I have a problem that I am hoping you can help me with. I am in the process of making a picnic table for my family so we can enjoy summer evenings out doors. I'm building it out of 2x4 lumber which is actually 1.5"x3.5" dimensional. The plans I have recommend cutting out a cardboard template and just 'fudging' the angles. Not my style!

So here is the problem: Imagine we have two crossing parralelograms in an X shape making up the legs at one end of the table. I know the width of the table top (w) and the height I want the table to be (h). Each parallelogram also has a known altitude of 3.5". Given just this information is it possible to figure out the length I need to cut each leg and the angle I need to cut the miters?

Thanks everyone! -Nate

  • $\begingroup$ Just to clarify, you know the height and width of the table and the length and thickness of the leg (before cutting it), right? $\endgroup$ – shardulc Apr 25 '16 at 5:24

I'm assuming that the height of the table $h$, the length of a leg $l$, and the thickness of a leg $t$ is known. As we shall see, the width of the table doesn't really matter. The length and thickness of the leg refer to those dimensions before the leg is cut into the proper shape, like the uncut leg $ABCD$ in the diagram below.

The line through $E$ parallel to the $x$-axis is the ground, and $CG$ is the top of the table. We take the point $C$ to be the origin with axes as shown. Now, $A$ is a point on the ground at a distance of $\sqrt{l^2+t^2}$ from $C$ (by Pythagoras' Theorem). To find $\angle BCG$ (or equivalently $\angle DAE$), we consider the blue angle $\angle CAE$ and the green angle $\angle CAB$.

View of the picnic table and leg

The sine of $\angle CAE$ is $\frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{\sqrt{l^2+t^2}}$ so $\angle CAE = \arcsin\Big(\frac{h}{\sqrt{l^2+t^2}}\Big)$. Similarly, the tangent of $\angle CAB$ is $\frac{\text{opposite}}{\text{adjacent}} = \frac{l}{t}$ and thus $\angle CAB = \arctan\Big(\frac{l}{t}\Big)$. We get $$ \begin{align} \angle DAE &= \frac{\pi}{2} - \angle EAB \\ &= \frac{\pi}{2} - (\angle CAB - \angle CAE) \\ &= \frac{\pi}{2} - \arctan\Big(\frac{l}{t}\Big) + \arcsin\Big(\frac{h}{\sqrt{l^2+t^2}}\Big) \end{align} $$ which, I think, is what is needed to build the legs for your table accurately!

You can play with the construction (made in GeoGebra) here.

  • $\begingroup$ I would like to know how can I upload the whole construction (not just the picture) so that everybody can play with it. $\endgroup$ – Mick Apr 25 '16 at 9:05
  • $\begingroup$ @Mick I don't think it's possible to embed GeoGebra constructions directly into the page, maybe someone on Mathematics Meta can help. $\endgroup$ – shardulc Apr 25 '16 at 9:22
  • $\begingroup$ Maybe that is the right place that I should go to but I am interested to know how did you incorporate your "toy" in your post? $\endgroup$ – Mick Apr 25 '16 at 13:40
  • $\begingroup$ @Mick It's just an image, uploaded with the standard SE imgur uploader. $\endgroup$ – shardulc Apr 25 '16 at 14:33
  • $\begingroup$ I think it is more than an image. It includes controllable (to some extend) buttons (with limited actions) on the left pane. $\endgroup$ – Mick Apr 25 '16 at 16:34

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