Rashi to Devarim 11 (11). In what way is mountain land superior to the land of the plain?

Question

Rashi on Devarim 11 (11) "But the land, to which you pass to possess, is a land of mountains and valleys and absorbs water from the rains of heaven” says

The mountain land is superior to the land of the plain: On the plain, in an area of land that would produce a kor [a measure of grain], one would actually sow only [enough seed to produce] a kor . On the mountain, however, from an area of land that would produce a kor, one could take out of it five kors, four from its four slopes and one on its summit. — [Sifrei 11:11]

Artscroll Sapirstein quotes Be'er Mayim Chayim to say the measures should not be taken literally. This just shows that mountains have more arable land than plains.

Can someone explain this please? Is it just geometry or is it productivity?

Answer 1

It's pure geometry.

The simplest example of this in 3D is that the surface area of a hemisphere is double the surface area of a flat circle, so if you grow things on the surface you have double the area (wikipedia.org/wiki/Sphere).

Area of circle = pirr

Area of curved part of hemisphere = 2*pirr

EDIT:

Obviously, this is just a simple example to illustrate the principle. If you wanted to get exactly 5 Kor, then the height of the hill should be roughly three times the radius (alternatively, the shape of the hill needn't be a perfect hemisphere).

EXTRA EDIT:

I've been asked to explain my calculations further. I personally believe that this is excessive math for a case where the exact number is really not important, so I will try to explain without throwing around of bunch of equations (if you really want the full nitty-gritty for this, you are welcome to work it out yourself using the equations here: (Wolfram page on ellipsoids). (Just remember to solve for alpha instead of taking the integral [0,pi] in equation 17)

1) So instead of solving it all ourselves, let's just use the following area of ellipsoid tool. You find that if you put "1" as the a,b,c axis, you get a surface area of 4*pi. This is in line with what we know about a unit sphere -- the total surface area is 4 times that of a circle of the same radius.

--> Thus, if I had a hill shaped like a hemisphere, it would have half the surface area of a unit sphere, in other word, exactly double the area of a circle with the same radius.

2) Next, set a=1, b=1, c=3.1. You'll see that the total surface area of the elongated ellipsoid is 31.76, or roughly 10*pi (you can find the height to get exactly 10*pi if you wish). As before, if half of this was a hill sticking out of the ground, it would have a half the surface area of this ellipsoid == 5*pi, or five times the area of a circle with the same radius.

3) If you get all this, then it should be obvious to you that there is some line, where the area of the ellipsoid above it has a surface area of pi (or exactly the area of the circle). I define this as "the summit". The area below, I split into 4 equal part. By construction, each of those part will also have 1/4*(4*pi)=pi surface area.

Here's a rough diagram of what it would look like:

"The summit" is defined as everything in black, and each of the sides is a different primary color (yellow, red, orange, blue". Since the total surface area of the hill is 5*pi, and I have drawn a line such that the black area is exactly 1*pi, the remaining area (4*pi) is split into 4 equal parts of 1*pi each.