Introduction

This article is concerned with the problems of dividing line and space into parts that are pleasing to the eye. In the article, I develop useful relationships between the parts or divisions. Aware that many, if not most, people are not interested in algebraic details, all of the results are accumulated in a table in the appendix.

Although this article is copyrighted, the reader may download and copy it for his own use. And it may be copied for any non-commercial educational use. Other webmasters also have my permission to freely place links to this article on their web sites.

The Problem And The Golden Mean

As far back in time as there are records, and almost certainly before, artists and craftsmen have been concerned with the problem of how to divide lines and spaces so that the resulting divisions are pleasing. Please refer to Figure 1. It shows a line that has been divided into two parts.

Figure 1. A line arbitrarily divided into two parts.

The length of the shorter division is ‘a’ and the length of the longer division is ‘b’. The length of the entire line is ‘a + b’. From these simple divisions, it is possible to form three ratios, which are generally unequal. The ratios are

and

One could also form the reciprocals of these three, but they are of no consequence here. As I said before, these ratios are usually unequal. For example if ‘a’ is 1 and ‘b’ is 3, then the ratios are

and

But now consider a second example. Suppose ‘a’ is 3 and ‘b’ is 5 so that ‘a+b’ is 8. Then the ratios are

and

Notice that two of the ratios are now close to being equal.

If we force the ratios and to be equal, the human mind finds the result pleasing and this ratio has a name. It is called the golden mean or the golden section.

Let us now proceed to determine the value of the golden mean.

To do that we set the ratios and equal.

Then we have this situation

where k is just some number that both ratios are equal to.

From this relationship, we can form two simple algebraic equations:

a = kb and b = k(a+b)

These can be solved by substituting the value kb in the equation b = k(a+b) to get

b = k(kb + b ) = k²b + kb = b( k² + k )

This reduces to

b = b( k² + k )

The b on each side of the equation can be divided out and we are left with

1) 1 = k² + k

which can be rewritten by factoring out a ‘k’ on the right as

1 = k( k + 1 )

1) can now be written as

2) k² + k – 1 = 0.

1) can also be written as

Equation 2) can be solved by a method known as the ‘quadratic formula’.

This is explained as follows:

Given an equation of the form Ak² + Bk +C = 0,

the solution is given by the formula

4) where the symbol means plus and minus

In equation 2), A = 1, B = 1 and C = -1.

Substituting these values into 4) above gives

The symbol is the square root of 5. That is the number that, when multiplied by itself gives 5. The square root of 5 is about 2.236. This is only an approximation because the decimal representation of the square root of 5 is infinitely long and it never repeats a pattern of digits.

Since negative lengths have no meaning, we can ignore the solution –1.618034 (but the magnitude of this number is significant, as we shall see later).

So, k = 0.618034 approximately. For use in woodwork, setting k = 0.618 is quite sufficient. Note that from equation 3) above we also have that 1/k = 1.618, an interesting symmetry. Note also that 1 – k = 0.382.

It is interesting to note that k is a little more than 3/5 and a little less than 5/8.

The Golden Rectangle

The Greeks were aware of this ratio and ascribed mystical properties to it. But the early Greeks were unaware of numbers like the square root of 5. That number is called an irrational number because it cannot be written as the ratio of two integers. i.e. cannot be written as a/b where ‘a’ and ‘b’ are one of the numbers 1, 2, 3, 4, 5, 6, … But the Greeks knew how to construct rectangles having this ratio. Here is how they did it. Refer to Figure 2.

Figure 2. A ‘Golden’ Rectangle

Start with a square, ABDC, where each side of the square is 2 units in length.

Find the midpoint of one side and label it as ‘m’. The two lines Am and mB are then each 1 unit in length. Using a compass, place the compass point at m and the pencil point at D. Then strike an arc from D crossing the extended line AF. Now let’s examine the ratios of the sides AB to AF. The length mB is 1 unit and the length BD is 2 units. From the theorem of Pythagoras, we know that the square of the side mD is equal to the sum of the squares of the other two sides namely mB and BD. In algebraic terms

So that

Since the length mD was used to strike the arc from D to F, the length mF is the same as mB, the square root of 5.

Now consider the ratio of AB to AF. AB is 2 units in length and AF is Am + mF.

So, AF is of length 1 + .

Hence the ratio AB to AF is the ratio 2 to 1 + . This ratio is 0.61803… which is the golden mean.

So if you need to mount a picture in a frame and have the frame be pleasing to the eye, then make the long side 1.618 times the short side. (Or make the short side 0.618 times the long side.)

How does this apply to woodturning? Lets consider the problem of designing a turned box that will have pleasing proportions. If we let the length of the body be 0.618 times the total length, then we should expect from the previous discussion that the box will have pleasing proportions. If we do that, then the lid will be 0.382 times the total length.

So, how do you start? One method is to make the box square. That is, let the total length of the box when it is closed equal the diameter of the box. So measure the diameter of the blank and multiply it by 0.618 and you have the length of the body of the box. Subtract the length of the body from the diameter of the box and you have the length of the box lid. All you have to do now is to decide how long the spigot will be and how much waste you will have due to chucking and parting off the box, and you have the total length of the blank required to make the box.

An example may help. It is much easier to work in millimeters when doing design than it is to work in inches and fractions of inches. So let us assume that we will make a box with an over-fitting lid from 2 inch or 50mm square stock and further assume that the finished box will be square.

The diameter of the box will be 50mm.

The body length will be 50 x 0.618 = 31mm or just under 1 1/4 inches.

The lid length will be 50 – 31 = 19mm or about 3/4 inch.

The total length or height of the box will be 31mm + 19mm = 50mm

or 2 inches.

Assume that we will allow 10mm or 3/8 inch for the spigot and that we will have 4mm of waste at each end for chucking and at the parting cut for parting off and cleaning up. We then have a waste of 12mm or 1/2 inch.

So, the blank length is 50mm + 10mm + 12mm = 72mm or just under 3 inches.

But this box will not look square. Because of the way we have divided the space, the box will look taller that it is wide. Try this for yourself and see!

How about a second example. This time lets make the height of the body the same as the diameter of the blank. This will produce a taller box. Lets also assume that we will make a box with an over-fitting lid and lets further assume that the spigot will be 10mm long and that there will be 12mm of waste. This is just as before, but the method of setting the height is different.

The diameter of the box will be 50mm.

The body length will also be 50mm.

How do we find the length of the lid and the total height?

Lets call the body length ‘b’ and the height ‘h’.

Then we know that

so that

So the height of the box is 81mm.

The length of the lid is 81mm – 50mm = 31mm.

The length of the blank is then 81mm + 10mm + 12mm = 102mm which is

4 inches.

It is a good idea to check the ratios, and here we have:

The ratio of the lid height to the body height is 31/50 = 0.62.

Perfection is 0.618, and we are close.

The ratio of the body height to the total height is 50/81 = 0.617.

Perfection is 0.618 and we are almost right on.

The ratio of the lid height to the total height is 31/81 = 0.383

Perfection is 0.382 and we are very close.

Approximations And The Fibonacci Sequence

Years ago a man named Leonardo Pisano Fibonacci (1170 to 1250) won a contest in mathematics to solve the ‘rabbits problem’. The problem is as follows:

You start out with two rabbits.

Any two rabbits can mate.

Any two rabbits that can mate will mate.

Each mating produces exactly one offspring.

At the end of ‘n’ mating cycles, how many rabbits will you have?

Leonardo started with the numbers 1 and 1 and then he added them to produce 2. Then to get the next number he added the previous two and so on forever.

His sequence of numbers is

1, 1, 2, 3, 5, 8, 13, 21, …

and it bears his name today. It is important in computer science and other scientific disciplines.

And, as you will see, it gives increasingly accurate approximations to the golden mean.

If you carry out the sequence above very far, the numbers become enormous.

So that sequence by itself is not useful to us. But consider the sequence of ratios:

1/1, 1/2, 2/3, 3/5, 5/8, 8/13, 13/21 … formed by dividing each number by its successor in the sequence.

These numbers are 1.000, 0.5000, 0.667, 0.600, 0.625, 0.6154, 0.619 …

The first two numbers are poor approximations to the golden mean, but the rest are quite good and are adequate for our use. If you look carefully at the ratio sequence, you can see that the numbers oscillate about the golden mean. 1.000 is larger, 0.5000 is smaller, 0.667 is larger but closer, 0.600 is smaller but closer still, 0.625 is larger and even closer and so on. See the chart below.

You may find it useful to remember this sequence and how to write it down starting with a pair of ones. You can then always reproduce the first few terms of the ratio sequence.

The sequence is also useful for dividing masses. For example, suppose you want to place a band of beads on the side of a turned bowl. Let the width of the band be 10 units. Then the width of the beads can be 2, 3 and 5 units wide. This technique will give a pleasing look to the beads. It is up to you whether to place the largest bead at the top or at the bottom of the band.

Figure 3. Fibonacci Ratio Sequence Chart

Other Ratios

Other ratios can be used to create boxes that are pleasing to the eye. If the lid is 1/3 of the total height, then the box will generally be perceived as pleasing and visually balanced. But if the lid is reduced in height any further, then the result starts to appear unbalanced and stingy.

Sometimes it is desirable to make the lid even larger than the approximately 40% figure of the golden mean. For example Richard Raffan makes his spillikan (pick-up sticks) boxes with a lid only slightly shorter than the base. I do the same in the stubby or ‘squat’ jewelry boxes that I make. It is usually unwise to make the lid and the base the same height for this causes most boxes to look top heavy. Again this is due to how our brains perceive images received from the eye. So if there is a need to make the top and base nearly equal, the proportions of 4 to 9 (0.444) or 3 to 7 (0.429) work well. That means the in the first case the lid is 4 units and the base is 5 units. In the second case, the lid is 3 units and the base is 4 units. An exception to the ‘avoid 50% rule’ is the mosque boxes that Richard Raffan makes. These boxes look fine when the lid and base are the same height. This is due to the fact that the primary mass in the lid is close to the join and it then rapidly tapers to a point. So, in this case, the eye is fooled and the result looks good.

Rulers And Micrometers

When learning to turn boxes, it is necessary to measure in order to accurately place the parting cut between the body and lid. The easiest way is to use a hardware store vernier caliper and measure the lid height from the tailstock end. It is much safer to turn the lathe off when doing this. After some experience is gained it is better to trust your eye and not measure. With experience your mind will know where the line should be and if you get it wrong by just a millimeter, your mind will scream at you. This has a down side in that when you want to work on a new design, your mind is uneasy with the placement of the parting cut. Just tell it to go away and leave you alone and it will. Pretty soon your mind will get used to the new design and will know where the cut should be made.

The ratios presented here are useful and you should try to remember them. The approximations to the golden mean are especially useful as are the first few members of the Fibonacci sequence, which are useful for dividing bands of beads. But one should not be slavish about numbers and ratios. It is fine to use the number 0.618 when designing boxes, but it is unnecessary to carry calculations out to several decimal places. I know that I cannot work to an accuracy any better than a millimeter and so I always round to the nearest millimeter. If one slavishly tries to work to numbers, the usual outcome is an artistically dead result. That is the piece may be technically correct but with no life; it is the little deviations and errors that give a piece life and artistic appeal. I advise you to read and understand the material presented here and to absorb it into your subconscious. Let it become a part of you. Use it when you are designing; be aware of it when turning; but make no attempt to adhere to numbers with mechanical precision. If you do this, then even when you use the same design repeatedly, the results will always be somewhat different.

Completeness

In mathematics and scientific circles, the Golden Section or the Golden Mean is assigned the Greek letter Phi, . In the discussion so far, I have not used this symbol. Instead I have used the letter k to represent the ratio a/b (See Figure 1). The Golden Mean is actually the ratio of the longer part to the shorter part. But I have used the reciprocal of this relationship. i.e. I have examined the ratio a/b, not the ratio b/a. I took the approach that I did because the numbers 0.618 and 0.382 relate directly to the height of the base and the height of the lid, respectively, of a box or other vessel whereas the ratio is the ratio of the total height to the height of the base, as we shall see.

To be complete and to satisfy the purists, I shall now discuss the ratio and show how it relates to what I have shown previously.

Consider Figure 1. If we say that the longer portion, b, is to the shorter portion, a, as the whole, a+b, is to the longer portion, b, then we have the equation