## Introduction

Is it faster to sail downwind a straight line or to gybe from one tack to the other? The answer, of course, depends very much on the boat performance and the prevailing winds. It is however possible to provide some practical guidance with a little use of theory.

I was recently tasked with conveying a catamaran accross the Atlantic ocean. The boat, wonderfully designed to cruise in the Caribbeans, is not rigged nor designed to sail downwind. Its stays are fixed roughly 40 degrees aft of the mast, meaning that the mainsail cannot be fully opened. The boat has no whisker pole nor a spinnaker, investments that would only make sense if the boat were to sail anything else than a beam reach.

Crossing the Atlantic gives one plenty of time to think. Nudged by the fact that the boat’s speed is slower downwind than at 120 degrees (see Figure 1 at the top), I spent the better part of two days figuring out the most efficient angle to the wind that would minimize the total sailing time.

Such endeavour may not be a desirable goal for everyone. I enjoy sailing as much as anyone and some of my passage plans focus much more on enjoyment than on time. But during a race, or when days become indistinguishable from one another, as in an Atlantic crossing, speeding things up may be a wortwhile endeavour.

Even if time is not a sensitive issue, you may discover – as I did – that the fastest route may also be the most comfortable one for the crew. It may not change your cruising plans, but it may comfort you knowing that you are not wasting time.

### Time Economics

In a race, in a cruise (or in our lives), time is a finite – thus scarce – ressource. The science specialized in managing scarce ressources is economics. At its core is the weighting of various trade-offs when considering options. It should thus not come as a surprise that an economists – yours truly – took a look at possible sailing paths to consider the trade-offs.

As established in antique greece, the shortest path between two points on earth is defined by the great circle passing through those points. Those are the spherical equivalent of a straight line. When the speed of the vessel does not depend on the wind, it is also the fastest path. Thus, great circles have been used as the aircrafts’ and tankers’ baseline for commercial routes.

Things are however different for sailboats. Speed depends greatly on the wind direction and how the rigging is configured. Therefore, the shortest route is not necesarily the fastest one. This opens the question as wheter a longer route can reduce the total sailing time.

To understand the key trade-off, it is perhaps best to imagine a straight line between two points (see Figure 2 above). A skipper must then choose a given speed and course. He may wonder if a change of course from the straight line, later corrected by a change of course towards the endpoint, would reduce the duration of the trip. In considering such alternative, three key factors are important to figure out if its worthwhile.

First and foremost, the deviation from the straight path must increase the boat speed. If not, there is no point in changing course, as it only increases the length of the route. This is the first practical rule of thumb and for the rest of this post, I will assume so.

The second factor is that the speed increase *parallel* to the original path is what matters. When changing course, the skipper increases the overall speed, but because of the course away from the destination, some of this speed is “spilled” away. Only the speed that is parallel to the original path matters, as this is is the speed that brings us closer to destination. This increased speed is the key factor reducing the total sailing time.

The third factor is the increased length of the path that is induced by the change of course. This means more time spent sailing and acts as a penalty to the total sailing time.

The trade is thus valuable if the speed increase parallel to the original line outweights the increase in distance. If it is the case, then a longer route requires less time.

This argument can be repeated for any additional deviation. As long as each degree brings more in terms of speed factor than in terms of length penalty, the total saling time will be reduced. In fact, the most efficient angle, minimizing the total sailing time, is when the two factors break even. When the time saved by the accrued speed is equal to the time lost by the acrrued distance, changing course further – meaning that more time is added by distance than reduced by speed- will only increase the sailing time.

In the next section, I will show that the condition for mimimizing sailing time while sailing downind corresponds to the equation:

\frac{\Delta V(\theta)}{V(\theta)}=\tan(\theta),

where \theta is the course deviation, V(\theta) is the boat speed at that course (the polar curve of Figure 1) and \Delta V(\theta) is the increase in speed due to change of course. The \tan function is nothing but the tangent trigonometric function.

The left hand side of this expression measures the speed increase in percentage, while the right hand side measures the length increase factor. When the course is chosen so that the two are equal, there is no more gains to be made from a course change. This expression, however, is a simplification and only works downwind (as in the trade winds). For an answer that works for all points of sail, some mathematical artillery is required.

### Now Comes the Math Part

If you are not comfortable with mathematics, you may want to skip up to the “practical application” section below. You will then have to accept the mathematical result of this section on its face. If you however have some knowledge of trigonometry and calculus, you may find this section helpful to understand how the trade-offs in changing course functions at any point of sail.

It pays to invest in the additional notation of Figure 2. Denote \theta_1 and \theta_2 respectively the departure angle and the arrival angle with respect to the shortest path route. The distance on the shortest path is called D. As no units are specified and the efficiency condition does not depend on it, it may very well be equal to one. The part of the shortest path associated with the first leg is named d_1 while the second part is the remainder, that is D-d_1. As in the previous paragraphs, V(\theta) is the polar curve of the boat with respect to *the boat deviation from the shortest path.*

It should be noted that while this problem presents itself with three variables (\theta_1, \theta_2 and d_1), only two are the choice of the skipper. The remaining one is implictly pinned down by geometry. Mathematics have their way to sort themselves and the problem is much easier to solve as if the skipper were to chose \theta_1 and \theta_2. It can be understood as if the skipper chose the course deviation and kept such course until a relative bearing to the destination gave him \theta_2. This is merely for mathematical convenience, as once the problem is solved, all variables are known, and how to communicate the solution is up to the navigator.

With this notation in hand, the total time of the A-C-B trip can be written as:

T(\theta_1, \theta_2)= \frac{d_1}{V(\theta_1)\cos(\theta_1)}+\frac{D-d_1}{V(\theta_2)\cos(\theta_2)}.

This expression is nothing but the “distance divided by speed” for each leg of the deviation, accounting for the length and the speed that is parallel to the shortest path.

Now, note that the dotted line in Figure 2 can be measured both from angle \theta_1 and \theta_2. Since both measures must be equal, it implies that d_1 \tan(\theta_1) = (D-d_1)\tan(\theta_2), which gives us, with little algebra:

T(\theta_1, \theta_2)=\frac{D}{\tan(\theta_1) + \tan(\theta_2)}\left[\frac{\tan(\theta_2)}{V(\theta_1)\cos(\theta_1)}+\frac{\tan(\theta_1)}{V(\theta_2)\cos(\theta_2)}\right],

which indeed only depends on the two angles \theta_1 and \theta_2.

#### Simple Case: Downwind

It is useful to begin the analyse when the function V(\theta) is symmetric, that is when V(\theta) = V(-\theta). This is a sensible restriction when the boat is running downwind. In such a case, the restriction boils down to stating that the speed gains are the same wheter the skipper deviates either to starboard or to port.

If this is so, the time function T(\theta_1, \theta_2) becomes symmetric, which implies that the time minimizing angles \theta_1 and \theta_2 must be equal. To understand why, assume the opposite, which would mean that the shortest of the two legs (say the first) has the highest speed. In such a case, it would make sense to increase the first leg and reduce the other one so as to diminish the total time travelled. This can only be done by reducing the difference between the two angles. As this argument holds for any angle difference, the time minimizing angles must be equal. Thus, \theta_1 = \theta_2 and the function boils down to:

T(\theta_1, \theta_1)=\frac{D}{V(\theta_1)\cos(\theta_1)}.

Such function can be minimized using unconstrained optimization techniques, which gives the following first order condition:

0=\frac{D}{V(\theta_1)\cos(\theta_1)}\left[\frac{V'(\theta_1)cos(\theta_1) - V(\theta_1)\sin(\theta_1)}{V(\theta_1)\cos(\theta_1)}\right].

Such expression defines the optimal angle \theta_1 that mimimizes sailing time.

Since neither the distance D, the cosine \cos(\theta_1) and the speed V(\theta_1) cannot be zero, such expression equals zero only if:

V'(\theta_1)\cos(\theta_1)= V(\theta_1)\sin(\theta_1).

The left-hand side measures the time reduction that is induced by the increase in speed from the deviation \theta_1 *paralel* to the original path. It is the speed factor described earlier. The right-hand side describes the increase in the path length *perpendicular* to the original path, which is a pure waste of travelling distance to reach point B. This is the distance penalty factor also described earlier. The condition states that the marginal gain of the deviation must equal the marginal time penalty. Rearraging such expression yields the equation stated in the beginning of the text.

#### At Any Point of Sail

In a general setup, the function V(\cdot) may not be symmetric. This may be so because the path from A to B is not downwind, but maybe also because the boat may have a different performance on one tack or on the other. In such a case, the time equation is a two variables unconstrained optimization problem whose first-order conditions satisfy:

\frac{1}{\tan(\theta_1) + \tan(\theta_2)}\frac{1}{\cos^2(\theta_1)}\left[\frac{V(\theta_2)\cos(\theta_2)-V(\theta_1)\cos(\theta_1)}{V(\theta_2)\cos(\theta_2)}\right]=\frac{V'(\theta_1)}{V(\theta_1)}-\tan(\theta_1),

\frac{1}{\tan(\theta_1) + \tan(\theta_2)}\frac{1}{\cos^2(\theta_2)}\left[\frac{V(\theta_2)\cos(\theta_2)-V(\theta_1)\cos(\theta_1)}{V(\theta_2)\cos(\theta_2)}\right]=\frac{V'(\theta_2)}{V(\theta_2)}-\tan(\theta_2),

Both equations have the same structure. I focus on the interpretation of the first one, while the second one as a similar interpretation for the angle \theta_2. The right hand side has a familiar face and reflects the trade off between accrued speed and accrued distance. In the symmetric case, this is the sole trade off to account for. However, in the general case, there is an additional trade-off. When the angle \theta_1 is increased, a portion of the path d_1 is reduced. As a result, this portion that was travelled at speed V(\theta_1) is now travelled at the speed V(\theta_2). This portion of distance is proportionnal to \frac{1}{\tan(\theta_1) + \tan(\theta_2)}\frac{1}{\cos^2(\theta_1)}, explaining why it pre-multiplies the speed difference on the left-hand side.

This substitution between speeds may increase or decrease the overall time, depending on if the speed *parallel* to the original path in the first leg (V(\theta_1)\cos(\theta_1)) is greater than the speed parallel to the original path in the second leg (V(\theta_2)\cos(\theta_2)). If the speed is greater in the second leg, then the right-hand side must be greater, which implies a greater angle. Conversely, if this substitution induces a loss of speed – in effect a time penalty -, the trade off on the right hand side should be made smaller by decreasing the angle. This is so because relatively speaking, it pays off to travel more along the first leg.

If one divides the two equations forming the first order conditions, we can further refine the analysis of the angle differences. A little algebra brings the following expression:

\cos(\theta_1)^2\tan(\theta_1) + \cos(\theta_2)^2\tan(\theta_2) =\cos(\theta_1)^2\frac{V'(\theta_1)}{V(\theta_1)} + \cos(\theta_2)^2\frac{V'(\theta_2)}{V(\theta_2)} .

This expression should look familiar by its simialarity to the downwind case. On the left-hand side, we can recognize the time cost factors that are induced by the extra distance on each leg, respectively \tan(\theta_1) and \tan(\theta_2). On the right hand side, we can further see the time benefits of the accrued speed on each leg, respectively V'(\theta_1)/V(\theta_1) and V'(\theta_2)/V(\theta_2). What is now new is that each of the costs and benefits are now weighted by the (square of the) distance travelled on each leg that is parallel to the straight path. If the penalty occurs on a longer leg, it thus weights down the benefits of increasing the angle on such leg (and *vice-versa*).

In summary, a time minimizing path should be such that the (weighted) net marginal time gains due to speed equal the (weighted) marginal time losses due to accrued distance.

## A Practical Application

Telling the skipper to “equate the weighted net marginal time gains” while underway is so abstract that it may induce a few expletives. It is best to make this theory practical prior to embarking in the cockpit. The expression in the beginning of the text can be approximated by:

\frac{\Delta V(\theta)}{V(\theta)} \approx 0.0174\theta.

The left hand side measures the speed increase in percentage. The right-hand side is simply an approximation of the tangent function. The approximation states that everytime that the boat speed increases by more than 1.74% with a one degree deviation, the extra mileage reduces the total sailing time. This is a good rule of thumb for *small deviations*.

For larger deviations, a bit of preparation pays off. Most polar curves provided by manufacturers do not behave as well as the function V(\theta) described in theory. Most are approximations drawn from empirical data and there is absolutely no guarantee that the function is always increasing for deviations between 0 to 90 degrees. The gist of the analysis however remains: *as long as the net gains drawn from speed are higher than the net losses induced by the longer path, it is worthwhile to deviate from the shortest path*. How much angle then becomes a matter of keeping tabs on the running gains while increasing the angle. The angle with the highest tab wins. This can be calculated in advance or, if you are in the middle of a long cruise, calculated while on a passage.

Some peole read books. Others scribble down mathematics on pieces of paper and write a blogpost about it. The truth of the matter is that I was genuinely interested in knowing what would be the best tactical plan to tackle an Atlantic crossing. Thus, I did compute the passage time per nautical mile travelled with the polar curve provided by the manufacturer (Figure 1).

I do not wish to argue that this curve is an accurate performance of the boat. Our days aboard goes to suggest that it clearly overstates its performance. However, the curve seems to match the *relative* performance with respect to the wind angle. The boat is faster on a broad reach than on a beam reach, and it is ridiculusly slow on a close reach. Thus, I kept the overall polar curve structure, but I did use the 8 knots figures for 15 knots winds. For a proper analysis, I used the symmetric, downwind case.

Figure 3 shows the travel time per *parallel* nautical mile for deviations between 1 to 41 degrees. The sweet spot is at 26 degrees off downwind, where the sailing time is minimal (it is also the round angle where the equation above is the closest to be satisfied). The time difference is of roughly 1 minute per nautical mile. It is not much, but it costs nothing and over 2600 nautical miles, it amounts to day less at sea. Not bad for a *free* course correction!

## Conclusion

One may wonder how this analysis relates to route planning algorithms found in most specialized softwares (e.g. *PredictWind*). It bears much of the same logic, with two notable exceptions. In a broader set-up, winds can vary over time (meaning that V() becomes a function of time) and multiple deviations from the straight path should be allowed. This broader setup is best served by Euler-Lagrange equations or, when you get down to programming, the Dijsktra algorithm. Computers can perform this on the fly.

If the mathematics are slightly more involved in a broader multi-path route, the basic underlying principles remain: each change from the straight path is analysed in terms of gains in speed versus accrued distance. Racing skippers may not do trigonometry and advanced calculus on the fly, but I would not be surprised that they weight the benefits of speed gains against longer routes. If they do not, they miss a basic concept that may help them win a race.

For others doing longer passages, the lesson here is that it pays off to look at the boat’s performance and to weight it against comfort. On the boat I had to convey on the eastern side of the Atlantic, I found that the comfortable point of sailing was also the fastest one. Paradoxically, with a lot of time to spend on my hands, I ended up saving a day. That may very well be the broader lesson.