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We have established the forces required to heel the boat under sail. Now we have to establish at what point or points these forces are acting on the mast. For simplicity at the moment we will assume that the vessel has just the mainsail set, no other sails. We will tackle more sails and two masts later on.
The simplest way of looking at this is to assume that all the forces acting on the sail were concentrated at the centre of pressure. We actually use the centre of geometric area of the sail as an approximation of the centre of pressure (see the article on Balance). This is called the Centre of Effort (CE).
On Design 152, with the rig we are considering for this article, we have the following sail areas and CE positions:
Mainsail Area (SAmain) | 19.000 | m^2 |
Mizzen Area (SAmizz) | 5.640 | m^2 |
Mainsail CE to Partner (Lmain) | 3.705 | m |
Mizzen CE to Partner (Lmizz) | 2.720 | m |
With only the mainsail set, the whole of the force to heel the vessel comes from the mainsail and the force (F) acting at the CE is found from the formula F = RM/L.
Fmax = RMmax/Lmain = 7,030.729/3.705 = 1,897.633 N
F30 = RM30/Lmain = 4,048.881/3.705 = 1,092.815 N
At this point we need to consider what material we are going to make the mast from and have a stab at the approximate size. When doing this for real, it is pretty simple to build a spreadsheet with appropriate formulae in it and then keep entering spar diameters and material data until the right safety factor emerges. For the purposes of this article. I will use the mast sizes that we derived from this for Design No. 152. Note that the ^ sign means "raised to the power of", so 123^3 means 123 cubed, that is the same as 123 * 123 * 123.
Wood – Douglas Fir, we need to know:
Modulus of Elasticity (E) | 13,400 | N/mm^2 |
Rupture strength (R) | 69 | N/mm^2 |
Weight (W) | 480 | kg/m^3 |
We decided on a outside diameter (D) of 150mm as a starting point. With hollow timber masts, the wall thickness is normally 20% of the diameter. This will give us a wall thickness of 150 x 0.2 = 30, and an internal diameter (d) of 150 - (2*30) = 90mm. We will also taper the mast so that the diameter at the head is 60% of the diameter at the partners. This doesn't affect the strength calculations (because we are only interested in the strength at the partners), but it does make the spar lighter, which will be of interest when considering which material to use on the basis of the weight of the spar. We will deal with the weight calculations later, once we have finalized the basic diameter.
The diameter of the spar gives us the distance to the neutral axis (y) – this is the axis about which the spar can be considered to bend. With a circular mast the neutra axis runs through the centre of the circle. So y = D/2. Now we can calculate the Moment of Inertia (I) and the Modulus of Section (Z) of the spar
The formulae are:
I = π(D^4-d^4)/64
Z = I/y or put another way, Z = π(D^4-d^4)/32D
Working these out for our Ø150mm spar, we have:
y = 150/2 = 75 mm
I = 3.142(150^4 - 90^4)/64 = 21,629,865.420 mm^4
Z = 21,629,865.420/75 = 288,398.206 mm^3
You will often find Moments of Inertia given in cm^4 because this simply gives more handleable numbers. To convert from mm^4 to cm^4 you have to divide by 10^4. Our figure above would thus be 2,162.987 which rounds off to 2163 cm^4. For our purposes here we will stick with mm^4.
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© George Whisstock. This article is for information only and may not be commercially reproduced in any form or used in any way without permission. Do not use this material as the basis for designing a mast without professional advice.
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