I have this question that I have no idea were to start and I would like a push in the right direction.
A boat has a sail surface of 10m^2 and traveling N40°E. And wind is 6m/s from N30°W. What is the best angle to sail at. We are assuming the keel of the boat is N40°E, the sail is flat and the wind is relative to the boat. We are also ignoring everything except for the wind.
the sail is fixed perpendicular to the axis of the boat, or mathematically, the boat axis is the normal of the plain the sail locates in.
The direction of boat(or steering the boat) is ruled by Rudder.
with a certain angle there exist two force components; one aligned with the boat axis, another perpendicular to axis.The first one drives the boat forward and the second is lateral and push the boat aside.
For NE the fourth quarter of an vertical ellipse is the path of the boat.
The equation of ellipse is:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
For the boat $a$ is the distance it moves to East and $b$ is the distance it moves to North. Suppose the fluid dynamic shape of the boat is such that the ratio of these parameters is:
$\frac{b}{a}=\frac{10}{1}$
that means a 10 meter motion forward is associated with 1 meter lateral motion.Then the equation of path, as shown in figure is:
$$y-d=-10\sqrt{1-x^2}$$
Where d is for transformation when the center of ellipse is moved from (0, 0) to (0, d). Let $y-d=Y$ for simplicity and the boat starts moving from origin O then the vector of the speed of the boat at a location like A must make $10^O$ with y axis that is we must have:
$$Y'=\frac{10 x}{\sqrt{1-x^2}}= \tan(80^o)≈5.67$$
⇒$x ≈0.5$
Plugging x in equation we get $Y=8.66$, so the boat has move from (0, 0) to (0.5, 8.66) and we have:
$tan(\alpha)=\frac{8.66}{0.5}=17.32$
$⇒ \alpha≈86.7^o$
This is the angle between the axis of the boat and x axis at point(location) A, that is the boat has turned $90^o-86.7^o=3.3^o$ clockwise.This angle plus $10^o$ will be total angle with y axis which is necessary for $N40^oE$.So the desired angle must be $13.3^o$.