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Very interesting! It's not clear to me why it's so obvious that doubling the height of a pyramid while keeping the side constant would double the volume of the

Very interesting! It's not clear to me why it's so obvious that doubling the height of a pyramid while keeping the side constant would double the volume of the pyramid. The reason it is not obvious to me is that the proportions of the pyramid would change. The author of the post seems to think this is so obvious that it doesn't bear explanation. Can anyone explain to me why this works?

ilovepyramids 7y ago
Okay, maybe I can accept this readily. Easy enough to prove that doubling the height of a convex 2-d shape will double the area if you stretch all of its vertical proportions to double. If that shape is on a graph, the x axis is what it normally is, and the internal height of the shape for a given value of x is given by f(x), when you double the height it will be given by 2*f(x). The integral of a function times a constant is that constant times the integral of the function, so the area inside the shape would double. I don't remember multivariable at all, would someone be willing to offer a 3-d proof?
ColinWright 7y ago
Imagine that the pyramid is made up of a very large number of very thin slices. Now take two copies of each slice ... they are thin enough that you really haven't changed the shape, except to double the height. You've doubled the height, and doubled the volume. It's a little like performing a riffle-shuffle on two identical packs of cards. If they aren't simply cuboids, but are identically interestingly shaped, riffling them together results in something that's the same shape, but twice the height. Does that help?
- ilovepyramids 7y ago
  Took me a moment. Interesting! So you're saying: Divide the pyramid into very thin slices, parallel to the base (horizontal). Above each slice, insert a duplicate of that slice. If using infinitely small slices, you have effectively doubled the height of the pyramid, and obviously you have doubled its volume. Yes, that makes sense, and I think it's very clever! Thank you.
chmod775 7y ago
Its not specific to pyramids - its true for stretching any shape in 2D and 3D or lines 1D. Stretch it to twice its former size across a single axis and you get double the length/area/volume. It becomes obvious in 2D if you look at how the area of a rectangle is calculated (x*y) and imagine your shape is just projected/drawn on that rectangle and stretches with it.
- ilovepyramids 7y ago
  I did consider this, and it seemed to make it feel a little clearer—but when I first posted, I didn't feel it was obvious that the fraction of the rectangle occupied by the triangle would remain constant if you stretched it. But I think the other replies prove that that is true. Thank you.
reidacdc 7y ago
I stumbled at that point too, but satisfied myself with a slightly more general argument than the first reply. Imagine the pyramid is made up of a bunch of small but finite-size cubical blocks. Extending a block along any single axis doubles its volume. Extending all the blocks vertically at the same time doubles the height and volume of the pyramid, because the volume is just the sum of the volume of all the blocks. Mentally pass to the limit of small blocks, and you're there, or at least, I convinced myself I was there.
- ilovepyramids 7y ago
  This feels like a perfectly intuitive 3d proof to me. Thank you.
deleted 7y ago
[deleted]

ilovepyramids 7y ago

Okay, maybe I can accept this readily. Easy enough to prove that doubling the height of a convex 2-d shape will double the area if you stretch all of its vertical proportions to double. If that shape is on a graph, the x axis is what it normally is, and the internal height of the shape for a given value of x is given by f(x), when you double the height it will be given by 2*f(x). The integral of a function times a constant is that constant times the integral of the function, so the area inside the shape would double. I don't remember multivariable at all, would someone be willing to offer a 3-d proof?

ColinWright 7y ago

Imagine that the pyramid is made up of a very large number of very thin slices. Now take two copies of each slice ... they are thin enough that you really haven't changed the shape, except to double the height. You've doubled the height, and doubled the volume. It's a little like performing a riffle-shuffle on two identical packs of cards. If they aren't simply cuboids, but are identically interestingly shaped, riffling them together results in something that's the same shape, but twice the height. Does that help?

ilovepyramids 7y ago
Took me a moment. Interesting! So you're saying: Divide the pyramid into very thin slices, parallel to the base (horizontal). Above each slice, insert a duplicate of that slice. If using infinitely small slices, you have effectively doubled the height of the pyramid, and obviously you have doubled its volume. Yes, that makes sense, and I think it's very clever! Thank you.

chmod775 7y ago

Its not specific to pyramids - its true for stretching any shape in 2D and 3D or lines 1D. Stretch it to twice its former size across a single axis and you get double the length/area/volume. It becomes obvious in 2D if you look at how the area of a rectangle is calculated (x*y) and imagine your shape is just projected/drawn on that rectangle and stretches with it.

ilovepyramids 7y ago
I did consider this, and it seemed to make it feel a little clearer—but when I first posted, I didn't feel it was obvious that the fraction of the rectangle occupied by the triangle would remain constant if you stretched it. But I think the other replies prove that that is true. Thank you.

reidacdc 7y ago

I stumbled at that point too, but satisfied myself with a slightly more general argument than the first reply. Imagine the pyramid is made up of a bunch of small but finite-size cubical blocks. Extending a block along any single axis doubles its volume. Extending all the blocks vertically at the same time doubles the height and volume of the pyramid, because the volume is just the sum of the volume of all the blocks. Mentally pass to the limit of small blocks, and you're there, or at least, I convinced myself I was there.

ilovepyramids 7y ago
This feels like a perfectly intuitive 3d proof to me. Thank you.

deleted 7y ago

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