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.

__WARNING!__

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!

I’ll explain.

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)

OK.

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.

Like this:-

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!

5
petalled flower tower crease pattern

7
petalled flower tower crease pattern

Or you could try creating
the crease patterns yourself!

If you have Visio, you could
use one of the following stencils:-

Flower
Tower Stencil for Visio 2002

Flower
Tower Stencil for Visio 5