Resulting force forward from a flat sail

  • Thread starter Thread starter cosy
  • Start date Start date
  • Tags Tags
    Flat Force
Click For Summary
SUMMARY

The discussion focuses on calculating the optimal angle for a flat sail on a reach, utilizing a simplified model with a 1m x 1m sail and wind pressure of 1N/m². The participant derived that the force forward is represented by the equation F push = F wind x (sail area x Cos Θ) x Sin Θ, leading to the conclusion that the maximum forward force occurs at an angle of 45°. The participant identified flaws in their initial calculations, particularly regarding the behavior of the function at extreme angles, revealing that the force available to move the boat forward is zero at both 0° and 90° angles.

PREREQUISITES
  • Understanding of basic physics principles related to forces and vectors.
  • Familiarity with trigonometric functions, specifically sine and cosine.
  • Knowledge of sailing dynamics, particularly the concepts of lift and drag.
  • Ability to interpret mathematical functions and their derivatives.
NEXT STEPS
  • Research the impact of sail shape on aerodynamic efficiency.
  • Study the principles of fluid dynamics as they apply to sailing.
  • Learn about the use of computational fluid dynamics (CFD) software for sail design.
  • Explore advanced sailing techniques for optimizing sail angles in varying wind conditions.
USEFUL FOR

Sail designers, physics enthusiasts, and anyone interested in optimizing sailing performance through mathematical modeling and analysis.

cosy
Messages
2
Reaction score
0
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.
955bf060b6044804b26090e7142e82d1.jpg

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.
234a08599582b3904a4a6f9e4a4a063e.jpg

The lift itself can be divided into sideways force and force forward. The force forward will be sin(x)*cos(x)^2.
2780becf93f8f77b111156942d5bc3da.jpg

This result however does not make for a satisfying answer as the graph for the function is repetitive and looks like this:
8e78d75b9f02b5b88a5fe0a8bfd030fc.jpg
1c36402d59e3f1318fc0a0ea9be517b9.jpg

I was expecting to get a curve looking something like this:
89779206b2dad17e54ff420f72838384.jpg

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!
 
Engineering news on Phys.org
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!
 
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.
 
Last edited:

Similar threads

Replies
11
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
17K
  • · Replies 21 ·
Replies
21
Views
11K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
11K
Replies
0
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 28 ·
Replies
28
Views
2K