Sail power and horse power?

[kjell]

If I know from calculation that my boat need 40 HP to reach hull speed, How big sails do I need to make hull speed with 15 knots wind?

[sharpii2]

About 112 square meters, asuming one two horsepower per ton to about 176 square meters, assuming one horsepower per ton under power.

[kjell]

Many thanks. How do you get this number and how many sails?

[sharpii2]

Many thanks. How do you get this number and how many sails?

Dear Kjell:

I used the following proceedure.

1.) Estimated your boat's displacement in tons, based on your horsepower requirements. For convience sake, I used Metric tons.

2.) Found the cube root of the number of metric tons I estimated that your boat displaces.

3.) Squared that number.

4.) Multiplied the new number I got from step 3.) by 15 to get the area in square meters.

The number of sails that this is to be divided up into and their placement on your boat is entirely dependent on the design of your boat.

For example, is your boat long and narrow or short and wide? Does it have any kind of keel projecting below the bottom? Is it a single hull, or is it a catamaran (two hulls of equal size connected by crossbeams), a single outrigger (one large hull connected by crossbeams to a float on one side), or is it a double outrigger (one large hull connected by crossbeams to a float on both sides)? If it is a single hull, does it have ballast?

Your boat most likely was designed to be a motorboat. If it's to be made into a sailboat it will need: some kind of keel to keep it from slipping sideways in a cross wind, a bigger rudder for when there is less wind and to make up for lack of prop wash, and relatively short mast or masts so the wind doesn't tip it too much. Tall masted rigs usually require much more ballast to keep them upright. And ballast cuts into your cargo capacity. Your boat could have one or two masts and have anywhere from two large sails on two masts to as many as four large and small sails one one mast to five large and small sails on two masts.

All of this depends on the proportions, shape, and intended purpose of your boat, as well as the number of crew.

I would have to know much more about your boat before making any additional comments.

Bob

[kjell]

The boat is a 15 meter motorsailer 17 ton.
I understand now how you get the area of the sail. Will this change if we are talking about 20 knot wind speed?

[sharpii2]

The boat is a 15 meter motorsailer 17 ton.
I understand now how you get the area of the sail. Will this change if we are talking about 20 knot wind speed?

Dear Kjell:

Yes. And substancially.

To figure out how much, we'll use the following proceedure:

1.) take the new wind speed and divide it by the old wind speed.
20kts/15kts=1.33

2.) Take that number and square it because the force of the wind increases with the square of the increase of its speed. (thats why huricanes are so deadly)
1.33*1.33=1.77

3.) Take the old sail area estimate and divide it by the new number, the increased force of the wind, to get the new sail area.
112sm/1.77= 64sm (I rounded it up just to be sure)

Bob

[kjell]

Dear Bob,

I am learning a lot. I have a program that calculates the power needed to move a displacement hull at different speeds. I introduce the weight in ton and the LWL and I got the result in HP. I know that sailors don’t think in HP.
If 112 sqr/m can produce 40 HP at 15 knot wind sped and 64 sqr/m give 40 HP at 20 knot wind speed how many HP will the 112 sqr/m produce at 20 knot wind speed?

Kjell

[Raggi_Thor]

One very old way to estimate sail area is to multiply thw wetted surface with 2 or more.. Old racing boats may have 3 or 4 times the wetted surface.
If you want to sail in light winds, multiply with 2.5, if you only want to sail in stronger winds you may need less.

[Tim B]

Have a look on the web for Hazen's Aerodynamic Model of a Sailing Rig. That should give you all you need, then just play with areas until you get a good solution. You will need to know what conditions you're sailing in. By which I mean, what is the apparrent wind and the apparrent wind angle.

Scribble that up into a bit of code (in C or whatever) and then you can see what the force from the sail is.

Of course, Force=Power / Boat speed

So, you know the wind, know the force, all you're after is the area. and with a little trial and error, starting at the aforementioned values, it shouldn't take too long.

Cheers,

Tim B.

[sharpii2]

Dear Bob,

I am learning a lot. I have a program that calculates the power needed to move a displacement hull at different speeds. I introduce the weight in ton and the LWL and I got the result in HP. I know that sailors don’t think in HP.
If 112 sqr/m can produce 40 HP at 15 knot wind sped and 64 sqr/m give 40 HP at 20 knot wind speed how many HP will the 112 sqr/m produce at 20 knot wind speed?

Kjell

Dear Kjell:

Wish I knew. Most of the power in a power boat does little or nothing to move the boat forward. Mostly, it just agravates the water. So out of the 40hp used to push your boat at hullspeed, an amount unknownt to me is just wasted. The trouble is tip vortices caused by the propellor. Only about 75% of the propellors diameter is actually pushing the boat forward. The rest is just stirring the water.

Sails too have their inefficiences. Most sails aren't perfect airfoils. The ones I grew up using didn't even try. they were flat cut. Also, your keel is never perfect either. Both sails and keels create tip vortices of their own. It's not that sail, keel, and propellor designers are lazy. They do do their best. But mean old Mr. entropy gets in the way and makes us waste energy, whether we like it or not. Improvement is always possible. Perfection is not.

Most of the formulas and rules quoted on this thread are 'rules of thumb'. Or, in other words, rules that have been found to work by trial an error. So, I can neither tell you how much sail is pushing the boat forward nor how much engine is doing the same. I can only tell you how much of either is likely to be needed.

It would be very tempting to multiply the 40hp by the increased force of the wind (1.77*40hp=71hp) but it could also be very dangerously misleading.

I have gone through a lot of trouble not answering your question. But I have done so with the most honorable of intentions :)

Bob

[kjell]

Tim B,
Cant find the “ Hazen's Aerodynamic Model of a Sailing Rig. “ in the Web
Is it the same difficulty to compare Wing sails power with HP ?
Why cant we have a table that show. X sqr/m sail at X AoA at X apparent wind speed = X HP?

[Tim B]

kjell,
It's just a little more complex than that. I took my version of Hazen's model from "Principles of yacht design". Typically, when comparing the "prime mover in a yacht, we concern ourselves not with power, but with force. basically, at a steady speed, force forward=force back. I strongly suggest that you obtain a copy of principles, as it is the one book that EVERYBODY keeps within grabbing distance.

Tim B.

[kjell]

Dear Kjell:

I have gone through a lot of trouble not answering your question. But I have done so with the most honorable of intentions :)

Bob


Dear Bob,
I am sorry if my questions have has produced any problems. My intension was to get a simple answer for a simple question. What I can see there is nobody who can give me a simple answer. I have many good books of yacht design and I have been consulting them all. If anybody is interested in to know how to calculate needed HP for displacement hulls I have very good formulas.