Geometric construction of a specific gothic rose window

Question

Christmas is coming up and I wanted to gift my boyfriend some nice laser cut gothic tracery. While browsing "The Power of Form applied to Geometric Tracery" by R. W. Billings, I came upon this construction diagram. I have some difficulty comprehending how one would find the centerpoint of the circle marked 1 on the left. Notably the centerpoint of the mirrored circle marked 2 lies slightly above the horizontal line.

Answer 1

Depends on what kind of tools you have available to you. The center of a circle can be found by drawing two chords, then constructing their perpendicular bisectors. The intersection of the perpendicular bisectors is the center of the circle. That's how you'd find it on this drawing; translating it to a physical model is harder. I'd ask the shop about whether they have those kind of geometric tools or something else. https://www.youtube.com/watch?v=kGz_7OcU4EM 2:12 - end

Answer 2

Your requirement of locating the center of the circle in question sounds more practical than theoretical to us. Therefore, in our exposition, we decided to include a practical solution as well as a theoretical solution. One needs not to fully comprehend the theoretical aspects of our solution to carry out the construction described in the practical solution, which needs the use of only circles and lines. On the other hand, in the theoretical solution, we have to use parabolas, ellipses, and hyperbolas in addition to circles and lines. To keep things short, we do not describe the theory behind the methods used. Let the radii of the circles $\Omega_0$, $\Omega_1$, $\Omega_2$, and $\Omega_3$ be $r_0$, $r_1$, $r_2$, and $r_3$ respectively.

$\underline{\text{Theoretical Solution:}}$

After constructing the circle $\Omega_0$ with center $O_0$ and radius $r_0$, we draw its vertical diameter $AB$ as shown in $\mathrm{Fig.\space 1}$. Next, draw the two radii $O_0P$ and $O_0Q$, each of which makes an angle of $60^o$ with $AB$. Construct the perpendicular at $A$ to $AB$ to intersect the extended $O_0Q$ at $C$. When we draw the angle bisector of $\angle O_0CA$, it meets $AB$ at $O_1$, which is the center of the circle $\Omega_1$. Since $O_0A$ is equal to the radius $r_1$ of this circle, we are now in a position to sketch it.

After marking the point $D$ on $O_0B$, such that $O_0D = O_0O_1$, draw the line through $D$ parallel to $O_0P$. Using $DE$ as the directrix and $O_1$ as the focus, construct a parabola shown in $\color{red}{\pmb{red}}$. Now, using $O_0$ and $O_1$ as its foci, we need to construct the ellipse shown in $\color{green}{\pmb{green}}$, so that it passe through $A$. The parabola and the ellipse meet at $O_2$ and that is the center of the circle $\Omega_2$. When we join $O_1$ and $O_2$ with a line segment, it cuts the circumference of the circle $\Omega_1$ at $G$. With that we can obtain the radius $r_2$ of the circle $\Omega_2$ as $O_2G$, which let us draw the circle.

Finally, we construct the hyperbola shown in $\color{blue}{\pmb{blue}}$ with foci at $O_1$ and $O_2$ and passing through $G$. It intersects $O_0P$ at $O_3$. Draw the line segment $O_2O_3$ to cut the circumference of the circle $\Omega_2$ at $H$. The radius of the circle $\Omega_3$ is equal to $O_3H$ and its center is located at $O_3$.

The above described construction yields the following results, which we need to methodize our practical solution given below. $$r_1 = 0.46410161514\times r_0\tag{1}$$ $$r_2 = 0.16401253984\times r_0\tag{2}$$ $$r_3 = 0.10808712968\times r_0\tag{3}$$

$\underline{\text{Practical Solution:}}$

We assume that $r_0$ is given. So, calculate the radii of the smaller circles, i.e., $r_1$, $r_2$, and $r_3$, using the equations (1), (2), and (3) respectively.

After constructing the circle $\Omega_0$ with center $O_0$ and radius $r_0$, we draw its vertical radius $O_0A$ as shown in $\mathrm{Fig.\space 2}$. Mark the point $O_1$, such that $O_1A = r_1$ and draw the circle $\Omega_1$ using $O_1$ and $O_1A$ as the center and the radius respectively.

Draw the arc-1 taking $O_0$ as its center and $r_0+r_2$ as its radius. In a similar vein, construct the arc-2 using $O_1$ as its center and $r_1+r_2$ as its radius to intersect the already drawn arc-1 at $O_2$. Now using $O_2$ as the center, draw the circle $\Omega_2$ with the radius $r_2$.

By drawing the arc-3 with center $O_1$ and radius $r_1+r_3$ to meet the segment $O_0P$, we can locate the center $O_3$ of $\Omega_3$. Finally, complete the diagram by drawing the circle $\Omega_3$ with the radius $r_3$.

$\underline{\text{A Request at OP:}}$

We would like to request OP to add an image of the laser cut gothic tracery to your post, if you were able to produce it using the method decribed in our text or any other method.