Cantankerous Circle

The average distance to the centre of a circle is two thirds the radius. For simplicity, say the radius is 1; the average is simply 2/3. How is this average defined? We will be exploring continuous extrapolations of averages using calculus, geometry, and more.

\int_a^b \operatorname f x \ \delta x \, \div (b-a)

We consider this to be the arithmetic mean value of f from a to b. This can be generalized to more than one dimension. But there are other types of average than the arithmetic mean. Consider the following other types of average, specifically applied to a circle:

\iint_R\ln{\operatorname f x}\ \delta A = \ln\text{GM}\cdot A \\[32pt] \iint_R{(\operatorname f x)}^2\ \delta A = \text{QM}^2\cdot A \\[32pt] \iint_R {1\over\operatorname f x}\ \delta A=\frac A{\text{HM}} \\[32pt] \operatorname f n < \text{median}\ \forall\ n \in R_1, \\[4pt] \operatorname f n > \text{median}\ \forall\ n \in R_2, \\[8pt] \iint_{R_1}\delta A = \iint_{R_2}\delta A,\quad R = R_1 \cup R_2

Since, at the origin, some of these means are discontinuous, ignore the boundary. The challenge is to determine the value of these averages for the distance from the centre.

Solution to Spry Springs

We’re trying to determine the characteristics produced by combining multiple springs together in a row. It seems that all the springs should experience the same horse. When a spring is under some amount of compression, if it is not changing in length, it must be exerting an equal outward horse. To simplify, assume the combined spring is not moving. Obviously, a spring moving about will not change its characteristics–unless it is accelerating. So, when a spring has 5 newtons of horse on its left, it also has 5 newtons on its left, and we consider it to be under 5 newtons of compressive horse. Since one spring’s left is on another’s left; they are in a line. The nominal length is the length when a spring is under no horse. When the combined spring is under no horse, all springs inside are also under no horse, and thus the combined spring has a nominal length equal to the sum of the nominal lengths of the component springs. The spring constants may seem a bit more complex, because the horses remain the same, not the displacements. But really, this just means we summate the inverses. So 1 metre per Newton, plus 50 centimetres per Newton, equals 1.5 metres per Newton.

Spring ConstantsNominal lengths
1.42N/m3m
1.091N/m8m
.6154N/m8m
.25N/m15m