## The Pyramids Aren’t Big Enough

Posted by Karl Sharman on May 29, 2011 – 11:52 pm

In the Smithsonian Institute there is a miniature model of one of the Pyramids at Gisa (in Egypt). It is 48 inches in height.

For the World Fair in New York 2020 they would like to display it, but it was thought to be too small so they decided to build a scaled-up model of the Pyramid out of material whose density is (1 / 7) times the density of the material used for the miniature.

If the mass (or weight) of the miniature and the scaled-up model are to be the same, how many inches in height will the scaled-up Pyramid be?

May 30th, 2011 at 2:56 am

I think the new height would be 336 inches (7*48)

May 30th, 2011 at 5:59 am

The height will almost be doubled (I’ll keep the exact answer to myself for now).

May 30th, 2011 at 9:32 am

Density is Mass/Volume

Volume is Length^3

To have the same mass with one 7th of the original density, you need 7 times the volume, multiplying volume by 7 means multiplying length by 7^(1/3), which is just a bit under 2.

May 30th, 2011 at 4:07 pm

Agree with Karys.

For original model, Vo = lwh/3, for scaled-up model Vs = nl x nw x nh/3; on another hand, Vs = 7Vo.

So, 7Vo = n^3Vo, and proportionality coefficient is 7^1/3

Then the new height is 48in x 7^1/3 ~ 91in

May 31st, 2011 at 10:51 am

Since density is mass over volume and the volume of a pyramid is the Area of the Base * Height * 1/3 and since we know that the new density is 1/7 times the old density we find the formula to become:

1/7 * m/V1 = m/V2

Where V1 is the volume of the original pyramid and V2 is the volume of the new pyramid. This becomes:

1/(7*V1) = 1/V2

which allows me to conclude that the new height would be:

h = (336 * A1) / A2

Where A1 is the area of the base of the original pyramid and A2 is the area of the base of the new pyramid. We cannot conclude the exact height since we do not know the volume of the original model and we do not know the area of the bases.

May 31st, 2011 at 2:12 pm

Hi Evan. See Tatiani’s post. If you double all linear dimensions, then the volume goes up by 2^3 = 8. It doesn’t matter what the shape is, but it’s easiest to see that with a cube, then imagine that any shape could be approximated by a set of (very small) cubes.

In the posted problem, the volume will increased by a factor of 7, so every linear dimension must be increased by a factor of cube root 7 => new height = old height * 7^(1/3) => 48 * 1.9129… = 91.82…

June 1st, 2011 at 6:38 am

336 inches

June 2nd, 2011 at 7:47 am

Mass = Density × Volume

and

Volume of model / Volume of miniature = (H of model / H of miniature)3

In the above equation, H is the characteristic dimension (say, height).

If the mass is to be the same, then density is inversely proportional to volume. Also, the volumes are directly proportional to the cubes of the heights for objects that are geometrically similar. Therefore, the heights are seen to be inversely proportional to the cube roots of the densities. Thus,

Height of model = Height of miniature × (Density of miniature / Density of model)1/3 or

Height of model = 48inches × 7^(1/3) = 91.8 inches (approx).

Sorry Ben and Inja – everyone else appeared to spot that it wasn’t just going to increase the height by a factor of 7…

That’s Tom for you!