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Anybody remember high school physics/
http://www.hobie.com/au/en/forums/viewtopic.php?f=14&t=11266		 Page 1 of 3
Author:  alfred_new [ Sun Nov 23, 2008 7:27 pm ]
Post subject:  Anybody remember high school physics/
I am working on a solo righting pole system. On the commercial one available, their photos show 2 people using it.
Here is the question (sorry for the low quality graphics):
Click to enlarge
Image :?: How long does X have to be for the force (moment, vector or whatever) that the single 200 lb. person applies to be the same as the force the 2 people on righting lines apply.
I am remembering how when my kids were on a see saw with me; they would sit on the end and I would move up towards the middle until we "balanced".
I just don't want to find out I have to be 10' out in the air to get the boat to right.
Author:  Karl Brogger [ Sun Nov 23, 2008 8:50 pm ]
Post subject: 
you need to be 1.5 times farther out than the stickman on the lower hull to get the same amount of righting moment. (I think)

edit: That has to be straight out too. Making the pole 1.5 longer won't be enough as the pole goes up at an angle. Being higher won't add any righting moment.
Author:  alfred_new [ Mon Nov 24, 2008 7:05 am ]
Post subject:  thanks
That was my first thought too, but I am wondering if it might be less since it seems to me more of the force on the righting lines is down vs. being out on the pole
Author:  srm [ Mon Nov 24, 2008 9:26 am ]
Post subject: 
It's all about torque=Force (F) times Distance (X).

The force in this case is the weight in pounds of the person or persons hanging off the righting pole.

So if a crew weighing 300lb can right the boat using a 5ft pole, then the torque is 300lb x 5ft = 1500lbft. (The 5ft of course is made up, you would have to input whatever your real value is.)

The length of righting pole required for a crew weight of 200lb can then be calculated by:
1500lbft / 200lb = 7.5ft.

For righting the boat, we only care about horizontal distance from the fulcrum (bottom hull). Vertical height above the water doesn't matter as long as you're not floating in the water.

sm
Author:  alfred_new [ Mon Nov 24, 2008 10:41 am ]
Post subject:  what is the average distance out when using a righting line
If you are 6' tall and standing on the lower hull, you'd have to have the rope at your head to reach 5' out from the hull. I have never righted a boat with anyone else, so I don't know how far out you have to get for it to tip.
In photos on the rightiing bag test post,
http://www.hobiecat.com/community/viewt ... g+bag+test
he stands on the lower hull and leans out with the bag on his back and it looks like only 2-3' to get to a tipping point.
Since the ratio looks like 1:1.5, even 3' out @ 300 lbs means 4.5' for 200. 4.5 plus the length from the center of the boat, plus length lost to the pole being at an angle, and I think you are talking about a pole that is too long to be practical.
Author:  srm [ Mon Nov 24, 2008 10:59 am ]
Post subject: 
Quote:
If you are 6' tall and standing on the lower hull, you'd have to have the rope at your head to reach 5' out from the hull.


The location where you hold the rope has zero bearing on how much torque you transmit to the boat. It only effects how comfortable it is hold onto the righting line. All the boat cares about is the horizontal distance of your center of mass from the lower hull (fulcrum). The 5ft was a number that I picked out of the air to keep the math simple, but you're right, for someone standing on the hull, the distance is probably more like 3 to 4 feet horizontal.

Quote:
Since the ratio looks like 1:1.5, even 3' out @ 300 lbs means 4.5' for 200.


There is no way to get around this. The physics is what it is. You either have to move out farther or add more weight.

sm
Author:  Karl Brogger [ Mon Nov 24, 2008 11:13 am ]
Post subject: 
Just going back to the bag thing. I can pull the FXone upright no problem. That mast is a 1' 3" longer than the HC16 mast, and I would think that with spreaders might be a bit heavier. The way I do it is I have a loop knotted off in the line for the bag that I hook into my harness. I have just enough left over to get the bag onto my shoulder. That way I don't have to worry about hanging onto the line, the bag, and myself. I just have to worry about the bag this way. I weigh about a 160lbs, and use the smaller bag from Murray's.

The downside of putting a loop around your trap hook is sometimes you can't get back on. :oops:

Image

Image
Author:  alfred_new [ Mon Nov 24, 2008 1:03 pm ]
Post subject: 
fx 0ne
How far do you have to lean out before the boat starts to right?
Author:  Karl Brogger [ Mon Nov 24, 2008 1:33 pm ]
Post subject: 
I'm about as flat as I can get, with the bottom of the righting bag maybe a foot out of the water. Its getting the mast clear of the water those first few inches that are tough, once you get some wind under the sail it almost does it itself. I probably stop trying to pull the boat up when the tip is at best 5 feet out of the water. I'm trying to perfect getting back on the boat with out having to swim.
Author:  alfred_new [ Mon Nov 24, 2008 5:22 pm ]
Post subject: 
I am just thinking out loud. When you are standing on the lower hull before grabbing the righting line, 100% of your weight is supported by the hull. At 45 degrees, is 1/2 of your weight still on the hull? If so, then of the 300 lb crew the effective weight is 150 lbs times the horizontal distance out. At full horizontal (which Karl says he gets close to) all the weight is force times the distance out.
If the tipping point is reached at 30 degrees above the horizontal then is 1/3 of the weight still on the hull? That puts the effective mass at 200 lbs and at 3' out you get 600 ft/lbs. So if I can get 3' out on the pole, I would match that.
I know physics is physics, I am just trying to estimate whether using a pole to get out and to use all your mass is worth trying for solo righting. I have plans to make a righting bag if this doesn't work, but I am trying to avoid adding 150 lbs to my back the additional pull on my arms.
Author:  BobMerrick [ Mon Nov 24, 2008 6:21 pm ]
Post subject: 
alfred_new wrote:
I am just thinking out loud. When you are standing on the lower hull before grabbing the righting line, 100% of your weight is supported by the hull. At 45 degrees, is 1/2 of your weight still on the hull? If so, then of the 300 lb crew the effective weight is 150 lbs times the horizontal distance out.


It doesn't work that way. The moment about your feet is your weight times the horizontal distance from your feet to your center of gravity. Don't fake yourself out with trying to reduce the weight for different angles.
Author:  srm [ Mon Nov 24, 2008 6:39 pm ]
Post subject: 
Quote:
When you are standing on the lower hull before grabbing the righting line, 100% of your weight is supported by the hull. At 45 degrees, is 1/2 of your weight still on the hull?


Unless your body is in the water, 100% of your weight is being supported by the hull regardless of what angle your body is at.

sm
Author:  BobMerrick [ Mon Nov 24, 2008 6:45 pm ]
Post subject: 
It's easier to picture if you think of you and the boat as the system under consideration instead of making yourself the system. The forces and moments on you relative to the boat will always net out to zero. The moments on the boat relative to the water will not (hopefully).
Author:  Karl Brogger [ Mon Nov 24, 2008 6:52 pm ]
Post subject: 
srm wrote:
Unless your body is in the water, 100% of your weight is being supported by the hull regardless of what angle your body is at.


But if the righting line is pulled over the top hull, and you're pulling on it, thats taking some weight.

mI brane iz starrttyng two hert...
Author:  srm [ Mon Nov 24, 2008 7:23 pm ]
Post subject: 
Quote:
But if the righting line is pulled over the top hull, and you're pulling on it, thats taking some weight.


Uh, yea, but what's supporting the top hull? The bottom hull.
Swinging your weight off the side of the boat by holding the righting line doesn't change the weight on the bottom hull. You could hang completely off the top hull, it wouldn't change the amount of weight on the bottom hull.

The critical issue is that the moment generated by the crew trying to right the boat (weight times horizontal distance) is greater than the moment generated by the weight of the rig (of course wind force in the rig can either help or hurt this as well).

sm
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