Constructing the Pattern on the Sala de la Barca Ceiling

I was flipping through Owen Jones’s Grammar of Ornament a couple months ago, and my eye was caught by this handsome pattern I had not noticed before. This is Jones’s Plate XLII, in the chapter on designs from the Alhambra in Granada, Spain. He calls it, “Part of the ceiling of the Portico of the Court of the Fish pond.”

Since then I’ve kept out a clipboard with a compass and straightedge, and, in the small triangles of time after meals or before working on something else, I’ve slowly figured out how one might construct this. It’s less complicated than it looks at first, and at the same time, weirdly complex.

This post discusses how to construct this figure with compass and straightedge; the next post is about the mathematics of it.

To get a bigger picture of what the original is, take a look at this beautiful photograph at the David Wade archive of the actual ceiling:

It turns out that it is constructed not on a flat surface, but on the inside of a barrel-vaulted ceiling. And it consists of multiple square repeat units, each holding a large 12-pointed star, arranged in regular rows and columns.

At Brian Wichmann’s Tiling Database (tilingserch.org), this pattern is called “Alhambra, Sala de la Barca,” and the notes say…

The shape of the ceiling of the Sala de la Barca is essentially cylindrical, with quarter spheres capping each end. Consequently the basically planar pattern has to be greatly modified where it covers the spherical surfaces at each end. The colours on the painted wooden ceiling are black, white, silver, brown and orange. The symmetry group of the tiling is *442 (p4m).
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And they have this rendering of the pattern, which shows that someone else has figured out how to construct it, even if they did not publish the method.

I found a photograph of the entire ceiling on the page for The Hall of the Boat at alhambragranada.org, so you can get a sense of the whole piece of work. The ceiling extends down a long, thin space.

The basic pattern, three full square repeat units with half a unit on each side, is rendered nine times down the main semi-cylindrical body of the vault. In the quarter-dome caps on each end, there is a single square repeat unit, with three half-units along the edge, and two triangular spaces cleverly filled in between these.

For now, I’m interested just on what’s going on with the regular square repeat units in the semi-cylindrical vault, the part we could actually unroll and lay flat. If you are interested in how they worked the pattern in the quarter-dome caps at each end, you can read more in Jay Bonner’s Islamic Geometric Patterns, p. 113 and p. 543.

Last, before going into the construction method, we should discuss where, exactly, this is in the Alhambra. We’ve seen it called “the ceiling of the Portico of the Court of the Fish pond,” “the ceiling of the Sala de la Barca” and the “Hall of the Boat.” Are these the same places?

A little research reveals that there is no Court of the Fish Pond, but there is a Court of the Myrtles that used to be called the Court of the Pond. At its northern end….

through a pointed arch of mocarabes, visitors may enter into the Hall of the Boat (Sala de la Barca). The origin of its name is the Arabic word baraka, which means blessing and which degenerated into the Spanish word barca, which means boat. This rectangular hall is 24 meters long and 4.35 meters wide, but was apparently smaller at the beginning and later extended by Mohammed V. A semicylindrical vault was destroyed by a fire in 1890 and it was replaced by a copy completed in 1964.
https://www.alhambradegranada.org/en/info/nasridpalaces/halloftheboat.asp

Interestingly, Jones’s drawing dates from before the 1890 fire, yet it differs slightly from what we see in photographs of the ceiling today. Tilingsearch.org cites Jones’s drawing as

Plate XLII+ (Alhambra, Sala de la Barca, Spain. Drawn incorrectly without a kink) from [jones]

I was a long way into constructing this figure before I found what I think is the “kink” — and I don’t know if Jones got it wrong, or the builders of the copy in 1964 got it wrong. But there is a difference.

In today’s ceiling, the line coming off one of the vertices of the eightfold stars is aligned with neither side of the star, but rather with the octagons that appear at the corners of the repeat unit. As a result, the arrowhead shaped piece off the end of one of the rays of the 12-pointed star is symmetrical. In Jones, that line is aligned with one side of the star, and not with the octagon side, so the arrowhead shaped piece is asymmetrical.

*Rendering from tilingsearch.org (left), compared with Jones’s 1856 drawing (right).*

My construction method goes with what we see on the ceiling today.

Technically, the repeat unit is just this much:

And there are much smaller repeat units of course: a quarter of the square would suffice. Even half a quarter of the square would suffice. But from the point of view of construction, it’s easier to visualize with the entire outer frame.

So let’s construct that.

Phase 1

Draw an initial circle and establish within it the square which frames the basic repeat unit of the pattern. Then draw the outer edges of the 12-pointed star that dominates the centre of the pattern

Draw a circle (“Circle A”).

2. Inscribe the master square within it.

2.1. Draw a horizontal line through the circle’s centre

2.2 Bisect it to make a vertical line through the same centre

2.3. Draw four arcs (same radius as Circle A) centred on the four points of intersection between Circle A and the two lines just drawn.

2.4. Draw diagonals connecting the points where the arcs intersect outside Circle A.

2.5. Note where the diagonals intersect Circle A, and join those points to form the master square. You should now have this:

3. Locate the quarter-centres of the square. These are the centroids of each quarter-square.

3.1 Locate the midpoint of each side of the square.

3.2. Connect adjacent midpoints with diagonal lines

3.3. The quarter-centres are where the diagonals from step 2 above intersect the lines connecting the side midpoints. You should have:

4. Draw a circle centred on the same point as Circle A, but through the square’s quarter centres (Circle B).

5. Draw the additional lines to split these circles into 24 equal segments. Getting these radiating lines evenly spaced will make the final pattern look much better, so it’s worth taking your time with this step.

5.1. So far we have already divided Circle A into 8 equal arcs using the horizontal, vertical and diagonal lines.

5.2. Eight more divisions can be made by connecting four pairs of points. They are the intersections between, on the one hand, the construction arcs we made for the master square, and, on the other, Circle A.

5.3. For the last eight points, there is no easy solution. However, we already have enough of the radial lines that we can just copy the interval between them onto the remaining arcs. You could also do this using the usual angle bisecting technique, but really it’s easier to just use a compass set to the span of one of the existing 1/24th arcs.

You should now have 24 radial lines, with 1/24th of the circle between each pair of them.

6. Observing where the radial lines intersect Circle B, construct a few lines between intersections that are four apart, extending them out to the next 1/24th radial. You really only need a few of these.

7. Now draw a circle (Circle C) going through these outer intersections.

8. Draw line segments alternating between circles B and C, starting at a quarter-centre of the square. You should get a 12-pointed star.

This is the first part of the pattern we will actually ink.

Phase 2

In the second phase, we establish the 8-pointed stars centred on the midpoint of each side of the master square, and then let them dictate the twelve pairs of parallel lines that make the star at the heart of the pattern. Also, we will draw the 8-pointed stars in the corners.

1. At the midpoint of each side of the master square, draw a circle centred on that midpoint whose radius just touches the nearest approach of the 12-pointed star we’ve drawn. We’ll call this Circle D. We’ll be drawing a lot of circles with the same radius as Circle D over the next few steps.

2. For that circle, draw circles of the same size centred on each of the four cardinal points.

3. Then draw diagonals through adjacent circle centres and intersections. This allows you to identify 8 evenly-spaced points around Circle D.

4. Trace the lines of the 8-pointed star. These are inkable.

5. Repeat to create 8-pointed stars on the other 3 sides of the master square.

6. Extend the 8-pointed star sides in parallel pairs, to create channels that run across the master square and along its sides.

7. Now we need to draw four more pairs of parallel lines crossing the 12-pointed star. Using the same radius as Circle D, mark out two points on either side of each vertex of the 12-pointed star. (No need to do the vertices that already have an eight-pointed star adjacent to them.)

8. Connect corresponding pairs of these marks to make four more pairs of parallel lines. Extend these lines until they hit the master square.

9. Ink these lines inward from the 12-pointed star, stopping where they meet partners that are perpendicular to them. You should have twelve channels within the central star, emanating from another 12-pointed star at their centre. Each channel has the same width, and this width is shared with the channels framing the master square.

10. At each of the small corner squares formed by outer frame channels (drawn in step 6, above), draw a circumscribed circle. (It’s the same radius as Circle D.) Extend the sides of the master square to intersect the far sides of this circle. Noting the cardinal points of the circle, connect adjacent ones as shown, so diagonal lines extend into the framing channel.

11. Ink in an 8-pointed star inside each circle.

12. And ink in the parallel lines extending from the stars in the middle of each side of the square, as shown.

Phase 3

In the final phase, we will fill in the corners.

1. Draw construction lines for the corners. As shown below, there are two lines in each corner. They both go from the tip of one of the mid-side-8-pointed stars to a specific intersection where the master square meets one side of a channel. (This is actually slightly wrong, but the error should be imperceptible. See Part 2 for a discussion of what’s going on here.)

2. Ink extensions of the channel lines out of the 12-pointed star and turning through a right angle onto one of these construction lines just drawn, as shown below.

3. Draw similar construction lines for the octagons. This is the only part of the pattern for which you need to draw a line without having two specific points to connect. You have one point, which is where a 1/24th radial, at either 30° or 60°, hits the master square. Through that point you need to draw a line parallel to one of the construction lines drawn in step 1.

4. Ink these in as shown.

And you’re done!

You now have a completed pattern which you can repeat above and below, as far as the space you need to fill. The frame provides an overlap for aligning adjacent tiles.

Now, having done all of that, you probably would like to know why we did some of the things we did. Why did they work?

If so, read on to Part 2.

The Mathematics of the Pattern on the Sala de la Barca Ceiling

(This is the math geek part about the Sala de la Barca ceiling. For instructions on constructing the pattern with compass and straightedge, go over to Part 1.)

In the process of figuring out how to draw this pattern, I ran into a lot of questions, and had to do more than a little math to answer them.

Two families of symmetry

One of the most interesting things to notice about this pattern is that it includes two distinctly different types of symmetry. There’s twelvefold symmetry inside the frame, but there’s eightfold symmetry on the frame.

They’re not incompatible, but eightfold and twelvefold patterns do things a bit differently. The eightfold stars have lines at 0° (horizontal), 45° and 90° (vertical), but the twelvefold stars send out lines at 0°, 30°, 60° and 90°. So we shouldn’t be surprised if there will be some weird little shapes in the corners where the 30° and 60° lines coming out of the twelve-fold stars meet and blend into the 0°, 45° and 90° lines coming out of the frame. Maybe a little hanky-panky to make it all work. More on that later.

Let’s look at the twelvefold part first. The major feature inside the frame is the 12-fold star. But this star is actually composed of two stars.

One is the {12/3} star that forms the outside of what we initially see as the star.

The notation “{12/3}” (a Schläfli symbol) tells us that there are 12 vertices equally spaced around a circle, and the star is generated by stepping around these vertices hitting every third one.

If you’re familiar with the stellations of a dodecagon (there’s a nice chart on the Wikipedia page), this one is recognizable as the {12/3} star because the internal angles of its vertices are 90°.

The other component is the {12/5} star that sits in the centre. It’s recognizable as a {12/5} star by its vertex internal angles of 30°.

The sides of the {12/5} star have been extended outwards in parallel pairs to the {12/3} star, forming 12 channels.

All of this twelvefold symmetry is set in a square frame based on eightfold symmetry. The frame is made from channels of parallel lines. The width of these channels is the same as that of the channels extending from the {12/5} star.

The frame has {8/3} stars at the midpoints of each side. The {8/3} stars are sized so the channels are extensions of their sides, as are the cardinal (North, East, South, West) channels coming out of the {12/5} star.

At the corners of the square are {8/2} stars, also sized to match the channels.

Dividing the circle in 24

The first challenge of drawing this pattern was divide a circle into 24 equal parts. (Twelve for the twelvefold symmetry, but then 24 because the halfway divisions are very useful too.)

By the time we got to phase 1, step 5.2, we already had the circle divided into eight parts (i.e., at 0°, 45° and 90° in each quadrant) simply by virtue of having drawn a square in it.

But how then to get the important lines at 30° and 60°? One can’t trisect an angle, so we can’t generate them by trisecting the right angle in each quadrant of the square. Instead, we got them by drawing through points A & B, as shown below.

But why does this work?

It works because A & B are intersections between circles that share the same radius and have their centres on each other’s edges.

Just as in the classic construction of a hexagon, equilateral triangles are made and it’s easy to show that the two angles have to be 30° and 60°.

Then, to finish up getting the full 24 divisions of the circle, we also needed lines at 15° and 75°. I have seen people suggest that in this case you can draw through the intersections of those same arcs with Circle B (points C & D, below).

But this doesn’t really work. The result is not 15° and 75°. This becomes apparent when we construct a triangle like this…

We’d be hoping that the two angles at the base of this isosceles triangle would be 75°, leaving a perfect 15° between point C and the x-axis.

But in fact for such an isosceles triangle, where the two equal legs are twice the length of the base, the peak angle is…

…leaving 75.5225° for the two base angles. The complementary angle is 14.4775°. It’s close to 15°, but not quite right. And visibly so: if you draw the actual 15° line next to it you can see the difference.

So we resorted to copying the arc length for 15° from other parts of the circle.

How many independent variables are there?

It does initially look like the width of the channels is more or less independent of the length of the side of the square. It appears one could, for example, narrow the channels and make a variant pattern where the {12/5}, {8/3} and {8/2} stars shrink, and the {12/3} star grows larger. Something like this…

However, in the Sala de la Barca ceiling there’s no such latitude. The proportions of the twelvefold pattern inside the square, and the eightfold pattern of the square frame itself, are locked together by the fact that the dimples of the {12/3} star at the semi-cardinal points (NE, SE, SW and NW), fall exactly on the quarter-centres of the square.

This is why we drew Circle B through the quarter-centre of the square, and then built the {12/3} star outside it.

This requirement fixes the size of the {12/3} star relative to the square. The size of the {8/3} stars then depends on the space between the {12/3} star and the square. And the size of the {8/2} stars, the width of the channels that frame the pattern, and the width of the channels that emanate from the {12/5} star all depend on the size of the {8/3} stars. The whole thing is connected. Once you draw one part, you’ve determined the size of all parts.

Building a {12/3} star on its inscribed circle

In Phase 1, step 6, we determined the radius of Circle C (which is the circle circumscribed around the {12/3} star) by drawing line segments between points on Circle B, and extending them out until they met. Circle B was divided into 24 equal arcs, and we joined points in a skip-4 pattern. The intersections of the extended lines told us where Circle C was. Why did this work?

As noted above, we knew that Circle B was the inscribed circle for a {12/3} star. Our problem was how to generate a {12/3} star on it. It’s a problem in what I would call outward step stellation (as opposed to either inward stellation or outward jump stellation).

Outward step stellation takes us from one stellated figure to the next by extending sides. For example, you can stellate a {12/1} star (aka a dodecagon) outward into a {12/2} star. You do it by extending sides until they meet.

*Dodecagon sides in black (left), extended (in red) to their first meeting. They form a {12/2} star (right). N*ote *that, in the process, the first figure’s circumscribed circle became the second figure’s inscribed circle*.

With the new {12/2} star, there’s still a vertex every 30° as you go around, but the vertices have been offset by 15°, or 1/24th of a circle.

The dodecagon had interior angles of 165°; the vertices of the new {12/2} star have interior angles of 120°.

Now, we draw a new circumscribed circle around the {12/2} star, and then we are ready to stellate outward one more step. We get a {12/3} star by extending the sides to their next meeting.

*The {12/2} star’s sides extended to make the {12/3} star. The {12/2} star’s circumscribed circle becomes the {12/3} star’s inscribed circle.*

The {12/2} star had interior angles of 120°; the new {12/3} star has interior angles of 90°. The vertices are back on the original 1/12th divisions.

It was this, the second outward step of stellation, that we called into play to get from Circle B to Circle C.

Circle B was the inscribed circle of a {12/3} star. Therefore it was the circumscribed circle of a {12/2} star. If we could draw just a few sides of that {12/2} star, and extend them to where they met, we would see where the {12/3} star’s vertices were going to be.

With a {12/2} star, sides join points on the circle that are 2/12ths apart.

That’s also 4/24ths apart.

Since we had our circle divided into 24, we could locate suitable pairs of points for a {12/2} star by skipping ahead four, and then extending them out.

This turned out to be a straightforward way to generate a {12/3} on Circle B.

The ratios

The math to calculate their relative sizes of the elements is not terrible. From a mathematical point of view, we should probably take the radius of Circle D as 1, and express the sizes of the other elements in terms of that. But from a design point of view, we are likely to need to base everything on the size of the square, so we’ll start with that.

If the side of the square is s units, then we know the radius of Circle A.

The radius of Circle B is then half of that.

As we go from the circle circumscribed around a {12/2} star (Circle B) to the circle circumscribed around a {12/3} star (Circle C), we can calculate the proportional amount by which the radius increases.

Hence…

And we can calculate the radius of Circle D because it is the space left between Circle C and the master square.

So r_D is about 0.066987 times s.

The width of the channels will be √2 times that:

…or about 0.094734 times s. So if your square is 100 cm on a side, your channels should be about 9.5 cm wide.

Now we can go backwards and express everything in terms of r_D. If r_D is 1, we get…

Drawing the channel lines

Drawing the channel lines that run North–South and East–West was easy because we were simply extending the sides of the {8/3} stars. But to generate channel lines at 30° and 60° we employed the technique of copying Circle D onto each vertex of the {12/3} star.

But those places where the channel lines will hit the {12/3} star: aren’t they just the midpoints of the sides of the {12/3} star?

They are very close. But not exactly. We can prove that.

Just as when they are coming into an {8/3} star, the channel lines hit the sides of the {12/3} star a distance of r_D down from the vertex point.

For this to be the midpoint, then each side of the {12/3} star would have to be twice r_D. We can figure out the side of the {12/3} star from the triangle we drew above.

Whereas…

So the length of a side of the {12/3} star is almost twice r_D, but not quite.

Constructing the lines in the corners

In the corners we need construction lines for what I think of as the “wings,” (in red below) and the octagon (in blue).

Both the pattern at tilingsearch.org, and the photograph from the Wade archive, suggest that these lines are extensions of lines in the adjacent stars.

This makes sense—everything in the pattern should be at 0, 30, 45, 60 0r 90°—and these lines falling on one of those headings makes them parallel to other arms of the star, and produces the pleasing square motifs that catch your eye.

Maybe it’s just me, but I kind of like these squares.

It also means the channel in the octagon is half the width of the main channels, which seems like a nice relationship.

But, I find it a little mathematically creepy that a line extended from a {12/5} star should hit the vertex of an {8/3} star on the border of the next pattern. Would it really?

Let’s construct the triangle XYZ as seen below, spanning two adjacent full patterns. We’re curious about the angle θ, which should be 30°. If we can determine the lengths of the two legs of this right triangle, h and t, we can compute the ratio t/h and compare it to the tangent of 30°. We should get:

First we’ll calculate h, the horizontal leg of the triangle between vertex Y and vertex Z. The distance between two pattern centres is s, so

Calculation of g is straightforward. Vertex X is at the point of the {8/3} star, so

Bearing in mind the definition of r_D we worked out above,

If s = 1, g is about 0.114237.

Calculating j is a bit more involved. We start by noting that vertex Z falls on the circle inscribed in the {12/5} star, which we can call Circle E. So j is the radius of Circle E (r_E) .

From the discussion of outward step stellation above, we know that if Circle E is the circle inscribed in the {12/5} star, it is also the circle circumscribing the {12/4} star. Because there are specific ratios between successive stellations, we can get the radius of Circle E (r_E) if we first get the radius of the circle inscribed in the {12/4} star, Circle F. We’ll call its radius r_F.

The {12/4} star was generated in the first place by the intersecting channels of width C, so we can see that r_F is closely related to C.

And then it remains to figure out the relationship between the inscribing and circumscribing circles for a {12/4} star. The key point is that a {12/4} star has inner vertex angles of 60°.

Hence

That’s a surprising and elegant relationship, that Circle E, which is never drawn but is readily apparent to the eye, has the channel width as its radius.

Now we can calculate h:

That’s ugly! But it’s usable. If s = 1, h is about 0.790911.

The other leg of the triangle is of length t:

So, bearing in mind the definition of C above, we have

If s = 1, t is about 0.452633.

If we divide t by h, we get

We were hoping for the tangent value of 30°, or 0.577350. Instead we get 0.572293, which turns out to be the tangent value for an angle of 29.78°. Just a hair shy of 30°.

In other words, if you extend the channel line from an adjacent star, a line that really is at 30° to the horizontal, it would come in a bit high for the vertex X in the diagram above: 3.5 mm high in a pattern where s, the side length, is 100 cm.

Although I was pretty picky above, on the difference between 14.4775° and 15°, here I’m going to say that this angle difference (29.78° vs. 30°) is close enough for practical purposes — given that in the real world materials are coarse, and geometric pattern lines typically have significant thickness. Introducing a tiny “course correction” as the construction line crosses the boundary between adjacent patterns, a correction of about half a degree, is not going to be noticeable. It helps that the channel line does not extend to the pattern boundary, so you don’t see the course correction.

Hence in Phase 3 I created the construction lines in the corners by connecting the tip of one of the mid-side {8/3} stars to a specific intersection where the master square meets one side of a channel.

Conclusion

Well, it’s probably time to wrap up this overly long post. If you are here, thanks for reading to the end! It might not hurt of someone double-checked my math…

Constructing Bourgoin’s Figure 171 – Part 2

Now that we know our way around the pattern (go back to Part 1), it should be fairly straightforward to construct with a compass and straightedge. But be aware: any pattern that requires you to construct a pentagon is an advanced challenge. They are trickier to make than squares or hexagons.

Here’s what we want to draw:

There are different scenarios for beginning. You might know where you want to site two rosette centres, and that will then determine the size of the master triangles and the rest of the layout. This is the scenario I’ll go through here. But, alternatively, you might want to scale the pattern so a certain number of rosettes will appear in the space you have; or you might have an exact size that you want the diameter of a rosette (or a pentagram) to be. Each of these is, in a sense, a different problem.

The first nine steps will take us from drawing one leg of a master triangle to having a compass set to the radius of a rosette circle.

1. Pick two points that will be adjacent rosette centres, and draw a line through them. We know that one of these will be at the apex of a master triangle, and the other will be one of the remaining corners (Figure 1). For the moment, we’ll call the triangle side that connects them the main axis.

2. Bisect the main axis between the rosette centres, and establish a midpoint (Figure 2). The midpoint will be useful later when we want to draw diagonals.

3. Create circles centred on the two rosette centres, each with a radius that takes it to the main axis midpoint. (Figure 3).

Note that you could actually draw these two circles with any radius. Our purpose in drawing them is simply to give us the ability to draw 20 divisions of a circle (i.e., at a spacing of 18°), and once we have those we won’t be using these circles any more. Drawing them to meet at the main axis midpoint has the advantage that these are large circles, which should make the 20-fold division more accurate.

Now we need to construct twenty evenly-spaced rays from each circle centre.

4. Construct a pentagon in one of the circles so that one vertex touches the axis midpoint (Figure 4). (You can find methods of constructing a pentagon within a circle at many places on the internet, including Wikipedia’s page on “Pentagon.”)

In the process of doing this, your compass will become set to the length of a pentagon side.

5. Without changing the span of the compass, use it to draw a pentagon in the other circle (Figure 5).

6. Continue using the same span to draw a second pentagon in each circle, with one vertex touching the place where the axis exits the circle. You now have a 10 pointed star, or 10/2 star, in each circle (Figure 6).

7. Divide each circle in 20 sections by drawing lines from the rosette centre through every point of the 10/2 star, and through each of its 10 dimples. You now have a line meeting the circle every 18°. Be sure to extend, outside the circles, the first rays adjacent to the main axis until they intersect (Figure 7).

8. Create a circle centred on this intersection, using a radius that will take it exactly to the main axis midpoint (Figure 8). This is the pentagram radius.

9. Note the point where the first 18° ray from one of your circles enters the pentagram circle (Figure 9). The radius of the rosette circle is the distance from the rosette centre to here. Draw a circle with this radius around each rosette centre.

You can even erase the initial circles you drew.

We’re now ready to extend the grid of master triangles, to locate other rosette centres and to put rosette circles around them. This occurs in the next three steps.

10. Using the appropriate rays from the two rosettes you’ve already sited, extend the grid of master triangles (Figure 10). Draw a rosette circle around each vertex, and construct the twenty evenly-spaced rays. Remember, you already know the rosette circle radius, and ray spacing can be copied from one of the other rosette circles. (E.g., place one leg of your compass on the point where one ray leaves the circle, and the other leg where the fourth next ray leaves the circle. Use this distance on a new circle to set up rays.)

11. Each pair of rosette centres allow you to construct a third. In my case, I have room for four, and all other possible centres are off my page (Figure 11).

12. You already know the distance from a rosette centre to the midpoints of the master triangle legs, so set your compass to that and add in leg midpoints. You can then add the diagonals that connect the midpoints (Figure 12). For triangles with missing vertices, you can still place a midpoint on a leg from the nearest rosette centre. Notice that even though you’ve never figured out where the midpoint of a master triangle base is, the intersecting diagonals will lead you to it.

Rosettes in place, it’s time to construct 10/4 stars in them, and extend the lines from these stars. This takes place in the next four steps.

13. Although you have twenty rays from each rosette centre, only ten of them are important from now on. These are shown above, by circling their intersections with the rosette circle (Figure 13).

14. Connect each to the vertex four along (Figure 14). This is the “10/4 star”.

15. You want to extend some of the 10/4 star lines outside the rosette circle. How far to extend each line is a bit weird. It’s okay if you extend a line too far, because you will wind up erasing a lot of construction lines anyway. But ideally, it looks like this (Figure 15, above).

Lines at 12:00 and 6 o’clock (a) go until they hit the next master triangle base. Lines going out parallel to master triangle legs (b: at 1:00, 5:00, 7:00 and 11:00) go out only as far as a midpoint bisector of that leg—at which point they meet identical lines coming from the next rosette. Lines that cross at 3:00 and 9:00 (c) go out just a short way: as far as the vertical line coming up or down from an adjacent rosette. But their opposite ends (d) go a long way: all the way to the midpoint of the next master triangle leg they encounter.

16. Having extended the 10/4 stars in each rosette you’ll have something like this (Figure 16).

17. As always, the final pattern is made by selecting some of the construction lines for inking, and the rest for erasure. We can get partial success with the construction lines we have so far (Figure 17).

In a virtual space, you can just go on creating master triangles and rosette centres as far as you like. But in the real world, you come to the edge of the page, or wall, and there are still areas in the corners where the adjacent rosette centres are off-page, and you do not have the lines coming out of them to guide you. This is where it gets doubly interesting, as the physical limitations of the space in which you are working create additional geometric problems.

The remainder of the process now is just working out how we can extend the pattern into these spaces.

This process will be different for every space. I’ll show how I completed this for the rectangular space I’m working in.

18. Drawing additional pentagram circles is quite handy. Their centres are a known distance outside the rosette rays, and their radius is known from way back in step 8. We can even locate those in the far corners because their centres lie on a line passing through other pentagram circle centres, and the spacing between centres can be measured with the compass elsewhere in the pattern (Figure 18).

19. Each of these circles can have a pentagram inscribed in it—we know the spacing between the vertices from other, existing pentagrams—and then we can extend the sides of that pentagram to form the guidelines that we need (Figure 19).

20. Some more inking and erasing, and we’re almost done. There are only four small areas near the corners (marked with question marks in Figure 20) yet to be finished. We know what should be there, but we don’t yet have the construction lines. At this point I’m inclined to use the compass to measure spaces and lengths out of completed parts of the pattern and sketch/copy them into the areas that need to be filled.

Whew, done!

Constructing Bourgoin’s Figure 171 – Part 1

Just veering off into geometry here….

In November I was watching Eric Broug, an Islamic geometric design guru, give a talk online at an Islamic art conference, and I noticed that behind him they were projecting an interesting pattern on the scrim. I froze the video and grabbed a screenshot…

What the heck is this? How do I look this pattern up? How do I find what pattern this is, and how to draw it?

I have a few books of geometric patterns, but this was not in Eric Broug’s book Islamic Geometric Patterns, nor in Daud Sutton’s Islamic Design. So I took the tack of searching Jules Bourgoin, the 19^th century Frenchman who catalogued Islamic geometric patterns, and whose 1879 book, Les Eléménts de l’Art Arabe, is available for free on the internet. Bourgoin’s book provides that useful function, like Köchel numbers for the works of Mozart, of giving us a handy identifier, albeit a random number, for many patterns.

After some swimming back and forth in a sea of patterns, I finally recognized it as his Figure 171.

Bourgoin gave some enigmatic and tense (and French) instructions about how to construct this pattern, and I couldn’t make head or tails of them. So next, a web search on “Bourgoin Figure 171.”

This did not yield instructions about how to construct the pattern, but it opened a number of satisfying rabbit holes. One was a talk by Lars Erickson at the 2021 Bridges conference, who was constructing this pattern (among others) using an extended girih tile site. He gave several references about this pattern, including a page at http://tilingsearch.org dedicated to it. Here I could see that it is fairly common, turning up in the great mosque in Damascus, the Kalyan mosque in Bukhara, the tomb of the Mughal emperor Akbar in Agra, India, and even on the spine of one of my favourite cookbooks, Taste of Persia by Naomi Duguide.

A video posted by Samira Mian was also helpful, even though it was not about this exact pattern. It clarified that the pattern grows out of five-fold symmetry, and that subdividing the circle into 20 equal portions is key.

After long period of staring at the pattern, I think I can do an analysis. We’ll do construction in part 2.

Analysis

Here is the pattern, both as a line drawing and as a tiled pattern (Figure A). Believe it or not, you can construct this with a compass and a straightedge. No measuring of lengths or angles required.

The first thing to notice is that we have a regular, repeating pattern of rosette centres with 10-fold symmetry. Each rosette has a 10-pointed star in its centre (yellow, in Figure B) surrounded by ten points (orange).

Although it initially might appear that the rosette centres are placed at the vertices of equilateral triangles, they are in fact, as Bourgoin notes, at the vertices of isosceles triangles whose angles are 72°-54°-54°. (Bourgoin’s text about Figure 171 says, “Plan isocèle ou losange. Le triangle isocèle a son angle de base égal aux 3/5 d’un droit.” Isosceles or diamond plan. The isosceles triangle has a base angle equal to 3/5 of a right angle. I.e., 54°.)

These isosceles master triangles alternate, apex up, apex down. Be sure to recognize which sides are the legs of the isosceles triangles (equal in length, meeting at the apex), as opposed to the third side which is the base (joins the two 54° angles, and is slightly longer). Notice that the head-to-head kites (red) always occur on the base, whereas the legs run through two opposite-facing petals (green).

In addition to the 54° and the 72° angles in the master triangles, the pattern is full of angles with measures like 18°, 36°, 108° and 144° (Figure C). These all reinforce the impression that the pattern will be constructed from pentagons and 5-fold symmetry. The full circle of 360°, divided by 5, 10 and 20, respectively, gives 72°, 36° and 18°. The angle 54° is, in turn, three 18’s, and 108° and 144° are doubles of 54° and 72°. So, all in a family. The “5” family.

Each rosette centre features a 10/4 star, which gives us the rosette points (orange). I’m calling them “10/4 stars” after Magnus J. Weninnger, who used this kind of expression in his 1971 book Polyhedron Models. A “10/4” polygon is a star formed by connecting each decagon vertex to the fourth next vertex. A “5/2” polygon, by the same logic, is the star formed by connecting each vertex of a pentagon to the vertex two along: in other words, a pentagram.

As shown above in Figure D, the majority of construction lines for the pattern are extensions of the 10/4 stars. So once we draw these stars, we’re going to have most of the lines we will need to draw the pattern.

These 10/4 stars themselves can be drawn if we can construct circles of the right radius around the rosette centres. It won’t do to use just any circle: the radius has to be just right so the extended lines from the 10/4 stars meet and form 5/2 stars, or pentagrams (purple, in Figure E above).

In fact, this is the key puzzle of drawing this pattern: the relationship between the length of a master triangle leg and the radius of the rosette circles.

But, we don’t need to determine the radius of the rosette circles first! Instead we should first determine the radius of the circles enclosing the pentagrams, the pentagram circles.

If we construct 20 equally spaced rays coming from each rosette centre, the rays will be 18° apart (Figure F). These rays alternate in function: one coincides with a point of the 10/4 star (orange), and the next coincides with the centre of a “petal” (green) between two points. We can think of these as point rays and petal rays.

Notice that each pentagram occupies the space between two petal rays, and it does so simultaneously for two different rosettes. In other words, it occupies 36° of arc from the point of view of two different rosettes.

One petal ray lies along the master triangle leg connecting these two adjacent rosette centres (let’s call it the main axis). The next ray, a point ray, comes out at 18° from the main axis. Note where this ray intersects the corresponding ray from the other rosette. This location, halfway between two petal rays—and this is true looking from either rosette centre—is the centre of a pentagram circle.

The radius of the pentagram circle, and of all pentagram circles (r, in Figure F) must be the distance from that centre to the midpoint of the main axis.

Once the pentagram circle is drawn, the radius for the inner circles (R, in Figure G) falls out. It is the distance from the rosette centre to where the first point ray meets the pentagram circle.

Now we see the the relationship between the length of a master triangle leg and the radius of the rosette circles. It’s governed by this 144°-18°-18° isosceles triangle, where the main axis is the base, and it involves subtracting the triangle’s height (r) from the length of one of its legs.

We can quantify this (it’s interesting, but not of practical value in constructing the pattern) and say that if the length of the main axis is 1, then

If we look at the relationships between the pentagram circles and the inner circles, we can see they pack nicely. While most of our construction lines will be based off those 10/4 stars in the rosette circles, the pentagram circles may come in handy as we reach the edges of the space we’re working in, in places where we cannot draw a rosette circle because its centre is off the canvas, board, wall… or whatever medium we are working on.

The kite pairs along the bases of the master triangles clue us in to a few construction lines which are not generated by the 10/4 stars at each rosette. These are diagonals which cut through the pattern. They form the edges of some pentagrams, and the “noses” of the kites.

These diagonals are actually a secondary grid, of the same spacing and direction as the master triangles. The symmetry of the features that lie along them tell us that they are based on connecting the midpoints of the legs of the master triangles.

Finally, what wallpaper group does this pattern belong to? It has two axes of reflection (blue, in Figure J above), and three 180° rotational centres (red diamonds). So it’s cmm, also known as 2*22. The basic unit of repeat is shown above, a 36°-54° right triangle, or half of a master triangle.

It’s kind of unexpected to find that a pattern built on 5- and 10-fold symmetry has a repeat that is basically rectangular. Maybe I should have seen that coming, though, from the underlying pattern being isosceles triangles arranged in rows where they alternate apex-up, apex-down.

On to part 2, Construction.

Octagons in Baku

Besides the window in the Divankhana, I saw a lot of geometric design in Baku based around octagons.

For example, consider this pattern in a window in the external courtyard wall at Baku’s Taza Pir mosque.

DSCF5870

This beautiful pattern, with its eight-pointed stars set within octagons, turns up on plate 67 in Jules Bourgoin’s 1879 Les Éléments de L’Art Arabe (which you can download from archive.org).

Bourgoin plate 67 3x3

It’s wallpaper group is the fairly common *442 (p4m) and it is generated by tessellating a square cell.

Bourgoin plate 67 single tile Construction of this pattern is straightforward. The eight-pointed star in the centre is inscribed in a circle whose radius is one quarter the side of the square. The vertices of the octagon are found by extending the sides of the star. The rest of the construction lines are extensions of the octagon sides, and lines connecting star dimples that are three apart.

bourgoin pattern 67 construction lines

But one enters the Taza Pir compound via a stairway from the street. The panels in the stairwell are related, but subtly different from the window design!

DSCF5873

Seen head-on…

taza pit entryway pattern 4-tile What did they do here? There is the same eight-pointed star in the centre, and the same enclosing octagon, but in this case they’ve trimmed back, to the borders of the octagon, the square that one repeats.

taza pit entryway pattern square2

As a result, the square tile borders stand out strongly as lines, and around each point where four tiles meet we get a big diamond holding four small diamonds.

taza pit entryway pattern 12-tile

(This also belongs to the *442 wallpaper group.)

Now, in the Old City, I came across a piece of octagon-based decoration that illustrates what happens if one doesn’t follow best practices, as explained by Eric Broug. This pattern involves starting with the same pattern as in the Taza Pir windows (above; a.k.a. Bourgoin’s Plate 67), but then repeating a somewhat random subset of it. In other words, incorrect tessellation.

DSCF5758

It would appear that the manufacturer of these pre-cast concrete blocks selected a piece out of the overall pattern that was not the all-important basic square, but rather a rectangle.

subset taken for pre-cast

Hence each of the concrete blocks looks like this:

single block

When you put them together, lines match up, but the effect of the original design is lost.

tiled pattern

The wallpaper group of this pattern would be*2222 (pmm).

Elsewhere in the Old City, there were pre-cast patterns that did tessellate pleasingly, again with octagons.

DSCF5747

But back at the Taza Pir mosque, I spotted this on an adjacent building, which I believe is Baku Islamic University:

Taza Pir

The grill pattern is octagons packed together, with squares in between; and eight radial lines emanating from the centre of each octagon. It’s basically the central column of this pattern:

Taza Pir end pattern

But look at what they did in the point of the arch. It’s beyond my knowledge to know whether this is best practice or not, but it is definitely creative.

A Little Geometric Creativity

There’s a nice old geometric pattern…

…that Eric Broug presents in his book Islamic Geometric Patterns as being from the Great Mosque of Herat.

Its construction is based on a hexagon, and the pattern repeats as additional hexagons are tiled around the first one. The underlying hexagons (which are not drawn in the finished pattern) are shown in red here:

This makes a fine star/wedge/triangle pattern that is quite satisfying to look at, and suitable for decorating things in your house.

The construction of one of the hexagon-based units of the pattern begins with drawing a circle, and then a hexagon within that circle. Once the hexagon is drawn, the lines of the pattern can be drawn within in it.

But in the real world, a medium on which we are drawing (or painting) always has edges, and those edges do not go along the edges of the hexagons. For example, let’s say we want to draw one of the pattern units, plus however much extra fits, on a square board or tile. Something like this:

Draw in the supporting hexagons and you will see that the one complete pattern is indeed surrounded by only parts of adjacent copies of the pattern.

It’s a bit of a construction conundrum. How will we fill in the portion of the pattern that lies outside the one central hexagon that we will have room to draw? How will we extend the pattern to the edges of the square when we cannot place a compass foot at the centre of any of the adjacent circles that would define the basic hexagons?

There is a way.

Let’s review how the pattern is made, and then examine how to extend it without being able to draw more circles. Follow along with your own piece of paper, pencil, compass and straightedge.

There is something magical in the fact that this pattern, with all of its exact proportions, can be constructed solely with a compass and a straightedge. You never need to measure an angle or a line.

To construct a hexagon, one begins by drawing a circle.

Within the circle inscribe a hexagon using the standard method of walking the legs of your compass (still set to the same radius) around the circle. We’ll call this Hexagon A. It is oriented so it is a point-up hexagon.

Draw radii through all the vertices of the hexagon. We’ll call these the ribs of hexagon A.

Now we need to construct a second hexagon of the same size, and on the same centre, but rotated 30°. To do this we will connect the midpoints of the six arcs from the hexagon already made.

If we had plenty of room, we would find those midpoints by first drawing six more circles (of the same radius) centred on the six vertices of hexagon A; and then draw lines from the centre of the original circle to the outer points of intersection of adjacent secondary circles. These ribs (purple, below) would show us the midpoints of the arcs.

But with the limited space we are working with, we need instead to bisect one side of the hexagon. This can be done with two compass arcs from adjacent vertices of hexagon A; we can then connect their intersection outside the circle with the circle’s centre.

This line bisects one of the arcs, and we can use that as a starting point for making a second hexagon. We’ll call this Hexagon B, and it is oriented so it is a side-up hexagon.

Draw ribs for hexagon B.

Note that the ribs of hexagon B (purple) go through the midpoints of the sides of hexagon A (red).

Connect the hexagon A midpoints to make a six-pointed star. We’ll call these star 1 lines (green in the next figure).

Similarly, we can note that the ribs of hexagon A pass through the midpoints of the sides of hexagon B. Connect the midpoints of hexagon B to make Star 2 — being sure, however, in this case to extend the star lines beyond hexagon B as far as the first edge of hexagon A you encounter. We’ll call these star 2 lines (blue in the next figure).

Finally, we ink certain portions of the Star 1 and Star 2 lines to make the final pattern. Note that the entire pattern lies within hexagon A.

The pattern that is inked consists only of portions of star 1 and star 2 lines. In fact we drew hexagons A and B, and their ribs, only to create the star 1 and 2 lines. And we drew the initial circle only to create hexagons A and B.

In order to extend the pattern we need to draw more star 1 and 2 lines outside our original hexagon. On a larger medium we could easily place the centres of new circles outside hexagon A, but on this limited surface where we are working we cannot.

It may help to look at what we need to find.

From our earlier construction of adjacent hexagonal cells, we know a few things about what should be happening. Star 1 lines leaving the hexagon (A in the diagram above) go a certain distance to a point Y (which we’re not really sure how to locate) and then turn through 120° to follow a line D.

Star 2 lines leaving the hexagon (B in the diagram above) go a certain distance to a point X (which we’re not really sure how to locate) where they cross a star 2 line (C) in the adjacent hexagons. At some farther point Z they also go through 120° turns.

As well, it’s valuable to make a few observations about what happens as these lines enter adjacent hexagons…

1. Hexagon A sides, when extended, become the ribs of other hexagons A, and vice versa.

2. Star 1 lines become some of the star 1 lines of adjacent hexagons.

3. Star 2 lines become some of the star 2 lines of adjacent hexagons.

This means that by extending these hexagon sides, ribs and star lines, we have much (but not all) of the construction information for the pattern outside the original hexagon.

What’s missing are lines C and D, and points X, Y and Z.

But now we can see that point X is the place where the extended hexagon A sides meet the extended star 2 lines. Drawing a line through them gives us line C.

Similarly, Point Y is the place where the extended hexagon A sides meet the extended star 1 lines. Drawing a line through pairs of points Y gives us line D.

At this point we have all the construction lines we need to ink the rest of the pattern outside the original hexagon…

…and then remove the construction lines.

It’s very satisfying to be able to construct this figure in a limited space, and to solve the problems associated. But now, as a bonus, it appear that the pattern presents us with a fascinating geometry problem!

As the pattern is extended, a new, larger hexagon has appeared, a hexagon that is similar to hexagon B, but is formed by star 1 lines that pass on the outside of the six small triangles. We’ll call it hexagon C. In the illustration below, hexagon B is purple, and hexagon C is blue.

It’s a bit of a puzzler, but I’ll just leave the problem here for the intrigued reader to solve. In terms of the side length of hexagon B, what is the side length of hexagon C?