Post II in what is planned to be a relatively short carpentry drawing thread, detailing a French carpentry problem and its 2D descriptive geometrical solution.

Here’s our problem, a situation in which 2 sticks are intersecting in the rough form of an ‘X’:

These sticks have different slopes from one another and are oriented differently to the floor plan from one another.

Let’s give this problem some parameters, so that those who wish to try this at home can follow along more easily. Taking the left side member, which is colored light brown, in the next illustration I show some of it’s particulars:

This piece is rotated in plan 60˚ from a line which connects our two sticks on the ground. I have labeled this piece as ‘A’ where it meets the ground, and it will be called piece ‘a’ from here on out. Notice that if you follow the arris of the stick from the ground, at ‘A’, up to the top, we reach a point marked with a small circle and which has a line running vertically through it down to the ground. This line is plumb to the floor and meets the floor 135 cm away from line A-B.

Point ‘B’ on the line, which is located 150 cm from ‘A’, defines the place at which the second piece, colored yellow, meets the floor:

This is therefore piece ‘b’, and it runs 45˚ to the plan. Again, following the arris of the stick from the ground to the top of the stick, we find a small circle and a plumb line back down to the ground. That plumb line meets the ground 150 cm from the A~B line.

Here are both sticks ‘a’ and ‘b’, shown in their positions relative to one another:

Note that the height of the two circles at the top of each stick is exactly the same for each piece. I chose 125 cm for that height, but it could of course be whatever height it needed to be.

Now let’s translate the previous 3D drawing into 2D. What we want to create first is the basic ground plan. Here’s what I come up with:

Hopefully that was straightforward enough for everyone to follow. If not, carfully compare the above sketch to the one previous, and you should see the common points.

In the next illustration, I bring in a little 3D once again, superimposing it directly on that 2D plan. I draw in a triangle representing piece ‘b’, from the point it meets the floor at ‘B’, up along its arris, to the circle at the top. This forms a triangle which represents the slope of that stick of wood:

That right-angled triangle has a rise of 125cm and a run of 150cm x √2, which is about 212.132cm. It’s not critical to know or understand why the run of the triangle is √2 times longer than the distance A~B. Those who have obtained my first two volumes of the carpentry drawing essay series should be quite savvy about this, but for completing this particular drawing problem it is not too important at all. Don’t sweat it if that √2 thing isn’t making sense.

Now let’s look at the triangle which defines the slope of piece ‘a’, which I’ll just add on to the previous sketch:

Again, piece ‘b’ runs along 45˚ to the line on the floor A~B, while piece ‘b’ sets off at a 60˚ from that same line. Both triangles rise the same amount, 125 cm.

Of course we are looking at a 3D illustration at this point, and, to follow traditional carpentry practice what we want to work with is a 2D drawing. You need to be able to visualize the 3D and translate it into 2D – that’s the key to this sort of drawing work, and it by no means comes easy, especially as the constructions gain complexity.

Those triangles which are sticking up in the 3D drawing need to be placed on the floor somehow. What we do is to treat those triangles as if they were a folded-up flap of paper attached to the ground. We simply fold them down, using the run of the triangle as if it is a long hinge. Let’s do just that with the light brown triangle representing piece ‘a’:

The illustration attempts to show, through the use of the multiple triangles, how the triangle for the slope of piece ‘a’ is rotated from a plumb position down to the ground. Maybe one day I’ll get all fancy and do some sort of animation for this, but for now the above will have to suffice I’m afraid. When the triangle is down, it would look like this in our 2D plan:

Note that nothing has changed in regards to the geometry of this triangle – it still has a height of 125cm, only now we’ve drawn that on the floor directly, and at a 90˚ angle to the run of the triangle. We’ll be doing the exact same thing with the other triangle which represents the slope of piece ‘b’ too, however we’ll stop here for today. We’ll do some more work on this light brown triangle next time. If you are finding the whole ‘rotate the triangle’ part of this a bit mystifying, I would suggest making a cardstock triangle to represent the slope of piece ‘a’ and rotate it on the drawn plan just as I have illustrated. It’s a simple 90˚ rotation downward. Up down, up down – it will be clear after staring at it long enough I hope.

Thanks for coming by the carpentry way today. –> on to post 3

G'Day Chris,

There's so much exciting development at the Carpentry Way at the moment – It's like a smorgasboard.

I'm really looking forward to seeing how the bell tower and the Ming table start to manifest, almost as much as I'm looking forward to the next volume of the carpentry drawing essay.

I shall follow along with this X-pose' at home, should make for some good carpentry drawing practice.

Regards

Derek

Hi Derek,

good to hear from you! How are ya travelin'?

Yes, there is a lot on the plate right now, project-wise, and to add to all of that there is the carpentry essay volume III on splicing joints, now past 120 pages, which soaks up a fair amount of time, and the upcoming presentation on Japanese architecture I'm giving in Boston in another week or so, and a woodworking machine on the way to me from Colorado, and the bubinga also in shipment this coming week, AND I'm moving house at the end of the month. A little busy, but at least my truck doesn't need any attention at the moment!

Glad to hear you are following along with the drawing exercise – maybe you are the only one out there doing so, but I'm glad nonetheless. One tip: that triangle we laid down on the plan today is going to get moved slightly in the next installment, so if you're drawing with pen and paper you might want to hold off drawing that triangle in for the minute.

~Chris

'maybe you are the only one out there' After giving sketchup a try in 2007, I thought that this exercise is a good opportunity to pick it up again. Thanks for this introduction to traditional geometric drawing.

..and now there are TWO!! :^)

Damien,

it's not really a fitting intro to descriptive geometry, as this exercise is not among the most basic sort of layout problems, but for anyone who wants to give it a try and follow along, it should be achievable, and, hopefully, my attempt at explanation will be digestible too.

~Chris

Hey Chris

Very interested to see how this progresses.. I work with a bunch of German and Swiss master & apprenticed carpenters who have often talked about this method of layout but have never shown me how it works. Thanks for doing it for them!

Hi Adam,

well you're welcome. The German method is similar to the French one in certain respects, and while I have no idea how the Swiss do their lay out exactly, I suspect it would be quite similar.

~Chris

Chris,

I've been following your development of the X rated problem, and I realized that I don't know how you arrived at the dimensions; Point B is 135 cm from A-B, and A is 150 cm from A-B. Clues? Mike Laine

hi Mike,

the dimensions are somewhat arbitrary – I set them out as given distances, rather hen determined them by some formula or other reason. I'm also faithfully following the example in the French carpentry text and using the exact dimensions given there for the problem. The distances used ensure that the pieces 'a' and 'b' do in fact intersect one another.

Since piece 'b', is rotated 45˚ in plan, it makes sense that its displacement along the plan, in both x- and y-axes, would be the same: 150 cm in this case. At that point that is 150 cm along the x-axis, and then 150cm up on the y-axis, I set the altitude of the piece at 125cm. Again, this dimension if simply arbitrary – the greater or lesser it is, then obviously the slope of the piece would vary as a result.

Piece 'a' on the other hand, is rotated 60˚ up from the baseline in plan, so it cannot have an x- and y-axis displacement that is the same. The y-displacement is set at 135cm. One could readily do the trig to ascertain the x-axis displacement if one wanted (that is, tan 30˚ x 135 = 77.9422..cm). However, if one can lay out a line along the 60˚ angle that would be all you really needed.

I could have set the y-axis displacement for that piece 'a', not at 135cm, but at 250 cm or any other number, however if that spot was too far away, and the altitude at that location was kept the same at 125cm, then piece 'a' would have too slight a slope and would pass completely below piece 'b' and there would be no intersections. No problem to solve in other words.

So, a value of 135cm up along the y-axis is a number that works well, though it could have been varied slightly and there would have been an intersection between the two pieces.

Did that explanation, long-winded as it was, make sense?

~Chris

The drawing exercises…….How do I get them to do..?

TAJCD essays are shown in the bar to the right of the page.

~C

Thanks. My computer skills are very poor. I am not at sure this msg will go through. But if it does: what is the algebra on : “in an equal sided right triangle the hypotenuse equals the side times the square root of two? I know it is true but in 53 yrs of messing with trig I have never seen it.

Robert,

thanks again. That question was posted by you in Part 1 of this series, and answered at length by me, so please refer again to post 1.

Comments are moderated so there is a delay between the time you post the comment and the time it appears on the page. I realize that can be a little confusing at first.

~C