Maxi sailing (Fuerteventura)
The purpose of the sails to make air flow smoothly in a curved fashion. Causing the air to go in a curve creates a centrifugal force on the sail.
For example, if the sail is 5m high and 2m wide at the foot and curved 20cm (10%) deep and affects a total of 5 cubic metres of air at a speed of 10 knots (Force 3-4 Beaufort scale), then the force may be calculated as follows:
Mass of air moving in curve = 5 cubic m x 1.225 kg/cubic.m = 6.125 kg
Speed of air, V = 10 knots = 10 x 0.5144 m/s = 5.144 m/s
The radius of the curved airflow may be calculated using this formula:
W = width of sail (2m) C = camber depth (0.2m)
Radius of curved airflowm R = C / 2 + (W x W) / (8 x C) = 0.2 / 2 + (2 x 2) / (8 x .2) = 2.6 m
Centrifugal force = M x V x V / R kg
This calculator below works out what centrifugal force applies to the sail and how this may be resolved into in-line driving force and sideways capsizing force.
Default values assume 10% camber and close hauled with sail 10 deg off the centreline.
The thickness of the air affected is assumed to be half the width of the sail along the foot, at the boom.
If you play with this a bit you will learn that the total force on the sail may be increased by increasing the camber from 10% to 12 %, or the total force decreased by reducing the camber from 10% to 5%.
Also, and rather important, that the more the sail is let out the greater the component of total sail force is in line with the boat. Don't keep the boom in too tight !
So, to trim the sail, let is out as far as possible, without allowing the smooth curved airflow to break down. Watch for smooth trailing of the tell-tails all down the leech (back edge) of the sail.
The camber may be adjusted up to about 12 deg, when the smooth airflow may start to break down. You may decrease the camber to reduce both driving and capsizing forces if you can't reef the sail or use smaller sail.
Note that the angle of the sail to the centre line of the boat varies as you go higher, due to twist in the sail. Use the angle about 1/3 of the way up.
Feedback welcome, click to email me firstname.lastname@example.org to say what you would like to add below...
If you can explain the difference between kg and N (Newtons) that would help. Do the results even approximate to reality ?
Page created 15 June 2018, amended 3 Nov 2018 ECJ (c) 2018 Copyright. All rights reserved.