Information on the Royal Cubit I found in my work (length 523.33mm)
There is an interesting relationship between the Royal Cubit and the foot. I found that 7 Royal Cubits is essentialy equal to 12 feet. (it is actually 12 feet plus .2238 inches )

(I left the first part of this page unchanged because it still has good info)

The Royal Cubit is re-discovered

I've been continuing a search for the Royal Cubit that fits my model. I had a break-thru on July 13, 1996 and I thought it important enough to add this page.

A quick review of what I was trying to do.

The perimeter of my model (from "the best set" in the paper) is:

which equals what I call the perimeter fraction

The height of my model is one. So the task was to find cubit expressions for both the perimeter and the height, that when divided would equal the above perimeter fraction.

I tried and tried through many hours of computer time to find these two cubit expressions but to no avail. I was ready to give up but thought I should at least try one more thing.

The break-thru

This time instead of using just any unit fractions. I would try using only Horus Eye unit fractions. These are fractions that are alway evenly divisible by 1/64 (I hadn't tried this before, because there are no good Horus Eye fractions in the perimeter fraction).

I thought that at least I could find a rough approximation to the perimeter fraction. I was wrong! The approximation I found was astoundedly close to the perimeter fraction.

How I did it, was to first set up this expression:

I would change the number indicated by the arrow and check the results. As the expression shows, when I tried 10 the results was 43.000.. I was surprised by this.

This indicated that the height and perimeter of the pyramid equal:
278 43/64 and 1755 10/64

__and here is where the error happened__

I got it into my head that 1755 was probably the integer part of the perimeter. And finding such a good approximation right away dissuaded me from looking any further.

But some really thoughtful email about how others were searching in the field, and finding, different cubits from mine, prompted me to take another look.

This time I looked at 16 other perimeters (1755 to 1771). And soon I found that the approximation I first reported on, was just the first in a cluster of good approximations. I should have known. I had come across clusters like this before and they are always elements in a type of series I call a CCI series.

The series I found was:

I found that when n equals 312, CCI equals:
(17835) / (112330)
which equals:
(64*278 + 43) / (64*1755 + 10)
dividing both top and bottom by 64 gives:
(278 + 43/64) / (1755 + 10/64)
which is the first approximation I found.

But more significantly, when n equals 313, 314 or 315 it produces three better approximations. They are summarized in the graph below. (the lengths are found by dividing height in cubits into146.302288m)



What does all this mean? Just maybe ... the Egyptians might have have known this CCI (it is a level of math they could handle). They might even have had (on purpose) four distinct lengths of the cubit.

So I'll leave it there. I guess all I could ask is for all you out there to go out and measure some ancient monument and see which of the above cubits fits it the best.

I did check one thing and that is that the Kings Chamber appears to be ten cubits of .523326m wide.

Let me know if you find anything.

As always, any one with math questions E-mail me. I will answer promptly.

To Pyramid page

Terry 4/11/99

©;1996,1997 Terrance G. Nevin