Resulting force forward from a flat sail

cosy · Oct 24, 2016

I wanted to calculate the best angle for a sail on a reach, but realized I had to make the calculation a bit unrealistic by using a flat sail and doing other simplifications to make it possible for me to solve. However it seems that wasn't enough ([emoji4]) since I ended up with a quite useless answer, so I was hoping that I could get some help on here! Here is my approach (not that I assumed all sideways force was counteracted by a centerboard):
I started by assuming we had a sail that was 1m*1m and the pressure from the wind was 1N/m^2 (I just needed a constant). I then defined the area catching wind as cos(x)* the area of the sail, so the area will be cos(x)m^2.

Handwritten sketch on graph paper of a line at angle x to a horizontal dotted line, with the segment labelled cos x · 1² and a second arrow labelled cos x.

That also means that the total force on the sail will be cos(x)N.
This vector can be divided into the drag (which I will ignore for this calculation) and the lift which will be cos(x)^2.

Handwritten free-body diagram of a flat sail at angle x to the wind: a sloped line with normal force vector cos(x)N = F_T pointing up, a lift component cos(x)·F_⊥ = cos(x)^2, and a forward component labeled cos x · 1^2 = cos x along the horizontal.

The lift itself can be divided into sideways force and force forward. The force forward will be sin(x)*cos(x)^2.

Handwritten diagram on squared paper: a right triangle with angle x at the origin, a force arrow labelled cos(x)^2 = F_L along the hypotenuse, and below it F_P = sin(x)·F_L = sin(x)·cos(x)^2.

This result however does not make for a satisfying answer as the graph for the function is repetitive and looks like this:

Screenshot of a plot graph app showing a wavy curve oscillating around zero along the horizontal axis, with y-axis ticks from -7.5 to 7.5 and x-axis labels 2.5, 5.1, 7.6, 10.2, 12.7.

Screenshot of an Android app's plot graph screen showing a nearly flat, noisy line along zero on a y-axis from -60.5 to 60.5 and an x-axis from 0 to about 90.

I was expecting to get a curve looking something like this:

Hand-drawn graph on squared paper: a vertical axis at the left and a horizontal axis, with a smooth arch curve rising from the origin to a peak and falling to a point labelled 90 on the right.

where I could find the point where the derivative was 0 to get the maximum value.
So my question is, of course: What did I do wrong? Thanks for any help in advance!

cosy · Oct 24, 2016

I forgot to point out that something was wrong about my funtion on a very deep level since the value of the function at x=90 was >0!

JBA · Oct 25, 2016

It has taken me a bit of of mental thrashing to work this out but the fact is that since the direction of travel is always perpendicular to the direction of the wind then the amount of force available to move the boat forward: F push = F wind x (sail area x Cos Θ) x Sin Θ = F wind x sail area x Tan Θ.
That formula will give you: Tan Θ = 0 at Θ = 0° and Θ = 90°; and, Tan Θ = 1 at Θ = 45°

At Θ = 0°, F sail = 100% (the face of the sail is perpendicular to the wind) but the perpendicular F push vector = 0% of that force and at Θ = 90° the F sail = 0% (the face of the sail is parallel to the wind), while the F push = 100% of that force.

Resulting force forward from a flat sail

Discussion

Similar threads