I am in the process of building a twin mast catamaran. The two sails will be under the 1400 sq inch total. I have two airtronics 945841 sailwinches and 1 Hitech hs 725 bb at home. Are these strong enough for the twin sails on a cat?
I was thinking of putting an airtronics arm in each hull to simplify rigging. What are your thoughts?


(Also for Jeff on servo needed for given sail size.)

An easy approach for a monohull with bulb and fin is to say that the heeling moment of the sails is in equilibrium with the righting moment of the keel. So the forces on the sail can be taken to be the same as the forces on the fin and bulb. If we have a 96 oz bulb on the end of a 20 inch fin (say) at a heel angle of 90 degrees (extreme case, boat is knocked down), then both keel and sails are experiencing a torque of 1920 oz-in. OK. The sails are giving their force through their centre of effort of the sail plan. To make life easy, imagine the CE of a (triangular) sailplan is around 3/8 up the mast, and let’s have an 80 inch mast. So the 1920 oz-in of torque means the sail force itself is around 64 oz.

If you don’t have a bulb and fin to make the calculations easier, then let’s calculate the forces on a sail directly. To do this, you need to know the sail area and the apparent wind speed, things we didn’t need to know before. Roughly,

Sail force = 0.004 * V^2 * A

where Force is in lb, V is apparent wind speed in mph, and A is sail area in sq ft.

Let’s sail the cat at 10 mph apparent. Force = 0.004 * 10^2 * 10 (assume sail area of 1440 sq. in, ie 10 sq ft) = 4 lb or 64 oz. OK, we can now pick up where we left off.

The sails are being held by a sheet control line. This line is controlling part of a sail force of 64 oz by controlling a boom attached to the sail clew. Let’s imagine the boom controls 3/8 of the sail force, and the other 5/8 is absorbed by the mast (I have no better basis to estimate this, so Will needs to guide us here). Then, of the part of the sail force being controlled by the boom, let’s say that the sail force CE is located about 3/8 of the sail foot away from the luff. Let’s take the foot of the sail and the length of the boom to be the same. If the foot is 18", the sheet is handling a torque of 162 oz-in (64 oz * 3/8 * 3/8 * 18"). Let’s attach the sheet at a radius of 12 inches from the boom pivot point. Hence, the sail winch servo needs to deliver 162 * 18/12 = 243 oz-in of torque. If the drum on the sail winch has a radius of 1/2", the servo must be rated to deliver at least 121.5 oz-in, but for safety double it to accomodate friction losses and unanticipated gusts. Around 250 oz-in then…

Does that seem right?

Lester Gilbert


thanks for your reply (I think)
As I believe in the principle of K.I.S.S (keep it simple stupid, no offense Lester)

Can you put this into plane english now for some of us really became lost in what you were explaining. Well ok I did. Guess I should have stayed in school longer then I did.

If I remember correctly the web site I had found then lost had it broken down to basicly
Servo X will hold XX amount of sail area
No Math involved at all. Nice and simple for me



You are perhaps the most qualified person I know to quantify the sail forces. In fact, I am rather surprised that you have not tried to actually measure sheet loads in one of you wind tunnel experiments.

As far as your breakdown of the total force to the sheet (5/8 to the mast vs 3/8 to the sheets) this is a bit tricky. The lift from camber will be centered chordwise (50% chord), but the lift from angle of attack will be centered at the 1/4 chord. Given a 10% camber airfoil at 10 degree angle of attack (near stall), you get about 40% lift from camber and 60% from angle of attack. Using the lever rule, this places the the center of lift (in the chordwise direction) at 35% chord. Now, given a triangular sail planform, the center of area (representing the average of the 50% chord locations on an area basis) is 1/3 of the foot length back from the luff. So, the 35% point will be 23% of the foot length back from the luff. If the sheet attaches to the boom at the 2/3 back location, then by the lever rule the sheet must absorb 35% of the sail force. Looks like your guess was pretty close, Lester.

Now, one simplification that you made that will drastically change the numbers is the assumption that the sheet is only pulling the boom toward centerline. Effectively, the sail force you calculated (which seems about right) is the total force to the side. So the reaction force breakdown between the mast and the sheets would also be in this direction. But as we know, the sheets generally also pull down. So the total sheet force will be the vector sum of the lateral force plus the downward force on the boom (untwisting the sail). From Geometry, it is fairly easy to divine what the totoal force is if you know the lateral force and the sheet angle. Generally, I would say that for closehauled sailing, you will need to at least double the lateral force to get the total force. but with a sheeting post and a good vang, you might be able to keep that under control.

The rest of your math seems about right. My only question is whether a drum with a 1/2" radius would have enough sheet travel for a 12" sheeting radius on the boom. With a 12" sheeting radius you need about 17" of sheet travel. With a 1/2" radius drum, that would be 5.5 revolutions of the drum. How many revs can you get out of the big winches (like Guyatt)?

  • Will

Will Gorgen

Jeff -

not familiar with the site, but seems to leave a lot of open loop-holes.

Do you have interior room for single arm, double arm winch - or can you only use a drum type?
How many part purchase will you use?
How fast (or slow) is the winch?
How far out on the boom are you attaching sheeting?
What kind and how many blocks are you using for guiding your sheets?
How long of sheet travel do you need - from full in to full out?
What voltage is the winch being rated at?
What is your budget?

Sail area is only a very small part of winch selection. In my case for the MultiONE and maybe for the F-48, I wanted a very fast, yet strong winch. 1100 sq. inches for the 1 Meter multi, and 1400 sq. inches for the F-48. I selected the HiTec 815 - it is big, heavy, but strong and fast. I need the speed for dumping the main if the boat starts to heel too much, and I wanted above deck sheeting (and steering) on the 1 Meter as it is too narrow to get hands inside of hull in case the drum winch has problems wtih elastic tensioner, or it comes off the turning block forward. If I build a 2 Meter multihull, I will probably use the same winch for the jib, but opt for a Guyatt winch for the main, using two channels for sail control.

Darrin - both the Airtronics and HiTec 725 are pretty slow by comparison to Guyatt drums - and may cause problems speed wise when sheeting in and out - especially on a cat without ballast.

Sure throw a wingnut into the pile.

Will have to work area out for the sails that will be controled (Main, Foresail and jibs) then see if I can find a servo with either a drum or go with a double arm to control them. Just rough guess the main looks like it is around 350 to 400 square inches of sail. lots of sail to pull in.


Jeff -

In reality (!!!) 300-400 square inches of sail area isn’t that big. Consider the 10 Raters, Marbleheads, and even IOM boats are running around 1000 - 1200 (or more) square inches of sail using only one winch. Even the 36/600 uses one winch, and sail area measurements do not include roach - so boat could have sails close to 800 square inches!

If those are truly the sail area of your main, and similar for multiple jibs, you probably won’t have too many problems - unless you are out there sailing in an (hee hee hee) ALBERTA CLIPPER ! [:D]

I am going to use my two(maybe one) Hitec hs-725bb servos. They are slow but fairly strong and cheap! I am toying with the idea of converting my Mistrals (2) to arm winchs and that leaves me with the two Hitecs. My schooners mast is only 48" tall (approx) so I don’t think I will have that much sail area. I don’t think winch speed is as important on a scale-type boat as the competition isn’t that intense. Something that came as a bit off a suprise is that there is less room in the schooner than the is in a Mistral. Even though the schooner’s hull is 47" long it is only 8" wide and there is only about 2 1/2" of usable depth under the deck. I’m sure everthing will fit but it won’t be as easy as the Mistral. Two winches might create problems.

Vancouver Island

I just did some very rough calculations of sail area for the schooner. The sail plan is approx. 4’ high and 5’ long. Thats 2880 sq. in. Its a gaff rig so the plan is sqareish but knock off 30% for corners I’m still in the 2000 range. If I don’t control the jibs that will knock of some more but I think I’m going to need both winches.

<blockquote id=“quote”><font size=“1” face=“Verdana, Arial, Helvetica” id=“quote”>quote:<hr height=“1” noshade id=“quote”>Originally posted by Albertaclipper

If I remember correctly the web site I had found then lost had it broken down to basicly
Servo X will hold XX amount of sail area
No Math involved at all. Nice and simple for me
<hr height=“1” noshade id=“quote”></blockquote id=“quote”></font id=“quote”>
Hi Jeff

I’ll do my best, but Servo X holding XX sail area is a little too simple for me, sorry!

Doug Lord commented to me, seeing my post, that he uses a rule of thumb that one sq foot of sail area yields about 1 lb of force. This is a good approximation that builds in a multiplier of around 100% for all those friction losses, gusts, and so on. So lets have your 10 sq.ft sails yielding 10 lbs of force.

Doug also suggested to divide the sail force half and half between the mast and the boom for simplicity (forget my previous 3/8ths), so you need to sheet 5 lb with your boom. That’ll do.

Now, let’s say that the sail force acts about 1/3 along the boom from the mast (I prefer 3/8ths again, but hey, it doesn’t really matter here). If your boom is 18" long, then the 5 lb force is acting around 6" along the boom. Turn 5 lb into oz (80 oz) and multiply by its lever arm (6") and you get the torque the sails are putting out to the boom – 480 oz-in.

OK. Now let’s attach your sheet to the boom at a radius of 12". Your servo needs to deliver a force of 480 divided by 12 = 40 oz.

The servo is sitting in the hull and will deliver its force either via a drum or via an arm. We’ll start by imagining you want to run an arm with a radius of 8" in order to pull in around 16" of sheeting from close hauled to running. The servo must therefore deliver 8 times 40 = 480 oz-in of torque. This is too much for any arm winches I know about.

Alternatively, let’s use a drum, and have the drum rotate 5 turns to pull 16". Each turn must pull 16/5 = 3.2" of line, which defines the circumference of your drum. The circumference of a circle is 2Pir, where r is the radius, and 2*Pi = about 6.3. So the radius of your drum needs to be 3.2 / 6.3 = .51". Great, the standard (smaller) drum of the RMG series has a radius of about 0.5". Using 0.5" as the radius of your drum, you are asking your winch to pull 40 oz with it, and it therefore needs to deliver 20 oz-in to do it. No sweat, the RMG 280 will give you 150 oz-in, much more than you need.

Let’s work backwards for fun. You want to use the RMG 280 and you want to use the RMG drum with a radius of 0.63". The idea is to minimise the number of drum turns, to obtain maximum speed, so where do you attach the sheet on the boom? Let’s try two turns. This sheets 2Pi0.63 * 2 = about 8", and your sheet attachment to the boom would probably be around 6" from the mast (say). The sail force is 80 oz at this point on the boom, and you are going to pull it on a drum radius of 0.63, giving a required torque from the servo of 0.63 * 80 = 50 oz-in. Still no problem at all for the RMG.

Lester Gilbert


Your arm winch math does not compute. The only time that the arm winch would have 480 oz-in or torque is when it was at 90 degrees to the sheet. The sail would be halfway out (presumably on a reach) and the loads would be lower than the 80 oz. You really only see those kind of loads when you are sailing close hauled. When you are trimed in near close hauled, then the “effective radius” of the arm gets much smaller (can even drop to 0 in my “lock off” scenario) so the torque load from the high sheet forces can in fact be quite low.

I’m running a futaba s3801 with 156 oz-in of torque in my boat. The Fairwind has about 600 square inches of sail (or a bit over 4 square feet) thus about 4.2 lbs of force (67 oz) by Doug’s rule of thumb. The boom length on the boat is 14" if we use your 1/3 rule, that give us a boom torque about the gooseneck of 312 oz-in which way exceeds my servo (it really does not matter what the sheeting radius needs to be becasue the sheet force and sheet travel will always be inverses of each other and will therefore cancel out to give the same force). But because my servo has a smaller effective radius near close hauled, the effective servo torque goes way up. I have never had a problem with my servo slipping (even in very strong winds).

Others who have used this same servo for their Fairwinds do have trouble with the servos slipping. They have set their arm on the servo such that there is still a rather large effective radius at close hauled and you will see their sails easing in heavy winds as the servo slips.

So the “strength” of a servo to hold xxx sail area can have a lot to do with how you set-up your sheeting geometries.

As a bit of supporting evidence, Futaba makes a smaller sail servo - the s3802 - that has 122 oz in of torque. I use this servo with a very similar sheeting setup in my second Fairwind boat. I have had that boat out pleasure sailing in 25 knots with no problems of servo slippage.

  • Will

Will Gorgen

For those not following, perhaps this photo will better explain Will’s theory/comment about “Locking Off” the winch.

In the photo, viewed from the port side above of my old 36/600 - I used a double arm winch. You will notice the the bottom sheet is for the jib, the top for the main. You will also notice the the arm of the winch is slightly past midpoint when the main is sheeted completly in. (fine sheet adjustments made at boom). In this position, the sheet is pulling at almost a straight line, and it would take major wind to cause the arm to come back past 180 degrees in order to slip.

When you sheet out, the arm rotate clockwise (arm on right moves toward bottom of photo) and when the arm is perpendicular (in photo) the main is about 1/2 way out. While it is possible for the winch to be overcome at that point by strong gusts, one would normally be playing the main or heading up slightly in gusts which would eliminate a lot of the force on the arm.

Obviously at 180 degrees from the photo, both sheets would be eased to maximum and jib and main out for a downwind run.

Hope the photo helps a bit.

<blockquote id=“quote”><font size=“1” face=“Verdana, Arial, Helvetica” id=“quote”>quote:<hr height=“1” noshade id=“quote”>Originally posted by wgorgen

The only time that the arm winch would have 480 oz-in of torque is when it was at 90 degrees to the sheet.<hr height=“1” noshade id=“quote”></blockquote id=“quote”></font id=“quote”>
Hi Will

Sure. One of the simplifications. Useful, because it provides a maximum requirement. Although that maximum might hardly ever be seen, IMHO it’s worth sizing the winch to suit, otherwise an unpleasant surprise might await.

Lester Gilbert

<blockquote id=“quote”><font size=“1” face=“Verdana, Arial, Helvetica” id=“quote”>quote:<hr height=“1” noshade id=“quote”>Originally posted by Dick Lemke

In reality (!!!) 300-400 square inches of sail area isn’t that big. Consider the 10 Raters, Marbleheads, and even IOM boats are running around 1000 - 1200 (or more) square inches of sail using only one winch.<hr height=“1” noshade id=“quote”></blockquote id=“quote”></font id=“quote”>
Hi Dick

For info, IOM No.1 rig is around 950, 6.63 sq.ft. I use the Hitec 5735 (260 oz-in, 3.5" arm with pulley to give double purchase, total sheet run around 14", sheet attachment on main boom approx 9" radius) as an arm winch, and find that at the top of the wind range for this rig (around 5 m/sec, 10 knots, 11 mph) it struggles to winch in from the run – but I have particularly high friction losses… The RMG 280, on the other hand, has no problems (150 oz-in, step-down drum, 4 turns). I have actually broken my boat when the sheet snagged and I didn’t realise it – the RMG simply pulled the sheeting post clean out of the hull and broke off bits of the deck as well.

Lester Gilbert

For some reason, I was thinking the IOM was a bit larger for the “A” rig. Seems when Kris Harig and Doug Lord were discussing the sail area for the MultiONE, they tried to keep rules on sail area at the IOM/1 Meter size, so those with monohulls could use their existing rig on a multihull. With the MultiONE limit at 1100 sq. inches, that is where I came up with the approximate size.

I am aware that Rob’s winches would work to almost pull a truck out of the mud !

Wouldn’t it be great if Hitec were to build a newer drum winch - something a bit faster than the current one, but with about the same torque. Seems it would be a “seller” for the boats with smaller sail area. FMA had one for a short time until they stopped selling. They wrote to advise they were going to try a new design, and then a short time later decided to specialize on aircraft servos. Haven’t checked lately to see if they have anything to offer.

I agree with you that the Guyatt really is the way to go if one wants a drum winch with few (if any) problems with performance.

Hi Guys

The standard HiTec drum winch has 4.5 rotations. It is plenty strong enough to pull in full size sail area in 15 knots apparent. If you are going with twin rigs, the sail area will be still the same but the overall force on the winch will be greater. I think that the drum winch will handle the job. The battery power however will be the only thing that you could have a problem with.

One of our skipper’s is using a Guyatt winch in his multihull, and after racing against him the boat seems to “slow” dramatically when sheeting on from a reach to windward. Possible cause of this is the spped of sheeting in is just"dumping" the power out of the sails.

The club travelled to Ballina in northern NSW a couple of weeks ago. We sailed there against a local boat and got soundly beaten.

The boat’s that we sail here in our club are just not on the pace. I am very surprised that they are as slow as they are. We are all going back to the drawing board and coming up with something else.

In contrast to what I have been saying in the past, weight is the most important part to these boat’s. Build light but not stupidly light. Try to design so that weight can be added easily.(for heavy conditions)


mmm, the Hitec HS725BB has 3.5 rotations, well, going by the box for mine it is.
And Dick, it seems that your picture doesn’t work so can you please fix it.

I see said the blind man to the crippled nudist who put his hands in his pockets & promtly walked away.


I have heard that the 725 will have different numbers of rotations with different radio gear. I can’t confirm that, but I do have an interesting tale to tell:

On my canting mast boat, I am using a 725 as the mast canter stepped down through a 4:1 gearbox to get me the correct amount of canting travel. I originally hooked it up to my “sheet” channel on my radio (4 channel Futaba FM) because it had a rachet so I would not have to hold the stick while I fiddled with some of the installation details. When I was done and everything was working correctly, I switched the servo to the other channel on the “sheet” stick and gave it a try. The servo moved significantly more on this channel than it had on the original channel - perhaps an entire extra turn. It was enough to cause the servo to rip off the temporary mounts. If was very frustrating and I have not figured out how I am going to fix it yet.

  • Will

Will Gorgen

You can wire a couple of pots in series with the transmitter pot on the channel you want to use to give you adjustable end points.

Vancouver Island

Thanks for the tip, Don. I was thinking about something along those lines at the servo. I had not thought of doing it in the transmitter. I guess that you could probaby figure out a way to drill some holes in the Tx box to stick the pots through so that you would have access to them…

Do you have any good detailed instructions on how to do this? I’d like to have a guide in front of me before I start snipping wires in my transmitter. I’d also like to have specifications for the pots so that I knew that the ones I put in were nto going to screw up the Tx.

  • Will

Will Gorgen

Adding resistance in the servo increases travel-adding it in the transmitter reduces travel.I used 2 5000 ohm pots. I beleive I put a pot on each side of the transmitter pot and left the wiper lead alone. If Steve Andre is around he knows this as it was his idea. I would have to take my transmitter apart to be sure of what I did so if Steve doesn’t respond I will do that for you. I remember Steve saying to use a ganged pot so that there is one adjustment that reduces that travel on both ends. I couldn’t find a ganged pot so I used individual pots. This gave me End Point Adjustment of sorts. Each pot affects one end or the other but it isn’t as precise as the factory EPA on my Futaba. It does work though-just a little fiddley. I did the mod on an Airtronics.

Vancouver Island