In cartesian space, the closest logarithmic fit is about y=0.22*ln(x)+0.22.

This is still not great though, it is a decent approximation for values with x<0.3, but is worse between 0.3<x<0.6. Also, the accuracy is nowhere close to 1ppm.

Ether,
are you just looking for “here’s my answer” with work shown, or could we start posting “here’s what I tried; here’s what I’ve ruled out; I’ve got some speculation on how/what needs to be done”?

To be clear, I’m fine with either, just wanted to know what parameters should be followed for posts.

The answer is: a dialog is most welcome. Please do post your thoughts and ideas.

The purpose of these quizzes is to start a conversation from which both participants and readers can gain insight. The journey is as important as the goal.

Ok, so here’s what i’ve got:
The problem statement says a coordinate system exists, but it doesn’t state if it’s Cartesian, or polar, or maybe even something else. The equation is “simple” but something simple in a polar coordinate system can be complex in a Cartesian one. It’s also undetermined what the origin of said system is.

I know it’s not a circle (polar, r=constant). The perpendicular bisectors of each adjacent point segment aren’t close to intersecting at a single point.

I tried rotating the points to get a good fit on a parabola, which would have a relatively simple equation, assuming the origin was on the vertex. The best I got was around 142.5 degrees clockwise, but that didn’t quite match up.

Also, the radius of curvature appears to be increasing from the starting point until the ending point.

Other observations:
The points are not equidistant in spacing. The minimum distance appears to be around the 5th or 6th point.
I couldn’t find a rotation which places the x or y coordinate in equal spacing in Cartesian coordinates.

I haven’t had time to work this out thoroughly, but if the points are equally spaced in the theta direction of a polar coordinate system, I think you could derive the coordinate system origin.

with playing around with a spreadsheet, I think it’s in the vicinity of 0.86,-0.51 and the theta interval is 0.025 radians.
as far as taking a swag at the final answer, since it’s supposedly “simple” I’m going to guess r=theta.

I had the same idea as MechEng83 last night, then started with his solution this evening (also in a spreadsheet, which I cleaned up just a bit and posted). My goal was to minimize the ratio of the standard deviation of the angle with its mean. (Actually, I used the sine of the angle, much easier to calculate with a cross product.) The point kept marching up and to the right. I eventually hit paydirt.

The center point where all of the angular displacements are the same is
(1.16, -0.3)
At this point, the angular displacements are all exactly 1 degree (to within the precision of the numbers as given). Using the mathematical convention (x axis is 0 degrees and the y axis is 90 degrees), the points span the range from 139 to 199 degrees, inclusive.

While the radial function increases monotonically as the points are given, it appears to be somewhat concave down. Of the stock excel fits, polynomial (quadratic by default) is better than the others, but is not an exact match.

An explanation of the spreadsheet is in order. The top two rows were used to create the “origin” for the polar coordinate system. Each cell to the right of C1 and C2 was added with another power of 1/10 to create C1 and C2. This let me manually migrate towards a minimum value for the scaled deviation. Row3 just has the results I was trying to optimize on. Row 4 is headers for the table. Below this, columns A and B are Ether’s data. Column C is the distance from the origin (not used). Columns D and E are the x and y values relative to the origin specified by (C1,C2). Column F is the distance of the point from the point (C1,C2). Column G is the change in the sine of the angle from (C1,C2) to this point and the next. Column H is the angle in degrees from (C1,C2) to Ether’s point, x towards y, in the range from 0 to 360.

OBTW, I’ve been working MQ9 for the cube extension off and on. I think I have all of the terms integrated on scratch paper, but I’m still writing them up and checking. I definitely have the mean distance from the origin to an arbitrary point complete, but the random line has eight times as many terms to start, and several of those are quite a bit nastier than the distance from the origin.

In any case, I found the answer a different way:
Assuming that r=θb, we know that dr/dθ = bθb-1, so that r/(dr/dθ) = θ/b. I therefore calculated the slope of r using the numbers given. The intercept of the line with r=0 is therefore (a good approximation to) θ[sub]0[/sub], and the slope is (approximately) 1/b. My spreadsheet gave a θ[sub]0[/sub] of 2.007223652 radians, or 115.0054438 degrees, and a b=0.499973651. These did not work out exactly, but a bit of tweaking found that 115 degrees and 0.5 worked “exactly”.

Using the mathematical polar coordinate system centered on 1.16, -0.3 and rotated 115 degrees counterclockwise, the equation is simply:

r = √θ

Going back to calculate the original series:
D = [139…199]
θ[sub]1[/sub] = D * π / 180
θ = (D-115) * π / 180
r = √θ
x = cosθ[sub]1[/sub] + 1.16
y = sinθ[sub]1[/sub] - 0.3

An updated spreadsheet with a “double check” tab is attached. The maximum error in the reconstructed x and y is less than 10-14.