<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

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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 2*Pi*r, 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 2*Pi*0.63 * 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

http://www.iomclass.org/

http://www.onemetre.net/