How do I calculate how much lift an outrigger needs?

I have been thinking of putting a sail on my canoe with will probably require the use of an outrigger. (related: Outrigger and/or leeboard when sailing a canoe?) I was thinking about using a section of PVC pipe as an outrigger, this reference indicates a 4 inch pipe will hold 0.653 gallons of water per foot weighing 5.44 pounds, a 5 inch pipe will hold 1.02 gallons at 8.50 pounds. Every pound of water displaced is a pound of lift. The farther out you place the outrigger the more leverage is applied, but at the same time it applies more strain to the arm that connects the outrigger to your vessel

The first image below could be considered two outriggers with a sail between them. At this point my mind spirals out of control with all the variables. If the base was wide enough they would not tip, but then it would require stronger cross bars, which would add more weight...

Many of the off the shelf out riggers for canoes, only have a couple feet long arms with volume of a couple of gallons, but if you see professional sea going outriggers they have much larger volumes with much longer arms (second image).

What are the basic things I need to consider and how do I calculate how much lift I need from an outrigger? I am planning on inland water (river, lake) sailing, so will not be at the extremes in the images, but I want to understand the principles required to keep the canoe from tipping over, and keep away from putting to much pressure on the outrigger.

• I think you might be overthinking this. First, you are comparing the sailing you will do with your canoe to that done on the open ocean - which is a completely different animal. Secondly, no matter how big you make your pontoons they will fail given enough sail and wind. But given the limitations of a canoe I'd imagine things would start breaking before you have any concerns for overcoming the stability of the pontoons. Also remember that as you increase the weight of the craft, you slow it down - increased mass to get moving and it'll sit lower in the water so greater drag. Commented Sep 10, 2015 at 14:25
• It's a canoe, I'm pretty sure you could get away with just using a couple of pool noodles duck taped to hockey sticks. Commented Sep 10, 2015 at 14:44
• @ThatIdiot & ShemSeger you are both pretty much on what is happening in my mind as i think through this, but would like some real science to back it up. Good research tends to keep you from appearing on TV. Commented Sep 10, 2015 at 16:13
• @njzk2 Ah, I see where you're confused now. They aren't filled with water, but you can measure how much float a container will provide by measuring the amount of water it can hold. Example: It would take 10kg of weight to submerge an outrigger with a volume of 10L just below the water's surface. Commented Sep 10, 2015 at 22:25
• @ShemSeger indeed. In this light this sentence now makes sense: Every pound of water displaced is a pound of lift. Commented Sep 10, 2015 at 22:29

(I realize my vocabulary about those things in english is very limited. Please edit if I use the wrong words)

My understanding of how that works is that the outrigger displaces the amount of force applied by the wind, like so:

|
|
WIND -> |
|
|
BOAT -> O

VS

|
|
WIND -> |
|
|
BOAT -> O----O

In the first case, there is absolutely no resistance to prevent the boat from tipping over.

In the second case, assuming the outrigger always floats (which is an approximation, but will do for now) the wind will need to apply enough torque to lift the boat.

Now, all that's left is to compare the opposing Torques of a/ the wind in the sail to b/ the weight of the boat. We want equilibrium between the 2.

The wind force can be integrated as a force applied to a single point at the barycenter of the sail. Assuming a triangular sail, the barycenter's distance to the base of the sail is 1/3 of the sail's height. The total heigh needs to take into consideration the offset of the base of the sail

boat_weight * outrigger_length = wind_force * barycenter_height

The wind force is (apparently in kg. The 0.063 constant is something I found somewhere, not sure if it is right)

0.063 * wind_speed^2 * sail_surface

Putting it all together, if your canoe with you in it is 100kg, the outrigger is 1m out, the wind is a moderate 10m/s, the sail is 5m^2, the mast 3m and the sail start .5m above water, it gives:

100 * 1 > 0.063 * 10*10 * 5 * (1+.5) (47.25)

However, if the wind rises the 15m/s, then the force would be 106kg, tipping your boat over.

The next part is to make sure the outrigger does not sink. I guess the same force applies on it, so you need it to float 100kg, so about 100L. In a 12.5cm pipe, that's about 8m. Using a larger 20cm pipe, only 3.2m. Considering that the floatation device can sink a little, and that you are not going to actually lift your canoe above the outrigger, the real value is probably quite less.

The same works approximately identically if your consider the outrigger to be a weight on the other side. (except you need weight instead of floatability)

I am not entirely sure of those computations, but it sounds about right. Overall, the wind applies a force depending on the square of its speed at the barycenter of the sail. The moment of that force needs to be compensated by an equivalent one. Lower winds, smaller sails, will result is less floatation required

• Great answer, clarified some of the language, hope it's still correct?
– user2766
Commented Sep 11, 2015 at 10:57
• @Liam thanks for that. Moment, however, is a specific term. It refers to how a force applies with respect to an axis. Torque is probably a better term, though. Commented Sep 11, 2015 at 13:30
• Sorry, @njzk2. This isn't my subject. Thanks for clarifying!
– user2766
Commented Sep 11, 2015 at 14:34
• I think this answer is close, but after thinking about it over the weekend. I think we need to consider the boat as fulcrum, the sail is a lever, and the wind is a force. The outrigger arm is another lever and the outrigger provides as much lift as the maximum amount of water it can displace. You need to come to a balance between the length of the levers and the force applied to each. Commented Sep 14, 2015 at 23:31
• This link comes up with answers very similar to yours but without the math. My brain is starting to hurt.... Commented Sep 15, 2015 at 14:49

I don't think it is possible to answer your question in such a general way.

Needed lift of your outrigger depends on your type of canoe, the way its built, how it is packed with people/goods (weight distribution), how big your sail is, how high your mast is, if you are sailing with/across/against the wind, etc, etc.

What we can observe is:

• If you choose a hollow outrigger (relying purely on lift from flotation), it will only prevent your craft capsizing in the direction of the outrigger.
• If you want stability towards both sides you'll need either two outriggers, or attach weight to the outrigger, such that this weight will counterbalance forces which roll your canoe away from the outrigger.
• Alternatively you could set up your mast/sail not directly over your canoe, but in the style of a catamaran somewhere between canoe and outrigger. This could be quite a bit more challenging to build.

I'd suggest either finding professionally built outriggers, then figure out how much lift they produce, or do some experimentation yourself.

But I think you're approaching the problem from the wrong side.

• I think you should first try to figure out how long your outrigger arm is. This will heavily impact how agile your craft will be. Only then would I try to find the right amount of lift/weight needed for the outrigger to keeping the craft stable.
• How would I figure out how long my outrigger arm is? I would think the length of the arm and the lift of the outrigger are dependent on each other. If I had an outrigger arm 30 feet long on my 17 foot canoe, it would only need enough lift to hold half the weight of the arm, there would be no amount of sail or speed that I would be able to obtain (assuming it does not sink on launch) that would strain the system. Commented Sep 10, 2015 at 16:09
• I don't think it is possible to answer how long the outrigger arm should be. Have you tried looking up similar projects on the web? Are there professional outrigger-kits available for canoes? If not then you are very much on your own and will just have to do your own experiments to see what works... Commented Sep 14, 2015 at 9:16

Here is what a sailor with 37Kmiles of blue water sailing thinks ... the out rigger needs to have two properties when used with a canoe, floating displacement, keel side area. with out keel side area, you wont be able to tack, unless the canoe has a centerboard or a dagger board. a good general rule of thumb is an out rigger needs to displace at least half of the amount the main hull does in volume. two pieces of structural foam or just 2 inch pieces of home insulating will do fine. make a 1-2 foot keel and floating flat part that is 3/4 as long as the canoe, or just start with 8 foot lengths. like a "T". use 2x4/2x3 studs and rope. take it out and trim it ... relocate the mast and width as needed. you might need to add foam ... to bond the Styrofoam to anything use gorilla glue. it foams, pushes parts out of place. so clamp it.

• How far out (arm lenght) would you put an outrigger that has 50% the displacement of the main hull? Thoughts on keel shape? The top image in the question has a short deep keel, but a canoe might want a longer shallower keel. Commented Sep 18, 2015 at 17:49