Given the Y-axis values of a set (n > 3) of 2D-points that are known to be on a circle with an unknown center and radius, is it possible to find these points their corresponding X-axis values, if you know that the points are sampled at a constant, but unknown angle? The points are discretely sampled.
So there will be 2 circles: 1 to the left, 1 to the right of the Y-axis, that will return 2 sets of X-axis-values, that correspond to the given set of y-values.
It is known that:
- The y-values are strictly > 0
- These circles touch the Y-axis in 1 place: the first y-value (closest to the origin), where they also touch each other.
- These circles pass through the X-axis twice, but never through the origin (0,0)
- The center(s) of the circles are above the X-axis, at the same y-value, and they have the same radius, so they are mirror images.
- The first (closest to the origin) x-value is always 0
Or simply put: more than half of the circles lies slightly above the horizontal axis, touching the vertical axis in 1 point above the origin.
In the image below, -only- the green dots are initially known, all else must be calculated somehow. The points of which the green dots are the vertical coordinates, are known to be on a circle (the orange circles), and the lowest green dot is never on the origin (0,0): that is the only place where the orange circle touches the black, dotted, vertical Y-axis. The green dotted segments go from the (known) green dots to the (unknown) orange points on the (unknown) circle(s), and turn into orange segments to the corresponding (unknown) red dots (X-axis coordinates).
So what is required are the (X-axis) values of the red dots, if you only know the green dots. It is important to realize that the orange dots on the circle happen to be spread out over the circle at a constant, but unknown angle. So naturally, the arc length between green dots is also constant.
For example: given the following y-values:
yval<-c(0.10000,0.19877,0.29388,0.38298,0.46388,0.53459,0.59337,0.63877,0.66967,0.68531)
The corresponding x-values would be:
xval<-c(0.00000,-0.01564,-0.04654,-0.09194,-0.15072,-0.22143,-0.30233,-0.39143,-0.48654,-0.58531)
The radius $r$: $0.63726$
The circle center at: $(-0.6353,0.05002)$ For the left circle, and inverse these values for the dotted orange right circle.
Given these data, the coordinates of the first (lowest) orange point on the circle will be (0,0.10000) (not so clear in the image because it obviously coincides with the first green dot).
If the given y-values would not come from points that are sampled at a constant angle, a unique solution would not exist, but they are, and 1 solution for the left circle, and 1 for the right should be possible to find. All help much appreciated!
ps: the values are discretely sampled cumulative values of a sinusoid with a period of 40, divided by 10, starting at phase = $pi/2$, so $90°$: in R:
startpoint<-90
n<-10
phaseshift<-9 #so period = 40
cumsum(round(cos((startpoint+((0:(n-1))*phaseshift))*pi/180)/n,5))
pps: I used this simple method to estimate a circle from 3 known points (x,y): small bit of R code included per illustration: I'm sure there are computationally cheaper ways to do it:
circleFromThreePoints<-function(x1,x2,x3,y1,y2,y3)
{
vara<-x1*(y2-y3)-y1*(x2-x3)+x2*y3-x3*y2;
varb<-(x1*x1+y1*y1)*(y3-y2)+(x2*x2+y2*y2)*(y1-y3)+(x3*x3+y3*y3)*(y2-y1);
varc<-(x1*x1+y1*y1)*(x2-x3)+(x2*x2+y2*y2)*(x3-x1)+(x3*x3+y3*y3)*(x1-x2);
vard<-(x1*x1+y1*y1)*(x3*y2-x2*y3)+(x2*x2+y2*y2)*(x1*y3-x3*y1)+(x3*x3+y3*y3)*(x2*y1-x1*y2)
varx<- -varb/(2*vara)
vary<- -varc/(2*vara)
varr<- (((varb*varb)+(varc*varc)-(4*vara*vard))/(4*vara*vara))^0.5
# x, y , r:
# (x-x1)^2+(y-y1)^2 = r^2
# h,k,r for equation: (x-h)^2+(y-k)^2 = r^2
# To plot: upp<-(((r^2)-((x-h)^2))^0.5)+k & dwn<--(((r^2)-((x-h)^2))^0.5)+k
return(c(round(varx,5),round(vary,5),round(varr,5)))
}
This question is a more specific version (sampling at a constant angle) than my previously asked, related question
There is some ambiguity in the problem statement. You say the circles are tangent at the first $y$ value, which therefore is also the $y$ coordinate of the center of each circle. But in your worked example, the first $y$ value is not the $y$ value of the center of the circle. In fact the circles in that solution would not be tangent to the $y$ axis or each other, but would intersect twice.
I did not assume that the circles are tangent at the first $y$ value. I assumed only a sequence of $y$ values of equally-spaced points along the circle that may or may not include the tangent point.
Consider four consecutive points $(x_1,y_1),$ $(x_2,y_2),$ $(x_3,y_3),$ and $(x_4,y_4),$ where initially only the $y$ values are known, with $y_1 < y_2 < y_3 < y_4.$ Choose points such that $y_4 - y_3 \neq y_2 - y_1,$ since otherwise the solution is not determined.
Since the central angles are equal, the distances between consecutive pairs of points are the same, and likewise the squares of the distances are the same, that is, $$ (x_2 - x_1)^2 + (y_2 - y_1)^2 = (x_3 - x_2)^2 + (y_3 - y_2)^2 = (x_4 - x_3)^2 + (y_4 - y_3)^2. $$
Let \begin{align} a &= \tfrac12(y_2 - y_1 - y_4 + y_3),\\ b &= \tfrac12(y_3 - y_2),\\ c &= \tfrac12(y_4 - y_1),\\ t &= \tfrac12(x_4 - x_3 - x_2 + x_1),\\ u &= \tfrac12(x_3 - x_2),\\ v &= \tfrac12(x_4 - x_1). \end{align}
Then $a,$ $b,$ and $c$ are known, whereas $t,$ $u,$ and $v$ are initially unknown.
We have the following facts: \begin{align} x_2 - x_1 &= v - u - t, & y_2 - y_1 &= c - b + a,\\ x_3 - x_2 &= 2u, & y_3 - y_2 &= 2b,\\ x_4 - x_3 &= v - u + t, & y_4 - y_3 &= c - b - a. \end{align}
Therefore $$ (v - u - t)^2 + (c - b + a)^2 = (2u)^2 + (2b)^2 = (v - u + t)^2 + (c - b - a)^2.\tag1 $$
Let $(x_m,y_m) = \left(\tfrac12(x_2+x_3), \tfrac12(y_2+y_3)\right)$ and $(x_n,y_n) = \left(\tfrac12(x_1+x_4), \tfrac12(y_1+y_4)\right).$ That is, $(x_m,y_m)$ is the midpoint of the chord from $(x_2,y_2)$ to $(x_3,y_3)$ and $(x_n,y_n)$ is the midpoint of the chord from $(x_1,y_1)$ to $(x_4,y_4).$ By symmetry of the trapezoid with vertices $(x_1,y_1),$ $(x_2,y_2),$ $(x_3,y_3),$ and $(x_4,y_4),$ the segment from $(x_m,y_m)$ to $(x_n,y_n)$ is perpendicular to the edge from $(x_2,y_2)$ to$(x_3,y_3).$ So $$\frac{x_3 - x_2}{y_3 - y_2} = -\frac{y_m - y_n}{x_m - x_n}. \tag2$$ (The condition $y_4 - y_3 \neq y_2 - y_1$ implies that neither the top nor bottom of either ratio is zero.)
But $y_m - y_n = a$ and $x_m - x_n = -t,$ so Equation $(2)$ can be rewritten $\frac bu = \frac ta,$ which implies that $$ u = \frac{ab}{t}. \tag3$$
The chord from $(x_1,y_1)$ to $(x_4,y_4)$ is parallel to the edge from $(x_2,y_2)$ to$(x_3,y_3),$ which implies that $\frac cv = \frac bu = \frac ta,$ so $v = \frac{ac}{t}$ and therefore $$ v - u = \frac{a(c - b)}{t}. \tag4 $$
Use Equations $(3)$ and $(4)$ to substitute for $u$ and for $v - u$ in Equation $(1)$. We can just look at the first equality, since symmetry ensures that the second equality will be true if the first one is true. So we have $$ \left(\frac{a(c - b)}{t} - t\right)^2 + (a + c - b)^2 = 4\left(b^2 + \frac{ab}{t}\right) . $$
This is equivalent to $$ t^4 + (a^2 + (c - b)^2 - 4b^2)t^2 - 4abt + a^2(c - b)^2 = 0. $$
Solve for $t.$ This is a quartic, so in principle it is solvable by radicals, but I would just do it numerically in practice. Once you have $t$ you can find $u$ and $v$ easily. Depending on the problem statement, it might take some additional work to set the $x$ coordinates so that the circle is tangent to the $y$ axis.
Note that the way I interpreted the problem, three $y$ values would not be enough. If there is a circle that passes through three equally spaced points with given $y$ coordinates and satisfies the other conditions, you can find another circle with a slightly larger or smaller radius that also will have equally spaced points with the given $y$ coordinates and that also will satisfy the other conditions. So you really do need four points under that interpretation.
Knowing that the first $y$ value is the tangent point, I think three $y$ values would be enough. One approach would be to label your first three $y$ values $y_2,$ $y_3,$ and $y_4,$ then set $y_1 = 2y_2 - y_3$ and proceed with the solution given above.