Analysing the Flower Tower
The Flower Tower is a wonderfully satisfying model by Chris Palmer. The geometry and beauty of the finished item is just captivating.
I first saw a Flower-Tower in 2000, but it was only at the BOS Bristol Convention that I finally learned how to fold one under the tutelage of John McKeever.
Being me, I immediately thought, ‘Can it be done with different numbers of petals?’ and this led me down the path that I am about to relate.
If you feel that mucking about with the geometry and the numbers will somehow undermine the beauty of the model then stop reading now. I understand how you feel, honest!
However if you feel that learning the structure may open up new opportunities for future flower tower type models, then read on.
We will concern ourselves eventually with the generic case of the n-petalled tower, but it helps to arrive there by looking at the familiar 8-petalled version.
Since the 8-petalled tower starts with a regular 8 sided polygon (it does, but with a square there’s a little extra to tuck away at the bottom!), it would seem reasonable to assume that an n-sided polygon will produce an n-petalled tower.
It doesn’t. As many of you have probably found already.
The 8-petalled version turns out to be a special case!
In the 8-petalled version, you will have noticed (during one of the interminable unfolding sequences) the following familiar ‘frog-base’ shape repeated at every edge of each of the concentric octagons,
and this appears at the edges of the polygon, flanked by triangles that are formed by folding the points of the polygon to the centre of the polygon. i.e. it looks like this (8-petalled version)
Let’s look at the angles now.
The 8-petalled version is awash with 45 and 22.5 degree angles, i.e. 360/8 and 360/16.
In particular, look at the frog-base part again, this time I’ve marked the angles.
Where a = 360/n.
For this arrangement to fall flat when folded, we can see that all the angles marked, MUST be the same. It also means that this is only a triangle when n = 8.
In general it requires a kite (or dart!) shape with angles equal to a, 2 a, a and (n-4) a, but when n = 8 one of these angles is 180 degrees. It will look a bit like this:
I will refer to this as the ‘petals kite’!
Going back to the polygon, we can now see that each vertex of the polygon has an internal angle of 3 a. In the case of the 8 petalled tower this gives an internal angle at each of the 8 vertices of 135 degrees, which gives an octagon.
So now you’ve folded the lowest layer and you’re folding in the points to the centre. Each time you do this, you create an extra little flap of paper that will become the next set of petals as well as the ‘height’ for the next layer.
But how much paper are we left with?
And how much do we need?
The paper we are left with is a rhombus with angles equal to 2a and 180 – 2a. I will refer to this as the 'height rhombus'.
To make the next set of petals we need the kite shape mentioned above.
The 2a angle is fine, the problem is with the 180-2a.
If 180-2a < a, then it is impossible to make another set of petals.
If 180-2a = a, then it is possible to make another set of petals, but there will be no height (i.e. a flat model)
If 180-2a > a, then all is well. We can make petals, and there’s some left over for height.
And now the moment you’ve all been waiting for.
How does this apply to other values of n?
For n=8, well we know what happens. We start with an octagon. To make petals we need a right angled triangle, and the ‘height rhombus’ is a square.
For n=7, the internal angle for 7 of the points is 1080/7. This is NOT a heptagon. It is a form of heptagonal star.
To make petals we need a kite shape of 102.856, 51.428, 154.288, 51.428 and we have a rhombus of 108.56, 71.44, 108.56, 71.44. Therefore it is possible to make a 7 petalled flower tower..
For n=6, the internal angle of the 6 point is 180 degrees (1080/6)! This is a flat line. The effect is that you start with a hexagon but you fold the edges into the centre.
To make petals we need a kite shape of 60, 120, 60, 120, and we have a rhombus shape of 60, 120, 60, 120. They are identical! This means that we can make petals, but the fold will be flat. Martin Gibbs has already done this.
For n = 5, it can be seen that the internal angles become reflex angles and make stars. It can also be seen that it is impossible to make another set of petals beyond the bottom layer due to the petals rhombus being larger than the ‘height rhombus’. The final result is actually a pretty star-based raised stand!
For n < 5, the angles become impossible to fold!
For n>8, it can be seen that an n-pointed star is required. And the petals kite will always fall within the height rhombus.
It won’t be much of a surprise to discover that the crease pattern is simply an n-agon with height rhombuses and petal kites attached to the corners. The next layer has the height rhombuses and petal kites placed where the previous layer’s rhombuses joined.
But what sizes should be used?
Here is a typical rhombus/kite:-
If we take C as the origin (0,0) and the length AB = BC = CD = DA = L
then the coordinates of the points are:-
The origin is then translated to the point (0, L), and the pattern is drawn. The coordinates are then rotated by a degrees about (0,0). Another n-2 rotations are required.
For the next level a new value of L is required.
The origin is then translated to the point (0, newL), a rotation of a/2 applied and the pattern is drawn. Another n-1 rotations of a are required.
Note. The extra a/2 rotation is only required on every second level.
This could be implemented as a program in almost any language, but the easiest way might be to use LOGO!
Or you could try creating the crease patterns yourself!
If you have Visio, you could use one of the following stencils:-