Updated: Jun 6
There are lots of interesting links between the metre, the Egyptian royal cubit and the imperial system. However, these depend on what values you give all these, as well as the values you accept as the ratios themselves. For example, the diameter of a circle being 1 metre and the circumference being 6 Egyptian royal cubits works best with a metre of 39.375 inches, or 54 digits of 0.72916666", pi as 22/7, and the Egyptian Royal cubit of 20.625". Alternatively the 20.6181818" Egyptian royal cubit corresponds to a circumference of 6 of these cubits, pi of 3.14181818, and a diameter of 1 metre of 39.375". (This conversion rate between metre and inch has never been an official one, but it fits very well, and some researchers into metrology, going back to the 19th century have used this, notably Mauss. It also yields little gems like 6.4 metres divided by 7 make one English yard.)
I also like this metre - royal cubit connection, that I first saw in another film about Ancient Egypt, the famous Revelation of the Pyramids by Patrice Pooyard, first put forward by Schwaller de Lubicz. It's: pi - Phi squared = Egyptian royal cubit in metres. And this works really well with the Fibonacci number approximations for Phi squared and pi, so ((55/21 x 6/5) - (55/21), obtaining the 20.625 inch Egyptian Royal cubit, or alternatively with (144/55 x 6/5) - (144/55), and to convert to inches, using the 39.375 rate again, 20.6181818.
On the GHMB forum, Holger Isenberg recently shared a link to a film by Fehmi Krasniqi, which looks at different interpretations of ancient Egyptian science and engineering, and mentioned that the film makes an interesting claim: the difference between the volume of a sphere and a cube gives the value of a royal cubit in metres. So the volume of a sphere with a diameter of 1 being 4/3 x π x (1/2)³ = 0.5235987756, this can be interpreted as also a length in metres, moving back down from the three dimensional to the two dimensional. There's no need to bring a cube into the equation in fact. All that you need is the volume of a sphere with a diameter of 1. Converted to inches this 0.5235987756 m gives 20.614125". Multiplied by 440 this does give a figure very close to Flinders Petrie's base side for the Great Pyramid, 9070.215, and multiplied instead by 280 for the height, it is again very close to Flinders Petrie's estimate, at 5771.955 inches. The height is even closer to Flinders Petrie's value simply by using pi: 0.5235987756 x 440 x 2/π x 39.3700787402 = 5774.2782152. This last number is very close to 10,000 / √3 = 5773.5026919.
With different values for Phi, using the volume of a sphere formula, you can arrive at the 20.625" and 20.6181818" Egyptian Royal Cubits, if you stick to a 39.375" conversion rate between inch and metre.
4/3 x 22/7 x (1/2)³ x 39.375 = 20.625 and 4/3 x 864/275 x (1/2)³ x 39.375 = 20.6181818.
Even more impressively, 0.5235987756 x 440 x 2/π = 146.66666 m, which is an important number for David Kenworthy.
So it is all the more curious that 0.5235987756 - π ≈ -2.617993978, so close to Phi squared. So that the volume of a sphere with a diameter of Phi² + π, would be 100.04182788, very close to 100.
(4/3 x π x ((Phi² + π)/2)³ ≈ 100.04182788
It's also curious that (0.5235987756 x 440 x 2/π x 39.3700787402 / √(Phi x 20,000,000) = 365.242199 / 354.36708, the ratio between a solar and lunar year in days.