[From Bruce Abbott (2018.04.02.1735 EST)]

Correction: The superscript code somehow got lost in the formula for R: it reads R = V3/D but should read V^{3}/D. I have corrected it in the text below.

Bruce

[From Bruce Abbott (2018.04.02.1725 EST)]

For those of you who may have been taken in by my post (reproduced at bottom below), April Fool!

But there was a serious purpose in posting it. I had presented the following equation:

V = 2A^{1/2}π^{1/2}/T

This form parallels Rick Marken’s analysis (Marken & Shaffer, 2017) of the formula for computing the radius of curvature at a given point as the point traces some curved path. That formula is as follows:

R = V^{3}/D, where V is the velocity of the point along the curve and D is computed from the velocities and accelerations of the point along the X and Y directions. Solving for V yields:

V = R^{1/3}/D^{1/3}, which is similar in form to my equation above relating velocity to circle area.

Converted to log form, my equation becomes:

log V = log 2 + ½ log A + ½ log π – log T

And Rick’s becomes:

log V = 1/3 log R - 1/3 log D

Empirically, a power-law relationship is found between V and R. This is obtained by doing a linear regression analysis in which log R is the predictor variable and log V is the criterion variable:

log V = a + b*log R,

where a is the intercept of the best fitting line relating log V to log R and b is the slope. Empirically, b has often been found to be close to 1/3, which would be the power exponent of R when the equation is expressed in its original (non-log) form. However, empirically the exponent can and often does depart from 1/3.

Rick notes that this regression analysis omits log D as a predictor variable. He asserts that it is this omission that allows the exponent to depart from 1/3, and that if log D is included, then the mathematics of the calculation of the radius of curvature necessarily impose a 1/3 exponent in the formula relating V to R.

In reaching this conclusion, Rick has fallen into a mathematical fallacy, because the radius of curvature itself depends on D. It is the division of V^{3} by D that makes R a pure measure of distance, uninfluenced by the velocity of the point that moves along the curve.

To better understand this assertion, consider my parallel construction “proving” that the velocity of a point traversing a circle depends on the circle’s area. Obviously this is nonsense – you can go around the circle at any velocity whatsoever no matter what the area of the circle may be. Yet my equation seems to indicate that the velocity of the point is mathematically determined by the circle’s area. How can this be? Well, let’s have a look at my original equation again:

V = 2A^{1/2}π^{1/2}/T

The key is to understand the role of T. Let’s move T to the other side of the equation:

VT = 2A^{1/2}π^{1/2}

V is the velocity of the point in, say, cm/second. T is the time in seconds required to traverse the circle at this particular velocity

V cm/s times T s = cm, a distance. In fact it is the distance around the circle ordinarily referred to as the circle’s circumference. I’ll call it C.

So the formula can now be rewritten as

C = 2A^{1/2}π^{1/2.}

I think you will agree that the area of a circle depends on its circumference. The two measured variables, V and T, were used as *a means* of determining the circumference. The faster you go around the circle, the sooner you will complete the circuit, so the two variables vary inversely in such a way that the circumference that you determine from V and T will be the same for a given circle no matter what speed you go around it at.

By exactly the same logic, in the formula for the radius of curvature, the denominator D of the formula compensates for the different velocities at which the point may be moving along the curve, so that when V^{3} is divided by D, the result, the radius, is purely a measure of distance that will be the same value no mater what the velocity of the point was observed to be when the curve was drawn. Because R as computed is independent of the velocity used to compute it, the empirically *observed* relationship between V and R could be any relationship whatever, and the empirically observed power-law relationship is not an artifact of the way in which R is computed.

It’s as simple as that.

Bruce

[From Bruce Abbott (2018.01.01.1145 EST)]

I have discovered a heretofore unsuspected relationship between the area of a circle and the speed with which the circumference of the circle is traced.

A point traces the circumference of a circle at velocity V, completing the transit in time T. Then the area A of the circle is given by

A = V^{2}T^{2}/4π

Solving for V:

V = 2A^{1/2}π^{1/2}/T

Notice that the velocity with which the circle is traced is a power-law function of the circle’s area, with an exponent of 1/2. Converting to logarithms:

Log V = log 2 + ½ log A + ½ log π – log T

Where it *appears* that velocity of the circumscribing point and the area of the circle are independent, this is only due to the influence of log T, which allows the relationship to depart from what otherwise would be a mathematically enforced power relationship.

It amazes me that professional mathematicians have overlooked this important fact!

Bruce