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.

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.

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

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!

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?

F-18 Infusion #626- SOLD it!

'Long Live the Legend of Chris Kyle'

Re: righting line position, engineering/physics needed
[Re: Todd_Sails]
#239365 10/26/1112:18 AM10/26/1112:18 AM

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.

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

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..

Re: righting line position, engineering/physics needed
[Re: Andinista]
#239371 10/26/1107:12 AM10/26/1107:12 AM

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.

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

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 didnīt 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 doesnīt 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. Thatīs momentum.

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

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

Re: righting line position, engineering/physics needed
[Re: Isotope235]
#239405 10/27/1106:55 AM10/27/1106:55 AM

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 itīs 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)

Last edited by Andinista; 10/27/1106:59 AM.

Re: righting line position, engineering/physics needed
[Re: presto]
#239406 10/27/1106:55 AM10/27/1106:55 AM

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

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.

Jay

Re: righting line position, engineering/physics needed
[Re: waterbug_wpb]
#239414 10/27/1109:20 AM10/27/1109:20 AM

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

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.