I recently received an email from a reader, a fellow named ‘Mo’ who recently graduated from the timber framing program of the American College of the Building Arts in Charleston, South Carolina. The instructor there is Bruno Sutter, who is French, and part of Les Compagnon’s Du Devoir. A craftsman. I’ve been in touch with Bruno on and off in the past few years and he comes across as a dedicated educator and experienced woodworker.

Apparently, as one of the parting gifts to program graduates, Bruno distributes a carpentry drawing with a problem to study. Mo passed one such drawing he had received on to me in the past few days. Hah! – truly the worst case of a spam attack I have ever suffered. Well, funny enough, I welcome that sort of mail very much indeed. Send me your poor, your wretched, your layout drawings….

I thought it would be fun to explore the drawing on the blog here, and, who knows, this might possibly of interest to some readers(?). Mo has given his permission to do so, providing I credit his teacher. So, I’m starting a micro build thread on this topic, to be nested in between the two other design build threads I’ve got going on. Hopefully for those readers who are interested in the topic of carpentry drawing it will be a nice sojourn, and for those who feel, well, ill at the sight of descriptive geometry, have no fear, as this will be a relatively short series and there will be other stuff to read about along the way. My aim is to demystify the French drawing technique and make it of interest to more people out there.

So, the problem exercise looks like this:

Hopefully the title of this post now will make a bit more sense. What you have in the above drawing are two square-section sticks of wood which cross one another. Each piece is rotated so as to be on a diagonal orientation to the floor, like a splayed sawhorse leg, and both sticks terminate at the same height from the ground. The challenge in this problem is to determine the lay out lines for the point of intersection between the two sticks, so that one could make such a construction if need be.

The first piece of information I will share, to help clarify things, is that each of the sticks is rotated, in plan, a different amount:

The green stick is 45˚ to the plan, and the light brown stick is 60˚to the plan.

I’ll share a couple more views of the problem:

Were those pictures ‘X-rated’ or what?

Now, this is not too tough a problem to draw in SketchUp, a 3D drafting program many readers are already familiar with. I’m sure it would be easy in other 3D modeling programs. One could easily draw the sticks, tilt them up into their respective slopes, and intersect their surfaces. It would then be a simple matter of measuring the pertinent angles and distances needed to go to the wood and start laying out. However, if one did that what results is next to zero in the way of practical carpentry skill, as one is entirely dependent upon the computer software to solve the geometry. Powerful as it may be, 3D drawing can be like a crutch. I think 3D is great for working out lots of kinds of problems and for communicating with non-carpentry specialists and clientele, however I think it is wise to develop the skill of working with the basic manual lay out tools and descriptive geometry techniques to solve such problems. Good old 2d was enough to build the cathedrals of Europe and the great temples of China and Japan after all. So, that is what we will do in this series, though I am going to take advantage of the benefits of 3D from time to time as a convenient means of explaining what is going on in 2D. 2D geometrical drawings, as I’m sure many will agree, aren’t always so easy to get your head wrapped around. I think the 3D will help with that visualization as we work though the drawing in 2D. Does that sound like a reasonable plan?

Forgive my presumption to be teaching anyone this material, however I do find it quite interesting and, I must confess, there is a bit of a selfish motivation at work too. You see, one of the best ways I know how to consolidate one’s understanding of something, especially something complex, is to try and teach it. I hope to learn a lot from the process too, and won’t embarrass myself too much!

Stay tuned. Those of you into carpentry drawing might want to have a crack at the problem on a sheet of paper. In the next installment we’ll start sketching in the elements of the basic plan view of these two sticks of wood.

Hasta la vista. –> on to post 2

Thank you for your blog. I am putting you on my Christmas list and on my Thanksgiving “Thank you” list. If you have time I would like to see the algebra on ………'when opposite sides of right triangle are equal the hypotenuse is equal to the length of one side times the square root of 2'.

Robert,

nice to hear from you and welcome to my blog!

The √2 issue is an important conceptual issue to grasp, especially for constructions involving rafters at a 45˚ plan angle – like a regular hipped roof for example. I'd be happy to go over this for you.

The Pythagorean Theorum tells us that the square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the sides.

That is commonly written out as a formula: c-squared = a-squared + b-squared,

where 'c' is the measure of the hypotenuse, and 'a' and 'b' denote the two sides of the triangle.

If we wanted to simplify the formula to solve for 'c' by itself, we could take the square root of each side of the '=' sign. If you take the square root of c-squared, you end up with 'c', and the full formula therefore becomes:

c = √(a-squared + b-squared)

I trust you followed me so far.

Now, let's take a right triangle where each of the sides measures '1', which would mean that the values for 'a' and 'b' are both equal to 1.

We plug that info into the formula to get:

c = √ (1-squared + 1-squared)

Now, 1-squared is simply equal to 1, so we can readily simplify the equation to:

c = √ (1+1)

One more step gives us:

c = √2

So, for a right triangle having equal sides of 1, the hypotenuse of that triangle is √2 in measure.

Imagine we now make a larger triangle, still with equal length sides, now each equal to 2. Since the triangle is proportional – or 'similar' as they term it – to the one having sides of 1, the fact that we doubled the length of the sides means we can do the same for the hypotenuse. A right triangle with sides both equaling 2 would therefore have a hypotenuse of 2√2 (meaning: 2-times root-two)

If the right triangle had equal sides measuring 30.23, then the hypotenuse would be 30.23√2, and so on and so forth. It doesn't matter – so long as the sides of the right triangle are equal to one another, the hypotenuse will always be √2 times longer than that side measurement.

I hope that explanation made good sense to you.

~C

I'll never calculate a half pitch common or jack length any other way than this way again. Very Nice! It has the same elegance as Sim Ayer's “Trigonometric Formulas Geometrically”

Just reread your “selfish motivation …..” An old maxim in learning a procedure in medicine: “See one Do one Teach one”.

Robert,

so glad you are getting a lot out of this series!

~C