Transferred forces on a sail

I can't find any explanation of how wind force on a sail is distributed.

E.g. a say 10m² square sail, fixed at a mast, a boom and a yawl.
Say the wind force is 400 kg total.
How much of this force is transferred to the sheet?

Solhi said:

I can't find any explanation of how wind force on a sail is distributed.

E.g. a say 10m² square sail, fixed at a mast, a boom and a yawl.

Click to expand...

I assume that is a typo and you mean a yard?

Solhi said:

Say the wind force is 400 kg total.
How much of this force is transferred to the sheet?

Click to expand...

That very much depends on the rig. If sail shape is maintained by the structure of the rig, say a wishbone boom or a balestron rig (one example of that is the Aerorig), and if the rig is balanced so that the centre of pressure is very close to the axis of rotation, as in a balestron rig or this split junk rig

then the load on the sheet may be a small fraction of sail force.

If you need the sheet to maintain the shape of a sail, say a jib, the force on the sheet may be a multiple of the sail force.

So you really need to give details of the rig, because that can change the answer by an order of magnitude or two.

Also, at what wind speed do you expect to get a force on a 10m² sail equivalent to 400 kg? Did you mean 400 Newton?

Anyway, seeing that you ask about a square sail, if the aspect ratio and shape are decent, say a sail like those on Maltese Falcon, then you can expect the centre of pressure to be about 25% from the leading edge. The load would be on the windward brace. Assuming that is at the leading edge, the horizontal load should be half that of the force on the sail. But then you have to take into account the angle of your brace to the horizontal. The load will be proportional to the horizontal load (which is half the sail force) divided by the cosine of the angle to the horizontal. And that angle to the horizontal will depend on the angle of the yard to the boat.

That calculation is for a brace that is, as seen from above, at a right angle to the yard. So next, divide the above result by the cosine of deviations from that right angle.

For example, say the brace is at 70 degrees to the horizontal, or 20 degrees from vertical (assuming the boat is level). Then 1/cos(70 degrees) = 2.92. Next assume that, in the horizontal plane, the brace is at 60 degrees from perpendicular to the yard. 1/cos(60 degrees) = 2. But if the brace is at the leading edge, then it has to counter only half the sail force, the other half being taken by the mast. The load on the brace should be 2.92 * sail force.

Yes I did mean yard. A bit lingo confused these days.
Spooky you immediately jump to my planned questions about the AJ .
I wanted first to have a "simple" idea of the distribution. So now I have to sit down and learn.
My AJ question was then to understand the notion of the force on the jib.
My eyeballing assumed that the jib negates ca. the same area of the main. Meaning "only" Main area minus Jib area account for the tension on the sheet, corrected of course with the formulas you just gave.

P.s. for a 14,6 meter trimaran (5-10° heel?)

View media item 30310

Haha, skjønner. Vi deler samme interesse.

Beauty

Robert Biegler said:

Also, at what wind speed do you expect to get a force on a 10m² sail equivalent to 400 kg? Did you mean 400 Newton?

Click to expand...

12,6 knots = 8.6 m/s = 8,6² * 0,6 * 10= 400 Kg.?

12,6 Knots

12,6 x 12,6 (Knots squared) x 1 (Coefficient) x 0,16 (Air Density plus knots-meter confusion) x 10 squared meters =

= 254 sir Isaac Newton

Isaac Newton - Wikipedia https://en.m.wikipedia.org/wiki/Isaac_Newton

ca. 25 kg

CarlosK2 said:

12,6 Knots

12,6 x 12,6 (Knots squared) x 1 (Coefficient) x 0,16 (Air Density plus knots-meter confusion) x 10 squared meters =

= 254 sir Isaac Newton

ca. 25 kg

Click to expand...

Thank you for the correction. I would wonder why the boat does not move

CarlosK2 said:

12,6 Knots

12,6 x 12,6 (Knots squared) x 1 (Coefficient) x 0,16 (Air Density plus knots-meter confusion) x 10 squared meters =

= 254 sir Isaac Newton

Isaac Newton - Wikipedia https://en.m.wikipedia.org/wiki/Isaac_Newton

ca. 25 kg

View attachment 202771

Click to expand...

Maybe the formula on the web are wrong? I got a suggestion when using a metric system, 12.6 kn = 6.48 m/s.
Then m/s² * 1,2 (air density at sea level) you get a result in N(ewton).
When I try so, the result is different from your formula, confusing.

Dynamic Pressure = 1/2 x V^2 x Density

6,48 m/s, 1,2 kg



x 10 m^2

= 250 sir Isaac Newton

knots x knots x 0,16 x m^2 =

m/s x m/s x 1/2 x 1,2 x m^2

Salty Sea Water: 1025 kg


x 10 m^2

= 215 K (!) Newtons

1 Sail

Force (sir Isaac Newtons) = Max Coefficient x Sail Trim (from 1 to 0,6) x knots^2 x 0,16 x m^2 (Sail)

Max Coefficient = 0,8 + (4 pi x arrow / chord-line-string) ...

AR: Aspect Ratio = h^2 / sail Area

---

0,8: High pressure: "Pressure"
4pi (arrow / chord): Low pressure: "Vacuum"



---

SailBoat

1 Wing constructed with 2 pieces (StaySail + MainSail)

Max Coefficient: 1,4

In some cases, it can be 1,5, in others 1,3, and in others 1,45 ... but this is too much detail for an introductory course on the subject

---

1 Wing built with 3 pieces (Jib + StaySail + MainSail)

I don't know, I'd like to know.

The maximum Coefficient could be high and it would be very useful for sailing in light winds.



---

Using the Sails as if they were a parachute, that is, Downwind with an Angle of Attack close to 90 degrees, the Coefficient would be the Drag Coefficient: 1,1

The Force is transmitted primarily through the mast and shrouds.

The aerodynamic center is usually in that area because the leading edge of the wing (the leading edge formed by the staysail and the jib) is more heavily loaded: x 1,7 or x 2

1 square meter of staysail is equivalent to 2 square meters of mainsail, but —Achtung— the mainsail works for the foresails: the staysail is worth twice the mainsail because the mainsail is there (!)

In the sheet of a triangular mainsail, there can be a force similar to the wind's force because a lot of force must be applied to maintain the sail's shape.

MainSail = 2 x StaySail Area

It is a very logical proportion