G3 - The Kings Chamber 3
Article first time online: 10 July 2009.
Reworked : March 2020.
Hypothetical calculations of a hydraulic press inside Cheops pyramid.
We have previously calculated that the volume of the small cylinder was 3075 cubit³. The kings chamber has a height of 11.15 cubit, the assumption is this room was originally 5 cubit high and that the walls and ceiling were pushed upwards over a distance of 6.15 cubit. If the upward movement of the large piston (the kings chamber) was indeed 6.15 cubit, then the theoretical area of that large piston must be 3075 cubit³/6.15 cubit = exactly 500 cubit².
It’s perhaps no coincidence that the calculation amounts to exactly 500 cubit², this could well indicate that it was originally planned this way. If the surface of the piston is exactly 500 cubit² then, purely theoretically, this piston could be pushed upwards exactly 6.15 cubit but this is without taking into account the inevitable losses.
However perfect the construction may have been, the designers of the pyramid had to take into account the inevitable loss of pressure, with water leaking away. For that reason they have undoubtedly calculated a certain safety margin, it’s of course impossible to trace how much reserve was provided. The theoretical maximum area of the piston was 500 cubit², with a reserve of, for example, 10%, the surface would have been 50 cubit² smaller, which would result in a base of 450 cubit².
KC45 – Top View of the King’s Chamber,
the walls have a thickness of 3.29 cubit.
If we only consider the walls of the kings chamber, its base has an area of (16.58 x 26.58) cubit - (10 x 20) cubit = (440.7 - 200) cubit² = 240.7 cubit². The large piston must have a surface of about 450 to 475 cubit², so the surface of the base alone is much too small.
KC46 – A base under the walls of the royal chamber.
In order to reach the required surface area for the large piston, the walls of the kings chamber must therefore stand on a base, which must have an area of approximately 450 to 475 cubit² (to maintain 5 to 10% reserve).
KC47 – The walls of the kings chamber are standing exactly
in the middle of the base.
To evenly distribute the upward pressure, the walls of the kings chamber had to be exactly in the middle of the base. If this were not the case, the base could be skewed or even broken, blocking the walls of the royal chamber in the shaft in which they stand. Because of those requirements not many options are left, the absolute maximum surface area is 500 cubit² and a certain margin had to be provided. On the other hand, the piston had to be kept as large as possible in order to obtain the maximum upward thrust.
It is absolutely impossible to determine the dimensions with certainty unless they can ever be measured. So it’s looking for a valid solution without ever having the certainty that these are indeed the right values.
What follows is a purely hypothetical approach to show that the kings chamber may well have been the large piston of a hydraulic press.
KC48 - Top view on the base and walls of the kings chamber (3.29 cubit).
Without you getting bored with the countless failed attempts, we take as an example a base with outside dimensions of 29.58 by 19.58 cubit, the wall thickness of the kings chamber is 3.29 cubit (3+2/7 cubit). For those walls to be exactly in the middle of the base, the opening inside it must have a size of 7 x 17 cubit. The base itself then has a wall width of 6.29 cubit. In this way the pedestal passes 1.5 cubit the two sides of the walls of the kings chamber.
Note: The dimensions such as 3.29, 6.29, 16.58, 19.58, etc. feel totally wrong. The values 0.29 and 0.58 after the comma stand for resp. 2/7 and 4/7 cubit. In case of such a large construction such as the kings chamber, one probably won’t have counted with palms (1/7 cubit). It looks like it actually has to be ¼ and ½ cubit. The correct values may be as follows: 3.25, 6.25, 16.5, 19.5, 26.5 and 29.5. But, in order not to have to recalculate everything, we continue to work with the first values found.
KC49 – Vertical section of the base.
Outer dimensions of the base 19.58 x 29.58 cubit.
Inside dimensions of the base 7 x 17 cubit.
Width of the base walls 6.29 cubit.
The ground surface of the base is then = (29.58 x 19.58) cubit² - (17 x 7) cubit²
= 579 Cubit² - 119 cubit² = 460 cubit²
460 cubit² is used on a theoretical maximum of 500 cubit²,
this is 460 cubit² / 500 cubit² = 0.92 or 92%.
So there is still a reserve of 8% to compensate for the losses.
KC50 – The kings chamber was originally 5 cubit high.
The height of the walls of the kings chamber is 16 cubit.
The base walls are 6.29 cubit wide and on average 5 cubit thick.
KC51 – The kings chamber was pushed up 6.15 cubit. Large stone blocks slid under the base to anchor the kings chamber in this position.
In total there was a volume of 3075 cubit³ of displaced hydraulic fluid. On a piston with a surface area of 460 cubit², this gives a theoretical maximum upward displacement of 3075 cubit³ / 460 cubit² = 6.68 cubit. Somewhere inside the pyramid, the upward displacement was limited to 6.15 cubit, the 8% difference was used to compensate for the pressure losses.
The small piston placed a pressure of 11.227 tons/cubit² on the hydraulic fluid, the large piston therefore experienced an upward thrust of 460 cubit² x 11.227 tons/cubit² = 5164.4 tons’. This force had to be sufficient to push up the weight of the base, the walls of the kings chamber and the four layers of granite beams above.
Not all dimensions are known, it’s not even clear how high the walls of the kings chamber are in total. If indeed there exists a base below the walls of the kings chamber then not a single dimension is known of it. It’s therefore impossible to determine the exact volume or weight of this whole. The only thing we can do is take this example to see whether this upwards thrust of 5164.4 ton’ theoretically would be sufficient.
1 – The Base.
KC52 – The pedestal external dimensions (29.58 x 19.58) cubit²
Inside dimensions of the base (7 x 17) cubit²
Ground surface of the base is 460 cubit².
Width of the base walls 6.29 cubit, average thickness 5 cubit.
In the example above, the base has a surface of 460 cubit² and an average thickness of 5 cubit. The volume is 460 cubit² x 5 cubit = 2300 cubit³, with a specific weight of 375 kg/cubit³ this gives a total weight of 2300 cubit³ x 375 kg/cubit³ = 862,500 kg or 862.5 ton.
2 – The walls of the kings chamber.
KC53 – The walls of the kings chamber 3.29 cubit thick and 16 cubit high.
The walls of the kings chamber are 3.29 cubit thick and are 11.15 cubit high, for the total height of the walls we assume 16 cubit. The bottom surface of the walls is (26.58 x 16.58) cubit² - (20 x 10) cubit² = 240.7 cubit², the volume becomes equal to 240.7 cubit² x 16 cubit = 3851.2 cubit³ and 3851.2 cubit³ x 375 kg/cubit³ = 1444,200 kg or 1444,2 ton.
3 – Four layers granite beams.
KC54 – The four layers of granite beams are on average 3.57 cubit thick.
The supporting granite blocks are on average 2 cubit thick.
The four layers of granite beams plus three layers of supporting blocks together are approximately 20.28 cubit high. The correct dimensions are not known and the calculations will therefore always remain an approximation. The assumption here is an average thickness of 3.57 cubit for the layers of granite beams and an average thickness of 2 cubit for the supporting blocks. The four layers of granite beams (4 x 3.57 cubit) = 14.28 cubit plus (3 x 2 cubit) = 6 cubit together makes 20.28 cubit. The layers have a respective width of 16.5, 15.5, 14.5 and 13.5 cubit, these four added together amounts to a total of 60 cubit.
KC55 – The four layers of granite beams have an average width of 23.3 cubit.
The four layers together have in total a length of 60 cubit. Each layer is approx. 23.3 cubit wide and has an average thickness of 3.57 cubit. The total volume is therefore (60 x 3.57 x 23.3) cubit³ = approx. 4991 cubit³, with an average specific weight of 375 kg/cubit³ this comes to a weight of 4991 cubit³ x 375 kg/cubit³ = 1,871,625 kg or approx. 1872 ton.
4 – The three layers of supporting granite blocks.
KC56 – The layers of supporting granite blocks are on average 2 cubic thick.
The supporting granite beams have a total width of (3.25 + 3.25 + 2.75 + 2.75 + 2.25 + 2.25) cubit = 16.5 cubit, a thickness of 2 cubit and a length of 23.3 cubit. The combined volume is then (16.5 x 2 x 23.3) cubit³ = 768.9 cubit³. Its weight is 768.9 cubit³ x 375 kg/cubit³ = 288,337.5 kg or approx. 289 ton.
5 - The limestone walls standing next to the granite beams.
KC57 - The limestone walls on both sides of the granite beams.
In addition to the granite beams, there are also limestone walls that extend to the outside of the walls of the kings chamber. Its surface area is [(16.5 + 15.5 + 14.5) cubit x 5.57 cubit] + (13.5 cubit x 3.57 cubit) = 259 cubit² + 48 cubit² = 307 cubit². Both walls together have a thickness of (26.58-23.3) cubit = 3.28 cubit. The volume is then 307 cubit² x 3.28 cubit = 1007 cubit³. The total weight of both walls is 1007 cubit³ x 375 kg/cubit³ = 377,625 kg or approx. 378 ton.
If we count all weights together, we obtain:
1 - The base: 862.5 ton
2 - The walls of the royal chamber: 1442.2 ton
3 - Four layers of granite beams: 1872.0 ton
4 - The supporting granite blocks: 289.0 ton
5 - Both limestone walls next to granite: 378.0 ton
In total 5164.4 ton was available, for this theoretical example we come to a weight of 4841.5 ton, so there remains (5164.4 - 4845.7) ton = 318.7 ton.
Of course we don’t know what the reality looks like, the base can be thicker or the walls of the kings chamber can be a little higher. This example only makes it clear that the upward force has been large enough to push the whole upwards. The "crown", consisting of the walls of the kings chamber, the base on which they stand together with the granite beams on top of these walls can indeed be the large piston of a hydraulic press inside the pyramid.