Resulting force forward from a flat sail

In summary, the conversation discusses the process of calculating the best angle for a sail on a reach. The speaker explains their approach, which involved using simplifications and assumptions to solve the calculation. However, the resulting answer is not satisfactory and does not match the expected curve. The speaker then asks for help in identifying the mistake in their calculation. Another speaker chimes in and provides a formula for calculating the maximum value, which involves taking into account the direction of travel and the force available to move the boat forward.
  • #1
cosy
2
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
  • #2
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!
 
  • #3
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:

FAQ: Resulting force forward from a flat sail

1. What is the resulting force forward from a flat sail?

The resulting force forward from a flat sail is the force that propels a sailboat forward when the sail catches the wind. It is the combination of the force of the wind on the sail and the resistance of the water against the hull of the boat.

2. How does the shape of the sail affect the resulting force forward?

The shape of the sail plays a crucial role in determining the resulting force forward. A flat sail is designed to catch the wind at a specific angle, creating a lift force that pulls the boat forward. A curved sail, on the other hand, generates a lift force perpendicular to the wind, allowing the boat to sail at different angles.

3. What factors influence the resulting force forward from a flat sail?

Several factors can affect the resulting force forward from a flat sail, including the wind speed and direction, the angle of the sail, the shape and size of the sail, and the weight and shape of the boat. The skill of the sailor in adjusting the sail also plays a significant role in maximizing the resulting force forward.

4. Can the resulting force forward be controlled by the sailor?

Yes, the resulting force forward from a flat sail can be controlled by the sailor. By adjusting the angle of the sail and the direction of the boat, the sailor can manipulate the amount and direction of the resulting force forward. This allows them to navigate the boat and reach their desired destination.

5. Is the resulting force forward always in the same direction as the wind?

No, the resulting force forward is not always in the same direction as the wind. The angle of the sail and the shape of the boat can cause the resulting force forward to be at an angle to the wind. This is what allows sailboats to sail against the wind, known as tacking or beating. However, the resulting force forward will always be in the same general direction as the wind, whether it is directly downwind or at an angle.

Similar threads

Back
Top