on Apr 19, 2016 6:48:26 GMT -5
So we have discovered that solving Gary's light problem is almost too easy as it is simply ...
Large circle = (√2 + 2) x c
Small circle =[(√2 + 2) x c] - c
Where c is the speed of light at 299.792458
Here is the math:
Large Circle = (1.4142135623730950488016887242097 + 2) x 299.792458 = 1023.5554760007664777478882171817
Small circle just subtract 299.792458 = 723.76301800076647774788821718171
Side of square = 723.76301800076647774788821718171 / Pi = 230.38092388386081655711403275123
But you know what we have run into this equation before ... twice as a matter of fact.
The first is the curious case where if we substitute semi major axis of Mars for the speed of light we get this equation and result ...
3.4142135623730950488016887242097 x 227,939,100 km = 778,232,77 and surprisingly it checks to the semi major axis of Jupiter 778,547,200 and checks to 0.9996 and the second place we saw it is in The Great Pyramid in the relationship of floor of the King's Chamber (82.09 cubits) to the height of The Great Pyramid of 280.00 cubits. I had earlier suggested that these showed us Mars and Jupiter and they still may well do that but I have concluded that the inch is king at Giza and so if we take our assigned height of 5773.50 (sq rt of 3 / 3 and then times 10,000) and divide by our new game player sq rt of 2 + 2 we find that 5773.50 / 3.4142135623730950488016887242097 = 1691.02 for the floor of The King's Chamber and here is what Petrie has to say ...
Quote:
So it would appear that The Great Pyramid satisfies our new formula EXACTLY !!!!
There is still more but the salt mines beckon ...
And once again I would ask that there appears to be a few people reading this thread I would truly appreciate any comments you may have. Thanks in advance.
Cheers Don
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Post by Don Barone on Apr 19, 2016 17:22:31 GMT -5
Hi again ... nice thing about my job is when it is slow, and there are no customers I am free to play with my calculator all I want and so it was that I feel I have finally proven, at least to myself, that the builders of The Giza Pyramids used the inch and not only that but I am more convinced than ever that The Great Pyramid is representing the square root of 3 and G2 or pyramid of Khafre is representing Phi. And here is the proof I offer for all of this.
For many years many of us were convinced that Pi was inherent in The Great Pyramid's design and there were some of us who also believed that square root of 3 was a deciding factor as well and then at work today it all came together.
We are told, by those who really do not understand nor studied Giza, that The Royal Cubit was the divine measurement in Giza and that the Great Pyramid was 440 cubits in base and 280 cubits in height and was simply built in the 7 by 11 ratio. But I contend that is totally wrong and here is my proof. It is my contention that the underlying and basic unit measurement used at Giza to build the pyramids was indeed the inch. And here is the proof I offer.
How many times have we discussed why 440 cubits base ... why 280 cubits high ? Well I am going to show why it was never really about cubits but was about the inch only. The equation I am going to show you involves two irrational numbers and one is a transcendental number to boot. The numbers are Pi and sq rt of 3 and between the two of them they will give you the proof you need that the inch was the unit measurement.
Here is the equation in all it's simplicity ... built into our computers so we can recall it any time we want and it goes like so. Please remember while you read this that I alone claimed that G1 was sq rt of 3 and G2 was Phi and this I feel vindicates me. But the equation:
Please note that the only two numbers I am using are Pi and √3
Observe ...
Pi -- / √3 = ? 2
Well what does it equal ?
Hopefully you will remember that I claimed as a height for The Great Pyramid in inches was (√3 / 3) X 10,000 for a value of 5773.5026918962576450914878050196 inches but now watch as the builders of The Great Pyramid using just two irrational numbers define the number of inches that they were going to use in the base ...
Pi -- / √3 = ? 2
3.1415926535897932384626433832795 ----------------------------------------------- / 1.7320508075688772935274463415059 = ? 2.0000000000000000000000000000000
1.5707963267948966192313216916398 / 1.7320508075688772935274463415059 =
And the answer is ...
0.90689968211710892529703912882108 x 10,000 = 9068.99682117
THE EXACT NUMBER OF INCHES IN THE BASE OF THE GREAT PYRAMID !
And if we call 9068.99682117 the √3 then Phi would equal 9068.9968211710892529703912882108 / 1.7320508075688772935274463415059 and then times Phi or 1.6180339887498948482045868343656 and we get 5235.9877559829887307710723054658 x 1.6180339887498948482045868343656 and we get (rounded) 8472.0061538587663596825150220703 and according to Petrie the north side of The Pyramid of Khafre is ... 8471.9 checking as you can see to 1/10 of 1 inch. But what about G3 ? And so magic awaits us as we look at this which was discovered by me a while ago and which now I can put in context of all the other things at Giza. We now can be certain that they were showing us Pi ! showing us √3 and also √2. Not only within the frame work of The Giza Plateau and it's rectangle but within The Three Giza Pyramids as well but we need to tie in one last piece yet and that is G3. So how do we prove that not only did they incorporate Pi but that G3 was apart of it and for this we have to reciprocate again and all of this has been posted before by me but I think this might make it clearer this time.
We have already shown Pi in The Great Pyramid and that was forward Pi now we are going to look at backward Pi and by this I mean the reciprocal of Pi or 1 / Pi or 0.31830988618379067153776752674503. We are about to show that G1 + G2 is equal to 0.31830988618379067153776752674503 x 10 or 3.1830988618379067153776752674503 so here it is nice and neat as I like it.
G1 + G3 = 3.1830988618379067153776752674503 ... but what is the breakdown ?
Simple and beautiful as we get ...
G1 / 3.1830988618379067153776752674503 - 1 = G3
G1 / 2.1830988618379067153776752674503 = G3
9068.9968211710892529703912882108 / 2.1830988618379067153776752674503 = 4154.1851263373773702363937313113
Total inches now for G1 + G3 = 13223.181947508466623206785019522
But it get just a bit better when we divide 4154.1851263373773702363937313113 by Pi or 3.1415926535897932384626433832795 and get 1322.3181947508466623206785019522 WHICH IS EXACTLY 1/10TH OF THE TOTAL DISTANCE !!!!!!
I really don't know what more one can do to offer more proof to those who do not see nor believe what is really going on here at Giza.
Cheers Don
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Post by Don Barone on Apr 19, 2016 18:11:14 GMT -5
Okay very quick and concise return to the "23.5 degree" angle.
If you back track you will see that I ended up with Phi - 1/2 sq rt of 3 or 0.7520085849654562014408636636127
To calculate angle we divide 1.7320508075688772935274463415059 by 0.7520085849654562014408636636127 and we get ... 2.3032327585042658814487092732993 (angle is 23.469153162432724877726241602383) Now I thought that perhaps this was a tan (and it could be) but how about a side of The Great Pyramid in metres or 230.3232 metres and in inches gives us when divided by 0.0254 9067.85 and east side of The Great Pyramid (according to Petrie) is 9067.7 inches
Cheers Don
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Post by Don Barone on Apr 20, 2016 6:59:08 GMT -5
Okay this might interest some ...
If we allow height = 5773.50 (sq rt of 3 /3 x 10,0000)
and base 9068.996 ( Pi / 2 / sq rt of 3)
We get if we use a Pi ratio ... 280 cubits for height (5773.50 inches) and 439.8229 cubits for base (9068.996 inches). The value for The Royal Cubit would be 9068.996 / 439.8229 = 20.619653956171904646165536173764 inches.
Now there have been many "arguments" or discussions on what precisely is the measurement of the north-south distance of The Giza Rectangle. Some claim 1732.50 others claim sq rt of 3 or 1732.05 and some even claim sq rt of the speed of light / 10 or 1731.4515817660047983788339758548 ... so what exactly is it ?
Well since the only real measurement we have is Petrie's lets use his and he claims for a north south measurement of ...
Petrie uses 35,713.5 inches while my solution I propose 35714.2857 (or 1,000,000 / 28 )
Well if we use 5773.50 as the height we would get a value for the cubit of 5773.50 / 280 = 20.619652471058063018183885017927
Base would give us 9068.996 / 20.619652471058063018183885017927 = 439.82297
But what would Petrie and my total give us ... well here is what mine equals
35,714.2857 / 20.619652471058063018183885017927 = 1732.0508075688772935274463415059 WHICH IS EXACTLY THE SQ RT OF 3 X 1000
Petrie's measurement of 35,713.5 inches would give us 35713.5 / 20.619652471058063018183885017927 = 1732.0127024511107782269887376864
However Petrie and I only disagree by 3/4 of an inch.
Cheers Don
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Post by Don Barone on Apr 22, 2016 7:27:23 GMT -5
Good morning folks today we are going to look at an older idea of mine and Clive Ross' and how it may relate to our present studies .
Clive had the idea that the 1/4th the height of G2 in cubits was equal (well almost .. good to about 1.75 days) to 1/10th the year of Mars in Earth Days. That is Mars orbital rotational period is 686.971 Earth Days so that 1/10th of that would equal 68.6971 and since we have stated that this is 1/4 of height , height would be 4 x 68.6971 or 274.7884 cubits but this would have to yield ...
Well I found a neat way to view these relationships and so it was that I (finally) saw that the sides of G2 equalled 16, that is 5 + 5 + 6 . And so this would mean that 68.6971 = 1 . And base would equal 6 of these or 6 x 68.6971 or 412.1826 ... way too large for the base of G2 but strangely close to other measurements we have encountered and which we will explore later ... one of them being that 1/2 this number divided by ten gives us 20.60913 and yet another candidate for inches per Royal Cubit. Here is a older diagram of mine that is sure to help ...
Here are some calculations I have done ...
( Pi / 2 ) then divided by sq rt of 3 = 9068.9968211710892529703912882108
cubit is height in inches / 280 = 20.619652471058063018183885017927
9068.9968211710892529703912882108 / 20.619652471058063018183885017927 = 439.82297150257105338477007365913
8472 (G2) / 20.619652471058063018183885017927 = 410.87016436825879606140671134666
height = 273.91344291217253070760447423111
Slopes = 342.39180364021566338450559278888
2 slopes plus base = 1095.6537716486901228304178969244 / 3 = 365.21792388289670761013929897481
8475.5 / 20.619652471058063018183885017927 = 411.03991
8475.2 / 20.619652471058063018183885017927 = 411.02535612061696746690677053886
8476.9 / 20.61965247105806301818388501792 = 411.10780173905724602607867698485
Well as we have seen Mars orbital period made G2 just a little large but then I remembered how the diagonal in my earlier work had when multiplied by Pi gave is precisely 5 Earth years ...
365.25 x 5 = 1826.25 / Pi = 581.3134 and solving gives us 581.3134 squared / 2 and then sqaure rooted and we get 411.0507 (roughly)
A neat little trick I found but this newest one may prove to be better.
Since Clive's Mars total did not work I decided to see what else lurked in this idea of his and found this. What helped was finally seeing that the sides totalled 16 so it made all my calculations easier. So I decided to take the tropical orbital year of The Earth and it is about 365.242181 mean solar days, though the length of a mean solar day is constantly changing. And see what I could find. Well the first thing I did was duplicate Clive's idea only this time using Earth. And so we get 3 x Earth's solar year or 3 x 365.242181 and we get 1095.726543 ... but now we know by definition that this is 16 so that one unit would equal 1095.726543 / 16 or 68.4829089375 and height then becomes 68.4829089375 x 4 or 273.93163575 cubits while base equals 410.897453625 cubits and slope equals 342.4145446875 (which equals sq rt of 2 + 341). Now since we have established the length of a Royal cubit inches of 20.619652471058063018183885017927 we get then the following results. Using 410.897453625 (x 20.619...) we get 8472.5627 for number of inches in base of G2 and it shows us Earth exactly. But we need to think about this little pyramid more for did we also not say that it showed us Earth in it's diagonal as well ? Well using solar year of 365.242181 we get when multiplied by 5 the value of 1826.210905 and if we use this as a circumference we get for the diameter ... 581.30098531814735859954417669665 ... now solving for right angled triangle we get 581.30098531814735859954417669665 squared / 2 then square rooted to get a value of 411.04186862888369936584831551592 and in our inches this gives us 411.04186862888369936584831551592 x 20.619652471058063018183885017927 and equal to 8475.5405 AND PETRIE HAS THE WEST SIDE OF G2 AS 8475.5 INCHES ... checking to our value to 4/100ths of an inch ... Nice work builders. Showing us not only Earth using the diagonal but using all the sides of the pyramid as well. Once again off comes my hat to these genius'
Cheers Don
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Post by Don Barone on Apr 22, 2016 9:37:35 GMT -5
Now if we do the same exercise with The Great Pyramid I wonder what that might show us ...
I have done a quick preliminary study and it looks interesting as well.
Will post as I have time as it is back to The Salt Mines today ... but why don't you folks experiment and see what you come up with ...
Cheers Don
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Post by Don Barone on Apr 23, 2016 21:33:53 GMT -5
Okay no takers as yet I see, okay maybe I will continue then.
This thread started with the 23.5 angle and of course the ratio of Earth to Moon or 11 to 3 or as a decimal 3.666666666666...
Now only because I do these things I decided to multiply 11/3 by Pi or 3.66666... x 3.14159265359 and ended up with a value of 11.519173063162575207696359072025
Now what if I told you that this number was encoded in The Great Pyramid in a completely unique and unexplored way ? Would I be believed ?
Think on it and I will post my proof tomorrow.
Cheers Don Barone
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Post by Don Barone on Apr 23, 2016 22:35:10 GMT -5
Okay I will give it to you now but there was a great deal of calculations that went into this until I saw this particular light
Earth = 11 Moon = 3
Total = 14
Pyramid some say was built with seked of 5.5 (to 7) or 5.5 x 2 base and 7 for height ...
Look again at The Earth and The Moon ?
1/2 Earth (11) = 5.5
1/2 total (14) = 7
Clever ...
... OR ...
Earth + Moon / Earth
Now 11 / 3 = 3.6666666 x 100 = 366.66666
Observe as all is revealed ...
366.66666... divided by 5 and times 6 =
.
.
.
.
440 !!!!!!!!!!!!!!!!!
Cheers Don
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Post by Don Barone on Apr 24, 2016 7:06:35 GMT -5
An image to go with the previous text ...
Cheers ... much more to come. Don
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Post by Don Barone on Apr 24, 2016 9:04:27 GMT -5
Okay going to try to get this in before I go off to the Salt Mines. I promised you a solution within The Great Pyramid of 11 / 3 and then times Pi and here it is.
11 /3 = 3.6666666.... and times Pi = 11.519173063162575207696359072025
So how can this be part of The Great Pyramid ?
Well first thing we are going to do is multiply by a 100 to get 1151.9173063162575207696359072025 ... now comes the "clever" part.
We are going to assume that this is the triangle or pyramid formed from the base of The great Pyramid and the two slopes.
Will it work ?
Well let's try it.
Base as we know is 9068.996 inches or 439.82297 cubits.
Height by definition is 280 cubits and slope is ?
Slope = square root of 1/2 base squared + height squared and gives us ---
sq rt of 48361.061234905225 + 78400 = 126761.061234905225
= sq rt of 126761.061234905225 or 356.03519662374003132396547295844
Okay simple now lets add these up.
356.0352 + 356.0352 + 439.82297 ============ 1151.89336 ============
Checks for 1151.89336 / 1151.9173063162575207696359072025 = 0.99998
NOW WHO WOULD HAVE THUNK IT ?
So the pyramid formed from the base of The Great Pyramid plus it's two slopes gives us (11 / 3) x 100 Pi
I do like this one !
Cheers Don Barone
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Post by Don Barone on Apr 25, 2016 6:54:29 GMT -5
Okay a bite size bit of brain food this morning. I was re-reading this thread and this occurred to me from an earlier post I made on this thread ...
We have been looking for a way to find Mars (well at least I have) on The Giza Plateau and this should make most smile.
Base of The Great Pyramid is 9068.996 while for G3 it is 4154 inches with a total of 13222.996 (rounded to) or 13223 inches.
Now if we allow this to be a circumference we get 13223 / Pi = 4209.01035 and let's call these miles so this would give us 6773.75 kilometres and if we assume this is the diameter then the radius = 3386.87. And this from a NASA web site on Mars - Earth comparison:
Equatorial radius (km) 3396.2 6378.1 0.532 Polar radius (km) 3376.2 6356.8 0.531
AVERAGE RADIUS OF MARS = 3386.2
Checks for 99.98%
Cheers Don Barone
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