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 $\ddot\theta + \dfrac{\phi}{I}\theta \simeq 0$                               

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  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.