Cubit with squares

Okay, REALLY last one for tonight. Found an approximation for the cubit using square roots. Not very good, in fact worst so far.

Formula is

Cubit in squares

Comparison table is now:

MethodValueDifference from π/6Abs difference
π/60.523598775598300.000000000000000.00000000000000
φ²/50.523606797749980.000008022151680.00000802215168
Ave Year0.523607797087850.000009021489560.00000902148956
π – φ²0.52355866483991-0.000040110758390.00004011075839
10²/φ²π²ⅇ²0.523764440990030.000165665391720.00016566539172
cube roots0.523600350171940.000001574573650.00000157457365
square roots0.523403736847960.000195038750330.00019503875033

Cubit with cubes

Okay, last one for tonight… since I upset √2 with the previous formula, I thought I’d have a go using square roots, but didn’t find anything yet. Maybe another day.

But… I did find something using cube roots… in fact, the cube roots of 2, 3, 5, and something derived from 7 …. and the accuracy is the best yet, even if the formula is horribly complex. Beauty and the beast, I suppose.

Cubit in cubes…

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Egyptians being annoying again

Was playing around with the calculator, still exploring exactly what that kink in the alignment of the three pyramids is.

The great pyramid has a designed height of 280 cubits.

The second pyramid has a designed height of 274 cubits.

That’s a difference of 6 cubits. Which is not very interesting until we turn that difference into metres (and you may guess where this is going …)

6 cubits = 6 x π/6 metres = π metres = 3.1415926+ metres ….

Am still pondering difference in height to the third pyramid …. nothing jumping out like π at the moment….

Some thoughts on the cubit (and foot)

According to the historians, the cubit is an ancient measure based on the length of the arm plus hand, like this:

Cubit and Royal cubit.

where the normal cubit is 6 palms long, and the Royal Cubit 7 palms long. Well, that’s one explanation of the origin. The other is that they got it from the gods, a long long time ago.

Now there are several problems with the common cubit as shown above. In the first place, it’s generally accepted as being 17.6 or 18 inches long. That’s 44.7 to 45.7 cm.

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Stonehenge and the Golden Ratio

So as I’m watching the video referenced in Metre, cubit, foot, megalithic yard, (https://www.youtube.com/watch?v=33dKFtCXEFA), and this image pops up on the screen:

Stonehenge 1

Which is a view of how the first version of Stonehenge looked. And I say, “Hang on a mo, that looks familiar …. ” and indeed it is … that’s the same arc from the Nebra Sky disc:

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Metre, cubit, foot, megalithic yard …..

So I’m watching a video on megalithic building, and they introduce the concept of a megalithic yard:

The length was apparently found via careful measurement of existing structures, as well as finding cross-referenced methods using astronomical means.

They claim it to be 2.72 feet or 0.83m, although the Wikipedia editors generally are sceptical of the whole idea.

Be that as it may… I just found the correlation between this 0.83 metre (or 0.8296 if you want to be more precise), and the sum of the cubit + foot of 0.8319m (they’re both 0.83 if you work to two decimals) as discussed on the Origin of the Foot page, to be rather curious.

The close similarity is hard to ignore, and cubit+foot may make a more compelling origin argument than arguing for a circle divided into 366 degrees instead of 360.

Another way of getting the cubit

[Note: I’m not entirely convinced that my “average year” calculations are correct. I was looking for a way to get 364.75 and since it is tantalizingly close to 365.25 I was looking for a way to get there from that. It may be better to just use 365.25 – 0.5 … now need a Reason Why.]

I was playing around with the calculator and stumbled across this curious sum.

Let’s start with how long a year is. As you should know, it’s 365.25 days for a solar year. If however, we measure against the background stars, it’s about 364.25 days.

From Wikipedia: “Both the stellar day and the sidereal day are shorter than the mean solar day by about 3 minutes 56 seconds. ”

If we take those 3 minutes 56 seconds == 236 seconds, and work out the difference over a year (x 365), we get 86140 seconds, which is 23.927777 hours, effectively one day.

So if we take the average of a solar year and stellar year, we get (365.25 + 364.25)/2 which is 364.75 days, or more precisely, 364.7436921 days.

Now we do this sum.

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The Irrational Mathematicians of Giza, Part 4

(continues from Part 1, Part 2 and Part 3)

When I started this exercise I was hoping to find “interesting” alignments between the centres of the three pyramids, in part to explain the curious “kink”. So in that regard I failed spectacularly (so far).

What I did find was a whole host of other interesting alignments. The table below summarizes the best ones, those that are within 0.5° of the correct angle.

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