Chapter 3 The “theory” needed to understand model yachts
In this title I have used the word theory in inverted commas. I need to explain why I have done so.
It is a sad fact that in the eyes of many people theory is regarded as very suspect and is often greeted with some comment like ‘theory is all very well but what happens in practice?’ People expect much too much from theory and are unwilling to accept that most things are too complicated for us to develop a complete theoretical explanation. This is made worse by a long process in which those who first attempted to explain the natural phenomena of this world in words and symbols have been misquoted. Their work has been separated from the context which produced it so that those who try to understand it from books and lectures find much of it incomprehensible. Any links that may have existed between theory and practice have been allowed to wither and now the two are frequently independent. One of the objects of this book is to try to bridge that gap and use the theory to help us in understanding our yachts.
The body of knowledge available to us is very formidable in its extent. We need only a part of it and we also need some ideas from mathematics. An attempt must be made to be selective and I shall try to pick only the knowledge which is essential to our subject of model yachts but allow myself the freedom to include items of general interest in the text.
Mathematical equations and Units
There is one aspect of science and mathematics that we must use. It is the idea of representing a quantity by a letter. The quantity may have many values. Take, for example, the expression for the area of a circle. It is A = p.r . (Note that the full stop means “multiplied by”) The area of the circle is represented by A and the radius of the circle by r. p is the well known ratio of the circumference of a circle to its diameter. The advantage of representing the areas of circles in this way is that, if the radius of a circle is known, only a very simple calculation is required to find the area of the circle. The alternative would be a table of figures for area in terms of radius. This idea of relating quantities by such expressions is an essential first step in mathematics and the idea, when exploited using a computer, is very powerful.
However when such methods are used by physicists and engineers the quantity represented by the letter has units e.g. pounds per square inch or feet per second. Probably the most famous example of this is Einstein’s which, when decoded, says that if a quantity of matter of mass, m, is destroyed the amount of energy released, E, is equal to the mass times the velocity of light, C, squared. This immediately shows the value of the letter notation as a form of shorthand.
If E is to be calculated numbers must be inserted for m and C and the actual numbers will depend on the units used. For example C could be in mph, km/hour, metres per second and so on. A free choice is not possible. The units for m and C must be such that they give comprehensible units for E. Scientists call such units consistent units.
Some decision must be taken about the units to be used in this text. Unhappily for us the attempt to introduce the SI version of the metric system to the UK about 30 years ago has not been an unqualified success. The trouble is that all consistent systems of units produce some outrageously large or outrageously small units for everyday quantities. The SI system of metres, kilograms and seconds produces some pretty silly units for some ordinary quantities and in fact has found little favour with the ordinary people of those countries who use the metric system. Most people who might read this book will think in Imperial units of feet, pounds and seconds and use a mixture of Imperial units and metric units (not SI). This may in fact give an extremely useful hybrid system but, in a text, mixing units is a disadvantage and this text will be mainly in Imperial units.
We need to know something about mechanics in order to understand the forces that are applied to yachts. These will be introduced under appropriate headings.
The triangle of forces
This is fundamental because yachts are mostly subjected to forces that are not acting in the directions that we would like. We must be able to split a force into two or more parts to find the magnitude of the useful part. Diagram 3-1a represents the standard school experiment to illustrate the triangle of forces. Three strings are tied together at a point and draped over three pulleys. Weights are attached to the free ends of the strings and the system released so that it comes to rest. The forces in the strings are equal to the weights attached to them and they all pull at the same point. It is possible to represent these forces on paper as arrows having lengths that are proportional to the magnitude of the forces and directions indicated by the angles of the arrows. It is found that, if the arrows are joined head to tail, they form a closed triangle as in Figure 3-1b.
This has importance when considering, say, the rigging of a yacht but there is another consequence of the triangle of forces that we shall find to be useful.
That is the idea that any single force can be replaced by two others in any direction provided that the three forces, when represented to scale, form a triangle. The two new forces are called component forces or components for short. A force and its components 1 and 2 are shown in Diagram 3-2a. Note that, as these two components are to replace the first force, the arrows point in the opposite direction to the forces that would balance the force.
The most common use for this representation of a force by two components is when the components are at right angles. Diagram 3-2b shows the same force and typical pairs of components at right angles. All the triangles lie in a semicircle. (In a sailing rig the force on the rig drives the yacht along its course and causes it to heel. The force can be divided into two components and the two-headed arrows could drive the yacht and the others cause it to heel.)
When a yacht is sailing on a lake the water is stationary and both the wind and the yacht are moving. We need some way of deciding the direction of the wind relative to the yacht. In some ways the sailor of a model yacht is at a disadvantage compared with a sailor of full sized yachts. Both yachts could be fitted with a device to indicate wind direction (a burgee) but the sailor of the model will not be able to see his at the distances involved in model yachting. In addition a burgee only gives direction to which the on-board sailor can add an assessment of wind speed. The sailor of a model yacht needs a picture in his mind of the way in which the two speeds are combining to act on the sails of the yacht. For this reason we need to understand relative velocities.
If we are to use an example we may as well use a yacht. Let us consider a yacht moving steadily along its course in a steady breeze. Suppose that a balloon is released from the yacht to be carried along by the wind. We can easily represent the position of the balloon after equal intervals of time by the circles in Figures 3-3 a, b & c where the balloon is shown moving up the page. The yacht can be represented by five little outlines of the deck to show its position after the same intervals of time. This has been done for three different courses. Someone on the yacht watching the balloon will see it moving in the directions indicated by the lines with arrowheads. For each direction of the yacht the eye looks continually in the same direction and all the lines are parallel. They show the direction of the wind relative to the yacht. These figures show us how to find the direction of the relative wind but not how to find its magnitude. For this we need a simple way of finding relative velocities.
Figures 3-3 a, b & c almost tell us what to do. In Diagrams 3-4 a, b & c the speed of the wind is represented to scale and in direction by the vertical arrows; the speed of the yacht is represented to scale and direction by the short arrows. In each case these arrows are joined tail to tail. The resulting arrows show the speed of the wind relative to the yacht in direction and to the same scale.
The standing rigging of a yacht makes extensive use of framed structures.
There are two forms of framed structures. The simple framed structures are made up of lots of loosely-jointed members that form triangles. The loose jointing ensures that these members of the structure are either in tension or compression but not in bending or in shear. The other form of structure incorporates members that can withstand bending and shear. The forces in simple structures are easily calculated but where members withstand bending calculations can be very troublesome. We are unlikely to make any important calculations which is just as well as the mast is a member subjected to bending. Nevertheless a knowledge of how forces are transmitted through rigging is desirable.
An example of a framed structure is shown in Figure 3-5. It is the arrangement of a simple crane. A load suspended from the joint D puts BD in tension so it is a tie and could be a wire rope. The load also puts CD (the crane jib) into compression so it is a strut and must be constructed so that it does not buckle. BD also puts AB into tension so it could be wire and AB and BD together put BC into compression. It too has to be designed as a strut. The movement of the joints of this frame is dependent on the stiffness of the members. If they were all to be just strong enough to support the load the members would extend or shorten under load and the frame would move quite noticeably. However, if all the members were to be very strong the deflection would be quite small. The builders of yachts tend to the view that some give under load is desirable.
There are hundreds of framed structures in use in power pylons, bridges, aerial masts, tents and so on all available to be examined.
Centre of gravity
We shall need to understand the idea of a centre of gravity. This is really quite an obvious concept but it has its difficulties. Not the least of these is that it is not fundamentally connected with gravity.
It is a matter of observation that the Earth exerts a force on all objects and this force seems to act towards the centre of the Earth. We have no explanation for this but no one has ever observed any exception so we just accept it as a fact of nature.
Suppose that we have a long rod. At all times gravity acts on every piece of the rod and, if one were to carry such a rod so that it was horizontal, it is likely that one would grip the rod at the midpoint because then the forces on the two halves balance out. Indeed if the rod were to be made of wood it could be replaced by a much shorter bar made of lead and the carrier would feel no difference unless an attempt were to be made to rotate the rod about its midpoint. The impression is that the gravitational force is actually exerted at the mid-point. It is just a small step to talk of a centre of gravity (CG) as the point through which the gravitational force appears to act. It is an idea that can be applied to solids of any shape.
The position of the CG of a solid body can be found by practical methods. The body can be suspended by a cord and the body will come to rest with the CG in line with the cord. Suspension from a different point will give another line through the CG and the intersection of the two lines will give the position of the CG. Diagram 3-6 shows a yacht suspended first from its mast head crane and then from a yoke round the bulb and the deck at the mast. The CG is clearly aft of the mast and some way down the fin. There is no reason why centres of gravity should be within the envelope of a solid. The CG of a deck quoit is somewhere in the hole.
Of course a force other than gravity can be exerted on a rigid body and the body will have a CG even in the absence of a gravitational field. (The suspension of the body in the gravitational field is just a convenient way of finding the position of the CG.) Scientists recognise this by calling the CG the centre of mass. A force other than gravity may act in any way on a rigid body. A force that does not act through the CG produces an acceleration and a rotation.
For example, when the balls used in ball games are struck they experience forces other than gravity. The behaviour of the ball after it is struck depends to some extent where the force is applied to it. If, say, a golf ball is struck by a lofted club the club exerts a force on the ball that can be divided into two component forces. One due to the compression of the ball acts more or less through the centre of the ball, which is also its CG. The other is a friction force that acts tangentially. These are shown in Figure 3-7 and the combination of these two forces is a force that does not pass through the centre of gravity. The result is that the ball flies up into the air and starts to spin backwards. We see evidence of this as the ball pitches when, if the surface is right, the ball rolls backwards. A tangential force exerted on a quantity of fluid also produces rotation as we shall see when we look at sails.
Centre of volume
There is another useful concept that is allied to the centre of gravity. It is the centre of volume. A model yacht is made up of components that are of different densities and may well have a centre of gravity that is somewhere down the fin. But some bodies are made of the same material throughout. Suppose that we had two cuboids (shaped like a box) of the same size, one made of expanded polystyrene and the other of lead. Both would have centres of gravity at the intersection of their diagonals as shown in Figure 3-8. The fact that one would be many times heavier than the other may be irrelevant and then one might say that the things that the cuboids shared were shape, size and uniform density. Then, for many purposes, the cuboids can be regarded simply as volumes and the point at the intersection of the diagonals called the centre of volume instead of the CG. This idea can be extended to a volume of any shape especially the volume of water displaced by a yacht
Centre of area
A natural extension of the concept of a centre of volume is the centre of area although it has no special use in this text. Suppose that any shape, like that in Figure 3-9, is cut from a sheet of perspex. The perspex is of uniform thickness and the same density throughout. The suspension method can be used to find its CG thatwill be at the intersection of the dotted lines. No matter how thin the perspex may be the position of the intersection is in the same place. Sometimes it is more useful to think of it as a centre of area.
In order to understand ship stability we shall need the concept of a second moment of area. This sounds very complicated but it is just a bit of mathematics that often crops up in engineering. What we want is the second moment of area of the water plane section of a yacht.
Figure 3-10 shows such a shape. It is just a surface. The surface can be divided into lots of little bits like a mosaic. One such bit is shown as having area dA, (d just stands for “a small part of”). It is x from the axis. Frequently, during from some piece of engineering analysis, the value of the total of all such quantities as dA. crop up. It is called the second moment of area. It would seem to be a tedious quantity to obtain but there are simple ways of reducing the tedium so it is a quantity that can be evaluated.
This is a very simple concept that is often misunderstood. Engineers often have to envisage a force that is spread in some unknown way over an area (eg a slick tyre in contact with a road). If the force is divided by the area, the result is an average value of the force per unit area. For example, if a force of 10 pounds acts on an area of 2 square inches the average value of the force per unit area (in this case 1 square inch) is 5 pounds per square inch (psi). This is the average pressure. Probably the force not does have this value all over the area but varies in some way so that in places it is greater than 2 psi and others smaller. Then an engineer is forced to think of the area being divided into many small areas each with a particular pressure on it so that the force of 10 pounds is made up of lots of small forces on the surface all adding up to 10 pounds. If, however, it was known that the pressure was the same all over then the pressure equals 2 psi and in general the pressure p equals force/area.
All the surfaces of a model yacht are exposed to uneven pressures that exert forces on the yacht, for example on a sail. Then we shall have to invert the above process and think of these forces as the sum of small forces exerted on lots of small areas by a pressure which is pretty nearly uniform on each small area.
This is really quite straightforward but somehow it gets involved with Archimedes who was not interested in floating bodies but in trying to decide whether a crown claimed to be made of gold actually contained silver.
We might like to start by thinking about a simple geometrical body. Figure 3-11 represents a cuboid floating upright in water. (A wooden cuboid would not float like this unless it was fitted with a weight at the bottom to make it stable.) Suppose that the cuboid floats to a depth d. All the surface area of the cuboid below the surface of the water must be subjected to pressure exerted by the water. The pressure exerted on the sides changes with depth and, if we look at the two small surfaces drawn as little strips on two opposite faces at the same depth, we see that the pressure is the same along the strips and the same for both strips. So the forces on the strips are equal on both sides and in opposite directions. This means that they balance out. This applies to all the strips that make up the vertical faces. So there is no net horizontal force exerted by the water on the cuboid. The upthrust that opposes the gravitational pull on the cuboid (its weight) must be exerted wholly on the bottom of the prism.
Let us look at the bottom face. The pressure on this face is the same all over and it can be calculated. Let us call it p. If the area of the bottom face is A then the upthrust is p.A which must equal the weight of the prism.
The floating cuboid is obviously a special case in which all the sides were vertical. We must now consider shapes with curved sides. Figure 3.12 shows a cylinder floating on its side with its axis horizontal. Part of the cylinder is immersed and therefore subjected to water pressure. From the cuboid above we know that the two end faces are subjected to water forces that are equal and opposite. Two identical strips of the curved surface are shown in Figure 3-12. Both are all at the same depth and so all at the same pressure. In each case the pressure acts radially. End views of the cylinder are shown in Figures 3-13a & b. In 3-13a the pressures are shown acting on the strips and in 3-13b the two arrows represent forces (each equal to the pressure times the area of the strip) also acting radially. The horizontal and vertical components of these forces are also shown in 3-13b. It is evident that the horizontal components are equal and opposite and produce no net force. The two vertical components add and exert an upward force on the cylinder. The whole of the immersed surface is made up of pairs of similar horizontal strips and their net effect is to produce a vertical force to float the cylinder and no horizontal force.
Before we leave the cylinder it is worth considering such a cylinder when it is fully immersed. This could be arranged if the cylinder were to have a density greater than that of water and be suspended from two strings with its axis horizontal as in Figure 3-14a. From our previous work we see that the net force on the ends is zero. Now we must imagine two pairs of strips along the cylinder and the water forces exerted on them. One strip is shown in 3-14a and the ends of all four strips are shown in the end view of the cylinder in 3-14b. As the strips are horizontal, the pressure along each one is the same and the force on it equal to the pressure times the area of the strip. The forces on the upper pair are equal and the forces on the lower pair are equal. They all act radially. The vertical and horizontal components of the four forces are also shown in 3-14b and it is evident that for each pair the horizontal components balance out. The vertical components on all four do not balance even though the components on the upper strips act downwards. There is a net upward force on these strips. The whole cylindrical surface can be represented by such strips and the sum of all the vertical components is the upthrust on the cylinder. It is this upthrust that Archimedes showed to be equal to the weight of water displaced by the cylinder. This is the upthrust that acts on lead weights, fins and rudders.
The idea of dividing an immersed surface into strips of area all at the same depth is valuable but not really a route to calculation. It is better to look at the floating body in a different way. Suppose that we had an object of any shape floating at rest and that, in some way, we could take the body out of the water leaving a hole in the water having the shape of the wetted surface of the object. All the pressure forces which had been acting on the object would now be un-resisted but, if the hole were to be completely filled with water they could be completely and exactly resisted at every point. We could find the weight of the water needed to fill the hole from its volume and density and this must be equal to the weight of the original floating object. (This is the basis of the statement in physics that the weight of a floating body is equal to the density of water times the volume of water displaced by body.)
There is another useful deduction from this idea. If the water that filled the hole were to be a solid it would have a centre of gravity through which the weight could be considered to act. The net force exerted by the water on the floating object must be equal and opposite to this and so we have a method of finding the upthrust from water without using lots of strips.
Stability of floating vessels
In order to make the inevitable compromises in the design of a model yacht we shall have to consider its stability. As we shall have to explore the well-established methods used for full-sized ships we might as well start with a body of a convenient shape. This is a simple cuboid such as one might find as the base for a floating crane. It is shown in Figure 3.15a. In order to decide why it is stable we must look at it when it is heeled through some angle as in Figure 3-15b. Do not worry about how the new position might have been achieved, it is the outcome that interests us. Clearly if the weight of the cuboid is unchanged the only change has been in the shape of the submerged volume. Somehow this change has produced forces which tend to right the cuboid.
We said that, if we had a boat that was floating at rest and took the boat out of the water leaving a hole, we could fill the hole with water and restore the equilibrium completely. The water used to fill the hole would have a weight that acts through its centre of gravity and the upthrust opposing this would be equal and opposite and act through the same point. But the centre of gravity of the water in the hole is also the centre of volume of the displaced volume. So the upthrust acts through the centre of volume of the water displaced by the floating body. By common usage the centre of volume is called the centre of buoyancy and denoted B. Let us apply this information to our floating cuboid. In Figure 3-16a the weight of the cuboid, W, is shown acting through the centre of gravity G of the cuboid. In addition the upthrust, U, is shown acting at the centre of buoyancy B. The forces are equal. Figure 3-16b shows the cuboid in a heeled position and the weight acts through G as before but the upthrust acts at the new position of . The important thing to note is that is to the right of G and between them and the weight and the upthrust try to push the cuboid back to an even keel. This combination of two parallel forces acting out of line is called a couple and the value of one force multiplied by the distance between them is the magnitude of the couple. It is called a righting couple.
Metacentre and metacentric height
We have seen that stability depends on the creation of a restoring couple when a floating body heels but this would be much more useful if we could put a value to the restoring couple in some way. This leads us to the concept of a metacentre and to metacentric height. Let us change to a ship. Figure 3-17a shows a ship in an upright position with G and B one above the other. In 3-17b the ship is shown in the heeled position with W acting downwards at G and U acting upwards at B¢. It is evident that, if a semblance of scale is to be retained, the small angles involved make it necessary to have an enlarged diagram. This has been done in Diagrams 3-18a & b. From 3-18a it can be seen that the distance between the lines of action of the weight and the upthrust is equal to pG.sin where is the angle of heel. Then the magnitude of the righting couple is W.pG.sin . (If you are not familiar with sines do not worry, we do not need this again.) The greater this couple the more stable the ship will be. The couple varies with it is mainly dependent on pG. So pG is a measure of the stability.
There is no reason to suppose that p is a fixed point relative to the ship and so we must ask what happens to point p when the angle of heel is reduced. In practice p moves but if we could actually draw our diagram for a very tiny angle of heel the upthrust would still intersect with the centre line and p would still have a position. We go on to suppose that there is a position for p when the angle of heel is zero. This point is called the metacentre and given the symbol M and MG is a measure of stability and is called the metacentric height. We can redraw our ship on an even keel as in 3-18b and show the positions of M, G and B.
In ship design and operation the metacentric height is very important. Experience has shown that for successful designs of ships the metacentric height must lie inside quite a small range of values if the ship is not to roll too slowly and too far, or, recover so quickly that the structure of the ship is strained. (For ocean going ships it is surprisingly small at about 5 feet.)
It is obviously desirable to be able either to measure or to calculate the value of the metacentric height. In fact it is possible to do both but, for the moment, let us concentrate on calculation. We cannot calculate MG directly but a glance at Diagram 3-18b shows that MG = BM - BG. It turns out that BM can be calculated quite easily. It is given by BM = I/V where I is the second moment of area of the water plane section of the hull (which is explained in chapter 2) and V is the displaced volume. It follows from this that BM is a dimension of the ship just like any other dimension. The position of the centre of buoyancy can be calculated for a ship before it is built and so the position of M can be found. But the position of the centre of gravity is not fixed. This means that if a ship is to sail with a “safe” value of MG it must be loaded so that its centre of gravity is in a suitable position. In the days of sail heavy cargoes were often loaded on timber laid criss-cross fashion to raise its centre of gravity.
Energy is a word that is used in all sorts of ways by the population at large. Mostly they mean energy in fuels that they use to drive cars, generate electricity and heat buildings. Scientists and engineers need a much better understanding of energy in order to think about devices like boats and so do we because we are not concerned with energy in fuels. The energy in fuels is, loosely speaking, called heat energy. We are concerned with the forms of mechanical energy that are relevant to the flow of air and water.
Potential energy and kinetic energy
There are many forms of mechanical energy but really only three forms concern us. Most people will have heard of two of them at least. They are potential energy and kinetic energy. The driver of a car has no direct knowledge of the potential energy or kinetic energy of his body even though both change continually as the car changes speed and goes up and down hill. The fact is that the concept of energy is totally in the mind although it is extraordinarily useful.
So what does the idea of energy help us to understand? Let us start with an obvious example. Suppose that a lead weight drops off the bench and lands on a hard floor. The result will be some evident damage because the lead is soft. What was the mechanical process involved? When the lead started its fall it had no velocity but it certainly had when it arrived and the damage occurred when the lead was brought to rest suddenly. By what process did it acquire the velocity? We have no real answer to this question; we have to accept the idea of a gravitational field. We imagine this gravitational field round the earth made up of radial lines giving the direction of the gravitational force and concentric spherical surfaces giving the value of the force at different heights. The force decreases with increasing height but the change is small and so we can ignore it when we think about yachts.
However we can see that the falling lead is acted on by a force as it falls because it accelerates. In some way it exchanges height for speed. So what changes? Nothing that can be identified in the lead. We have to give some name to what happens and try to use it. We say that the lead starts off with potential energy by virtue of its position in the gravitational field and, as it falls, gives up this potential energy and acquires kinetic energy, by reason of its increased velocity, in exchange.
It will help us later if we stop for a moment to consider a pendulum. This, in its most simple form, is just a heavy weight on the end of a string tethered to a fixed point. The rest position for the pendulum is when the weight is stationary and the string vertical. In this position the weight is in equilibrium in the gravitational field. If the weight is now pulled from its rest position and released it will swing backwards and forwards for quite a long time. What goes on? When the weight is released it has no speed but it is at a height above the lowest point of swing. As it swings towards the lowest point it gains speed acquiring kinetic energy in exchange for potential energy. As it passes through the lowest point it starts to lose speed as it moves upwards in the gravitational field and now kinetic energy is lost and regained as potential energy. The weight comes to rest and then a reverse swing starts and the oscillations go on and on. In the absence of friction the pendulum would swing for ever. At every instant it has the same energy either as kinetic energy or potential energy. This energy must be lost if the weight is to return to equilibrium. Evidently these two forms of mechanical energy are completely interchangeable in either direction. (This is not true of heat energy and mechanical energy. If you exchange mechanical energy for heat energy you cannot have all of the mechanical energy back.)
Fluids when they are in motion continually exchange potential and kinetic energy.
When it comes to mechanical energy liquids cause us a special problem. Figure 3-19a shows water in a tank and flowing out through a small hole. The jet has velocity and therefore the water of the jet has kinetic energy. Where did this come from and how did the water get it? Let us look at the drop of water at the level of the jet and heading very slowly towards the hole. It is certain to become part of the jet at some time and when it does it will have acquired kinetic energy during its sideways movement without giving up potential energy.
This is a bit of a puzzle and to resolve it we have to envisage another form of mechanical energy that is special to liquids and gases. It is called pressure energy. The explanation is started by observing that, if we just let the water continue to flow, the tank would empty. However if we arranged to spray water on to the surface so that the level remained steady we should be supplying water with potential energy relative to the level of the jet and at the same time losing water with kinetic energy at the same rate through the hole in the tank. It turns out that the kinetic energy in the jet is equal to the potential energy of the water sprayed on the surface. This is a step forward but does not help us with the drop of water acquiring kinetic energy without a change in level.
Suppose now that we follow the path of a drop of water that is sprayed on to the surface and eventually leaves as part of the jet. When it is in the surface it has potential energy by virtue of its distance above the jet. A possible path for the drop is shown in Figure 3-19b. As the drop moves downwards its potential energy falls; it moves very slowly so it does not acquire kinetic energy but its pressure does rise. This continues until it reaches the vicinity of the hole in the wall of the tank. Then, in a fairly short distance the drop of water accelerates to become part of the jet and the pressure becomes atmospheric. It is hard to see any other process than the steady storing of potential energy in a form of pressure energy as the drop falls and then the swift conversion of this pressure energy to kinetic energy in the region of the jet.
Scientists and engineers use the idea of pressure energy when dealing with liquids and gases and say that, in a liquid or a gas, potential energy, pressure energy and kinetic energy are interchangeable. They then take it one stage further and say that, in the absence of internal friction the sum of the three forms of energy for a fixed quantity of a liquid or gas remains unchanged. This was first put forward by Bernoulli and the best known application of it is the venturi. The venturi is a simple device that is just a pipe which first converges and then diverges. The point at which the diameter is least is called the throat. If a liquid or gas is made to flow through the venturi the pressure at the throat is lower that the pressure upstream or downstream of the throat. The application most often quoted is the carburettor.
This is the information which we have available to adapt to model yachting
 For example 1 psi = 6,895 Newtons per square metre in the SI system.
 Yachtsmen call this the apparent wind or the relative wind although they would turn through 180° and look into the wind.
 At first sight figure 3.16a looks to be unstable because G is above B but in fact this is the normal case for seagoing vessels. For a given ship the value of the righting couple can be changed by altering the position of G and this is what is done to satisfy safety requirements.