
Post by Don Barone on Mar 28, 2018 10:04:00 GMT 5
Okay then ... shall we take a stab at trying to find Venus ? I agree ... let's do it ! ... where we left off Now I know I have always said step out of the box ... well this time we are going inside the box and we will draw the enclosing square ... ... then we will draw the green line as shown creating the distance "square root of 5" ... next we will draw a circle of r = 1 again but from the bottom of our first one ... ... next we will extend our square root of 5 line to meet this new circle ... This new line becomes square root of 5 + 1 (obviously) ... next we will draw a circle with radius our blue square root of 3 line ... ... taking special note that this blue circle meets the point where green line has been extended to EXACTLY ! Next we will join this intersection point with the center of the blue circle next we will make our green line bolder ... Okay so what do we have at the moment ... well we have an orange line that is square root of 3 in length and we have a green line that is square root of 5 + 1 units in length or 2.2360679774997896964091736687313 + 1 or 3.2360679774997896964091736687313 and as the creator created all of this he deemed that Phi to be 1.6180339887498948482045868343656 WHICH IS EXACTLY 1 /2 OF SQUARE ROOT OF 5 + 1 ! And so we have the ratio of square root of 3 and TWICE PHI ! ... and we get this ... and the final image in this series ... labelled ! And we have accomplished our goal of designing the beginnings or our solar system using just 1 and 2 Cheers Don Barone



Post by Don Barone on Mar 28, 2018 13:40:40 GMT 5
Well since we are on such a roll why don't we go for Earth ... yeah let's do it !! But a word of caution ... I have found other amazing things as well that I had overlooked in the past so be prepared to be mesmerized with the geometric wizardry about to unfold ... Here is where we left off ... What we will do next is with point "Venus" as an origin or center of a circle we will again draw a circle with radius of 1 So now we will color "Venus length  1" purple and see that this is simply the square root of 5 Now with radius our purple or sq rt of 5 line we will draw the appropriate circle .. Now with diameter of twice sq rt of 5 ( or twice the diameter we have just created) will draw another circle with center on our Sun point Now we will draw a radius line to the left (it could be to any point on the circle) So now let's circle the point on the circle ... AND LABEL IT ... EARTH !!!!!!!!! Okay there you have it laid out for you and remember we started with but a unit of 1 !!!!! ... and all we used was a compass, straightedge and marker. Could The Greeks have done any better ... well The Ancient Egyptians sure could have as I have proved many times at Giza. We will take a short break and I will be back to show you some astonishing geometrical and mathematical wizardry within our solar system ... Cheers Don Barone Just remember I am only the messenger



Post by Don Barone on Mar 28, 2018 14:13:07 GMT 5
Okay then ... inspiration comes at the strangest times and in the strangest places for example I was sitting waiting in the waiting room at the ultra sound clinic about to go in and see my "grandchild" (now 21 weeks) and since I had shown that Earth was sq rt of 5 x 2 when Mercury is sq rt of 3 Quick proof: 149598261 / 57909050 = 2.5833313 x 1.7320508 (sqq rt of 3) = 4.474461052 and divided by 2 = 2.237230526 this number squared gives us 5.0052 instead of a perfect 5 ... 0.999 correct.
But that is not what came to me in the waiting room. Since I had decided the semi major axis of The Earth was now to be 2 times sq rt of 5 I took it in my head to divide the semi major axis of The Earth in 2 and got 74,799,130.5 (149,598,261 / 2). Next I figured I would see how many times the planets would go into this and I got this amazing result ...
74,799,130.5 / 57,909,050 (Mercury) = 1.2916656463886041991709413295504
... now if you have been following my work at all you would immediately have recognized that this number is almost identical to the circumference of The Pyramid of Khafre (G2) divided by 1000 !
1.2916656463886041991709413295504 x 1000 = 1291.6656464
Now in order for this to be a perfect match the base of G2 would have to be 1291.6656464 or 411.15 and times 20.62 would give us 8477.9 and Petrie has the south side of G2 as 8476.9 or within ONE INCH
Now that would have been a quaint coincidence but then I decided to see what Venus would show me and so I dutifully divided 74,799,130.5 / 108,208,000 and got 0.69125324 now again anyone familiar with my work would realize that the is very close to 1/2 the base of G1 x Pi / 1000.
So 74,799,130.5 / 108,208,000 = 0.69125324 and times a 1000 gives us 691.25 (and divided by Pi gives us 220.03 within 3/100ths of a cubit or within 3/5ths of an inch.
So where does that leave us now ? Well I am going to have to dig up some old images where I showed 691.15 (220 x Pi) and 1291.4145 / Pi = 411.07
But all I can say is ... THE PLOT HAS THICKENED QUITE A BIT !!!!!!!!
Cheers Don Barone



Post by Don Barone on Mar 30, 2018 9:01:09 GMT 5
Hi all I am still trying to figure out why the base of G2 x Pi (inner circumference) is equal to 1000 times the number of times Mercury's semi major axis (57,909,050) will divide into 1/2 of Earth's semi major axis (149,598,261 / 2 ) of 74799130.5 giving us 74799130.5 / 57,909,050 = 1.291666 while 411.10 (south side per Petrie 8476.9 inches and divided by 20.62 inches per cubit } times Pi gives us 1291.51 and checking to 1291.51 / 1291.666 = 0.9999 (rounded) So I continues to play with these numbers and came up with this. Is it just a coincidence or is there some higher math at work here ? I honestly don't know for as I am apt to say I am only the messenger. It troubled me a bit that the numbers of a planet I had claimed was Venus was being used in conjuration with Earth and Mercury so I decided to try and go a little deeper into the numbers. As I said what I found may well just be a coincidence but it is, I think, very interesting. Observe the tale of some numbers. Firstly let's start with the diameter of Mercury which becasue it is virtually circular can be said to be a circle of diameter of 4879.4 kilometers. en.wikipedia.org/wiki/Mercury_(planet) and from NASA pds.jpl.nasa.gov/planets/special/mercury.htm giving us a circumference of 4879.4 x Pi or 15329.0872 kilometers. Now to get 1.291666 we had used 1/2 of Earth's semi major axis or 74799130.5 so I decided to divide 74799130.5 / 15329.0872 and was somewhat taken aback when the result showed this. 74799130.5 / 15329.0872 = 4879.55 OR THE DIAMETER OF MERCURY TO WITHIN 1.5 KILOMETERS OR WITHIN 0.99997 (4879.4 / 4879.55) ... Most interesting but any guesses what it could mean ? ... mathematically that is ? Still working on a viable 691.253 solution ... Some imagery ... So is there hidden math at work here or is it just a quaint coincidence ? Cheers Don Barone



Post by Don Barone on Mar 30, 2018 10:28:46 GMT 5
To close exactly we would need diameter of Mercury to equal 4879.4777 kilometers and we would have (4879.4777 X Pi) = 15329.3313 and times 4879.4777 = 74799130.2 and times 2 gives us 149,598,260.4 and actual is 149,598,261.
And from Wiki we have:
Mean radius
2,439.7 Â± 1.0 km[6][7] 0.3829 Earths
So our figures and calculations could be 100% Correct ... but why is it correct ?
Once again we see Mercury being the messenger of "The Gods" and used as a base in yet another measurement.
Cheers Don Barone



Post by Don Barone on Mar 30, 2018 11:10:20 GMT 5
The base of The Red Pyramid is 420 cubits. The question is why ? Well let's do some math ... 420 cubits is equal to 420 x 20.62 = 8660.4 inches divided by 10,000 is equal to .86604 and is virtually exactly 1/2 the square root of 3 ! and the perimeter of this pyramid would be 4 x 8660.4 = 34641.6 / 10,00 or 3.46416 or TWICE THE SQ ROOT OF 3 OR SQ ROOT OF 12 ! and the circle enclosing this pyramid is ... get ready for it .... 2.720699046 and The Megalithic Yard !!!!!! The picture in lieu of the thousand words ... More to come most likely Cheers Don Barone



Post by Don Barone on Mar 30, 2018 11:25:39 GMT 5
and then we have this finale ... The diagonal is equal to the square root of 0.86604 squared then times 2 and we get 1.2247655 (sq rt of our 1.5) and 1.22447655 x sq rt of 2 or 1. 41421536 and it equals, not surprisingly I should think ... 1.7308 or the square root of 3 and lending more and more support of the sq rt of 3 by sq rt of 2 theory for The Giza Rectangle ... Note as well that the diagonal appears to be simply the square root of the height when used in a ratio to the diagonal ... Cheers Don Barone



Post by Don Barone on Mar 30, 2018 11:39:45 GMT 5
Okay next I am going to revisit The Bent Pyramid with a renewed feeling that the ratio WAS NOT 9 x 11 but was in fact 9 by 11.0227 or ratio being sq rt of 3 to sq rt of 2 or 1.2247448714 so that 9 x 1.2247448 = 11.0227
So the new ratio will become 9 x 11.0227 and that's what we will go in search of and we will start with the bent ...
Cheers Don Barone



Post by Darkpiano on Mar 30, 2018 13:20:48 GMT 5
From your post: the base of G2 x Pi (inner circumference) is equal to 1000 times the number of times Mercury's semi major axis (57,909,050) will divide into 1/2 of Earth's semi major axis (149,598,261 / 2 ) of 74799130.5 giving us 74799130.5 / 57,909,050 = 1.291666 while 411.10 (south side per Petrie 8476.9 inches and divided by 20.62 inches per cubit } times Pi gives us 1291.51 and checking to 1291.51 / 1291.666 = 0.9999 (rounded) Read more: ahatmose2002.proboards.com/thread/697/answersgizasolarsystem?page=1#ixzz5BFwMoT3dCould the math have something to do with how the planets were formed by gravity or how they go around the sun. I know that when venus and Earth move around the sun they form a pentarose pattern. I also read an article on the phi relationships in the solar system. 420 cubits is equal to 420 x 20.62 = 8660.4 inches divided by 10,000 is equal to .86604 and is virtually exactly 1/2 the square root of 3 ! and the perimeter of this pyramid would be 4 x 8660.4 = 34641.6 / 10,00 or 3.46416 or TWICE THE SQ ROOT OF 3 OR SQ ROOT OF 12 ! and the circle enclosing this pyramid is ... get ready for it .... 2.720699046 and The Megalithic Yard !!!!!! Read more: ahatmose2002.proboards.com/thread/697/answersgizasolarsystem?page=1#ixzz5BFyRYsIiIf the pyramid with 440 cubits at the base gives us the speed of light, maybe the other pyramids are giving us other numbers. Speed of light = 299,792458 m/sec If we draw a circle around the base of the pyramid and a smaller circle inside the base and subtract the 2 numbers we get a number similar to the speed of light. Circumference of outer circle = 1023.463491 meters Circumference of inner circle = 723.697975 meters Difference = 299.765516 meters The pyramid is also located at the coordinates: 29.9792458 N, 31.1344 E 2 different ways to get the speed of light means it is unlikely to be a coincidence in my opinion. Read more: ahatmose2002.proboards.com/thread/699/speedlightgreatpyramidgiza#ixzz5BFveLTjYAlso I read an article on how Stonehenge gives us an equation: (pi * phi)/ (e^phi) = 1.00793907077 which is similar to the atomic mass of Hydrogen. (1.00794)



Post by Don Barone on Apr 1, 2018 11:50:37 GMT 5
Moving on with a couple of new images before I correct the extremely large error I have made in one of my earlier drawings. (I have no idea how I could have made such a serious error) For now though ... forward. In this image you will notice that instead of 200 I have chosen as the height 197.9898 cubits. The reason I have done this is to keep in tune with our theme of all things square root of 2 or square root of 3 with 140 x sq rt of 2 = 197.989899. This distance is also the distance between where the shafts of the so called King's Chamber exit the Great Pyramid. As shown below ... but before we go there I have to post the horrible error I made ... Oops just double checked my drawings as I am apt to say ... "I thought I was wrong .. but I was mistaken" so let's carry on. Now what you are about to see is possibly what Pythagoras was playing at when the soldiers stormed his home and he cried ... do what you will with me but leave my circles alone for I think I may have inadvertently solved for The Pyramids but most importantly I think I have shown in the series of some 16 images I am going to be posting that all the pyramids are tied into each other and can be shown to do so from simple, and I repeat here, simple geometry ! And the cherry on top will of course be when I beautifully tie it all into our solar system and the planets for "The Creator" was a very simple creator and used only the simplest of ratios and such and thus it was easy for The Ancient Builders to emulate "it's" work for remember what is written in Enoch where "The Creator" said he would make it so simple anyone could understand it ... and so we begin ... We will start with Gary Osborn's copyrighted drawing of The Great Pyramid to scale ... Our starting point will be the fact that like the height of The Red Pyramid the distance between where The King's shafts exit the Great Pyramid IS ALSO 197.989899 cubits (most have it at 198 cubits exactly) ... so let's show that ... Now before we move on I just would like you to think upon the fact that using 197.989899 as the diameter gives us a circumference of 622.00 and is precisely the diagonal in a pyramid that yields the Pi base for The Great Pyramid. We will examine that more later, for now just think about it ... However from earlier work on The Red Pyramid I discovered that viewed on the 45 degree angle the base to height was in the relationship 1 to 1.5 so to get 1/2 the base we simply add 1/2 of 197.989899 to it and this is the easiest way to accomplish this ... Now let's draw in the line and set the height properly ... NOW THE THING TO KEEP IN MIND IS ALL WE ARE USING IS A COMPASS, A STRAIGHTEDGE AND A MARKER ! Okay now let's draw in The Red Pyramid as seen on edge ! Okay so we have nicely drawn The Red Pyramid precisely to scale and is shown by our blue triangle ! Now how easy was that ? So let's draw another pyramid. Let's do The Pyramid of Khafre which is in the sizes 3, 4 and 5 and happily our 1/2 base of our blue triangle of //2 base of The Red Pyramid viewed on angle of 45 and since that can be said to be 3 x .5 the height would simply need to be 4 x .5 and is simply 2 and give us the following series of diagrams ... and drawing in the triangle which is in the precise proportions to The Pyramid of Khafre or G2 ... However there will be those who will claim that this is not what was intended so I decided to go a little further with this drawing and drew a circle with diameter of "6" or diagonal of The Red Pyramid ... and then decided to draw the square using 297 ( 1/2 base of blue triangle) and arrived here ... Now there has been much speculation on why the "Queen's" shaft to the left ends where it does. It has been speculated that it ends at 43.3 degree and matches The Red Pyramids angle and I thought that made a lot of sense but I now think the makes more sense for since we have embarked on a path of sq rt of 3 and sq rt of 2 it seemed logical that the pyramid builders would stick with their theme of sq rt of 2 and make the angle 45 degrees and so I moved the diamond (now a triangle) down ... and got this ... ... and noted that it would appear that it 'COULD" be that the angle of our black pyramid hits the end of the shaft of The "Queen's Chamber" as shown in this diagram where I have projected the line ... ... and then I got the intersection of the two other shafts and got this diagram ... note here that so fat that intersection point has not yielded me anything yet ... ... and last but certainly not least I drew in our 45 degree triangle but with base of The Great Pyramid and got the image below ... I believe I promised to tie this all into our solar system and so I shall ... For that we have to turn 90 degrees and we will begin anew ... but that will be our next post ... Cheers Don Barone



Post by Don Barone on Apr 1, 2018 13:58:41 GMT 5
To find any correlation to our solar system I have to once again show the solar system and put the distances on it ... For now we will stick with Mars and Earth ... observe ... and now the distances using the semi major axis of both Mars and our Earth So here are the ratios we have ... 227,939,100 / 149,598,261 = 1.5236747972625163069241827617234 227,939,100 / 78,340,839 = 2.9095820635773379960865622079947 and lastly we have 149,598,261 / 78,340,839 = 1.9095820635773379960865622079947 Now I have already shown that a meter at 39.37007874015748031496062992126 inches (10,000 / 254 ) divided by the cubit at 20.62 inches gives us 1.9093151668359592781261217226605 and checking to within 0.99986 But this is old hat my regulars cry out and you are right but I need to set it up for those who have not traveled here with us before. Now according to me the correct base length for The Great Pyramid based on 9068.9968 inches is 9068.9968 / 20.62 = 439.816 while using 280 as a confirmed and fixed height and using The Pi Angle we get 439.823 and using 197.989899 x Pi as a diagonal we get also exactly the same 439.823 (rounded). So we will try both ... Now it is further my contention that the base of The Red Pyramid WAS NOT meant to be exactly 420 cubits but was in fact meant to be sq rt of 3 / 2 then times 10,000 and is 8660.254 and using 20.62 inches as a cubit this gives us 419.993 cubits. So where is all of this going you must be asking yourselves by now and I will tell you. Firstly I noted using our new 1 to 1.5 ratio of The Red Pyramid that if we allowed 220 cubits to represent the base of The Red Pyramid then the height would be 220 / 3 and then times 2 or 220/3 or 73.33333 x 2 or 146.6666666 and there it might have stood if I had not decided to divide it into the height of The Great Pyramid and got this ... 280 / 146.66666 = 1.90909091 or 21 / 11 and close to our above ratios of 1.9095820635 and 1.9093151668 but I felt it might be closer than this so I decided to use the exact measurements I felt were their and those were 419.993 for base of The Red Pyramid and 439.823 for base of The Great Pyramid ... what would these values yield ? These values were not very good so I went back to the drawing board and decided to reverse engineer it and start with 419.993 for the Red and divide in by our solar system ratio of 1.90958206358 and got 219.94 or base of 439.88 and times 20.62 = 9070.315 inches and equaling 230.386 meters all very extremely close so let's put up one last diagram and maybe we can all see it visually ... So what are our final conclusions after all of this ... well they are simply this ... The base of The Red Pyramid represents the distance from The Sun to The Earth ! The base of The Red Pyramid represents one meter ! One half the base of The Great Pyramid represents the distance between Earth and Mars ! One half the base of The Great Pyramid represents one cubit of 20.62 inches Wow is all I can say ... Cheers Don Barone



Post by Don Barone on Apr 1, 2018 15:00:41 GMT 5
Well since I seem to be on an Easter roll I might as well keep going ... I was thinking what could the ratio be between Mars and Ceres and knowing I had come to a dead end on this many times before I tried yet again to divide Ceres semi major axis distance of 414,001,000 by Mars semi major axis distance at 227,939,100 and got 1.816279. Now only because I had just recently been fooling around with square root of 1.5 which is in fact simply the sq rt of 3 / the sq rt of 2 and it comes to 1.2247448714 but the reciprocal is the key for it gives us 0.816496581 and of course I immediately noticed the close similarities to 1.81628 so I decided to press on and got this ... I decided that the distance from The Sun to Ceres shall equal sq rt of 2 / sq rt of 3 then plus 1 or simply 1.81649658092772603273242802490196. This should make Sun to Mars equal to 1 and Mars to Ceres equal to 0.816496581 ( sq rt of 2 / sq rt of 3) or thereabouts ... shall we see how we have done ? Okie dokie ... The most up to date figure we have for the semi major axis of Ceres (remember they just sent a probe there ) is 414,001,000 kilometers and Mars as we all know (or should know) by now is 227,939,100 so let's do the math ... 414,001,000 / 1.81649658092772603273242802490196 (remember this should give us Mars) = 227,911,797 and divided by actual is 227,911,797 / 227,939,100 and checks for 0.99988 ... I guess close enough for my tastes and remembering after all the planets are probably not where they were at "The Creation" So how about another diagram or two ... ? Now I don't know about you but I didn't know that ! Cheers Don Barone



Post by Don Barone on Apr 1, 2018 16:11:59 GMT 5
Okay well the fact that Mars = 1 ties in so very nicely with other things I had noted. So now we have .... The Sun to Mars = 1 The Sun to Ceres = 1 + sq rt of 2 divided by sq rt of 3 AND FROM PREVIOUSLY DISCOVERED THINGS Jupiter = 1 + (sq rt of 1 + 1 or 2.41421356 ) = sq rt of 2 + 2 or 3.4142135623731 Here is an image that is worth way more than a thousand words ... Link to a very large image So it would seem that the solar system is reasonably simple ... at least up to Jupiter .. Cheers Don Barone



Post by Don Barone on Apr 1, 2018 17:46:13 GMT 5
Well shall we keep on going ? Why not ? Next stop Saturn and amazingly the distance from The Sun to Saturn is the semi major axis of Mars x 2 x Pi or 1 432,183,604 kilometers and here is the image I have made ... So in this image the red of The Martian Circle is equal to the straight path from The Sun to Saturn ... And from the peanut gallery I hear ... why ? Well I am still working on that But keep all this in mind for we will be getting back to the pyramids and our Giza Rectangle eventually ... Cheers Don Barone



Post by Don Barone on Apr 1, 2018 20:35:36 GMT 5
Okay got another one for you. In this diagram the greyblue line is equal to the circumference of Saturn (Saturn x Pi) but it is also equal to (Mars x 2 x Pi) then times Pi again or simply Mars x 2 x Pi squared or 227,939,100 x 2 x 9.8696044011 = 4,499,337,489 = Neptune semi Major axis ... And of course the image ... Uranus is Mars x 2 x 2Pi = 227,939,100 x 2 x 6.28318530718 = 2,864,367,208 (image will follow later, much later) So it would appear that Mars is yet another unit of measure in our solar system. Cheers Don Barone



Post by Don Barone on Apr 1, 2018 20:39:15 GMT 5
But the real question is ... why ?
db



Post by Darkpiano on Apr 2, 2018 0:19:45 GMT 5
39.37 inches plus 20.62 inches = 59.99 inches so basically 5 feet.
e1 feet (1.71828183 feet to inches = 20.61938196) number e = 2.7182818284590452353602874713527..
(5  (e  1)) feet to inches 39.3806181 inches



Post by Darkpiano on Apr 2, 2018 1:47:20 GMT 5
5/(5(e1) = 1.52359213639



Post by Don Barone on Apr 2, 2018 7:05:49 GMT 5
39.37 inches plus 20.62 inches = 59.99 inches so basically 5 feet. e1 feet (1.71828183 feet to inches = 20.61938196) number e = 2.7182818284590452353602874713527.. (5  (e  1)) feet to inches 39.3806181 inches Hi DP ... I have been reluctant to use " e" but from you posting above we have ... e1 = 1.7182818284590452353602874713527 x 12 = 20.619381941508542824323449656232 While ... 5  ( 2.7182818284590452353602874713527  1) 5  ( 1.7182818284590452353602874713527) = 3.2817181715409547646397125286473 feet x 12 = 39.380618058491457175676550343768 39.380618058491457175676550343768 + 20.619381941508542824323449656232 ============================================= 60.000000000000000000000000000000 or precisely 5 FEET =============================================Ratio is: 39.380618058491457175676550343768 / 20.619381941508542824323449656232 = 1.9098835343466321219250100255451 However how is this on the topic of this thread DP ? cheers db



Post by Don Barone on Apr 2, 2018 8:40:35 GMT 5
Okay I will start with this image: And we will now list the latest NASA listings for semi major axis distances of the planets of our solar system. Neptune = 4,498,396,441 km Link to NASA NeptuneUranus = 2,870,658,186 km Link to NASA UranusSaturn = 1,426,666,422 km Link to NASA SaturnJupiter = 778,340,821 km Link to NASA JupiterCeres = 413,690,250 km Link to NASA CeresMars = 227,943,824 km Link to NASA MarsEarth = 149,598,262 km Link to NASA Earth Venus = 108,209,475 km Link to NASA VenusMercury = 57,909,227 km Link to NASA MercuryOkay let's do some math. Not sure how this is going to work out as a lot of these numbers have changed from what was and is posted at Wiki. Okay so we claim Neptune = Mars x 2 x Pi squared = 227,943,824 x 2 x 9.8696044011 = 4,499,430,737 kilometers Actual is 4,498,396,441 km so 4,498,396,441km / 4,499,430,737 = 0.99977 ... well not our best work but not bad. Okay let's do Saturn we claim that ... Saturn x Pi = Neptune 1,426,666,422 x Pi = 4,482,004,751 (actual = 4,498,396,441) Not very good at all Okay Mars then = Mars x 2 x Pi = Saturn = 227,943,824 x 2 x 3.1415926535897932384626433832795 = 1 432 213 286 (actual = 1,426,666,422 ) Not very good again ... hmmmm



Post by Don Barone on Apr 2, 2018 10:14:51 GMT 5
Well thanks to DP and the video he mentioned I have found a few new things to play around with ... but first I am going to post an old picture of mine totally independently arrived at ... It is in my blog "What The Dormouse Said" and here is the link: Teaching a 9 Year Old To Understand The Solar Systemso in this exercise for all intents and purposes ... semi major axis of Mercury is ONE FOOT OR UNITY as he calls it ... And here is the image in question ... and the end result ... More later ... db



Post by Don Barone on Apr 2, 2018 10:58:49 GMT 5
Well I have to add this before I go ... According to Wiki the closest approach of our Earth to The Sun is 147,095,000 kilometers Now if we call Mercury 12 inches (OR ONE FOOT) Our Earth's Perihelion (closest approach) is in inches 147,095,000 / 57,909,050 then times 12 and we get .... 147,095,000 / 57,909,050 = 2.5401038352382 and times 12 is 30.481246023 inches and as we all know there are 30.48 inches in a meter (2.54 x 12) so now we can revise our diagram slightly and do this ... (30.48 / 30.481246023 = 0.99996 ! ) Cheers Don Barone



Post by Don Barone on Apr 9, 2018 13:03:24 GMT 5
Okay we are going to take yet another look at The Bent Pyramid and try to figure out what is going on here. As if you need it. Here is a look at The Bent Pyramid. Then I drew the projected path of the sloping pyramid at the top: and then I drew the projected bottom pyramid ... Now the question becomes what was the base of The Bent Pyramid, what was the height and what were the intended angles of the top and bottom pyramids. Here is another "solution" to ponder on. I have to admit that this time around I am going to agree with John Legon and call the height at 200 cubits and then instead of 9 and 11 we will use square root of 2 divided by square root of 3 and we get 0.8165 and so we have 200 / 1.8165 to get 110.102 and 89.898 so 110.102 = 1 unit and 89.898 = .8165 units. Does this look familiar ? It should ... Now for those who might doubt this analysis this time around I would remind you that ... The pyramid at the top is at the exact same angle as the Red Pyramid and it was called the red because of the red blocks used and who can forget that they call Mars the angry red planet and so it is not at all surprising that the top portion of The Bent Pyramid and The Red Pyramid for that matter represented Mars ! Cheers for now ... Don Barone



Post by Don Barone on Apr 9, 2018 14:24:13 GMT 5
Okay but what were the angles and the measurements ... well ... I am going to start with the assumption that Petrie was spot on when he measured the west side of The Bent Pyramid and got: Here is a comparison that John Legon posted on his site: As you can see Petrie has it at 361.79 and I have determined that it should have been 361.8034 based on the fact that The Pyramid at Meidum had also used this value as the height in inches. So the base (at least in this one solution, there may be several) is 361.8034 cubits and times 20.62 = 7460.386108 inches. So base of the Bent is 7460.386 and the height is 4124 inches. Now please keep in mind that I think there is more than one correct solution here. Okay so I am going to start the proceedings by calling the 1/2 base of the top pyramid 116.6726 and this is based on a square of 82.5 cubits and this is the diagonal. The reason I chose 82.5 is that comparing G1 and Meidum we get 440  275 = 165 and divided by 2 is 82.5 ... the diagonal then becomes what I posted above. and we get this at Dashur for The Bent ... Okay so now it is a simple matter to figure out the rest as observe: Blue base is 1/2 361.8034 or 180.9017. We have used 116.6726 of it so we have left 180.9017  116.6726 and we get 64.2291 so our new triangle becomes 64.2291 by 89.898 ... as shown: The angle now calculates to the tan angle of 89.898 / 64.2291 = 1.3997 (virtually 1.4 ! ) and we get an angle of 54.4555 and ... here is what Petrie got ... You will note that Petrie has it at ... 54 degrees 48 minutes which equals 54.80 and Lauer had it at 54 degrees 44.8 minutes so it appears that there are a variety of correct solutions. Mine is 54 degrees 27.3 and appears to check spot on to a rise of 7 on base 5. So the last distance we need is the height of the blue pyramid and we get this by using tan of 54.455468 x 1 / 2 base of 3730.193054 and we get 5220.95 inches and equaling 253.198333 cubits ... And that's the end of SOLUTION #1 Cheers Don Barone



Post by Don Barone on Apr 9, 2018 16:29:12 GMT 5
And solution #2 ... Tan 0.7 db



Post by Don Barone on Apr 11, 2018 11:09:17 GMT 5
Hmmm ... just when I think I have something solved I come up with something else. In my other thread I showed nicely how Meidum was apparently the start of it all but here is something I stumbled onto and I am putting it here for reference ... This time we will start with Ceres which we had determined was Mars x 1.8165 or 1. sq rt of 3 / sq rt of 2 and I got this ... Semi major axis of Ceres is 414,001,200 km as of the latest posting by NASA ( but on this site it is 413,690,250 km) For now I will use the former ... I also decided to reintroduce one of the early players ... the sq rt of 3 into the mix and this is what I got ... 414,001,200 x sq rt of 3 x 2 x Pi and this is what we get 414,001,200 x 1.7320508 = 717071112.80 x 2 = 1,434,142,226 Saturn (Wiki = 1,433,430,000) and times Pi and we get ... 4,505,490,680 km (4,503,443,661 km based on Wiki original entry) and they check for the latter is 0.9995 and the former is also 0.9995 ... if we use the old figure still posted on most sights of 413,690,250 km we get 1,433,065,063 for Saturn and 4502106675 for Neptune and checking for 0.9997 for both of them ... to be precise Ceres would have to equal 413,813,103 km ... Cheers Don barone



Post by Don Barone on Apr 15, 2018 18:58:23 GMT 5
This is interesting and I am not sure why it works so close but I will work on it. Here is what I have discovered ... In this image below The solution to Mars was 51 / Pi to get 16.2338042 and added onto 31 gave us our distance to Mars of 47.2338042 ... But why 51 ? Well just now I was playing with the numbers and noticed that if we have 20.62 CENTIMETERS ( as opposed to 20.62 inches in a cubit) it will translate into 8.118110236 ( x 2 x Pi = 51.007591 ) inches and if use this as a radius we get as a diameter 16.2343030351 and added to 31 gives us 47.2343030351 and divided by 31 gives us a ratio of 1.5236872 and times 149,598,261 gives us 227,940,955 and the agreed upon distance for Mars is 227,943,824 km and checking to 227,940.955 / 227,943,824 = 0.999987 or 13 one millionths ! If we use 227,939,100 the number I have long used it closes to an accuracy of 0.999992 or 8 ten millionths ! Strange to be sure ... Cheers Don Barone



Post by Don Barone on Apr 16, 2018 11:15:09 GMT 5
Hi all ... just a bit of a recap here today to start things off. A question once posed to the ancients was what is the ratio of the volume of a sphere to the cube that encloses it. That is what is the ratio of volume that a sphere takes out of the cube's volume. Here is an image to make things clear ... The diameter of the sphere is the same as the sides of the cube so what is the relationship ? Well the diagram is showing you the result but not how we get there so I will try to explain ... We will use 12 just for clarity. So the cube is 12 x 12 x 12 gives us volume of 1728 cubic units ... Length x height x depth ... pretty straight forward ... The volume of a sphere is of the formula ... and using that formula above we get for a radius of 6 (it is one half the base of our cube) 6 cubed for 216 x 4.18888 and we get 904.79808 Actual is below ... So the ratio is 1728 divided by 904.778684 and we get ... 1.9098593176 .... but why bring this up on our solar system thread ... well ... Its back to this diagram we go ... and we find that 149,598,261 / 78,340,839 = ... 0.52367479726251630692418276172341 and reciprocal we get 1.90958206358 and checking to 0.99985 So on top of this ... and this one ... we can now add this one ... and now changing our first image to this ... So we have : Semi major axis of Earth = the meter = volume of The Cube = The Red Pyramid Semi major axis' of Earth to Mars = the cubit = The Enclosed Sphere = half base of The Great Pyramid Now the question one has to ask themselves is .... why ? Cheers Don Barone and just a little addendum to previous posts ... average distance to The Moon is 384,399 kmsq rt of 2 x "e" = 1.4142135623731 x 2.71828182845904523536 = 3.84423
384,399 / 3.84423 = 99994 or 0.99994 correct ... wonder why ?



Post by Don Barone on Apr 16, 2018 11:48:39 GMT 5
Okay this is unbelievable ... even to me ... and so it was that I was going to check the volume of the pyramids to see if my diagram of Red to 1/2 Great Pyramid would yield the ratio and so I went ot The Red Pyramid first and found that the volume was ... 1,694,000 cubic meters ... BUT ... here is what I got according to Wiki for the Great Pyramid ... this is truly mind boggling and goes a giant step forward in proving my theories ... the volume is .... 2,583,283 cubic meters (91,227,778 cu ft)Now unless you have been sleep walking through these posts the last week or so you will of course recognize this number as the number I claim is the number of inches from The Great Pyramid to The Pyramid at Meidum ... It is also basically 31 / 12 and checks to the Mercury earth ratio to ... a degree of accuracy of ... 1495,598,261 / 57,909,050 = 2.583331292777 / 2,583,283 = .99998 ... simply unbelievable ... So now do the volumes of the other pyramids show us the other planets ? I will look for us This is amazing ! Another piece of the puzzle to prove once and for all that The Great Pyramid was indeed meant to represent Earth and also Mercury ... and their relationship in our solar system. Cheers Don Barone



Post by Don Barone on Apr 16, 2018 11:56:46 GMT 5
Oh dear ... I am afraid Wiki is totally in error volume is not as stated ...

