Rudder Loads on the Modern Cruiser

To do their jobs, rudders must cope with enormous forces. So must emergency rudders.

“Look! All I have is two fingers on the wheel! It doesn’t take anything at all to steer this boat!”

So why the big problem when the rudder falls off?

Let’s take a close look at how those two fingers turn the rudder. As an example, we’ll take a stock, 42-foot cruising sailboat from the pages of Cruising World and see what’s involved in steering it.


The fingers rest on the rim of a 30-inch-diameter wheel. Turning the 30-inch wheel turns a 3-inch-diameter sprocket on which rides a chain connected to the steering cables. That’s a power ratio of 30:3 or 10:1, so 1 pound of force applied to the wheel rim delivers 10 pounds of force to the steering cable.

The rudder is a partially balanced spade–its stock is set back a short distance from the rudder’s leading edge. When the airfoil-shaped rudder is turned a small amount, its center of lift (the point at which the average of the forces on it act) is a little aft of the stock, in this case about 2 inches. The cable turns the rudderstock via a quadrant that has a radius of 16 inches, so the power ratio of the quadrant to this lever arm is 16:2 or 8:1. Combine this with the 10:1 power ratio of the wheel to the chain/cable, and the total power ratio is 80:1. Therefore, two fingers applying a force of 1 pound to the wheel rim deliver 80 pounds of force to the rudder (or counter a force of 80 pounds acting on the rudder).

And this is the easy case, with the sails well trimmed and the boat sailing to windward at about 5 knots in flat seas and 10 knots of breeze.


Let’s say that we have an emergency rudder deployed and that we want to generate the same steering force. We made the rudder by lashing a floorboard to the spinnaker pole, which is 14.5 feet long (see diagram below). The boat has a regular “scoop” transom, at a 45-degree angle to the vertical, and our fulcrum for the “tiller” is by necessity at the deck. Because of the boat’s freeboard, to immerse our 3-foot-long floorboard, this fulcrum is 9 feet from the immersed end of the pole, putting the center of our rudder 7.5 feet from the fulcrum. The inboard end of the “tiller” is thus 5.5 feet from the fulcrum.

The ratio of these levers gives us our power ratio, and it’s not in our favor: 5.5 on our end, 7.5 on the rudder’s end, or 1:1.36. Therefore, to generate our 80 pounds of steering force (or to counter a load of 80 pounds on the rudder), we need to apply a force of 80 x 1.36 = 109 pounds to the end of the tiller. That’s a lot of fingers. In reality, we’re going to have to work even harder than that because our floorboard is nowhere near as efficient a “control surface” as our rudder was.

And we haven’t yet begun to take into account the boat’s motion in a seaway.


How Does a Drogue Steer?
Basically, all a drogue can give you is drag. By moving the towline to one quarter or the other, you can move the drag to one side and hope to steer the boat toward the side you’re applying the drag. How much drag do you need? That depends on many things, but applying it effectively depends on your available lever arm.

If we tow the drogue from a corner of the 10-foot-wide transom, that gives us a lever arm of 5 feet off the centerline. We can reduce that a little if we again drum our 14.5-foot spinnaker pole into service, center it across the transom, and tow our drogue from one end or the other. However, that entrains the practicalities of actually adjusting the drogue’s position to vary its effect (and whether or not the spinnaker pole was designed to support loads applied in this way). As Serengeti’s crew discovered, a drogue’s behavior even at slow speeds can be capricious, and they weren’t trying to steer with it.

Why Do Rudders Fail?
We’ve already seen that even in the easiest of steering situations, the rudder is subject to quite high loads. Because the load on the rudder is proportional to the square of the boat’s speed, it doesn’t take a radical change in conditions for that load to get markedly higher. An increase in boat speed from 5 to 7 knots doubles the dynamic pressure on the rudder, and at 10 knots it’s quadrupled. These loads are cyclical, and they sometimes create vibrations, both of which contribute to fatigue–inducing strains on the blade, the stock, and the bearings and the structure supporting them within the boat.


What would we expect the worst-case load to be on our rudder? Let’s examine a wipeout at 16 knots, with the rudder hard over and stalled out. Let’s assume maximum dynamic loading and the simplest case.

The dynamic load on a flat plate in a stream is expressed as q = dv2A/2g, where d is the density of the fluid, in this case seawater at 64 pounds/cubic foot; v is the velocity in feet per second, so multiply by 1.69 for speed in knots; A is the area of the plate; and g is gravity, conveniently 32 ft./sec.2. This all boils down to q = v2A.

Our rudder only goes to 35 degrees, so the frontal area it presents is A(sin35) or .57A.

At 16 knots, with our 9-square-foot rudder, q = (16 x 1.69) 2 x 9 x .57 = 3,751 pounds. The rudderstock has to be able to support the equivalent weight of a J/24 and its four crew.

The helmsman would feel 3,751/80 = 46.9 pounds on his fingers. Sound about right?

Spade or Skeg?
It’s an old debate. There’s no question that in terms of lift versus drag, the spade rudder is the most efficient type of control surface used in steering a sailboat. And although experimenters have played for decades with rudders mounted forward, it’s no accident that rudders today are mounted about as far aft as it’s physically possible to put them. It didn’t take primitive man long to figure out that his arrows flew straighter with a couple of feathers stuck on their back ends. But when a spade rudder falls off a boat, we effectively are left with an arrow with no flights. As we’ve learned from those who’ve tried to do it (see “An Oar to Steer Her By,” March 2003), balancing the ever-changing forces acting on sails and a fin keel in the absence of a rudder is nigh impossible. Many rudderless boats have been abandoned.

Would a skeg improve the odds of being able to steer the boat? If it wasn’t damaged by whatever caused the rudder to fail, I think its surface area located well aft would contribute somewhat to directional stability. I believe yacht designers could devote some effort to studying this possibility. Forget efficiency for a moment and examine contingency. Serengeti is designed to race, and when doing so, she has a large crew of strong sailors to handle her. The spade rudder serves her best. Mr. and Mrs. Retiree aboard their contemporary production performance cruiser don’t have quite as many options should they find themselves in extremis. Maybe boat designers should give them some.