Sure, human scaled patches of the Earth’s surface can be approximated by a similarly sized patch of a plane, but if we’re talking about tiling the entire surface of the Earth with buildings, it can actually be done using twelve pentagons or twenty isosceles triangles. We just need buildings whose footprints are roughly 1/12th and 1/20th the Earth’s surface respectively.
For the pentagon, that’d be around 510.07 × 10^12 m² divided by 12 = 42.505 × 10^12 m². With the Pentagon building having seven floors, one such building would have roughly 297.541 × 10^12 m² of floor space.
For the triangle, that’d be around 510.07 × 10^12 m² divided by 20 = 25.503 × 10^12 m². Assuming this building has seven floors like the Pentagon building does, it’d have roughly 178.524 × 10^12 m² of floor space.
The good thing about dividing into triangles, however is that it can be subdivided into four similar isosceles triangles, which can be applied recursively down to a far more realistic scale.
Doing that, we can subdivide the original triangles sixteen times yielding the following:
25.503 × 10^12 m² / (4^16) = 5.937 × 10^3 m²
And since the area of an isosceles triangle is equal to s²(√3)/4 we can rearrange things to find the side length of a compound with area of 5 937m²
s = √(4A/(√3)) = 117.103 m
I think that’s a human-enough scale for buildings.
In total, there’s 85 899 345 920 such buildings, covering the Earth.
If one such building has 7 floors, it’d have at most 41 559 m² of floor space.
EDIT:
Hit enter too soon. Additional proofreading.
Damn, I discovered a small mistake in the calculations partway through. Corrected.
That makes me think of the last episode of Severance, where two characters who have lost their memory of the external world are debating whether the equator is a continent or a continent-sized building.
But the Earth isn’t a plane.
Sure, human scaled patches of the Earth’s surface can be approximated by a similarly sized patch of a plane, but if we’re talking about tiling the entire surface of the Earth with buildings, it can actually be done using twelve pentagons or twenty isosceles triangles. We just need buildings whose footprints are roughly 1/12th and 1/20th the Earth’s surface respectively.
For the pentagon, that’d be around 510.07 × 10^12 m² divided by 12 = 42.505 × 10^12 m². With the Pentagon building having seven floors, one such building would have roughly 297.541 × 10^12 m² of floor space.
For the triangle, that’d be around 510.07 × 10^12 m² divided by 20 = 25.503 × 10^12 m². Assuming this building has seven floors like the Pentagon building does, it’d have roughly 178.524 × 10^12 m² of floor space.
The good thing about dividing into triangles, however is that it can be subdivided into four similar isosceles triangles, which can be applied recursively down to a far more realistic scale.
Doing that, we can subdivide the original triangles sixteen times yielding the following:
25.503 × 10^12 m² / (4^16) = 5.937 × 10^3 m²
And since the area of an isosceles triangle is equal to s²(√3)/4 we can rearrange things to find the side length of a compound with area of 5 937m²
s = √(4A/(√3)) = 117.103 m
I think that’s a human-enough scale for buildings.
In total, there’s 85 899 345 920 such buildings, covering the Earth.
If one such building has 7 floors, it’d have at most 41 559 m² of floor space.
EDIT:
Hit
entertoo soon. Additional proofreading.Damn, I discovered a small mistake in the calculations partway through. Corrected.
That makes me think of the last episode of Severance, where two characters who have lost their memory of the external world are debating whether the equator is a continent or a continent-sized building.