
Post by Don Barone on Apr 3, 2018 16:09:10 GMT 5
Hi all ... Well I have studied the following pyramids in depth: The Great Pyramid The Pyramid of Khafre The Pyramid of Menkarre The Bent Pyramid at Dashur The Red Pyramid at Dashur ... but there is one pyramid that I have stayed away from mainly because I thought that the data needed was lacking ... and that last pyramid is of course ... The Pyramid at Meidum ! I have looked at it but have studied it from scaled off drawings and photographs which let's face it is not too reliable. The main thing I knew is that it was supposed to be a true pyramid ( I have never believed that ! ) in the proportions Height 175 cubits base 275 cubits. Now what most Egyptologists would never tell you, at least not unless you asked is that Petrie, yes our man Petrie again, based this 275 x 175 ON A CUBIT OF 20.66 INCHES ? Now help me out here for as far as I know and have been "taught" the same builder was responsible for the Pyramids of Dashur and this one and that was Sneferu. So perhaps someone might want to explain to me why he would use 20.62 at Dashur and 20.66 at Meidum ? Something to think on as we carry forward.I have to go out in a little while so I really can't go into this in too much depth now but suffice it to say that I managed to stumble onto a treasure trough of data that Petrie left us in one of his books that I have neglected for far too long. I will leave a couple of pictures to whet our appetite ... Cheers Don Barone



Post by Don Barone on Apr 4, 2018 6:35:00 GMT 5
Okay just a quick post before my morning coffee. What most if not all, including me, failed to realize or think about is that The Pyramid at Meidum is at a ratio of 7 and 11, the exact same as The Great Pyramid. What this means is that anything that was found for The Great Pyramid in regards to Pi, Phi, sq rt of 2 and all the other things WILL ALSO BE FOUND AT MEIDUM ! THE TWO PYRAMIDS ARE IN A RATIO OF 1 TO 1.6 (175 TO 280 FOR HEIGHT  280 / 175 = 1.6 AND 275 TO 440 FOR BASE  440 / 275 = 1.6) As a quick check for Pi we get base perimeter as 4 x 275 or 1100 while 2 x height of 175 x Pi gives us 1099.56 ... just didn't consider this before. So we shall go off and see what else this pyramid can tell us ... after my coffee Regards Don Barone



Post by Don Barone on Apr 4, 2018 7:16:28 GMT 5
Okay let's look again at the pyramid. The first thing I always wondered is what slope is the pyramid on. I had guessed at it but could I f1nd the actual measurements and then I stumbled onto Petrie's book entitled Meidum and was very pleasantly surprised that all of the measurements were there. Here is what he wrote ... So we have an angle of the sloping face of between 73 degrees 20 minutes (73.33 degrees) and 73 degrees 54 minutes (73.90 degrees) and these angle yield their tan at between 3.33952548 to 3.46458127. A note here that the reciprocal angles are 16.67 for 73.33 and 16.1 for 73.90 and the tans are for the former 0.2994437402218 and 0.28863516876 for the latter. Now in keeping with my theme of square root of 3 being a controlling interest in everything Ancient Egyptian may I suggest an angle of 73.897886248. Now why would I pick that angle ? Well it is because the tan is 3.46410161514 and is precisely 2 times the square root of 3 ! (2 x 1.73205080757) ... Is this the angle that was used ? Is this the reason why ? ... As always food for thought and as a little diversion we note that 2 x 149,598,261 (semi major axis of Earth) as a simple decimal is 0.299196522 and is the tam of the angle 16.657 or 73.343 ... close enough ? And lastly let's try the speed of light at .299792458 and gives us 73.312 ... something to think about I guess. Regards Don Barone



Post by Don Barone on Apr 4, 2018 9:53:49 GMT 5
Okay now in our earlier work on The Megalithic yard I suggested that the Megalithic Foot was 1.6666666666 with the yard being 2.7206609 feet. In keeping with this theme I thought well what if the angle opposite the large sloping angle was 16.6666666666 and that would leave our angle as 73.3333333333 and definitely in contention for a possibility ... but let's leave that for the moment and concentrate on the pyramid itself. Here are Petrie's notes on both the pyramid and the wall that encloses it ... ... Petrie suggest that he saw no reasonable hypothesis for the discrepancy well I have taken a stab at it and here is my "solution" below ... Petrie saw that the average of the pyramid in inches was 5682 inches and divide by 4 and times 6 should give us the length of the wall if he was correct so let's do the math ... 5682 / 4 = 1420.5 and times 6 = 8523 inches and the average of the north side of the wall and the south is 8561 + 8479 = 17040 and divide by 2 gives us 8520 inches and within 3 inches and close enough I feel to say that Petrie got this correct. But what of the distance going north ... it is out as Petrie states almost 780 inches ... why could this be ? Well as you can see in my diagram above I have suggested that this could be 1.55 and as a check we go 1420.5 x 1.55 and we get 2201.775 and close to within just over a single inch ... a possibility for sure ... but there is another possibility ... and it's back to Mars and Ceres and their ratio of 1 + sq rt 2 / sq rt 3 or 1 + 0.81649658092772603273242802490196 or 1.8165 ... so let's do the math ... average distance is 1420.5 ... difference is 2203  1420.5 = 782.5 inches ... so now again let's do some math ... average is 1420.5 and divide by 1.8165 and we get 782.00 and checks to within half an inch ... so 1420.5 is one and 782.5 may be the sq rt of 2 divided by the sq rt of 3 ... I wonder ? Anther possibility ( I like this one) could be that the distance to the wall on the north and the south could have wanted to equal 3600 (1393 + 2203 = 3596 ) This would then give us 3600 and with the east side of the wall 9300 the ratio equals PRECISELY the ratio between the semi major axis of Mercury as compared to that of Earth's is nearly also precisely 2.58333333 or 31 / 12 ! I wonder if the uneven distances of the spacing could be giving us the elliptical of Mercury ? Closest approach as 1393 and furthest approach as 2203 ... Hmmm will have to do the math. Well I decided to simply divide 2203 by 1393 and lo and behold I got 1.5814788 and the ratio of the distance between semi major axis of Mercury and the distance between the distance from Mercury to Earth is (149.598.261  57,909,50 = 91,689,211) so ratio of 91,689,211 / 57,909,050 = 1.5833313 and checking for 99.88 not as close as we usually are but again definitely food for though ... so let's draw our solar system at Meidum shall we ... ... and one last observation for now ... ... and in the last ratio of the image we notice nicely that M 1 + M 2 + Y = 2.583 and precisely what to should be as it represents distance to Earth ... Amazing ! Cheers Don Barone



Post by Don Barone on Apr 4, 2018 14:23:08 GMT 5
Okay ... forward. Now either I am the best at finding meaningless coincidences or these ancient builders knew the fundamental ratios of Phi, Pi, sq rt of 2 and sq rt of 3 and further to this they used it within their pyramid complexes to tell the story of our solar system and maybe even creation. The fact that we found the planet ratios in two separate complexes separated by about 65 kilometers would lead one to speculate that maybe they were all related in some way. Now so far I have beautifully shown how both Giza and now Meidum shown us the ratio of the inner planets, namely Mercury and Earth but once again Venus seems to be hiding on us so let's try to find "her". We must remember that if Mercury = 1/2 sq rt of 3 then Venus = Phi or 1.618034 and this was shown beautifully at Giza by using 1/2 base of G1 as Mercury and giving us Venus as base of G2 and after having shown that if G1 is 1/2 sq rt of 3 then G2 is Phi. Now at Meidum we have manged to eloquently show Mercury and Earth and their proper ratios but no Venus for the time being. As I was pondering this situation I was wondering how I might tie The Giza Plateau pyramids and Meidum into one nice little package and maybe find Venus along the way so I started thinking. Remember what Pooh told us ... THINK ! THINK ! THINK ! I decided to see if I could find Venus at Meidum and so I started with what we had determined was "Earth" and that was 9300 or east side of the "wall". The ratio of Venus to Earth is 108,208,000 / 149,598,261 and gives us 0.723324 so if east side of "wall" is Earth and is 9300 inches then Venus would be 9300 x 0.723324 and be equal to 6726.91 inches but try as I might I could not find that number at Meidum. So I thought maybe I should look elsewhere ... and I did. You have to remember that I was thinking in Phi and sq rt of 3 and so it was that I decided to invite G3 to the party to see what it had to say for itself. First I multiplied the base of G3 by sq rt of 3 and got 7194.94 and failed to give me anything I could use and then next I decided to multiply G3 by Phi (1.618034) to see what I could get, and to be honest wasn't expecting to see anything but you never know and that is why you should never leave home without your calculator. So I multiplied 4154 (inches in base of G3) x 1.618034 and got this result ... 6721.31 inches or within about 5 inches. So now, using the two different sites I had determined that G3 x Phi = Venus when used in the scale of Meidum. Awesome to be sure ! But can we close the gap of this 5 inches ... Well Petrie has the bases for G3 as: North = undetermined East = 4149.2 in x 1.618033988 = 6713.55 in South = 4157.8 in x 1.618033988 = 6727.46 in and closes to within half an inch. West = 4153.9 in x 1.618033988 = 6721.15 in ====================== Mean = 4153.6 in =============== But we have to remember that we also claimed if Venus equals Phi then Mercury is equal to 1/2 of sq rt of 3 or 0.8660254 ... so ....... 6727.46 = Venus also equal 1.618034 or Phi Mercury = 6727.46 / 1.618034 = 4157.8 (of course) and times sq rt of 3 = 7201.57 but Mercury is 1/2 sq rt so we get 3600.79 and basically it all ties in beautifully with the measurements at Meidum ... And finally G3 or 4157.8 divided into our east side of 9300 inches gives us 2.23676 and basically is the sq rt of 5 ! (0.9994) So we have G3 as sq rt of 5 and times 1.618034 gives us 1.80902 and doubling we get 3.618034 and times 1000 gives us 3618.034 and now what do you suppose the height in inches is of The Pyramid of Meidum ... well let's let Petrie tell us ... Very nice and neat ! Cheers don Barone



Post by Don Barone on Apr 4, 2018 14:33:52 GMT 5
Just as I was proof reading the post the light went on again and I suddenly realized that this statement of mine ..." ... First I multiplied the base of G3 by sq rt of 3 and got 7194.94 and failed to give me anything I could use ..." was giving me a very very important number for adding the north boundary and the south boundary of the wall gives us 3596 AND DOUBLED GIVES US 7192 ! ... and now back tracking 7192 / sq rt of 3 = 4152.30 and perfectly acceptable for the base of G3 ... I mean really ... just how clever were these master builders ?
Cheers Don Barone



Post by Don Barone on Apr 4, 2018 16:10:13 GMT 5
Okay another very, very interesting tie in to Giza from Meidum.
Here are the sides of The Pyramid at Meidum:
N = 5677.2 E = 5694.5 S = 5681.3 W = 5675.0 =========== Total = 22728 ===========
Now the base of G1 equals 440 cubits ... and inverse of this is 0.002272727272 and times 10,000,000 = 22727.3 and checks to within 7/10ths of an inch ... inverse of 22728 is 4.3998592045054558254135867652235e5
A fluke ? Maybe but there are an awful lot of these showing up ...
Cheers Don Barone



Post by Don Barone on Apr 4, 2018 17:03:00 GMT 5
Okay here is our solar system image again ... What I am going to do now is remove the space occupied by the pyramid in the vertical ... Here is that image ... Okay the overall "height" was 9300 inches and the average pyramid side was 5682 inches so we have this ... And base of pyramid was average of 5286 and height is 3618 and ratio is 1.570481 and doubled is 3.140962 and to Pi value is correct to 0.9998 ... so very nice ... Cheers Don Barone



Post by Don Barone on Apr 5, 2018 5:57:31 GMT 5
And we tie in all together nicely ... Cheers Don Barone



Post by Don Barone on Apr 5, 2018 20:08:23 GMT 5
Hi all not the post I thought I would be making but I am finding nothing of significance so far with the inner workings of this pyramid but this should keep you amused ... Let us assume that the height of The Pyramid of Meidum which is 3618 (1000 x Phi) inches + or  is equal to the semi major axis of The Earth or 149,598,261 in order to find the other planets we would need to divide 3618 by 149,598,261 and then multiply by the semi major axis' of the other planets. So let's do Venus first. And I am going to go out on a limb here and predict nobody will be able to guess what Venus will equal let alone Mercury. Okay let's do Venus first ... this is so nice ... 3618.033988 (Phi + 2 or Phi sqrd + 1) / 149,598,261 = 2.4184773110430742239710928190536e5 and times 108,208,930 gives us .... *drum roll is heard* ... 2617 ! or 1000 times Phi squared ... amazing ! And using the same scale we get 2.4184773110430742239710928190536e5 x 57,909,050 or 1400.5 for Mercury so let's list them for clarity ... EARTH = Phi + 2 or Phi squared + 1 or 3618 inches Venus = Phi squared or Phi + 1 or 2617 inches Mercury = 1400.5 inches or 5 times height of G1 ? Well I never ... in all my born days I never figured I would be able to show Phi as a base for our solar system ... but here it is ... again all I can say is very nice . Cheers Don Barone



Post by Don Barone on Apr 5, 2018 20:29:55 GMT 5
But how about this for yet another cherry on top of another nice topic ... Mercury is equal to 1400.5 inches and one could say that in absolute numbers this is 5 x 280 (height of G1 in cubits.) or one could say that it is 280 x 10 (2800) then divided by 2. But then when we subtract 1400 from 3600 we are left with 2200 and of course is 10 times 1/2 base of G1 or again simply base times 5. All the numbers seem to be playing together nicely for us so looking at this I am thinking that if all the pyramids are playing together and are playing the same game then it seems that The Bent Pyramid would have to be either 360(0) cubits OR 361.8 cubits ... My gut feeling is that I may have been correct a while ago when I suggested Phi squared + 1 times 100 for the base of this Bent Pyramid in cubits. It is late and I am tired and besides we have to save something for another day Now since it appears that a cubit of 20.66 or even a bit longer may have been used here we will have to look at The Bent Pyramid and The Red Pyramid with this measurement in mind. ... So until the next post ... Cheers Don Barone



Post by Don Barone on Apr 6, 2018 20:08:20 GMT 5
Okay as promised I have looked at The Bent Pyramid again and even though there is tons more to add at Meidum we will jump to Dashur. Firstly allow me to post the diagram of The Bent Pyramid ... next I will add the measurements I am using assuming the base of The Bent Pyramid IS NOT 360 AND IS NOT 362 BUT IS IN FACT, LIKE AT MEIDUM, 361.8034 CUBITS ! You will also see that here I agree with John Legon and have added the height of the two sections as 110.102 and 89.898 obtained by adding together sq rt 2 and sq rt of 3 and then dividing the sum into 200. From recent work you should remember that sq rt 2 / sq rt 3 is 0.8165 and you will see that in the second image down but the finale I find most satisfying as it all comes full circle and takes us back to our solar system as we get the final diagram. Enjoy ! and a comparison ... Cheers and much more coming ... Don Barone



Post by Don Barone on Apr 6, 2018 20:59:07 GMT 5
Okay I am sure there are some in the audience who are doubting what I am presenting. That is okay for only because of all my other work and the fact that it is all tying together now makes me a believer. But let's just take a look at one of a zillion coincidences I have found. This one is brand new and I almost missed it. Momentarily we will return to Meidum where the north distance to the wall was 2203 inches and the south distance was 1393 and it worked out that the total was 3596 and divided by 1393 gave us 3596 / 1393 or 2.5814 and we will all remember that I claimed Mercury = 12 and Earth = 31 and this is why this works out is becasue 31 x 300 = 9300 and east wall length and 12 x 300 is 360 or a ratio of 2.5833333 ... However 3596 / 1393 is 2.5814 and very close so now let's fly over the Dashur where after allowing the base of the Bent Pyramid to equal 361.8034 cubits and the angle of 55.007 (tan of 1/0.7 or 1.42857142857) we get a height of 361.0834 / 2 (to get 1/2 base) = 180.9017 x 1.42857142857 = 258.431 and very close to our 2.5814 from Dashur (0.9988 not our best but not bad) I have to assume we will have to fine tune it. To get it exactly to 258.33 as a height we would need an angle of 54.9978 degrees and extremely close to 55.007 degrees.
NOTE HERE: If we use 3600 instead of 3596 and divide 3600 / 1393 we get 258.435 and checks almost precisely to height of pyramid I got of 258.431 (0.99998) To me there can be absolutely no doubt that all the pyramids sties in together mathematically !
And there is still quite a bit I have found tying in Meidum and Dashur ... and Giza which I will present over the weekend ..
Cheers Don Barone



Post by Don Barone on Apr 7, 2018 17:54:05 GMT 5
Hi all ... I really don't what direction to go in so I will just post some bullet points for now ... Observe this: This is how some get the average length of the sides of Meidum ... John Legon is our guest here. So apparently to get the average length of the sides the plan is to ignore one of the sides ... Wow can you believe it. Okay here once again is the four sides added together ... 22728 inches Now Petrie told us his best INFORMED opinion of the height of the pyramid was 3619 I will use 3618.034 inches Now 36180.034 x 2 x Pi equals 22732.8 inches and checking to Pi to within 1 inch per side. Now no offense John but doesn't this seem a bit more logical and probable ... Okay now let's check our 7/11 template ... 3618.034 / 7 = 516.862 x 11 = 5685.482 The base lengths again: So it does fit but not even as close as we fit for the four sides added togehter ... And now something to ponder from DUNE who did the checking for me .. DUNE posts over at graham hancock's board and has done some great research at Giza and is somewhat of an expert when it comes to using Google Earth as a measuring tool. For the longest time I have wondered what is the distance between The Bent Pyramid and The Read and according to Petrie this is what we get .. This is Jim Allison's solution using Petrie's erroneous data ... a beautiful neat and concise master plan however there is just one slight problem according to Google Earth Petrie is not correct in this measurement of 6702 feet or 3900 cubits. Since I had found so much new solar system data at Meidum I was trying to figure out the link or bond between all the 3rd Dynasty and early IYth Dynasty pyramids, especially those credited to Sneferu so I asked DUNE to check Petrie and see what he scaled for this measurement and this is what he got. Neither of us had any preconceived notion on what we might find in fact we were expecting 6702 feet but what we got was far different ... Here is DUNE's image: and try as we might we could not figure out what it might mean ... 6743.18 x 12 = 80918.16 inches and this looked vaguely familiar so I multiplied by 2 and got 161836.32 and realized of course it was showing us 1/2 the value of Phi . The exact value would be 1/2 Phi = 0.809017 x 10,000 = 80901.7 and if this is inches then distance would be 80901.7 / 12 or 6741.81 and checks to within 0.9998 ... so in my opinion the distance was meant to be 6741.81 and it was that way for when we divide the height of Meidum into this value we get PRECISELY 22.36068 WHICH IS EXACTLY THE SQ RT OF 5 X 10 OR THE SQ RT OF 500 ... Much more to come Cheers Don Barone



Post by Don Barone on Apr 7, 2018 20:02:43 GMT 5
Okay I have to admit this is all getting just a bit confusing ...
We have semi major axis of The Earth = 31, also we have it at 3618.034 (Phi squared + 1), and then for good measure we have it at 2.583333 and another one is 9300 (inches) as illustrated at Meidum and our final one comes from the fact that we have labelled 1/2 base of G1 Mercury and thus G2 is Phi but if G2 is Phi then ... but wait we have also said that G2 is Venus and thus equals 2.618034 if and only if G1 equals 3.618034 ... so if G2 is Venus and Phi then Earth will equal 1.618034 / 108,208,930 then times 149,598,261 for earth and we get ... 2.236923246 and accurate to sq rt of 5 (2.2360679775) to 0.9996 ... pretty good actually and for those who have been here since the beginning THIS IS PRECISELY WHAT THE GIZA RECTANGLE WAS SHOWING US ALL ALONG !!!!!!!!!!!!!!
Earth: 31 3618.034 2.58333 93000 2.237
I will do Mercury and Venus tomorrow ... why don't you see if you can make your own list ...
Cheers Don Barone
Still quite a bit to post and I have DUNE hopefully getting me some accurate distances between the pyramids ... (amazing that it has not already been done and listed somewhere ! )



Post by Don Barone on Apr 8, 2018 7:06:13 GMT 5
Okay another very, very interesting tie in to Giza from Meidum. Here are the sides of The Pyramid at Meidum: N = 5677.2 E = 5694.5 S = 5681.3 W = 5675.0 =========== Total = 22728=========== Now the base of G1 equals 440 cubits ... and inverse of this is 0.002272727272 and times 10,000,000 = 22727.3 and checks to within 7/10ths of an inch ... inverse of 22728 is 4.3998592045054558254135867652235e5 A fluke ? Maybe but there are an awful lot of these showing up ... Cheers Don Barone Okay just a very quick post here ... If we accept that The Pyramid at Medium was at the ratio 7 to 11 and that Petrie was correct when he suggested that the height of this pyramid was 3618 inches then 3618.034 / 7 gives us 516.862 inches. Now this number can yield all kinds of things but this is the one that totally floored me. If you look above the perimeter is 22728 and if the height is 3618.034 then a circle with that radius gives us a circumference of 3618.034 x 2 x Pi and we get 22732.78 and as I had pointed out showing Pi to within 1 inch per side but there was an even better surprise waiting ... as I said if 3618.034 is the height and equal to 7 then 1 unit = 3618.034 / 7 or 516.862 and then I did , sometimes I am not sure why I do these things, but I took the square root of this number and guess what it gave me ? Well square root of 516.862 is 22.7346 and times 1000 gives us 22,734.6 and showing us Phi in the perimeter to an accuracy of 22728 / 22734.6 or 0.9997 ... nice ! But let's revisit G3 for just a moment. Here are what Petrie has for the sides ... North  undefined East  4149.2 inches South  4157.8 inches West = 4153.9 inches Now for some interesting mathematical diversions. As I have shown 3618.034 / 7 = 516.862 and times 8, or what 13 units will equal is 13 x 516.862 = 6719.206 inches . Now we had that 6719.206 divided by Phi or 1.618034 gave us 4152.7 inches and quite within the realms of possibilities for it to equal one of the sides of G3. However Venus shows us in it's movements through the sky the tracing of a pentagram in an 8 year to 5 year cycle. And an even better example And from this site ... "This is just to say that if Venus emerged in the east as morningstar on, say, Christmas, then Venus would again emerge on Christmas eight years later. 13 of these Sun/Venus cycles equal one Venus Round. Another interesting fact related to this is that, because of this 8:5 ratio, Venus traces a fivepointed star around the ecliptic every 8 years. The occult symbol of the pentagram probably has its roots in the discovery of this fact in ancient Mesopotamia. ..." So we have Venus/Earth in an 8/5 year cycle and 13 of these cycles equal a Venus Round. So now let's go back to G3 and our 6719.206 inches. We have shown that 6719.206 is 13 when Phi squared + 1 (3618.034) is equal to 7 (height of Pyramid at Miedum ) . and so now let us see what 8 will give us ... well 516.862 is 1 so 8 is simply 8 times this or 4134.896 and not really all that close to G3 BUT 13/8 is not really exactly correct for Phi so what we really have is 13 / Phi (1.618034) and we find that we get 8.0344418 instead of exactly 8 so now let's multiply our 1 unit of 516.862 times 8.0344418 and we get 4152.70 and giving us the same answer as above when we simply divided 6719.206+ by Phi (1.618034) so in a very neat way showing us The Earth / Venus pentagram cycle in the numbers and most importantly in the pyramids. So now let's list the choices for Venus as we had done in a previous post for Earth. sq rt of 505 2618.034 1.868585 6726.97 (6719.206 ?) (9300 / 149,598,261 then x 108,208,930) 1.6174 (1.618034 ?) and of course 108,208,930 kilometers 67,237,911.8 miles = 6726.97 ? Semimajor axis 0.723332 AUOkay a minor light bulb had gone off in my head and I think I need to switch to miles for a minute ... Above I had noted that (9300 / 149,598,261 then x 108,208,930) = 6726.97 and very nearly the correct distance IN MILES for the distance for Venus to The Sun and I was surprised theat it was that close and then the little light bulb turned on for staring me in the face was a plan drawn not only in the metric system and not only in inches BUT WAS SHOWING US MILES AS WELL for is it not almost correct to say that The Earth is 93 MILLION MILES from The Sun ? so I came to the conclusion that our 9300 inches was in fact logically showing us miles x our usual scale of 10,000 to 1 so 9300 x 10,000 gives us 93,000,000 miles. I t was an interesting insight and I was quite pleased with myself The actual agreed upon distance is 92,956,050 and divided by 10,000 and we get 9295.6 and accurate to 9295.6 / 9300 = 0.9995 and only about 4.4 inches away from showing it perfectly at Meidum ... coincidence yet again or did these ancient builders know what they were doing ? Cheers Don Barone



Post by Don Barone on Apr 8, 2018 7:33:50 GMT 5
Here is an image that shows the concept visually ... Nice ... Cheers Don Barone



Post by Don Barone on Apr 8, 2018 9:51:14 GMT 5
Okay we have this ...
Height of Pyramid at Meidum is 3618.034
3618.034 / 8 = 452.25425 and times 13 = 5879.30525 (remember The Venus Round)
Now since we are dealing with inches let's mark this as 5879.31 inches and this equals in 20.62 inches per cubit cubits as 285.126 cubits. Now again because I have worked with these numbers for many years this number immediately looked familiar and so I squared it and got 81297.033 and then I divided by 2 and got 40648.52 and then I took the square root of this and got 201.6147726 ... now all I have really done is called 285.126 the diagonal and calculated the sides which we get as 201.615 cubits and in inches gives us 4157.30 and tying into the south base of G3 nicely.
Cheers Don Barone
PS: It would definitely appear that as Giza was predominated by Phi and the sq rt of 3 The Meidum Complex seems to be using Phi exclusively and just a little humor ... isn't it nice that Phi sqrd + 1 has in it the Phi angle (36) and the angle who's sine gives 1/2 Phi (18 = 0.309017) [3618] very cute I thought ...
Cheers Don Barone



Post by Don Barone on Apr 8, 2018 14:25:04 GMT 5
Okay another very interesting result from Meidum ...
Let's list again the sides of the enclosing wall.
North  8561 inches East  9307 inches South  8479 inches West  9300 inches (Just noticed my diagrams are in error as I have flipped east and west ... nothing changes it's just that on the east should be 9307) ================ Total = 35647 inches ================
Now maybe it's because I have done this so often but that number looked familiar so I decided to multiply by Pi and I got 111988.35 and for the moment not ringing any bells and then I decided to divide by four as if assuming this number represented a perimeter and I got 35647 / 4 and arrived at 8911.75. Now again I guess it is because I have played with certain numbers so much I recognized this as maybe being important so I decided to assume a diameter here and multiplied by Pi and got 27997.088. Now probably because the height of G1 is 280 cubits and times Pi gives us the orbital period in Earth days of Mercury I had often divided when I should have multiplied and recognized this and so I simply assumed 28,000 inches as circumference and dividing by Pi got a diameter of 8912.6768 and times 4 to get a perimeter and we get ... 4 x 8912.6768 and getting 35650.7 inches and matching to the overall perimeter of the rectangle (better I guess simply a four sided figure) at Meidum to within less than 4 inches in over 35000 and accuracy of 35647 / 35650.7 = 0.999896 and close enough I feel to claim that this was their intent !
Cheers Don barone



Post by Don Barone on Apr 8, 2018 20:56:45 GMT 5
Okay not much this time just an updated and corrected version of The Meidum Pyramid Plan with enclosure. Now I have been trying, in vain I might add, to calculate the distances between the major pyramids of The 3rd and IVth Dynasties. The reason I mention in vain is that there are several different coordinates listed for the pyramids in question. This makes it impossible to get the correct distance between them because the coordinates keep changing and it is impossible to know which one is right. Allow me as an example to list or show the various readings I found for The Great Pyramid . Now with all these different coordinates it is virtually impossible to find a correct measurement to any of the other pyramids ... very frustrating so I will simply use what I think it should be and carry on with some previous work I had done in this area ... I wonder though who will have the correct coordinates ? Cheers Don Barone



Post by Don Barone on Apr 8, 2018 21:58:04 GMT 5
Okay taking a line from 1970 I believe ... Houston we have a problem ! And here is our dilemma and it was all tying in so nicely ... damn Okay back to our plan diagram: In Petrie's notes he said he took the measurements to the wall from the center of each side of the pyramid with North side being 2203 and south side being 1393 and I went on to find all kinds of seemingly brilliant things but there is one major problem ... the measurements can't be 2203 and 1393. Why ? Because the longest side is 5694.5 inches and 5694.5 + 2203 + 1393 = 9290.5 so even using the longest side (when we should be using the center of the pyramid and thus making it much worse) we can't get the pyramid and the measurements to add up to 9300.00 . Even using the longest side of the pyramid we are out about 10 inches and if we take the center and average 9300 + 9307 and get 9303.5 and for the center east west at the center we get ... 5694.5 + 5675 and average of (5694.5 + 5677 = 11371.5 / 2 =) 5685.75 then plus 2203 and + 1393 gives us 9281.75 and short to the shortest side of 9300 by 18.25 inches so maybe I am missing something here but I do not see how Petrie's measurements can be correct. Don Barone Damn ! In order for it to work we need 2203 + 1393  9300 and that equals 5704 and that is longer than any of our pyramid sides ... now what ?



Post by Don Barone on Apr 8, 2018 22:17:59 GMT 5
Okay I think I have it ... all the other things I have suggested dealing with 3618.034 and such seems to be still fine. It is only 2203 + 1393 that is the error. All things being equal the only possibility is that, like The Earth itself the center of this pyramid BULGES OUT. Impossible ? Well maybe not for we still have out solution of 3600 and 9300 being Mercury and Earth and it fits so well and after all the bulge would be on THE EQUATOR of our pyramid. If not I am open for suggestions ... (bulge would take it to 5707)
Okay how's this ...
Equatorial diameter of Earth = 12756.28 and we are going to call this 5707 Polar diameter of Earth is 12713.6 kilometers so if 5707 = 12756.28 then 12713.6 would equal 5707 / 12756.28 and then times 12713.6 and this gives us ... wait for it ... 5687.9 and within 6 inches of the average of 5682 and within Petrie's own tolerances which he set at + or  6 inches ... Well it works ...
Don Barone



Post by Don Barone on Apr 9, 2018 0:09:59 GMT 5
Actually it was 5704 and redoing the math get's polar diameter of 5684.9



Post by Don Barone on Apr 9, 2018 7:26:27 GMT 5
An image for clarity ... note that 9303.5 is average of 9307 and 9300 and 5684.75 is average of 5694.5 and 5675 Interesting to note however is that 5684.75 + 3618.034 = 9302.8 and within 7/10ths of an inch of 9305.5 So how does one explain the 2203 and 1393 ? I am not sure other than the argument I have put forth. We know that G1 and G3 had indentations and were actually an 8 sides figure perhaps Meidum was the same only with an outward bulge instead of an inner recess. Anyone have any ideas I would like to hear them ... NOTE: If we use Petrie's best guess of 3619 for the height of Meidum we get 3619 + 5684.75 = 9303.75 and checks for 25/100ths of an inch ... Cheers Don Barone



Post by Don Barone on Apr 10, 2018 9:12:26 GMT 5
Hi all ... I have been thinking about the curious fact that the perimeter of The Pyramid at Meidum of 22728 is virtually the identical number as the inverse of 440.00 (0.00227272727). Now I have been trying to find a relationship on why this is and so far I have come up with this ... 440 / 10,000,000 = 0.000044 and reciprocal of this is 22727.27 which I find fascinating for some reason and shows a possible mathematical tie in from Meidum to Giza and The Great Pyramid. So I decided to do a bit more playing around and found that 22727.27 / 1760 (perimeter of G1) gives us 12.9132 and if you have followed my work you will immediately recognize this number. 12.9132 x 100 = 1291.32 and if we use this as a circumference gives us a diameter of ... wait for it ... 1291.32 / Pi = 411.04 and precisely the base distance needed at G2 to meet the criteria of the ratio we established between G1 and G2 of sq rt of 3 (G1) to Phi (G2) Ratio being 1.7320508 / 1.618034 = 1.0704662693192697958259095291383 and times 411.0406 gives us 440.0052 and checking to accuracy of 440.00 / 440.0052 = 0.999988 or accuracy of 12/1,000,000 ths (twelve / one millionths) ! (Actually just thought of another way 5682 / 440 gives us the same result) Now isn't that interesting ... And from an early thread we found that half the semi major axis of Earth (149,598,261 / 2) equaled 74799130.5 and then we noticed that 74799130.5 divided by semi major axis of Mercury of 57,909,050 divided into this gave us 74799130.5 / 57,909,050 and equaling 1.2916657 and times 1000 gave us 1291.6657 and divide by Pi to give us 411.15 (and times 20.62 = 8477.9) Too big but only by 1 single little inch ... Petrie has the south side at 8476.9 inches) There is probably more here but I can't see it at the moment but regardless they always all seem to tie in mathematically if only we look. Cheers Don Barone



Post by Don Barone on Apr 10, 2018 9:37:34 GMT 5
Okay this is a gem ...
Since we had already established G1 as square root of 3 and G2 as Phi I starting trying to figure out this result using just these numbers and honestly you are going to love this ...
I started with G2 and came up with this. The side of G2 is Phi so the diagonal would be the square root of Phi squared x 2 and we get sq rt of 2.618034 x 2 = 2.2882456 ... so this number is representing or really equal to our value of 12.9132. Next I moved to G1 where we had determined that this was equal to the square root of 3 so using the value of Mercury (and base of The Red Pyramid) we got (1/2 sq rt of 3) x 4 and we get ... 3.464102 and following the above example multiplying the two of them together would give us Meidum and we get this ... I really, really, really like this result. We get 2.2288246 x 3.464102 and we get ... 7.9267153178348392327669819027139 ... Now anyone who has done any research on The Earth or anything will immediately recognize this number ... maybe if I multiply by 1000 it might help ... now we have 7926.71532 and do you see it yet ? Maybe if I multiply by 1.609344 (ratio of kilometer to mile) it might be just a bit clearer ... x 1.609344 gives us 12756.812 AND THE EQUATORIAL DIAMETER OF THE EARTH TO AN ACCURACY OF 12756.28 / 12756.812 = 0.99996 ... now isn't that nice ...
So this now means that the perimeter of Meidum represents Earth and G1 does as well in it's own way as it is 1 / Earth times 10,000,000 ... very, very interesting ...
Cheers Don Barone



Post by Don Barone on Apr 10, 2018 10:00:52 GMT 5
So now back to Meidum we go where we get a perimeter of 22728 ( shall we use 22727.2727 ? ) Now we have used the perimeter but what of half the perimeter or 5682 x 2 ... well we get 11363.6363 now again not a number that many of you will recognize but I recognize it as the number I needed to make my 9 by 11 diagram work. If we divide by 10 we get 1136.363 and if we divide by base of G1 or 440 we get 1136.363 / 440 = 2.58264 and times 57,909,050 (semi of Mercury) gives us 149,558,497 and checks to Earth at 149,598,261 to 0.9997 ... All of it proves I think that The Pyramid at Meidum is indeed, like The Great Pyramid meant to represent ... THE EARTH ! So, so nice



Post by Don Barone on Apr 10, 2018 11:47:55 GMT 5
Okay I have learned that when you are on a roll ... keep playing so here we go ... Average of perimeter of Meidum is 5682 and times Pi gives us a circumference of 17850.529457697205180944739703794 and times 2 gives us 35701.0589 ... now Petrie has the distance from top to bottom of rectangle at Giza as 35713 inches ... so let's keep this in mind. So 35701.0589 in cubits is when we divide by 20.62 we get 1731.3801607853739263767933757317 and squared gives us 2997677.26 and in relation to the speed of light is 299792.458 / 2997677.26 = 9.9992 or 99.992 % accurate. Interesting ... 1731.4516 would give us a precise value and that is within 0.07 of a cubit or about 1.5 inches. Difference happens to be 1731.451582  1731.380161 and gives us 0.071421 and reciprocal is ... what else could it be but ... 14.001 Okay so that one might be a stretch but the main point here is that I think we have discovered that ALL THE PYRAMIDS were interrelated and they all either showed the solar system or are showing us the ratios that the solar system uses . So what do I think about all of this. Well if it all started at Meidum and I think it did then The Pyramid at Meidum I feel must be the more accurate. So using Petrie's value of 35713 and divide by 2 and divide by Pi we get 5683.9 and well within the bounds of possibilities for base at Meidum. I really feel that this is a break through discovery at Meidum and sadly like all my other research will probably wallow in the land of who gives a heck forever ... So basically what we have found is that the base length of The Pyramid at Meidum when used as a radius will give us the vertical height of The Giza Rectangle and that folks is simply amazing ... but another alarm bell is ringing somewhere for isn't "X" times 2 times Pi another ratio we have found in our solar system ? Yup ! And that will be our next post Cheers Don Barone



Post by Don Barone on Apr 10, 2018 12:30:06 GMT 5
Okay the plot is thickening ... Time for some more diagrams [ I love diagrams ] First using the base of Meidum as a radius ... And now our old friend The Giza Rectangle Okay let's combine them ... And label it ... Now isn't that about the most fascinating thing you have ever seen ? Well hang on because it is going to get a whole lot better very soon. Cheers Don Barone



Post by Don Barone on Apr 10, 2018 12:31:43 GMT 5
Yikes I hit 5 instead of 3 in 35,713 ... damn I will have to change all the diagrams ... db EDIT: All images corrected

