righting line position, engineering/physics needed

A question being debated on the F16 forum. Is it easier to right a cat if the righting line is thrown over the upper hull than if the line is tied to the front beam just inside the hull? everything else is the same. assume the line is tied to your trap hook.
I think it would be easier to right it over the top hull.

For starters, I'm not a fan of connecting the rope to my harness from a safety standpoint. But, if you're going to do that, where the righting line connects to either the inside (top) end of the main beam or around the hull makes no difference. It should only affect how easy it is to hang onto manually and would be easier to hang onto if it was thrown over the hull and carrying more of a downward angle to you.

From a physics perspective, the only thing that matters is how far away from the center of rotation (the hull in the water) your righting mass (your body) is. The further your body is away from the hull, the easier it is to right. Laying back flat gets your upper body further away from the center of rotation.

Jake,
Are you sure you're right about this?
From experience it seems easier over the hull. It would put the righting line a foot and a half higher and 2 feet farther out( it's around the hull). I'm certainly no engineer but that's got to change the leverage in relation to the fulcrum.

If this image is what we're talking about ... yeah. No weights have shifted to give it more righting leverage between these two images. You can consider the person and the boat as one fixed system here. It might be easier to hang on with it over the hull - and you might be able to get lower to the water (getting more leverage) if the line is over the hull...but which way the line runs doesn't add anything to the equation with all other things being equal.

Sigh..I called this crazy talk in the other thread but it's right. As long as the weight is fixed and in the same place, the righting force is the same. Move the line farther down and in reality all sorts of things would break, but the righting wouldn't change.

Same way having feet in hiking straps or torso hooked to trapeze doesn't change anything provided you're in the same spot both ways.

This is absolutely correct. The boat doesn't "care" whether the line is over the hull, under the hull, or if you use a pole, etc. The only thing that matters is the location of your center of mass relative to the axis of rotation. Different line positions may make it easier to hold onto the rope, but they have NO affect on whether the boat comes up or not. It's also important to consider wind and wave conditions as these will greatly help or hinder your righting success.

As for using your trap harness to right the boat, I like to throw two wraps of righting line around my hook and then just hand hold it. This provides enough friction to easily hold my body up with one hand, yet as soon as I release the line, it will easily slide out of the hook.

sm

So by that logic, you could put the line anywhere up or down the crossbeam and the righting moment would be the same? Might be right, but I ain't buyin' it. Put the line at mid beam or 3/4 and it will be harder to right, so why wouldn't that change at the top also? You have a longer lever.

The 'lever' is how far you get your CG, or CM away form the fulcrum, or axis, which is somwhere in the bottom hull.

You could hang from the top board, and b/c it gets your CG out from the fulcrum, the boat will right, you don't need a 'line'.
I do use a line however, kinda hard to get up to the top board, etc.

Not anywhere. All the way down the crossbeam won't work.
Assuming you can keep the body straight (using the harness for instance), the effect is exactly the same, as said. But the lower the attachment, the higher the rope tension, so you it will be harder actually. Not harder to right the boat but to stay straight.

Not just that, the angle of your body matters too. It can be seen as how far is the projection of your CG on the water. If you were able to stay horizontal, just above water, it would be the most effective position.
The force vector is gravity, vertical. The lever is the distance between the vector line and the center of rotation. That distance is measured by projecting the perpedicular line from the center of rotation towards the vector line. That is the horizontal distance between the lower hull and the projection of your CG on the water.

If I walked out to the tip of those super-long daggar boards, I wouldn't need a righting line at all, would I?

Same thing for hiking, by the way.

I once talked a writer into including a scene where a sailor was being drowned slowly and horribly because the monster attached the sailors harness to a righting bar. It was a proposed low budget horror movie. It was so low budget they wanted to borrow catamarans and sailors. The scene didn't make it through editing and they never found anyone to back the movie.

It would have been a great clip to show all new sailors.

The problem with this sketch is vector Fy(b) is on the wrong side of the C.B. at 'A'?

I can't really follow the detail that you added, but aside wind effect, the only forces that matter here are the weight of the person and the boat, and their correspondent lever arms. The rest cancels out.. You could also take the person out of the system you analyze and think of the rope tension as an external force, but it's worthless, the forces A, B and C that I think you are detailing (rope attached to three different positions right?) would have a different cos(anlge) but also a different magnitude, which gives the same result if the person is on the same position (except the rope has different tensions on each case).

I see. For some reason you are thinking about the effect of the feet at A and the rope at B and C.
Or maybe you are talking about what happens if you attach the rope at A. Well, the component is not reverted but goes to infinity as the attachment approaches to the actual center of rotation. Y component also goes to 0. If the rope and your arms are strong enough (and you are still a bit above), it would work too.

Comparing attaching the rope at B or C.
The lever arm includes the vertical vector?
Assuming vector Fy(b) and vector Fy(c) are in the same vertical plane, and on the same side of the C.B., I AGREE with you. >...approaches to the actual center of rotation<
As drawn vector Fy(b) and vector Fy(c) are on opposite sides of the C.B. at A.

You're putting too much effort into it and the solution is getting lost in the analysis. The system we are evaluating is a static system where the sailor's position (weight) and the boat (including mast) are not moving in relation to each other and can be looked at as one fixed system at any instant in time. While this relationship does change as the boat is righting, we're only looking at the system in a fixed point in time so the forces at play will start to right the boat.

With a static system, it simply breaks down to where the center of mass of the combined system is in relation to the center of buoyancy (rotation). The sailor isn't moving in relation to the boat in this analysis so there's no point in looking at the individual loads on the line related to it's angle. If the line angle does get lower, the line carries more load to support the same weight of the sailor (because the angle is sharper) but that is offset by the beam it's connected to and the amount of force the sailor is having to use to hang onto the line - it cancels itself out of the equation. The line isn't doing anything but supporting the sailor and keeping the system static....so it doesn't matter if the line is higher or lower when you look at the system as a whole. Only if you were trying to minimize the amount of effort you needed to hold yourself up or minimize the amount of loading a fixture on the boat has to hold does the angle of that line come into play. On the system as a whole, the relationship is fixed, non-moving, and the details about the line do not matter.

If having the line over the hull vs. having it attached lower to the beam really did make a difference, all keel boat's keels would look like upside down "A"s since they would also want the support to the keel from the outer most point of the hull....but they don't because it doesn't matter. Only the amount of moment the weight applies to the fixed system comes into play - not how it's supported in the fixed system. Hence, typical keel boats use a slender fin to hold the bulb as far away from the center of buoyancy/rotation to get the most righting moment.

In the diagram below, all that matters to get the boat righted is that Fa+b is on the sailor side of the center of buoyancy marked with a circle and cross. Once it passes that center point, the boat is righting. Note, however, that my formula below is wrong...Fa+b in the formula should read Ma+b to represent the righting moment...weight times distance from the center of rotation. Or it should read Fa+b(Dc) where Dc is the distance the combined sailor and boat weight are from the center of buoyancy.

Can't help but think:
What's your Vector Victor?


Roger Murdock: Flight 2-0-9'er, you are cleared for take-off.
Captain Oveur: Roger!
Roger Murdock: Huh?
Tower voice: L.A. departure frequency, 123 point 9'er.
Captain Oveur: Roger!
Roger Murdock: Huh?
Victor Basta: Request vector, over.
Captain Oveur: What?
Tower voice: Flight 2-0-9'er cleared for vector 324.
Roger Murdock: We have clearance, Clarence.
Captain Oveur: Roger, Roger. What's our vector, Victor?
Tower voice: Tower's radio clearance, over!
Captain Oveur: That's Clarence Oveur. Over.
Tower voice: Over.
Captain Oveur: Roger.
Roger Murdock: Huh?
Tower voice: Roger, over!
Roger Murdock: What?
Captain Oveur: Huh?
Victor Basta: Who?

We are thinking!
http://www.youtube.com/watch?v=KLSdOY-6R_U

You could choose as lever arm the body of the sailor instead of the horizontal. Then you'd have to calculate force component perpendicular to that lever arm.

The result would be the same as using the full weight and calculating the horizontal lever arm, that is perpendicular to it.

No reason to use the line angle though.

"You're putting too much effort into it and the solution is getting lost in the analysis."


yes. you should put boobs on the sailor

Again I agree with you as stated above, it seems like you disagree, that I agreed with you above.
Again, I am NOT disagreeing with Mrig =< Mrighting, as Jake drew above.

Simply, as drawn the FBD appears to show the location of tension members with Fy(b) and Fy(c) are on opposite sides of the C.B.
Can you edit the FBD of the tension member(s) whose vertical component(s) are on the same side of the C.B.? (not sys diag as Jake drew)

I disagree that I disagree that you agreed with me above...

I also agree that the vertical component that you mention does act in the opposite way. But there is also a bigger horizontal component that compensates it. I think we would bore the audience if we try to draw it..

Ok, here it is....
The horizontal components produce more torque because they are farther from the axis.

By the way, if you want to consider the sailor external to the system, you should also include the force that she applies with her feet. As it doesn't produce much torque, no problem here.

hahaha, nice.

engineers are funny!

The only things that matter in this system are:

The CG location of the boat (somewhere to the left of the hulls)
The CB location of the boat (also somewhere to the left of the hulls since the mast in the water contributes to it)
The CG location of the person righting (approximately lower chest)

As soon as the location of the combined CG(boat)+CG(person) is to the right of the CB(boat), the boat will start to rotate clockwise. Moving the combined CG out further will just increase the rotational speed / and counteract the movement of the CB to the right (when the mast clears the water).

The only thing that pulling the line from the top of the hull does is make it easier for the person to hold on to. Makes no difference in righting force or leverage.

Ok that's solved, now... which came first the chicken or the egg?

I'm still not really convinced, but I'm ignorant and prone to stubbornness. Damn that 10th grade education.

I'm with ya, but I think this is like politics,Chevy vs. Ford, and which boat is best. No winners.

Disagree. What center of balance you mean? There is an axis of rotation that is on the hull, it changes somehow as the boat rotates but is always on the hull. The center of balance that you are probably referring is what you then call the combined CG of the boat plus the person which will never be to the right or to the left because it´s the same thing..

Oh, but it's clear. Somebody gets their boat righted.

CB= Center of Buoyancy. It's the force counteracting gravity.

In a stable configuration, the horizontal position of the CB = horizontal position of the CG, otherwise the boat will shift until they do.

As the person leans out (to the right), the combined CG shifts to the right, as does the CB as the mast comes out of the water. The further the CG moves away from the CB, the faster the boat will move in an attempt to restore equilibrium.

The bottom line is that only the horizontal positions of the Center of Buoyancy and Center of Gravity matter. How they get to where they are makes no difference at all.

Ah.. got it. Acronyms..

At least now I'm looking at the diagram!

And I would venture that the boat is not perpendicular to the horizontal plane when you are attempting to right the boat. When you are at that point (trampoline is vertical), you've already cleared most of the main from the water, and righting should be almost automatic at that point (if the boat is facing somewhat windward)

We simplify man! (that's why it never works at the first time..)

Actually, that's at it's worst point from a pure weight balancing perspective. The center of weight for the boat extends further away from the center of rotation when the mast is horizontal. In the real world scenario, however, wind is starting to get under the sail, providing lift, and helping move the combined weight center move further toward the sailor side of the equation.

Come on, don't make me disagree everytime...
I think the the weight of the upper hulls are a significant factor and it's not the same whether they are in one side or the other of the vertical. The worst point to me is just when the mast left the water and none of it is floating. If you can move it up from there you are fine.

Aaahah...you are correct sir!

Agree, this is what I said in my last post, I thought.

Ya know, I'm glad you posted that Jake. 'Cause I was going to have to agree with Andy on this one.

Have you ever had it where you could get the mast out , but not righted? Don't ask me how I know. remember that thread?

You did?.. I thought you were talking about just getting away from the axis, like up on the boards. So I had to start making all these drawings instead of getting back to work..

Sorry to take away from your work.
If you think it's any different than Mass x radius squared = righting moment, then lets see a drawing on why it's not.
I realize this equation doesn't allow for all the other factors that influence this equation however.
HOwever, I always thought that righting moment, while sailing, righting the boat, etc. is the mass times the square of the radius from the axis. Is that not correct?

The mass has to be exerting a force thru some kind of lever, like a line, a board, a righting pole, etc.

Sorry if I was mean, I was actually but was only kidding..
I thought I had deleted that comment, obviously it didn´t work.
It think that you are mixing concepts.
like this: http://en.wikipedia.org/wiki/Angular_momentum
And this: http://en.wikipedia.org/wiki/Torque
And the square probably comes from a formula of angular momentum of something else than a punctual mass (like a cilinder maybe..)

If you make the analogy between linear and rotational systems:
torque (or moment) is equivalent to force
Angular momentum (or just momentum) is equivalent to mass.

Force vector = Mass * Acceleration vector
Torque vector = Momentum * Angular acceleration vector

Force makes a mass move (accelerate)
Torque makes a rotational body move (accelerate rotation)
The stability or balance equation or whatever it is called, is done for forces and for torques, not for angular momentum.
The righting moment is torque, not angular momentum. All the geometry stuff (why it matters whether more or less horizontal)is related to the vector part.. On the other hand, mass and momentum are just scalar values (the angle doesn´t matter but just the distance to the center, as you noticed with respect to momentum)

Watch how the artistic skaters control spinning speed by extending or retracting their arms. The closer the arms to the body, the faster he/she spins. That´s momentum.

So what you do is trying to make net torque more than 0 towards the side you want to turn the boat. When it´s more than 0 it starts accelerating and you adjust your body position a bit forward to go back to zero torque so that it doesn´t happen too quickly and also to adjust for the vector changes as it rotates.

Mi brane hertz

Someone cursed engineers earlier, a true Engineer would have assumed that the person was a sphere just to make the math easier.

Well we did.. you just didn´t notice

Mine too, I stopped reading stuff after the 3rd post. I just happened to go here and see yours, so I thought I would jack this topic up some more.

I think everyone now agrees that in terms of righting moment, the point of attachment doesn't matter. The original question, however, was "which way is 'easier'". I submit that the "easier" way is whichever one requires less pull on the righting line. I'm too lazy to do the math, but my intuition is that a right-angle vector to the body will require the least magnitude pull.

Therefore, you need to determine the height of the sailor righting the boat, the attachment point (harness or arms), the angle that the sailor leans, and the beam of the boat before you can answer the question.

A tall sailor that hauls by hand and leans way back on a narrow cat will prefer a line thrown over the hull. A short sailor who pulls from the waist and stands up straight to right a wide cat will want to attach to the underside of the crossarm.

Regards,
Eric

You are right.
Here is another drawing that shows that.
The boobs drawing is wrong by the way.. I never said I was a true mechanical engineer.. It would work only if the person was horizontal.

On the picture the vertical vector is not exactly the weight of the sailor, except if his CG is exactly at the harness hook. Otherwise it´s in proportion of the distance between feet-harness and feet-CG. That is because there is a lever arm effect with respect to the feet.If CG is above harness, the blue force would be higher than weitht (and the difference would be compensated by the feet)

we reduced them to an arrow.

Well, thankfully, Andanista reduced the drawing to TWO spheres I wasn't really looking at the rest of the drawing.

And the only engineering I'm particularly concerned with is the thermodynamic and volumetric analysis of the rum & coke I intend on destroying my liver with at Hiram's this weekend.

Gotta have at least one Margarita. To ward off scurvy.

sh*t, I thought that was what the lime was for....

Thanks for the health tip!