Kick-Up Rudder

The original rudder blade on my micro-cruiser was a trapezoidal slab angled back from the rudder head. The leading edge was radiused and the trailing edge was beveled across the last 1-1/2". It definitely was not a foil shape. This large area located far behind the pintle and gudgeon put a heavy load on the tiller. The slab rudder always wanted to kick-up either due to drag or just due to buoyancy.

The top of the slab was recut to allow the blade to hang with the trailing edge vertical. The planform was cut to give the leading edge an elliptical shape. The section was machined to a NACA 0012 with a straight section between the leading and trailing sections (the slab was only 3/4" thick).

Now, the foil seems to stay down almost by itself.

I'm wondering if the lift being so far forward of the kick-up pivot point actually causes the blade to stay hard forward. Is the lift vector sufficiently greater than the drag vector that the rudder tries to push forward against the stop in the rudder head?

Have I explained this properly?

 

lurking

 

Yes, it is possible.

No, you didn't explain it properly

 

So let me try again.

Does the lift created by the rudder when angled to the flow of water, actually cause the rudder to pivot DOWN rather than kick UP? The kick-up pivot is near 50% of the root chord.

 

I don't see how it could be the case. The drag force will always want to kick it up, and the lift force creates only a sideways bending moment. There is no forward-pointing force which could push the rudder down.
IMO, the most plausible explanation is that by machining the blade, (streamlining it and taking away some volume) you have decreased both the drag and the buoyancy so much that the kick-up moment has become very small, so the blade now remains in place.

 

The total force can be forward of the pivot axis of the kickup pivot. Lift is defined wrt the free stream direction. If there is an angle of attack on the rudder, say 6 degrees, then as long as the drag causes the total force vector to be less than 6 degrees aft wrt the free stream, the force will tend to hold the board down.


Since drag increases faster than linearly, and lift increases linearly as AoA increases, The total force vector starts out dead aft at zero lift, then rotates forward for a while until a certain angle of attack, then it starts to swing aft again as more helm is added because drag is increasing proportionally faster than lift at that point.

One minor point. Since the question is about a blade kicking up, one must consider the moment arm as well as the force. In general, the vertical center of drag effort would be lower than the vertical center of lift. So you can't just blindly combine the total lift and total drag vectors in 2D. You have to work out the moments for each section and integrate along the span. This would be a small difference for typical kick-up geometries though.

If you were to take your favorite L/D vs AoA plot, say for NACA 0012 with AR = 2, and add the line 1/TAN(AoA) to the plot, you would see the angles where the L/D lies above the line you drew.

 

I did not blindly combine the lift and drag together. I have considered them separately, if you read through my previous reply.

Quite honestly, I am having hard time trying to imagine a rudder which creates a forward-directed force, or thrust. If you could upload a picture of what you mean, it would probably make my life easier.

 

Daiquiri, The "blindly" comment, and the issue of vertical centers in general, wasn't directed at you in any way. It was a separate issue from the other ones, and probably more of a red herring than anything else.

 

SuperPiper, the fore/aft position of the pivot doesn't matter (much). It is just a matter of the direction of the total force compared to the cord's normal vector. When working out the actual moments that a particular section generates about the pivot due to HYDRO FORCES, the fore/aft pivot location is irrelevant. The fore/aft position of the pivot does affect the moment due to buoyancy.

 

Phil, now I see what you meant and you are right - there is a possibility for the total force to be tilted forward with respect to the blade plane.
But I believe that it is a pretty academic issue, becuase in practice the effect you have described is marginal (if any at all) for rudders of common aspect ratio (between 2 and 4). We are talking about (L/D)max = 10-11 at 5° AoA approximately. It means that hydrodynamic force would be tilted backwards by 5° respect to the free stream, and the rudder blade is turned by 5° too. Giving at best a 0° angle to the pivot axis. See the attached drawing.
The resultant hydrodynamic force could be tilted forward of pivot axis for rudders of much higher AR, but IMO even then the kick-down effect remains very weak because of very small angles in play.
Cheers

P.S.
I have just noticed that I have used a notation tg^-1(x) instead of arctan(x). That's an error which would have cost me a degree many years ago...

 

Instead of a degree, the error only cost you a minute. Needed to write the apology.
Just avoid 60 such errors and you are safe. Losing 60 minutes would equal losing one degree.

 

LOL!!
And how many degrees make one academic circle?
Ok, ok, I'm leaving...

 

While the rudder could never apply a forward force to the boat in the boats frame of reference I could envision with my stupid human brain the rudder developing some net forward forces on itself in its own frame of reference. Good topic.

 

philSweet and daiquiri, excellent presentations. I can see that under certain circumstances, the rudder will seat itself in the DOWN position.

Now that we're aware of this phenomenon, is there anything that could be done to take advantage of it (other than keeping barnacles off the appendage)?

 

I think that you have misunderstood my words.
What I said is that the possibility of hydrodynamic force pushing the rudder forward does exist, but is limited to a very narrow range of angles of attack (+/- 3°, to be precise) around the point of maximum L/D, and only for rudders of rather high Aspect Ratio (which is roughly the Depth/Chord ratio). By "rather high" I intend AR=4 or more. Both requirements have to be satisfied in order for the phenomena to happen and be appreciable, which makes it nearly useless in practice.
For example, on a straight course the tiller angle is zero and the only hydro force acting on the rudder is drag, which is directed backwards and wants to rotate the blade up. So you need a mechanical mean to prevent this (a bearing friction, a securing pin, cleats, weight or whatever). That example alone shows that you can't count on the above phenomena for keeping the rudder down in every situation.
Cheers