**THE THRONE OF INIQUITY**

**Durt Fibo
**

**May 26, 2024**

Everyone from hut dwellers to office workers knows that the obligatory five-legged chair is one of the worst impediments to civilization. It blocks both movement and creative labor, and angers bureaucrats to such a degree that they feel compelled to steal money from the citizenry. But few people know the where and whyfore of its excogitation. –I can state from personal observation that it was instituted in Germany, to prevent the natives from keeling over after their traditional beer-lubricated lunch hours.

The exact rationale for the fifth wheel atrocity is explained by Herr Professor Doktor Wasserkopf, flounder and Oberdada of Der Koolschrank, as follows:

“Assume that the chair has to support a constant weight W, and that we want a constant stability. Stability is determined by the shortest “tipping distance” D. For a radial distance R, a chair with n legs has D=Rcosπn

So we can define a “stability factor” S=1Rcosπn

Thus, for constant S we get R∝1cosπn(1)

Next, we look at the stress on each leg. The stress will be greatest when the tipping torque Γ

is directly in line with just one leg. At that point, Γ=W⋅R

Now we want to calculate the shape (section) of the leg that can support this torque. The maximum stress σ for a rectangular beam of width w and height h is proportional to wh2, and the mass of the leg of length R is whRρ; if we assume a constant aspect ratio wh, then mass is proportional to area times length:

m∝h2R(2)

where the first term is a function of the strength, and the second term a function of the stability.

Similarly, for given torque W⋅R we can write the bending stress as

σ=MyI where M is the bending stress, y is the perpendicular distance to the neutral axis, and Ix is the second moment of area about the neutral axis x. For a rectangular section, y∝h4

For constant σ, the maximum will occur at the outer edge of the beam where y=h2, leading to h3∝W⋅R

For given weight W, it follows that h∝R1/3(3). Substituting (3) into (2) we get m∝R5/3

For constant breaking strength, we get the total mass of n legs:

M=n⋅m∝nR5/3

For constant stability, we use (1) to obtain M∝ncos53πn

We can evaluate this for n between 3 and 7, and obtain M as a function of the number of legs:

n=3: 9.524

n=4: 7.127

n=5: 7.118 <— lowest value

n=6: 7.625

n=7: 8.329

This shows that indeed the structure with five legs needs the lowest mass to support a certain torque – if we can equate “mass” with “cost”, and stability is indeed the main driver, this proves that a chair with five legs is optimal, and can now support any amount of dopplebock and a Räucherkäse schnapps.”