# Generalising a formula that calculates the area of a four-centred arch given only its width and height.

Problem: To generalise a formula that calculates the area of a four-centred arch given only its width and height.

Root of the problem: As you all know better than I, the area of a circle, an ellipse, or a segment of both, as well as the arc lengths of these shapes, can be calculated with only the width and height known. While I was reading something about arches, I came across different kinds of arches: four-centred, ogee, and gothic. There is material on how to draw certain arch types, but I couldn't locate anything on how to use parametric equations to compute their length and area. If these arches can be drawn by parameterizing, I reasoned that they might be expressed to determine their area and length given their width and height, similar to an ellipse.

My background on the subject: I am not a student, architect, mathematician, or any other type of professional. I graduated from university 17 years ago, and I appear to have forgotten a lot about circles, triangles, and trigonometry. For me, this was simply a challenge, similar to learning how to write LATEX commands for this question, however it has now turned into an obsession:)

My attempt to solve the problem: First, I draw a 4-centered arch by following this example. Below is the drawing.

From the figure we know that: $$\overset{\frown}{AL}=\overset{\frown}{BM}$$, $$\overset{\frown}{LN}=\overset{\frown}{MN}$$, $$|AC|=|CQ|=|QE|=|EB|$$, $$|LK|=|KN|=|JM|$$. Now, let the area of 4-centred arch be $$\boldsymbol{A}$$, total arch lenght be $$\boldsymbol{L}$$, $$|AC|=\boldsymbol{r}$$, $$|KL|=\boldsymbol{R}$$, $$|NQ|=\boldsymbol{h}$$, $$|QP|=\boldsymbol{x}$$, $$\angle ACL=\theta$$, and $$\angle LKN=\alpha$$ and $$|KE|=\boldsymbol{H}$$. Then the area and arch length can be calculated using the formulas given below, respectively.

\begin{align} A & = 2\bigg(\pi r^{2}\frac{\theta}{360}\bigg)+2\bigg(\pi R^{2}\frac{\alpha}{360}\bigg)-2\bigg(\frac{H(r+x)}{2}\bigg)-\frac{2xh}{2} \\ & = \pi r^{2}\frac{\theta}{180}+\pi R^{2}\frac{\alpha}{180}-H(r+x)-xh \\ L & = 2\bigg(2\pi r\frac{\theta}{360}\bigg)+2\bigg(2\pi R\frac{\alpha}{360}\bigg)=\frac{\pi}{90}(r\theta+R\alpha) \end{align}

Consider $$\triangle GCE$$, and line $$|GQ|$$, since $$\angle GCE$$ and $$\angle GEC$$ are equal, and $$|GQ|$$ line bisects $$|CE|$$ and $$\angle CEG$$ into two equal sizes, we can conclude that $$\triangle GCE$$ is an equilateral triangle. Then $$\theta=60^{\circ}$$, and $$H=r2\sqrt{3}$$ and $$R=5r$$ . The only unknow terms are $$\alpha$$ and $$\boldsymbol{x}$$. By using law of sines for $$\triangle CPK$$, we can state that:

$$\frac{\sqrt{12r^{2}+(r-x)^{2}}}{\sin(60)}=\frac{4r}{\sin(120-\alpha)}=\frac{r+x}{\sin(\alpha)}$$

I tried to get $$\alpha$$ or $$\boldsymbol{x}$$ from equation above but I could not get any further, because it got too much complicated to draw anything meaningfull.

What dis I do before asking the question: didn't just use the sine law, I tried to use the information I know about triangles and trigonometry or the information I found from the sources, but I got mixed results from these attempts. By complex, I mean quadratic trigonometric expressions and impractical results like 3-degree variables. Of course, I may not have seen the connection between the equations or variables, as it is possible that I did not use theorems or expressions correctly. I searched the web using terms that could describe the shape, such as 4-centred arch, Tudor arch, and Persian arch. I searched for a book or article on the subject in libraries with open content. I searched this site using the same terms. The only resources I could find were architectural drawings, a few resources on how to draw a 4-center arc, and videos that explained how to draw a 4-center arc using various software.I may have searched using the wrong terms, but I don't know what the correct term is because it is a subject far from my area of expertise.

A little help: I want to calculate not only the area of a 4-centred arc but also the areas of shapes such as 3 and 5 centred arcs and Gothic arcs, using the height and width of the shapes. Therefore, considering my level in this subject, could you recommend a resource that can help me?

Thank you very much in advance.

P.S.: My native language is not English.

Update (2022-6-9): By using @Suzu Hirose's hint, we can obtain $$\alpha$$ and $$\boldsymbol{x}$$. Define $$\angle NKE$$ as $$\beta$$ and assume we start at point $$\boldsymbol{N}$$ and drop perpendicular to a line in the $$|KE|$$ direction. This perpendicular is equal to $$\boldsymbol{r}$$. Then $$\sin(\beta)=1/5$$ and $$\beta=\arcsin(1/5)^{\circ}$$. For $$\triangle KEP$$,

$$\sin(\beta)=\frac{r-x}{\sqrt{(r-x)^{2}+(2r\sqrt{3})^{2}}}=\frac{1}{5}$$

From this equation, we get that $$x=\frac{r}{2+\sqrt{2}}$$. Or, from @Suzu Hirose's equation, $$x=r-2r\sqrt{3}\tan(\arcsin(1/5))$$. From $$\triangle KEC$$, we know that $$\alpha+\beta=30^{\circ}$$, then $$\alpha=30-\arcsin(1/5)$$. Another way to express $$\alpha$$ is using $$\tan(\beta)=\frac{1}{2\sqrt{6}}$$ in

\begin{align} \tan(\alpha+\beta) & = \frac{1}{\sqrt{3}} \\ \tan(\alpha)\sqrt{3}+\tan(\beta)\sqrt{3} & = 1+\tan(\alpha)\tan(\beta) \\ \tan(\alpha) & = \frac{1-\tan(\beta)\sqrt{3}}{\sqrt{3}-\tan(\beta)}\\ \tan(\alpha) & = \frac{2\sqrt{6}-\sqrt{3}}{6\sqrt{2}-1}\\ \Rightarrow \alpha & = \arctan\left(\frac{2\sqrt{6}-\sqrt{3}}{6\sqrt{2}-1}\right) \end{align}

Finally we can expess equations [eq:area] and [eq:arch] only with $$\boldsymbol{r}$$ and $$\boldsymbol{h}$$.

\begin{align} A & = \frac{\pi r^{2}}{3}+\frac{(4\pi r^{2})(30-\arcsin(1/5)}{45}-\frac{(2r\sqrt{3})(3r+r\sqrt{2}}{2+\sqrt{2}}-\frac{rh}{2+\sqrt{2}} \\ L & = \frac{2\pi r}{15}\left[45-\arcsin(1/5)\right] \end{align}

• It looks in the Wikipedia diagram that they are using $4r$ for the radius of the inner part of the arch where the "this is carpentry" web site uses $5r$. Wikipedia doesn't explain anything about the differences, I guess you would have to compare to real arches to work it out. But the arch would have to be quite high to extend an actual piece of string as far as the $5r$ radius to measure it. Commented Jun 5, 2022 at 0:47
• I came across at least four different diagrams that show how to draw a four-centred arc. I didn't measure them by drawing, but I expect all of them to have a different radius. I think the radius of each circle will change based on how "depressed" the arch itself is. Commented Jun 5, 2022 at 13:26

Relabelling your $$a$$ as $$\alpha$$ (alpha), define an angle $$\beta$$ (beta) which is $$\angle NKE$$ in your diagram's labelling, then
$$\tan(\alpha+\beta)=|CE|/|EK|={2r\over 4r{\sqrt 3}/2}=1/{\sqrt 3},$$ and $$\sin\beta=r/|NK|=r/5r=1/5.$$ You can then solve for $$\alpha$$ from these two, and $$x=r-|EP|$$, $$|EP|=|EK|\tan\beta$$.
It might be possible to write all of this out with square roots and fractions and so on, but unless you are very keen, I would suggest solving for $$\beta$$ and $$\alpha$$ using a numerical solution, that is substituting the numbers like $$\sin^{-1}(1/5)=11.53...$$ you get from a calculator or a computer into the equations.