A friend recently asked if I was interested in “a woodworking ‘problem’ involving math.” I couldn’t help but take that kind of bait, and he continued with a picture:

Photo of the end of a wooden table
Photo from Jason King Lee via Facebook

and a simple question: How do you find the cut angles and length of these cross members?

This ought to be relatively simple geometry but it’s been a while since I’ve thought about a problem quite like this, so it took me a few tries to work it out fully. My friend had worked out the solution in a spreadsheet, so we were able to compare notes while I worked toward a solution. Here’s a diagram we used for reference:

Diagram of layout highlights

Starting from this diagram, we assume that we’ll know the measurement for width (which we’ll call $w_o$ as in “width of the opening”), and for height (which we’ll similarly call $h_o$). We’ll also know the width of the crossmember, which isn’t well marked in this diagram. Let’s call that $w_c$. So the measurements we’re after are $\angle{EBD}$ and the length of $\overline{BE}$. Let’s call that angle $\theta_{cut}$, and the length $l$.

It doesn’t seem as though either of these is simple to calculate directly, so I started to look at what I could calculate directly. The easiest seems to be the hypotenuse of the opening, $\overline{BC}$. This is just plugging into the pythagorean theorem:

$$ \begin{equation} \overline{BC} = \sqrt{{w_o}^2+{h_o}^2} \end{equation} $$

I want to calculate $\angle{EBD}$, and I still don’t think I have enough information to do that. It looks like $\angle{CBD}$ is a hefty part of that, though, and I do have enough information to calculate that using the sin law of triangles:

$$ \begin{equation} {h_o \over sin{\left (\angle{CBD}\right )}} = {\overline{BE} \over sin{\left (90\right )}} \end{equation} $$ $$ \begin{equation} {h_o \over sin{\left (\angle{CBD}\right )}} = {\overline{BE} \over 1} \end{equation} $$ $$ \begin{equation} {h_o \over sin{\left (\angle{CBD}\right )}} = \overline{BE} \end{equation} $$ $$ \begin{equation} {sin{\left (\angle{CBD}\right )}} = {{h_o} \over \overline{BE}} \end{equation} $$ $$ \begin{equation} \angle{CBD} = arcsin{\left ({h_o} \over \overline{BE}\right )} \end{equation} $$

and just to keep things as close as reasonable to the original set of known measurements:

$$ \begin{equation} \angle{CBD} = arcsin{\left ({h_o} \over \sqrt{{w_o}^2+{h_o}^2}\right )} \end{equation} $$

Now I know $\angle{CBD}$ but that’s only part of the angle that I need. More specifically I know that

$$ \angle{EBD} = \angle{EBC} + \angle{CBD} $$

If I can figure out $\angle{EBC}$, then, I can figure out $\theta_{cut}$. I can construct a right triangle within the crossmember using $\overline{BE}$ and $W_c$. At this point I can use the sin law again on this new right triangle:

$$ \begin{equation} {\overline{BE} \over sin{\left (90\right )}} = {w_c \over sin{\left (\angle{EBC}\right )}} \end{equation} $$ $$ \begin{equation} {sin{\left (\angle{EBC}\right )} \over sin{\left (90\right )}} = {w_c \over \overline{BE}} \end{equation} $$ $$ \begin{equation} {sin{\left (\angle{EBC}\right )} \over 1} = {w_c \over \overline{BE}} \end{equation} $$ $$ \begin{equation} {\angle{EBC}} = arcsin{\left (w_c \over \overline{BE}\right )} \end{equation} $$

And substituting back in like before: $$ \begin{equation} {\angle{EBC}} = arcsin{\left (w_c \over \sqrt{{w_o}^2+{h_o}^2}\right )} \end{equation} $$

Since we have defined $\theta_{cut}$ as $$ \begin{equation} \theta_{cut} = \angle{EBD} \end{equation} $$

we’ve got the answer to the first part of our problem:

$$ \begin{equation} \theta_{cut} = \angle{EBC} + \angle{CBD} \end{equation} $$ $$ \begin{equation} \theta_{cut} = arcsin{\left (w_c \over \sqrt{{w_o}^2+{h_o}^2}\right )} + arcsin{\left ({h_o} \over \sqrt{{w_o}^2+{h_o}^2}\right )} \end{equation} $$

Ok! I’m part way there. I know what to set the saw to, so I could make the first cut. Now I need to figure out how far apart to make the cuts. For this, I leaned on my friend sin law once again. I considered $\overline{BE}$ as one side of $\triangle{BEC}$, so sin law gives me this relationship:

$$ \begin{equation} {\overline{BC} \over sin{\left (\angle{BEC}\right )}} = {\overline{BE} \over sin{\left (\angle{BCE}\right )}} \end{equation} $$

and solving that for $\angle{BE}$:

$$ \begin{equation} \overline{BE} = {{\overline{BC} \times sin{\left (\angle{BCE}\right )}} \over sin{\left (\angle{BEC}\right )}} \end{equation} $$

Since $\overline{BE}$ is a transversal of the parallel lines $\overline{BD}$ and $\overline{EC}$, it follows that $$ \angle{BEC} = 180^{\circ} - \angle{EBD} $$ and $$ \angle{BCE} = \angle{CBD} $$

Applying these to (16) gives:

$$ \begin{equation} \overline{BE} = {{\overline{BC} \times sin{\left (\angle{CBD}\right )}} \over sin{\left (180^{\circ}-\angle{EBD}\right )}} \end{equation} $$

Substituting values from (1), (7), and (13) simplifies this to

$$ \begin{equation} \overline{BE} = {{\sqrt{{w_o}^2+{h_o}^2} \times sin{\left (arcsin{\left ({h_o \over \sqrt{{w_o}^2+{h_o}^2}}\right )}\right )}} \over sin{\left (180^{\circ} - \theta_{cut}\right )}} \end{equation} $$ $$ \begin{equation} \overline{BE} = {{\sqrt{{w_o}^2+{h_o}^2} \times {h_o \over \sqrt{{w_o}^2+{h_o}^2}}} \over sin{\left (180^{\circ} - \theta_{cut}\right )}} \end{equation} $$ $$ \begin{equation} \overline{BE} = {h_o \over sin{\left (180^{\circ} - \theta_{cut}\right )}} \end{equation} $$ $$ \begin{equation} l = {h_o \over sin{\left (180^{\circ} - \theta_{cut}\right )}} \end{equation} $$

And there we have it! Given $l$ and $\theta_{cut}$ I could now get out the speedsquare and circular saw and cut myself a crossmember. Oh, but one more thing! What length of board would I need to cut the crossmember from? Depending on how you look at it, this could also be the length that you’d set a stop block when making the cuts. Let’s call this measuremnt $l_{overall}$. It turns out this is even easier to calculate than $l$. The first thing I considered is that $l_{overall}$, $w_c$, and $\overline{BC}$ form a right triangle. I drew a diagram to visualize but went a much lower-tech route than my friend did with the diagram above.

Hand drawn diagram of the larger right triangle

Don’t worry about the specific numbers here, or the extra labels (especially that $\alpha$ - that never happened). Since I know the hypotenuse and the height, I only need to calculate the base of this triangle. Plugging everything into the pythagorean theorem, I get:

$$ \begin{equation} {l_{overall}}^2 + {w_c}^2 = \overline{BC}^2 \end{equation} $$

which I can then solve for $l_{overall}$:

$$ \begin{equation} {l_{overall}}^2 = \overline{BC}^2 - {w_c}^2 \end{equation} $$ $$ \begin{equation} {l_{overall}}^2 = \left (\sqrt{{w_o}^2+{h_o}^2}\right )^2 - {w_c}^2 \end{equation} $$ $$ \begin{equation} {l_{overall}}^2 = {w_o}^2+{h_o}^2 - {w_c}^2 \end{equation} $$ $$ \begin{equation} l_{overall} = \sqrt{{w_o}^2+{h_o}^2 - {w_c}^2} \end{equation} $$

And that’s it! Now I can easily measure and cut, whichever way I’m thinking about it. Now I just need to find an excuse to actually build this table…

To recap, the solution to the measurements I’d need to build the table are:

$$ \begin{equation} \theta_{cut} = arcsin{\left (h_o \over \sqrt{{w_o}^2+{h_o}^2} \right )} + arcsin{\left ( w_c \over \sqrt{{w_o}^2+{h_o}^2} \right )} \end{equation} $$ $$ \begin{equation} l = {h \over sin(180 - \theta_{cut})} \end{equation} $$ $$ \begin{equation} l_{overall} = \sqrt{{w_o}^2+h^2-{w_c}^2} \end{equation} $$

Update: My friend also compiled a very nice spreadsheet to do these calculations, which you can find here. It includes more complete diagrams, and the ability to round to a precision that matches your equipment.