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Post by Don Barone on Apr 28, 2016 22:36:50 GMT -5
Hi all ... this really is a review lesson. I was going to continue on the thread I stared entitled "The square root of the sqaure root of Pi" but then realized I had already sort of explained it all here. Please read the last several posts dealing with The 104th course ... many interesting things in there. And just to explain ... I use THE SQUARE ROOT OF 23313.35 inches as the height to course 104 and note here please that this number, 23313.35 is the sq rt of the sq rt of Pi plus 1 then times 10,000.
So we have this equation:
The height to the bottom of course course 104 is ...
The square root of (the square root of the square root of Pi ) + 1 or
sq root of 13313.35 + 1
sq root of 23313.35
or
152.687 cubits
which equals
152.687 x 20.619654 inches and equals
3148.3554 AND THE BOTTOM OF COURSE 104 IS MEASURED BY PETRIE TO BE 3148.4
"The Square root of the square root of Pi as found in The Great Pyramid"
Cheers Don Barone
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Post by Don Barone on Apr 30, 2016 19:33:39 GMT -5
There is talk on Facebook about the exit of the shafts being at the 103rd course ...
But let's check the true measurements.
Petrie has height at 3119.1 inches to the bottom of the 103rd ...
If we assume 311.91 "cubits" we get if we divide by my cubit of 20.619652471058063018183885017927 inches a value of 15.126831086886638465476082074615 on the surface a meaningless figure until we reciprocate and we get 1 / 15.126831086886638465476082074615 and it yields 0.06610769924355763847963798857981. So what you may say ... well how about this
0.06610769924355763847963798857981 x 2 = 0.13221539848711527695927597715961 and times 100,000 = ...
But before I put the answer down let's recap base of G1 + G3 in inches ... 9068.966 + 4153.4 = 13222.4
0.13221539848711527695927597715961 x 100,000 = 13221.5
This seems to check for 13221.5 / 13222.4 = 0.99993
Just saying ... Not even sure if this means anything .. will look again tomorrow after the beer has worn off )
Best db
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Post by Don Barone on May 12, 2016 15:30:26 GMT -5
Hi all a strange thing occurred on Facebook the other night. Gary Osborn accused me of falsifying data to prove his theory on the speed of light wrong.
I still can't believe he said it but needless to say whatever shaky relationship we had has ended. To accuse me of falsifying data is beyond ludicrous and neither he nor his work hold any interest for me any longer.
Regards Don Barone
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Post by Charlotte on May 13, 2016 7:25:38 GMT -5
Hi all a strange thing occurred on Facebook the other night. Gary Osborn accused me of falsifying data to prove his theory on the speed of light wrong. I still can't believe he said it but needless to say whatever shaky relationship we had has ended. To accuse me of falsifying data is beyond ludicrous and neither he nor his work hold any interest for me any longer. Regards Don Barone
Hi Don,
I saw and in scrolling past the posts on facebook I noticed the name Gary Osborn. Not that you need my defence, I must say that "falsifying anyone's findings to prove them wrong is not your practice, rather, as far as I know, give your own findings on a subject to compare, are quick to correct yourself, and always give credit to an author, as you did on this board, including Mr. Osborn. Shame these disagreements happen.
Cheers Don
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Post by hackjack. on Jun 6, 2016 16:26:48 GMT -5
I have read your thread with great interest, whilst I'm ashamed to say my math is is quite lacking you explained it in a way i could just about keep up with so thanks for that There is more to the Giza complex than meets the eye for sure. I also just read your recent posts on HoM and am astonished at the closed mindedness of some alleged 'open minded' posters, surely we are all here to the same ends...to find answers. Numeracy has to be the ONLY universal message that would last over eons. I can't let myself believe that all that effort purely in just the construction was just to satisfy the ego of whomever was in charge at the time. You may be on the right track... you may be miles off, but so many coincidences??? If the hall of records is to be found in Rostau it makes perfect sense to me that rather than being a hidden chamber the "hall" is in plain sight. Only the adepts have access to this knowledge... The right minds needed to decode this information are few in number but (and i hope) very close indeed.
Please keep up the search Regards
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Post by Don Barone on Jun 7, 2016 4:41:03 GMT -5
Thanks Hackjack I appreciate the vote of confidence. There is much to be discovered on this message board so enjoy and thanks again ! And yes the HoM can be difficult at times but that's okay I have been fighting them since 1999 and I am very secure in my relief that my research is correct..
Best Don Barone
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Post by Darkpiano on Apr 3, 2018 20:53:17 GMT -5
Hi DP sorry I have deleted this post becasue it is not at all of the caliber I had expected. Please read through other files on this board where you will find your answers. All the questions you have asked have been answered in full on other threads ... please search there before posting non relevant posts. Consider this your first warning ... You will not get a second one.
Don Barone Owner and Moderator
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geoff simmons (geoffss)
Guest
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Post by geoff simmons (geoffss) on Jul 23, 2019 4:51:42 GMT -5
Hi Don. I revisited Atalanta Fugiens. I found that - if you accept the Maier "constraint" (that one end of the dividers rests on the very centre of the larger circle) then the angle is just over 22.5 degrees and it ain't Mercury (by 100 or so miles). If you don't accept the Maier constraint it gets worse: using Gary's figures - and letting the scale be 1 : 720 miles - Mercury will be 4.21111r (using yours 4.211). Half of this + I.5 (1/2 moon) gives us a triangle we can quantify. Divide by our (assumed) 23.5 tan and we get 8.29221 (8.292082). So we now have a side for a larger similar triangle in 8.29221 (8.292082) + 4.211111 (4.211) = 12.50332 (12.50308). We can now work out our triangle base to a bottom corner of the Maier triangle by X 23.5 tan = 5.4366* (5.4364939).
Next, we turn to the larger circle centre. It's 8.5 down from the Moon top, 5.5 from the Earth top. Our 12.50332 (12.50308) minus this = 4.00332 (4.00308). So we now have opp and adj of a triangle based on the model centre and a base triangle corner : 4.00332 (4.00308) sq + 3.9366* (3.9364939) - we took off the lunar 1.5 to find this triangle side length - gives us 5.6145695 (5.61432397) hypotenuse from the Earth centre. And there's the problem: this box does not fit in that circle. How do I know? The Earth is 11. It's radius is 5.5.
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Post by geoff simmons on Jul 23, 2019 10:02:45 GMT -5
So you removed my post? It's already messaged to Gary and on GHMB.
The 23.5/Mercury idea doesn't work. It's not that it doesn't have legs, it's that they are too long!
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Post by geoff simmons on Jul 23, 2019 10:11:31 GMT -5
Apologies - there it is. The triangle sides can NEVER fit to that circle given that angle and that construction. Don't know if you know, but this was all fairly originally plotted out on a 14 X 11 grid (each grid box side being 720 miles). Makes it much easier to see why there's a problem. A downward vlaue of about 12.5 can NEVER produce a base that will fit into the Maier outer circle.
Geoffss geoffss.weebly.com[.url] contact c/o GHMB Mysteries Board thread
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Post by geoff simmons on Jul 26, 2019 10:20:11 GMT -5
Where's the come-back? >22.55-ish degrees >2931 miles diameter.
It is the ONLY outcome that works (and could ever work). The rest is so much ?drivel? But you don't publish - why? I know you have the skill to check it!
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Post by Don Barone on Dec 8, 2019 17:52:43 GMT -5
Hi Geoff ... On re-evaluating with my most recent tsunami I have realized it has nothing to do with The Solar System but everything to do with my recent revelation of what The Secret Sign is and how it relates to this diagram. I am presently going through a very disturbing crisis in my life and if I can i will post the true solution to The Philosopher's Stone when and if I can
Don
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Post by Don Barone on Oct 24, 2023 13:57:12 GMT -5
Hi all some new and interesting data on our old friend 23.5 degrees. For starting reference please refer to this thread: The Philosopher's Stone Solved I showed the above thread to a very old friend and partner of mine and he asked innocently enough what is the angle of the triangle. Is it 18 degrees ? I suddenly realized I had never actually calculated the angle and the result left me stunned. For the triangle used in the solving of the Philosopher's Stone is none other than the 23.5 triangle that Gary Osborn had used way back in 2016 to search for a solution to our solar system. Here are the two images to compare:
First Gary's:
and now the image of The Philosopher's Stone from 1617:
and now side by side:
Now that we have solved for the Philosopher's Stone we now know the exact measurements and we can solve for Gary's solar system as long as we know Mercury's diameter ... and we of course do:
Mean diameter = 4,880 km
Mean radius = 2,439.7±1.0 km or 0.3829 Earths (Diameter = 4879.4 km)
Flattening 0.0009
Well as you can see below the two diagrams do not seem to agree with each other so I will have to do a bit of study to see which one is correct. I would think Gary's is correct. Pretty sloppy work from 1617 if you ask me.
db
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Post by Don Barone on Oct 24, 2023 14:07:44 GMT -5
And this image adjusted to fit and shows a much closer match
Still not exact and since it is 45 degrees at the top I would have thought that the one from 1617 should have been spot on. All one needs is a compass, straight edge and marker to draw this exact.
db
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Post by Don Barone on Oct 24, 2023 14:50:00 GMT -5
Hi I have posted what I think is a simplified image on how it was all calculated. Any questions please ask. I'm sorry 6.03554 is the radius of the light orange circle
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Post by Don Barone on Oct 24, 2023 21:32:28 GMT -5
Gary proposed that the square which when representing Mercury meant that the large circle was Earth. Let's check his work.
Diameter of Mercury is 4880 km Diameter of Earth is 12756.28 km
Ratio is 12756.28 / 4880 =2.6139918032786885245901639344262 (Which is ALMOST The Silver Ratio which is 2.6131259297527530557132863468544) This is an astounding revelation and proves yet again the geometric design of our solar system and that "The Creator" loved geometry CHECKS 0.9997
Okay let's leave this and check on Gary's call that if the centre square was Mercury the large circle was Earth.
1/2 the large circle is 6.774 and twice that gives us the diameter of 13.548 units. Mercury equals 5 so ratio is 13.548 /5 = 2.7096 and times 4880 gives us 13222.848 NOT 12756.28 SO IT IS NOT EVEN CLOSE. HOWEVER 13222.848 is exactly the number of inches in G1 added to G2 (9068.996821 + 4153.851179) Not sure why yet.
Fixing my brain freeze we have this. Diameter of circle is 6.774 x 2 = 13.548 Mercury equals 5.0 thus Earth should be approx. 2.613126 x 5 or 13.06563 closer but still not nearly a match
db
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Post by Don Barone on Oct 24, 2023 21:53:43 GMT -5
Thinking on 13222.85 we can get this gem ...
Compare to ...
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Post by Don Barone on Oct 25, 2023 7:58:21 GMT -5
Firstly it appears Gary Osborn's website is no longer active. I hope he is okay. John Legon is also off the air and I can find no trace of either one of them. I am afraid they may have passed.
A side line:
And the penny has just dropped. The angles in The Silver Ratio triangle are 67.5 and 22.5 degrees !!!!!
And so it would appear that The Philosopher's Stone incorporates both The Silver Ratio and The Golden Ratio.
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Post by Don Barone on Oct 25, 2023 12:30:34 GMT -5
It would appear that Gary was closer than I gave him credit for as this image shows it was very close. The blue line is 5.0 (Diameter of Mercury) times "The Silver Ratio" or 2.613126 and give us 13.06563 as opposed to what the diagram gives us at 13.54912.
Here is the image:
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Post by Don Barone on Oct 25, 2023 20:31:34 GMT -5
Hi okay here are some final conclusions on this 406 year old puzzle.
From vertical to where compass hits the circle = 15 degrees
From horizontal to where hand is holding the compass = 15 degrees
From horizontal and where hand is holding the compass to top point where compass touches the circle = 45 degrees
Remaining angles = 60 degrees and are ready to form 6 equilateral triangles with side lengths of radius of circle
Pretty neat drawing !
And the Silver Ratio front and centre to boot.
cheers db
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Post by Don Barone on Oct 26, 2023 18:43:14 GMT -5
Hi I keep trying to understand this image. There is so much hidden within Here is an image that shows what the diameter should be and what it is. Remarkably the true Diameter when using 4880 km for Mercury gives us 13222.84 which is precisely the number of inches in G1 + G3
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Post by Don Barone on Oct 26, 2023 20:01:07 GMT -5
Okay I have found a much easier way to lay this geometric figure out. No calculator is needed. All we need is a compass, a marker and a straight edge. this makes it very, very simple.
Diagram A: Our base drawing:
Some preliminary measurements:
and now the key to unlock this image. Not sure how I missed it before but it solves it so easily you might just yell ... "Doh !"
And the easy solution to a 406 year old puzzle.
and a bit clearer trying to show scribing the arc:
Wow not bad it only took 406 years to figure this out.
Cheers db
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Post by Don Barone on Oct 26, 2023 21:44:44 GMT -5
Hi and one last point to solve and I think it will be complete.
So the last point is, in our 5 unit square model, 5.0 x 2.614 = 13.07 - 5 = 8.07 approx (rounded off) units left of center.
Cheers db
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Post by Don Barone on Oct 26, 2023 22:01:23 GMT -5
Hi I was thinking ... there had to be a way to plot this point without a calculator and guess what ? there is. This diagram gets more beautiful by the hour.
And so it appears that this diagram can be drawn with only a compass, a straightedge and a marker.
I am amazed. cheers db
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Post by Don Barone on Oct 26, 2023 23:25:05 GMT -5
Strange I just noticed that I was using 67.5 and 22.5 and Gary was using 66.5 and 23.5.
Weird that it just hit me in the head now.
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Post by Don Barone on Oct 27, 2023 6:57:07 GMT -5
Well I have just used a program I have to check the angle at the top of the triangle and it is indeed 45.0 degrees precisely so that means that the angles are as I stated 67.5 and 22.5 degrees. Wonder how Gary got 66.6 and 23.5 degrees ? I will have to do a study of this diagram using 66.5 and 23.5.
db
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Post by Don Barone on Oct 27, 2023 9:27:23 GMT -5
Okay in the immortal words of Tom Hanks from Apollo 13, "Houston, we have a problem" (sigh again)
I drew my image to scale and ... well just have a look.
This is what it looks like ...
and the ruler shows where it should be and where the original image places the top of the triangle.
Not sure what is going on but I am working on it ... It appears that the error comes from my sloppiness in drawing the figure. I will have to find my compass and do it properly. The angle shows about 64.5 degrees instead of 67.5 degrees
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Post by Don Barone on Oct 27, 2023 19:03:00 GMT -5
Hi and one last point to solve and I think it will be complete.
So the last point is, in our 5 unit square model, 5.0 x 2.614 = 13.07 - 5 = 8.07 approx (rounded off) units left of center.
Cheers db
I could have just deleted it but this image is completely wrong.
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