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Following on from the last technical post on induced drag, this week i’ll give an overview of e, that quantity i left provisionally defined as 1 in the equation for induced drag;

.                                                $C_{D_i} = \dfrac{{C_L}^2}{\pi{e\Lambda}}$

$\Lambda$ is the aspect ratio.

Planform is the word that describes the shape of a foil when looking at it’s broad side, for instance profile view of a boat’s keel.

As you cannot have discontinuities in the fluid medium , the lift produced does not stop at the end of the wing abruptly, rather it tapers off gradually. It follows that if the planform is rectangular (leading and trailing edges parallel to each other) the pressure will be less at the tip than over the root of the foil. In other words the tip will be underloaded due to the fact that the lift must taper off but the chord retains it full width all the way to the end.

Lift distribution along span of foils with varying taper ratios.

Similarly, for each planform with its varying rate of taper, there will be a corresponding lift distribution curve. For a triangular wing, as expected, the lift distribution tapers down more quickly than the lift distribution for a rectangular wing. However, it does not taper down as quickly as the surface does, meaning the tips are relatively overloaded.

One way to even out the lift coefficient along the entire span is to twist the wing along its length so that each section of wing is at a different angle of attack. In the case of a rectangular wing, one would have to twist it such that the underloaded tips are twisted to greater angle of attack than the root, in such a way that each section of wing is loaded the same.

For triangular wings, the tips are relatively overloaded, so the wing would have to be twisted the other way, with the angle of attack reducing towards the ends.

The problem with this is that the amount of twist needed depends on the overall coefficient of lift, which varies. Thus there is no way to create a twist that will correctly compensate for the variations of loading at all global lift coefficients.

Furthermore, Max Munk  determined that for the least possible induced drag for a given span, the downwash angle has to be constant across the whole span, so that the air stream immediately behind the wing is deflected in a perfectly uniform way.

Without getting too far into it, for uniform planar flows, an elliptical lift distribution curve will result in a constant downwash angle across the whole span. Also it will give the best possible value of $e$

It turns out that there is a planform somewhere between a rectangle (big tips) and a triangle (vanishing tips) that has a lift curve that matches the area distribution curve, thus making each piece of the wing work at the same coefficient of lift, or loading if you will, and that, at all global coefficients of  lift. In this case the local and global lift coefficient would in fact be equal at every position along the span.

An untwisted elliptical planform will produce the required elliptical lift distribution.

Another factor of planform is sweep; the foil can be swept back, or forwards, rather than being at right angles to the flow. This will also affect the value of e. Note that what is important in determining the sweep back angle of a foil is neither the chord midline, nor the leading edge, nor the trailing edge. It is the quarter chord line, about which one can consider as being the aerodynamic “center of lift” of any foil. Explaining the rather complex effects of sweepback and sweepforward will have to wait for another post though.

For now suffice to say, that in general, any sweep is a deviation from the optimum.

An early example of all this being put into practice is the wing of the british Spitfire. Observe not only the eliptical planform but also the practically straight quarter chord line.

The famous elliptical wings of the spitfire

There is however another issue to take into consideration and one which may not evident at first. A pure elliptical wing, despite having the optimal area distribution, has a trailing edge that blends smoothly to the tip and on to the leading edge. Where one starts and the other ends is quite ambiguous, and this has the effect of pulling the tip vortex in towards the wing root. The tip vortex tends to follow the radius of the tip around towards the trailing edge until the angle becomes too great, forcing the vortex to break away. This makes the vortex separation unnescessarily messy and means the tip vortices are closer together than the actual extremity of the foil, thus reducing the effective span of the wing.

This would force a substitution of $\Lambda_e$ for $\Lambda$ with $\Lambda_e \leq \Lambda$

In order to get the tip vortex to peel away as far outboard as possible and neatly, requires a sharp corner between the tip and the trailing edge.  This then requires a little manipulation of the quarter chord line near the tip.

Elliptical area distribution with straight quarter chord line modified near tip so trailing edge is straight and tip has vortex shedding corner.

In the above image i have manipulated the quarter chord line in such a way that the trailing edge straightens out at 0.8 of the span, this is close to being an optimum planform for planar (non twisted) foils. It loses a tiny bit of $e$  but maintains $\Lambda_e = \Lambda$, ie , as high as possible.

The 1988 US catamaran Stars and Stripes with a vortex shedding tip planform designed by Burt Rutan.
Image from François Chevalier

Notice too that in all this, that the triangular planform is about the worst possible area distribution. There exist empirical tables for the value of $e$ and the further one deviates from an elliptical area distribution the lower $e$ becomes, increasing induced drag. Yet it has been generally considered over the last fifty years or so that the bermudian (triangular) is naturally the best shape for going to windward etc. Marchaj did a number of wind tunnel tests on this and confirmed that the triangular planform is actually quite poor. This belief mainly stems from whatever is the current trend in raceboats, setting general ‘idealizations’ about what makes a boat ‘fast’, when in fact raceboats have to optimize to arbitrary rules just as much as actually go fast. This conflict inevitably produces design distortions that are completely innapropriate for sailboats that do not have to conform to any race rule.

To be fair, sailboat rig span loading is actually considerably more complex than that due to the fact that a sailboat rig is not span (height) constrained but rather heeling moment constrained, which imposes optimization along somewhat different lines than as described above. But we’ll come back to the finer points of optimum rig lift distribution in due course.

The foot of the jib is in close contact with foredeck, effectively eliminating one of the airfoil tips.

The value for b or span is taken as the distance separating the separation points of the tip vortices. But what if the lower tip is closed off completely as pictured on the jib of the boat above? This would effectively eliminate the lower tip vortex. How to measure b? In that case, mirror theory states that one can model the three dimensional flow as being one half of the aerodynamic geometry and its mirror image as reflected through the fluid boundary, which in this case is the surface of the water. So b becomes from the upper tip vortex down to its mirror image (underwater).

In simple terms this means that closing off the gap doubles the effective span and thus also doubles the effective aspect ratio, halving induced drag.

This is a vast increase in aspect ratio and one that comes at no cost in heeling moment, so eliminating this gap is the single most effective way of improving a sailboat’s rig performance. This applies to not just jibs, but every sail. Of course, there are plenty of other reasons that may make it impractical to close off the gap completely, but if performance is high on the list of priorities, every effort should me made to reduce the gap as much as possible, for even if the gap is not reduced to the point of having a significant effect on induced drag, it still increases aspect ratio and sail effectiveness at no cost.

Stationary smoke wall shows the tip vortices as well as the downwash created by the passage of a lifting foil.

Induced drag , or vortex induced drag is a topic one often hears mentioned at the technical end of sailing enthusiast circles.

Induced drag is the energy cost of lift.

In this post I’ll elaborate a bit on the post of ‘How is lift made‘ of two weeks back as well as to lay some groundwork in place that will be useful in future posts that will attempt to correct some commonly held fallacies as to some of the implications of induced drag.

As i explained in the penultimate post, lift is produced by creating a pair of counter rotating vortices; one that is shed at the starting point, and one that is bound by the wing and travelling along with it. These vortices are created by the act of moving an inclined plane or streamlined plane through a fluid.

What i neglected to mention is that a free flying wing must have ends; i.e. it cannot be infinitely long. So what happens at the ends of the wing?

Since there is by definition a pressure difference across the thickness of any lifting wing (otherwise it would not be lifting!) , when one gets close to the ends, the higher pressure underneath tends to push the streamlines out towards the free area beyond the wing tip, and similarly, the lower pressure above the wing tends to pull in air from beyond the tip. Both of these together make the air curl around the end of the lifting foil, at the same time that the fluid particles are flowing aft. When looking at the streamlines it looks like they get twisted up around the tip of the wing. They carry on twisting behind the wing under the newly acquired momentum, forming a long contiuous vortex going back along the path of where the wing came.

Tip vortex streamlines

Kutta and Joukowski both discovered and defined the concept of circulation;

$\Gamma = -\oint_C V\cdot{ds}$

Now, Helmholtz’s theorems, state that  a vortex can never start or end within the fluid, apart from instantaneously when being formed; instead any vortex must either join up with itself forming a loop without ends, or terminate at a fluid boundary.

Putting this together, we realize that the circulation vortex that bounds the wing and the vortices being shed off the wings are one and the same. Then, remembering that a vortex cannot have an end, we follow these tip vortices back all the way to where the wing began its journey,… and where it left behind its starting vortex, and there, the tip vortices join up with the ends of the starting vortex.

So we see that there is no ‘end’, it is all actually one continuous rectangular vortex. One short side is where the trailing edge of the wing was at the beginning, the two longer, and continually lengthening, sides are the lines the wingtips traced out in space, and finally the other short side goes through the wing from tip to tip and moves along with it. This last is the “bound” vortex part of the total rectangular vortex ring, and the part that is actually creating the lift which is the always manifested by the moving of a bound vortex through the fluid.

Around the outside of this rectangle air is moving up, and within it is moving down. This downwards moving air is the air that the passage of the foil forced downwards, called the downwash.

Not only do the tip vortices not contribute any useful lift they took energy to create so represent lost energy for the plane. Also, the ‘spillage’ of air from the tip makes the foil less effective at accelerating air downwards, so represents a loss of lift too.

Lift distribution and trailing vortices
source; Olivier Cleynen

The anatomy of a wing rectangular vortex wake .

So how does this tip vortex create the extra resistance?
It actually does not, at least not directly, what it does instead is induce the resistance by affecting the overall airflow over the wing. Think about this; the extra downwash attributable to the tip vortices does not contribute to lift, yet is causing air to sink more than if there were no tips ‘leaking’. This in turn means that the foil is continually forced to climb out of the hole it is creating for itself. Since lift is produced at right angles to the airflow and this airflow is pivoted downwards it means the lift force is rotated aft as compared to the direction normal to the axis of travel. This creates a component of the lift force to be adding itself to the rest of the drag.

Downwash induced by the tip vortices negatively affect the orientation of the lift vector.

Now to quantize some of these concepts…

I will spare readers the full derivation of these equations, and just jump straight to the conclusions. For those interested in knowing more, wikipedia has a surprisingly thorough amount of information on this.

The Coefficient of lift of a foil is defined in the following dimensionless manner;

.                                                 $C_L = \dfrac{F}{\frac{1}{2}\rho{SV^2}}$

Where F is the lift force produced, $\rho$ is fluid density, S is foil surface area, V is flow velocity, all in SI (standard international) units. The  $C_L$ is a measure of how effective the foil is at producing lift.

And the coefficient of induced drag works out to be;

.                                                $C_{D_i} = \dfrac{{C_L}^2}{\pi{e\Lambda}}$

Where $\Lambda$   is the aspect ratio, the measure of how long and skinny the foil is (when looking down from and above on an airplane) and is;

.                                                $\Lambda = \dfrac{b^2}{S}$

where b is the span of the foil from tip to tip. $e$ is the planform efficiency factor, which i will explain later but for now can be taken as 1 for the perfect elliptical surface area distribution.

So we see that the longer and skinnier (higher aspect ratio) the lower the induced drag. Also the lower the $C_L$ the lower the aspect ratio, but squared.

Downwash wake
Photographer Paul Bowen, courtesy of Cessna Aircraft, Co.

I remember being on a neighbour’s boat when i was nine or ten and the owner of the boat showed me a boomerang he had sculpted. I had never seen one of these things so he satisfied my curiosity by explaining what it was for and how it worked. He then went on to explain how lift is produced by an airfoil; the air gets split into two parts, the part that goes over the top of the wing and the part that passes below. Since the top is curved (and therefore longer than the bottom) and the two flows need to meet back up again at the same time at the back the top flow needs to go faster. Further, since the faster the stream the lower the pressure, the pressure over the top of the wing decreases, thus “sucking” the top of the foil up.

Unfortunately, this often repeated explanation for lift is rather misleading, so i’ll set out here to give a more thorough explanation.

Basic aerodynamics 101;

Lift is defined as being the total force experienced by an object in a fluid flow orthogonal to the far field flow direction. In other words, the total force at ninety degrees to the relative motion between the undisturbed fluid and the object.

In order to have a force one must have a counter force (Newton’s third law; action and reaction) so the air that is pushing the foil up, means that the foil is also pushing the air down.
Furthermore, we have by Newton’s second law,

$F = \dfrac{d(mv)}{dt}$
Force is equal to the time rate of change of momentum.
This means that as the foil moves through the fluid it is constantly deflecting the flow downwards, imparting new momentum to it continually and it is this that creates the lift force.
So whatever happened to the longer path and the pressure drop?

Upper and lower flow past lifting foil;
the fluid particles never meet up again

Well, this is also happening, however it is entirely false to assume that the fluid particles have to meet up again behind the wing. They never do, unless there is no lift and no viscocity and no turbulence. In fact the flow over the top of the wing is much faster than necessary to make up the difference in upper and lower path length.
This actually makes sense when you think that the passage of the wing has deflected air downwards behind it, thus pulling back over the top of the wing the required extra air and similarly, slowing down the air that is passing underneath.

Now we also have Bernoulli’s equation for incompressible flow;

$P + \frac{1}{2}\rho{U^2} = const$

where $\rho$ is fluid density, U is flow velocity and P is pressure.

This is a conservation of energy equation since pressure is a form of potential energy and the second term is the so called “dynamic pressure”, which is the kinetic energy. Therefore, it is indeed true that there is a lowering of pressure over the top of the wing, although there is also a raise in pressure over most of the underside as well.

Typical pressure distribution upper and lower surface

In the above image the fractional distance along the chord (line connecting leading and trailing edges) is on the x axis and the y axis is negative coefficient of pressure. I generated a Naca 64209 foil and ran Javafoil with the foil at ten degrees to obtain this graph.

Of course the total force experienced by any object in a fluid flow is the integral of the pressure (normal) and shear (tangential) forces upon the wetted skin.

——–*———

How it all works in practice.

We need to introduce the concept of circulation. Circulation is just what it sounds like; the air circulates around, in this case the foil. It is a key concept to understand. Imagine a vortex spinning around the foil such that the axis of rotation of the vortex runs from wingtip to wingtip. Now imagine adding, superimposing, the straight line flow to this vortex; what happens is that the flow is accelerated on one side, decellerated on the other, and deflected one way ahead of the foil, the other way behind.

Mathematically this is;

$\Gamma = -\oint_C V\cdot{ds}$

Which is the closed line integral (enclosing the foil), V is the velocity of the fluid element and ds is the differential length along the loop.

from Marchaj “Aero/HydroDynamics of Sailing”

One may object that this seems to imply that the upwash (air being sucked upwards towards the low pressure zone above the wing) cancels out the downwash behind the wing thus leaving no overall momentum change to the fluid stream, but in fact the downwash is always twice as great as the upwash.

To understand why this is so, one must go back to the beginning of the path of motion of the foil.

The following images show the initial conditions when the wing commences its path;

from Marchaj “Aero/HydroDynamics of Sailing”

What happens is that as the foil begins to move forward at an angle of attack, the air tries to make up the short path difference by curling around the trailing edge, initiating a vortex which quickly gets left behind but which is the exact reciprocal of the wing’s bounding vortex and which forces the establishment of the bounding vortex itself. Only like so can the overall circulation in the total airmass remain zero, as it must be according to the law of conservation of angular momentum.

Therefore, the downwash is actually due to the sum of the two vortices, whereas the upwash ahead of the wing is proper only to the wing’s bounding vortex. And when tacking a sailboat, the circulation decays to zero, only to re-establish itself in the opposite direction by shedding a new starting vortex behind.

——*——

So, to recap;

The lift created by a foil or other lifting body always involves imparting by means of an effective angle of attack of the lifting body, new momentum to the fluid flow, in  a direction orthogonal to the stream and opposite to the direction of the lift force.

The mechanism for achieving this invariably involves circulation.

This circulation imposes a pressure distribution on the foil; low pressure above and high pressure below.

—-*—-

Some astute readers may object to me using an equation for incompressible flows (Bernoulli’s eqn) when air is in fact compressible. This is a very good point, but the reason that it is valid is because for small mach numbers (the ratio of flow speed to the speed of sound) , less than ~0.3, the amount of compression or expansion air undergoes is so small that it can be taken out of the equations with negligible loss of accuracy, greatly simplifying the mathematics.

As promised, here is a short article about scaling laws. Scaling is a topic that does not get much coverage outside of specialized engineering forums so i think it is worthwhile to explain the concept here more generally.

The basic concept is that you cannot simply take an existing design and scale it up or down without affecting the design in other ways than the linear dimensions. That may seem obvious, but apparently it wasn’t obvious to the not insignificant number of people who have attempted to do just this and ended up with boats that were floating disasters.

We will introduce a variable here, the linear scaling factor; $a$
Such that                                         $a = \dfrac {L_{OA_n}}{L_{OA_o}}$

Where   $L_{OA_n}$   is   the new linear dimension    Length Over All new (say)  and   $L_{OA_o}$ is the old corresponding linear dimension Length Over All old, and similarly for all other linear dimensions.

In other words ;                            $L_{OA_n} = aL_{OA_o}$

Now, the sail area is $a$ times longer and $a$ times taller so $a \times a = a^2$ times as much

so                                              $[SA]_n = a^2[SA]_o$
And, the volume of displaced water is $a$  times longer, $a$ wider and $a$ deeper so  $a \times a \times a = a^3$ more voluminous

so, displacement;                                      $\triangledown_n = a^3\triangledown_o$

Therefore, sail area increases as the square of the linear scaling factor, whereas displacement increases as the cube of a.

———–*———–

Relative stability.

Furthermore, righting arm;        $GZ_n = aGZ_o$                       since GZ is linear
and righting moment;                 $RM = GZgm$                      where m is total boat mass (same materials means same densities but times volume, which scales to the cube, so mass also scales to the cube)
So                                                   $RM_n = aGZ_o\times{a^3gm_o}$

i.e.                                                  $RM_n = a^4RM_o$

On the other hand, heeling moment is heeling arm times heeling force; HA x equation of lift, V is wind speed

.                                                                    $HM = HA\times(0.5\rho[SA]{C_L}V^2)$

([SA] is sail area) or                                           $= HA\times{[SA]\times(0.5\rho{C_L}V^2)}$

So                                                     $HM_n = aHA_o\times{a^2[SA]_o(0.5\rho{C_L}V^2)}$

i.e.                                                       $HM_n = a^3HM_o$

So we see that righting moment increases to the fourth power of the linear scaling factor, whereas the heeling moment increases to the cube. This means that in a given wind a boat which is scaled up by a factor of two is sixteen times as stiff, but is subject to only eight times as much heeling force; so would heel approximately (since the righting moment curves are not linear but at small angles can be taken as approximately linear) half as much, for a given wind.

This also means that when scaling down, boats become excessively tender, so draft needs to be proportionately increased, and/or greater ballast ratios used and/or other means employed to offset the disproportionate loss of stability.

In the case of functional model sailboats – as opposed to scale model sailboats – this translates to what appears to the novice as absurdly disproportionate keels as well as extremely high ballast ratios. On the free sailing models i used to make standard dimensions were; 30 cM Loa , 15 cM draft and a ballast ratio that was around 95%.

————–*—————

Speed to length ratio.

The equation for the Froude number is  $F_n = \dfrac{U}{\sqrt{gL}}$                    (the Froude number is the dimensionless speed/length ratio) where U is boat speed, g is the acceleration due to gravity on the surface of Terra and L is waterline length.

So in order for the boat to operate at the same speed to length ratio, the speed needs to scale too, in such a way that the Froude number remains the same.

i.e.                                                              $\dfrac{F_{n_n}}{F_{n_o}} = 1$

so                                                      $\dfrac{\dfrac{U_n}{\sqrt{gL_n}}}{\dfrac{U_o}{\sqrt{gL_o}}} = 1$

but                                                         $L_n = aL_o$

so                                                          $U_n = U_o\times{\sqrt{\dfrac{gL_n}{gL_o}}}$

Thus                                                      $U_n = \sqrt{a}U_o$

——————*——————

Period of rolling and period of pitching.

The mass moment of inertia ( $I$ ) is dependant on mass – which is proportional to volume – and distance squared so scales to the fifth power.

The equivalent equation for   $F = ma$    in rotation is    $T = I\ddot\theta$    where  $T$   is torque and  $\theta$  is heel or pitch angle (the corresponding values of RM and I need to be used depending on whether one is looking at pitch or roll motion).

Substituting $T = RM$ we get

.                                                           $\ddot\theta = \dfrac{RM}{I}$

.                                                         $\ddot\theta_n = \dfrac{a^4RM_o}{a^5I_o}$

so                                                      $\ddot\theta_n = a^{-1}\ddot\theta_o$

i.e.  the angular acceleration scales to the reciprocal of a or the larger the boat the lesser the g forces and vice versa.

For small angles,                            $RM \simeq -\phi \theta$       where $\phi$ is the slope of the righting moment per heel angle curve near the origin.

so                                                  $-\phi \theta \simeq I\ddot\theta$

or                                                  $\ddot\theta + \dfrac{\phi}{I}\theta \simeq 0$                               [1]

which we recognize as the standard differential equation for simple harmonic motion and which can be solved

by putting                                 $\theta = Asin(\sqrt{\frac{\phi}{I}}t + t_1)$

which gives                              $\dot\theta = A\sqrt{\frac{\phi}{I}}cos(\sqrt{\frac{\phi}{I}}t + t_1)$

and                                           $\ddot\theta = -A\frac{\phi}{I}sin(\sqrt{\frac{\phi}{I}}t + t_1)$    which, by inspection can be seen to indeed satisfy eqn [1] and gives an arbitrary starting time point $t_1$ and a roll frequency of   $\frac{\phi}{I}$   and an amplitude A. Both A and $t_1$ depend on initial conditions.

Now $\phi$ scales to the $a^4$, while $I$ scales to the $a^5$

so                                       $\theta_n = Asin(\sqrt{\frac{a^4\phi}{a^5I}}t + t_1)$

giving                                $\theta_n = Asin(a^{-0.5}\sqrt{\frac{\phi}{I}}t + t_1)$

and                                    $\ddot\theta_n = -A\frac{a^4\phi}{a^5I}sin(a^{-0.5}\sqrt{\frac{\phi}{I}}t + t_1)$

giving                                $\ddot\theta_n = -A\frac{\phi}{aI}sin(a^{-0.5}\sqrt{\frac{\phi}{I}}t + t_1)$

So we can see, that, as determined earlier, the magnitude of the acceleration is inversely proportional to the scaling factor a, and we see that the frequency scales as the inverse root of a. Or that the period scales as the root of a, since period is the inverse of frequency.

At larger angles the approximation of linearity is no longer valid. However, the dimensional analysis still holds up the same.

————–*—————

Apparent speed

Time taken to travel one boat length is $\tau = \dfrac{L_{OA}}{U}$

so                                                                 $\dfrac{\tau_n}{\tau_o} = \dfrac{\dfrac{L_{OA_n}}{U_n}}{\dfrac{L_{OA_o}}{U_o}}$

thus, substituting appropriately;   $\tau_n = \tau_o{\dfrac{\dfrac{aL_{OA_o}}{\sqrt{a}U_o}}{\dfrac{L_{OA_o}}{U_o}}}$

giving                                                $\tau_n = \tau_o\times{\dfrac{a}{\sqrt{a}}}$

which is                                            $\tau_n = \sqrt{a}\tau_o$

so if a video of the $a$ times scale version of a boat is replayed at $\sqrt{a}$ times the original speed it will simulate the appearance of speed of the original boat. And, happily, since roll and pitch period scale at a rate of $a^{0.5}$ as well, the same time stretch factor will also give the correct visual impression of the boat’s oscillatory movements.

Film producers everywhere; you’re welcome.

———————–*————————

Scale wind.

This brings us to the potential utility of scale models for prototyping full scale boats. There is certainly immense value in these kinds of trials which can be done at a fraction of the cost of finding a flaw in the full sized boat. However, as i hope is now pretty clear, considerable attention needs to be given to the subtleties of scaling laws .

The first consideration is to establish the scale wind; the amount of wind that will affect the model in a comparable way to the original boat.

Wind force is proportional to wind speed squared, as given by the equation of lift;

.                                                                                     $L = 0.5\rho{SC_L}V^2$

We need to find the scaling relation between the new wind speed $V_n$ and the original wind speed $V_o$ such that

.                                                                  $\dfrac{HM_n}{RM_n} = \dfrac{HM_o}{RM_o}$

heeling arm is $HA$

so                              $\dfrac{aHA_o(0.5\rho(a^2SA_o)C_L)V_n^2}{a^4RM_o} = \dfrac{HA_o(0.5\rho(SA_o)C_L)V_o^2}{RM_o}$

eliminating                                                   $V_n^2 = aV_o^2$

thus                                                               $V_n = a^{0.5}V_o$

So we see that in order for the boat to behave similarly the wind speed needs to be scaled by root $a$.  This explains why large sailboats have greater difficulties in light winds. If a is 10, say, then the wind needs to blow ~3.16 times as fast to have the same effect.

————–*—————-

Reynolds number.

There is still the issue of the Reynolds number, which is the dimensionless ratio of inertial to viscous forces in the fluid stream. This number is important to distinguish what type of fluid flow regime the boat is operating at and which governing fluid flow simplifications can be reasonably applied.

.                                                            $\mathbb{R}_e = \dfrac{UL}{\nu}$       where $\nu$ is the fluid’s kinematic viscocity

Which can be seen to scale to the  $a^{1.5}$   so it is impossible to keep both the Froude number and the Reynolds number correct as one scales a boat, unless different special fluids are used, but this topic is beyond the scope of the present essay…

————————————————*————————————————-

Feedback on if the maths symbols do not parse, along with which formula is giving problems would be highly appreciated , this is the first time i try “Latex” mathematical notation typesetting in wordpress.