Image by Ryan Briggs

This is the famous Bent Pyramid. The lower part of the pyramid has an angle of 54° and the upper part is at 43°. Why is it bent? Really, who knows. The two likely reasons are:

• Time or money (well isn’t time = money). Basically this idea says that they either didn’t have the time or money to finish the pyramid at the initial slope. To cut costs (or time) they changed the angle.
• Building the pyramid at the original slope caused structural instabilities. Either the foundation couldn’t take the weight or the building material itself started to crack.

I don’t really have anything to add to the debate about which theory is more likely (though I do find it quite interesting). Oh, then there is the theory that the aliens that give the Egyptians pyramid building technology played a practical joke on them causing the pyramid to end up bent.

The second reason is interesting to me. How tall of a pyramid can you build? What is the best angle? Let me assume that indeed there are structural problems with the material and look at two ways of thinking about the limiting height.

How tall can I make a column of stone?

What happens if you keep stacking stones on top of stones to build a column or pillar? If you are very careful so that it won’t tip over, you still can’t keep adding stones on top of stones. Eventually the pressure on the lower stones will be large enough to crush them. This property is typically called the compressive strength and is measured in units of pressure. I am not really sure of the common symbol to represent the compressive strength, so I will just use ?.

Let me pretend to build a stack of blocks. Here is a diagram showing the forces on one of the blocks.

Each block has a height of h, cross sectional area A and density ?. The net force on the block shown must be zero (vector) so that in the y-direction:

I guess I didn’t need that. All I really need is F-down (not F’ed-up). This will simply be:

Here, n is the number of blocks above the block of interest. Oh, I guess you can see that this is just the weight of all the blocks above – where hA is the volume of each block. But what about the pressure on this block? It would be this force divided by the cross sectional area:

The more blocks that are stacked, the greater the pressure. The greatest pressure will be on the bottom block. Ok, so if these blocks have a compressive strength of ? (the pressure at which they crack – crack under pressure, get it?) how tall can it be? I will call the total height H not to be confused with the height of each block (h):

Notice that in this model, it doesn’t depend on the horizontal dimensions of the blocks. The Engineering Toolbox lists the compressive strength of limestone at 60 MPa. Of course, there are all types of limestone. Maybe you are going to use some some better stuff. Let’s say the compressive strength is about 80 MPa. I will also use a density of about 2500 kg/m³. This would give a maximum column height of (remember, 1 Pascal = 1 Newton/m²):

That is quite a bit higher than I expected. I guess I should compare this with something else. What about bricks? Wikipedia lists the density of bricks around 2000 kg/m³ with a compressive strength around 30 MPa (but can be much higher too). Using these values, you could stack bricks in a column that would be 1500 meters.

Hmmm. Well, it just takes one bad brick to break the whole bunch. I suspect in real life the effective compressive strength is a bit lower. If I knock the compressive strength of limestone to down around 40 MPa, I still get a max height of about 1500 meters.

Pause: Honestly, this is not going how I expected. Here is what I thought would happen. I would calculate the maximum height of a column of limestone and find that it was shorter the the height of a typical pyramid. However, this could be used to get an estimate for the slope of the side of the pyramid. I would then point out that for rocks in the middle of the pyramid, the compressive strength is higher. Since the middle rocks can’t expand on the side, this makes them stronger. The last step would be calculate the average pressure as a function of height in a pyramid and use this to calculate the angle.

Since that doesn’t seem to be working (1500 meters is taller than a pyramid), I will just proceed with a lower value for ?. I know, it does seem like cheating. But maybe not. The tallest chimney is 420 meters tall. This is not a straight “column” but rather wider at the bottom. Also, I am not sure what this is made from – probably brick or cement. So, let me just pretend like the the tallest straight brick column is 200 meters. If this was at the point where it is about to break, this would give a compressive strength of about 4 MPa. So, that must be it. My compressive strength was perhaps too high. Unpause

If height is all that matters, what angle should I make my pyramid?

Maybe I should start with a diagram of a pyramid. Here it is.

Just to be clear, this pyramid has a square base of length s and a height of b. I am really interested in the slope of the side (?). If the pyramid is limited by some absolute height (as I estimated above), then the slope angle will depend on the length of the side. Using simple trig, I can write:

Now suppose b is a constant value. This would mean that if you want a bigger base for your epic pyramid, you would need a smaller sloped side. Here is a plot of the slope angle as a function of the width of the base (assuming you have a constant height):

Ok, this clearly is not the way to go. If this model were true, why wouldn’t ever pharaoh on the block build the tallest pyramid. Then the cool pharaohs would just make the base bigger. That doesn’t happen. Oh, maybe some just didn’t have enough money. Well, here is a distribution of the heights of different pyramids in Egypt (from Wikipedia’s list of Egyptian pyramids).

So it seems that most of the pyramids aren’t all that tall anyway. Probably the limitation on height was the amount of money. Or maybe there was an inverse proportional relationship between the height of the pyramid and the size of part of the pharaoh’s body. You know what they say about big pyramids?

What if it’s not just about height?

Let me move on. What if it isn’t about the height of the pyramid, but rather the average pressure at the bottom of the pyramid. This might seem to be reasonable. A stone block on the inside of a pyramid will likely behave different than a free standing block. As a block is squeezed vertically, it should slightly expand horizontally. For interior blocks, they don’t expand horizontally the same because of interactions with the blocks next to them.

Just to be clear, I am making the assumption that the pressure at a given level in a pyramid is the same on the edges as it is in the middle. Maybe this is unrealistic, but I am going to do it anyway.

First, what is the volume of a pyramid? I will need this to calculate the weight of the rock (if I know the density of rock). Off hand, I don’t know the volume of a pyramid. Oh, sure, I could look it up – but I don’t want to do that. That would be like saying:

“hey, let’s hike to the top of this mountain! Oh wait, you have a picture of what it looks like from the top? Oh the internet? That will do. Cancel the trip.”

It is the journey that I enjoy, not the destination.

Pyramids are sort of a strange shape. How will I calculate the volume? What if I take horizontal slices of the pyramid and find the area of each slice. Then I just have to add up all of these areas. Here is a picture of what I mean.

As I move closer to the top of the pyramid, the area of this thin slice gets smaller. If I can find the area of this slice as a function of height, it will be easy to add up an infinite number of infinitely thin slices. This is, after all, the key idea in an integration.

But how do I get the area of the slice? Let me draw a picture looking at the pyramid from the top down.

Here, I lined up the edges of the pyramid’s slopes with the x- and y-axes. I am calling a the distance from the center of the pyramid to the corner. I will need this later. The dotted line square represents some arbitrary slice. How big is that slice? Well, if I know thee x value for that slice, then the area will be the length of that diagonal squared. This would be:

The square root of 2 comes in from the 45-45-90 triangle that is formed. The length of one side of the slice is the hypotenuse of this triangle. Fine, but I need this area in terms of y, not x. There is a relationship between these two variables. The line the forms the slope of the edge of the pyramid is just the equation of a line. Here is a side view of just one of those edges.

I have add the equation of the line that forms the edge of the pyramid. Remember that a is not the side of the pyramid, but rather the distance from the center to the corner. Now, let me solve that equation for x:

This means that I can get the area of my slice in terms of y:

From that, I can get the volume of that thin slice by just multiplying by its height (dy) to get:

And to find the total volume, I just need to add up all these slices. This would be the integral:

Now, I just need to change back from a to s, this would be:

Now that I am at the top of the mountain, let me check that picture to see if I am at the same peak. Yep, the same.

Back to real pyramids. How do I calculate the pressure in the rocks as a function of height? It will be the volume of the pyramid above that point (times the density and gravitational field to get weight) divided by the area at that height. I already have the area as a function of height from above. So, the pressure will be:

I made up some notation here. I am calling V(y+) the volume of the pyramid above the value y. The volume of pyramid above level y will be the area at that level multiplied by (1/3)(b-y) where (b-y) is the height of this part of the pyramid (which is itself also a pyramid). So, I can write the pressure as a function of y:

I really didn’t need the pressure as a function of height, but I did it anyway. A couple of quick checks:

• Are the units correct? Yes. Remember the pressure due to depth in water is ?gh – so this is the same.
• What is the pressure at the top? If I put in y = b, I get zero. Great.
• There is a problem though. This model says the pressure at the bottom is independent of the size of the base. So, you could just build a super skinny pyramid and be just as tall as your neighbor’s wide-base one. That just doesn’t seem right.

Obviously the greatest pressure will be at the bottom, but something doesn’t seem right.

Back to the Bent Pyramid

Just to be clear, the bent pyramid does have a name. It is called The Southern Shining Pyramid (or so Wikipedia says). If indeed the angle on this was changed because of crushing rock, then I can assume the original angle is beyond the compressive strength of rock. That pyramid had a base length of 188 meters and a height of 105 meters – but it is bent. The angle on the lower part is 54.84°. If they had continued with this angle, the height would be 133.5 meters. What is the pressure at the bottom of this pyramid? Let me use a density of limestone at 2500 kg/m³.

This pyramid is attributed to the pharaoh Sneferu. It turns out that there was a similar pyramid built by Sneferu. It is just as tall (105 meters) but has a bigger base. Actually, it has the same slope as top of the bent pyramid. If the pressure model I calculated is correct, then he could have built a pyramid just as tall with the steeper angle. Maybe there is some aesthetic reason to have a bigger base – but maybe it is a structural reason.

What if the steeper angle of 54.84° wouldn’t work, but 43.37° does? This would mean the size of the base does matter. How about I introduce an extra factor? What if the pressure at the bottom is something like this:

I am not happy with this. But what can I do? How about another graph. Here is a plot of the height vs. the base length for all the Egyptian pyramids.

Looks pretty linear – shouldn’t I add a linear regression line here? No. Why? Because I am still upset with my failure. Also, this would only be useful if I assumed that all these pyramids were built as tall as they could possibly be.

I guess I never answered the question

How tall can you build a pyramid? Based on my assumptions, it looks like around 140 meters. How wide would it have to be? It doesn’t matter. I now have a bad taste in my mouth. Surely, I did something wrong. I guess it is a good thing I am not a structural engineer.

It still seems like I am missing something.  I just seems like the pressure at the bottom should depend on the size of the base.