I like making things in wood, so it is worth working through the design of a wooden mast.

A mast needs to be as light and thin as possible so large defects in the wood such as knots cannot be allowed. This means most wood in the retail market in the UK isn’t suitable. I eventually tracked down two suppliers: Robbins Timber and Sykes Timber.

There are a few types of timber that could be used, but when availability of the necessary grade is taken into account it comes down to two:

The figures for Douglas Fir are as follows:

- Sitka Spruce. This has the highest strength to weight ratio of any wood so is the traditional wood for boat spars. However it is expensive, a bit soft and not as strong as…
- Douglas Fir. A bit heavier for the same strength but cheaper, harder (less chafe from yard and battens) and stronger for the same size.

Density | 510kg/m3 |
---|---|

Modulus of Rupture | 86.2MPa |

Modulus of Elasticity | 12.2GPa |

Crushing Strength | 47.9MPa |

Density allows us to calculate how heavy a given volume will be.

The Modulus of Rupture is the stress (force for a given area) where the fibres will break when subjected to a bending load. The Crushing Strength is the stress where the fibres will begin to lose their integrity under compressive load. When a mast bends one side is under tension and the other is under compression, so since the Crushing Strength is lower than the Modulus of Rupture we’ll use the Crushing Strength in our calculations. If you are feeling brave then the Modulus of Rupture will give a much thinner, lighter mast but a much weaker mast too!

The Modulus of Elasticity tells us how much the mast will bend under a given load. Useful information but not vital for a junk rig. We’ll calculate this if everything else holds up.

Note that all these figures will be different for every piece of timber. Sometimes very different – a factor of two or more is quite possible.

## Mast Design

We know that the maximum stress on a cantilever beam is at the fixed end, so we only need to consider the load at the partners (where the mast is fixed at the deck). Thus even if the mast is tapered we’ll only consider the diameter and stress at the partners. We can work out the maximum taper later, if everything else works out.

We’ve got to fit our mast through the slot in the deck, which restricts the mast diameter to ~70mm. Let’s calculate the strength of a mast of around 70mm diameter made of Douglas Fir.

[latexpage] [ M = S × sigma ] where:- M is the righting moment our mast can withstand;
- S is the section modulus of our material;
- $sigma$ is the yield stress of our material; we’ll use 47.9MPa as the figure for Douglas Fir.

We can find our section modulus S from Wikipedia:

We can use this formula in a spreadsheet to generate a table of M against a set of interesting diameters:

Diameter (mm) | Maximum moment (Nm) |
---|---|

65 | 1,273 |

70 | 1,589 |

75 | 1,955 |

80 | 2,373 |

85 | 2,846 |

Thus we want a diameter of around 85mm to give us a strength of over 2500Nm. It is also worth emphasising that Douglas Fir can vary widely in physical properties, so a margin is needed to allow for this.

We have got an issue with this diameter in that the gap in the deck is only 70mm, but we can probably figure out a way round that limitation. Let’s calculate the weight.

The top section of our mast will be tapered. Lets assume that the top diameter is 50% of the bottom diameter; i.e. 42.5mm. A tapered mast with a circular cross-section is a truncated cone, so let’s find a formula for the volume of a truncated cone: [ V = frac{pi h}{3} (R^2 + Rr + r^2) ] Where:

- h is the height of the cone, or the length of the tapered bit of our mast, which is 5m for my mast;
- R is the radius of the base of the mast, or half the diameter of the base of the mast, which is 0.0425m;
- r is the radius of the top of the mast, or half the diameter of the top of the mast, which is 0.02125m.

Thus [ V = frac{pi times 5}{3} (0.0425^2 + 0.0425 times 0.02125 + 0.02125^2) ] Thus [ V = 0.01655m^3 ] and weight W: [ W = rho V ]

Where:

- $rho$ is the density of the material, which is $510kg/m^3$

Thus: [ W = 510 times 0.01655 = 8.44kg ]

A weight of 8.44kg isn’t bad. This is only the tapered section but the bottom section shouldn’t add too much more and is low down so we don’t have to worry about weight here so much – it won’t affect the stability of the boat too much.

We do have a problem with this mast – it won’t fit through the 70mm slot in the deck. One solution is to use a rectangular section through the deck – it could be up to 160mm front to back even if it can only be 70mm side to side. Let’s work out the yield moment for that section.

Revisiting the Wikipedia page on section modulus, we discover that the section modulus for a rectangular section is:

[ S = frac{bw^2}{6} ] Where:- S is the section modulus for our rectangular section;
- b is the breadth of the rectangular section; in our case the front-to-back distance of 160mm;
- w is the width of the rectangular section; in our case the side-to-side distance of 70mm.

We’re calculating the section modulus for the side-to-side loading as this is likely to be the weakest point.

Adding the values in we get: [ M = frac{47.9 times 10^6 times 0.160 times .070^2} {6} ] [ M = 6259Nm ]This is plenty of support – far above the 2500Nm required – so we can probably reduce the front-to-back dimension somewhat. Playing with these formulas in a spreadsheet indicates that 100mm is probably enough, although we do need to be careful about the diagonal loading of the mast where the section modulus is substantially less.

I made a prototype mast base out of an old fence post to see how it might fit.

It certainly fits ok and seems very strong. Heavy though – it weighs about 3.5kg. I suspect that the wood is Douglas Fir or Larch.

The trickiest bit to make would be the transition between the rectangular foot of the mast and the circular section required at the level of the sail. I can envisage the rectangular section blending into the circular section, but I’m not sure about how easy it would be to make it accurately.

Cost must also be considered. As mentioned above most wood in the UK has far too many knots to be considered. Even clear timber (knot free) tends to be fast-grown with only a few rings per inch; we want at least 9 rings per inch for decent strength. Spar-grade wood is expensive and phone calls indicate that the cost would be around £300 + VAT and delivery.

Construction would probably involve getting four pieces of timber of around 100×50 x 4m long. They would need to be scarfed to make two pieces of timber 5.6m long, then joined to create a single piece of about 100x100x5.6m. This would then be planed down to the final size. I’d need to factor in the cost of adhesive (epoxy or Resorcinol). Once everything was done I’d need to varnish it. Varnishing isn’t too hard but takes ages (one coat per day or less) and the varnish isn’t cheap either.

## Conclusions

A wooden mast is viable.

- It could be made strong enough, although the final strength would depend on the strength of the wood used.
- It would be light enough. The ability to taper the mast means that the top part would probably be lighter than the existing aluminium stayed mast.
- It would be reasonably thin at the top. The ability to taper the mast means that the top of the mast would be thinner than the existing aluminium stayed mast.
- It could fit through the existing hole in the deck, although this would complicate the mast design and manufacture.
- Cost is probably OK – certainly better than the carbon fibre mast.
- With my tools and experience it would be straightforward to make. The problem would be that my shed is somewhat waterlogged so I’d need to be careful about leaving timber in there too long. Varnishing couldn’t happen until the weather warmed up and the shed dried out.