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Taking a look at a satelite picture I noticed that the truncated top of Khaphre's pyramid is a lot smaller than that of Cheops. Also it seems tha the casing stones are in place not only on the top part of the pyramid but also all the way up to this truncated top. I don't know if you can confirm this. The current height is about 136.4 meters or 260.5 royal cubits. Thus it seems that you might be correct in saying that Khaphre's pyramid doesn't have one uniform slope. Now if at the bottom it has the maximum slope and going up this slope decreases to a minimun at the top then the slope you give, in other words 52.75 or 52.2 degrees might even be too big. I think it's not important to compute the slope at the top. What's important is to compute the original height and the base slope. From this we can also compute the geometric virtual slope. There are some clues that lead to a hight of 265.1 or 265.9 or 266.2 royal cubits. In the case of 265.9 rc the virtual angle whould be 52.3 degrees. I'd like to know what you think of these values and which you you think is more probable.

With the third value I give (266.2 rc) the volume of the pyramid is:

V = (411.154 rc)^2 x (266.2 rc) /3 = 15000158 rc^3 or a theoretical 15000000 rc^3

But at the same time this is:

V = 15000000 rc^3 = 15 x (100 rc)^3 = 2152698 m^3

But the mean base length of this pyramid accourding to Petrie is 215.26246 meters !

With the third value I give (266.2 rc) the volume of the pyramid is:

V = (411.154 rc)^2 x (266.2 rc) /3 = 15000158 rc^3 or a theoretical 15000000 rc^3

But at the same time this is:

V = 15000000 rc^3 = 15 x (100 rc)^3 = 2152698 m^3

But the mean base length of this pyramid accourding to Petrie is 215.26246 meters !

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