I always find this Siberian diamond mine to be quite impressive. According to Wikipedia, the Mir mine is 525 meters deep with a radius of 600 meters (at the top). It's not the largest excavated hole in the Earth, but it's a nice cone shape.
There are some cool questions to consider with this mine. What if they want to make it 10 meters deeper? How much dirt would they have to remove?
Before I answer any questions, let me derive the formula for the volume of a cone. Why? Why not. Well, a previous post used the formula of a cone as well, so I thought I should derive it.
Volume of a Cone
Warning: Calculus required. You have been warned.
Here is a my cone. It has a radius of R with a height of h.
In order to find the volume of this shape, I am going to break it into many different pieces. I want to choose the shapes of the pieces such that I can find the volume of each little piece. In this case, I will break the cone into really thin disks. Each of these disks will have a volume (since it is just part of the total volume, I will call it dV).
I put the height of these disks as dy - in case that wasn't clear. Now, the next step will be to add up all these thin horizontal slices of the cone as the thickness size goes to zero (this is the essence of integration). The problem is it that the radius of the disks changes as the slice gets higher in y value. I can easily solve this problem by writing the radius of the disk in terms of the variable y. You can see that I already drew a line in that shows the edge of the cone. From this function, I can get a value of y in terms of x. Since the cone has its vertex at the origin, the radius of each horizontal slice will be the x value of that function. That means I can write the volume of the slice as:
Now that I have dV in terms of just y, I can add up all these super-thin slices of the cone. This becomes the integral:
There you go. That's the same answer that you will find in your table of volume formulas. See, it wasn't so hard. Now, we should still check some things. Does it have units of volume (m3)? Yes. What happens when h gets smaller? The volume gets smaller - that is good. The same is true for R. One other thing - this formula does not depend on the orientation of the cone. That is what we would expect.
Speaking of cone orientation. What if I had put the base of the cone in the x-z plane and at the origin (so the pointy part was pointing up)? In this case, my method would be very similar. The biggest difference would be with the equation I wrote to define the edge of the cone. If the vertex is at the origin, this y equation would have a zero y-intercept. The other way, there would be a different slope for the equation with a non-zero intercept. IN the end, you would arrive at the same formula, but it would be a little more algebra.
Volume of a Mine.
Back to the Mir mine. If I use the dimensions listed, how much dirt had to be removed to dig this thing? All I need to do is to put a radius of 600 meters with a height of 525 meters and I get a volume of 1.98 x 108 m3. Surely that is a lot of dirt. However, that isn't a very interesting question.
Digging Deeper
Suppose there was a standard cylindrical shaped well that you were digging that was 5 meters deep with a radius of 1 meter. It would be simple to calculate the volume of dirt needed to dig this well since it would just be a cylinder shape. With these values, I get a dirt volume of 15.7 m3. Now, what if I wanted to make it twice as deep (10 meters)? Well, I would just need to dig out another 15.7 m3 of dirt. Not a problem.
Why is the Mir mine a cone instead of a cubic rectangle or a cylinder? A 10 meter deep cylinder might be difficult to dig, but I suspect it's at least possible. What about a 500 meter deep cylinder? Again, maybe possible. But there is a problem. What if you want to get a truck down there to carry out the dirt? You can't really drive a truck down a vertical wall. The Mir mine is sloped to accommodate a spiraling road down to the bottom.
There could be another problem with a vertical wall - stability. Depending on the type of dirt, a vertical wall might collapse. Next time you are at the beach, try digging a vertical shaft in the sand. Doesn't work too well, does it? So, I assume the Mir mine has a particular wall slope to allow the trucks to get to the bottom and to prevent wall collapses.
Does this mean it has to be a cone shape? No. My guess is that the cone shape gives the shortest path of road to get to the bottom. That's just a guess though.
Now for the fun question: If they want to dig the Mir mine just 10 meters deeper, how much dirt would they have to remove?
Let me assume that the slope of the cone must be the same value as it is right now. That means that the ratio of the depth of the cone (which oddly I am calling h) to the radius at the top (R) is constant.
Where k is just some constant. If I use the numbers for the Mir mine, I get a k of 525/600 = 0.875 (with no units). Now I will re-write my cone volume formula so that it only depends on the depth.
Here is a simple way to answer the question. If I want to make the mine 10 meters deeper, I can just subtract the volume for a 525 meter deep mine from that of a 535 meter mine.
Check that out. In order to go 10 meters deeper, you would need to remove almost as much dirt as you did to get to 525 meters. This is because the volume is proportional to the cube of the depth (which is a result of a constant slope side). Here is a plot of the volume of dirt as a function of depth.
You can see the volume of dirt you need to remove really gets big for really big mines.
What About the Size of the Hole At the Top?
Let's say you want to make the mine twice as deep - say 1050 meters. First, this would require removing 1.4 x 109 m3 of dirt. That's a lot of dirt. But what about the size of the hole? If it's a circle, then it would have an area of:
So, if you double the depth you would increase the area of the top by a factor of 4. Here is what that would look like. I have taken an image from Google Maps and added circles for a mine twice as deep and half as deep.
Last question: if you filled up the current Mir mine with water, how long would it take you to drink it all?
