mast climbing

[Dan H]

When is a boat too light or a mast too narrow to climb? Is there a rule of thumb or a ballast/climber weight ratio? What if you find out at the top of the mast that your boat is too light? A bosun's chair may not offer quick release.



djhhan@aol.com

[Tom Dale]

Good question - I look forward to some responses on that one! Especially since I weigh 230lbs.



choice@nternet.net

[M. R. Bober]

When is a boat too light or a mast too narrow to climb? Is there a rule of thumb or a ballast/climber weight ratio? What if you find out at the top of the mast that your boat is too light? A bosun's chair may not offer quick release.

Great questions. I don't think the mast could be too narrow to climb. Too weak, perhaps. If you are in still conditions, your weight either on the mast on in the cockpit will have the same effect on the waterline. When the boat is rolling side to side, the dynamics clearly change. Consider a centerboarder with a tall mast and a heavy climber. What you are looking for is the "righting moment" of the vessel.

One thing is certain (IMHO) if things are getting too exciting, you will know way before you get to the masthead. As a general rule send the lightest crew member up the stick (for two reasons: 1) Easier on the rig and those hoisting and 2) least likely crew member to win the "I don't want to go up there" argument.

I always wear long pants and a lifevest (Class III zip up type) to prevent bruises when I go aloft.
Good luck.
Mitchell Bober
RESPITE
CD330

[Larry DeMers]

Mitch is right..it's 'righting moment' that will determine the effect on your boat. Fortunately, the math is not complicated.

Lets set up an example, and see if this makes sense.
Using my Cd30 for example. Keel weight is 4000 lb, and the center of this mass is lets say, 4 ft below the waterline. Since the waterline is the fulcrum in this equation, the weight on one side of the fulcrum must exceed the weight on the other side before the fulcrum will pivot.

So on the boats lower portion, we have a 4000 lb weight at 4 ft. from the waterline, or 16000 lb of torque when the boat is lying at +90 degrees to the vertical.

The mast is say 40 ft. tall, and I weigh 200 lbs (I wish, heh)..so my lever arm if at the top of the mast would be 40 * 200 or 8000 lbs. when the boat is heeled over at 90 degrees to the vertical.

The reason I mention being over at 90 degrees to the vertical is that is the strongest effect either the keel or mast weight will exert on the boat. Less than 90 degrees over, and the vector addition will reveal that the effect grows considerably less quite fast.

In our example then, we have an 8000 lb. torque or force on the mast, counteracted by a 16000 lb torque on the keel. Ergo, the keel would win, and will continue to win as long as it is greater than the mast torque. Lets say I weighed 400 lbs (jeesh, better get the diet going there bud..) and then climbed (or was lifted) to the mast top. At 40 ft. long and positioned at +90 deg. from the vertical, that would be 40 * 400 or 16,000 lbs torque, the same as the keel exerts on the boat. Now if the boat were heeled to 90 degrees, it would be in balance and would not right itself automatically. This balance between the weight at the top of the mast and the keel would also be mostly true from the perfectly vertical position all the way to the the horizontal position, as each weight moves the same amount off of vertical. However, it is true that hull shape, beam of the boat, and other secondary factors confuse the equation somewhat, making an exact number something less straightfoward. I do believe that the basic idea presented here is however, workable and true to a first order approximation.

Hope this helps..

Cheers!

Larry DeMers
s/v DeLaMer
Cape Dory 30

You need to plug your own numbers in here to see what your boat's torque is. Now this exercise is not out of a book or reference source..just from common science. So, critiques are welcome.

When is a boat too light or a mast too narrow to climb? Is there a rule of thumb or a ballast/climber weight ratio? What if you find out at the top of the mast that your boat is too light? A bosun's chair may not offer quick release.

demers@sgi.com

[John M]

Hi Dan,
I can go to the masthead of my CD28 with no problem. I don't know if there is a rule of thumb. The boat displacement and beam should be taken into account, along with size of the crew member going aloft. I have been to the top of a CD22 mast, at the time I weighed about 200 lbs. To go up the mast of a small boat you must take certain precautions. Make sure the boat is tied securely in a slip or dock. Hoist the MIDDLE of an anchor line to the masthead. Lead the two ends out abeam and secure them. You will find that the mast section half way between the deck/spreaders and the spreaders/masthead will FLEX. If the flex is too much, come down.

OR, buy some beer, and get 2 or 3 friends to help you lower the mast.

Good Luck,
John Martin
CD28 #346 Intrepid

When is a boat too light or a mast too narrow to climb? Is there a rule of thumb or a ballast/climber weight ratio? What if you find out at the top of the mast that your boat is too light? A bosun's chair may not offer quick release.

johnmartin55@hotmail.com

[Neil Gordon]

You got the physics right, I think. The practical problem in actually going aloft is knowing that the rig will support the climber, allowing for the wake from some passing $&*#$@, and remembering that you don't have to be over 90 degrees before the water forgets to stay on the outside.

That said, there are practical solutions, too. If you start going aloft, you only change the weight distribution a little at a time... if you're four feet up and the boat's over 15 degrees, stop climbing. If you get to the spreaders and you're still in trim, keep going.

I watched a fellow on a Catalina 25 go up so where he was just over the spreaders. All was fine until his soon to be ex-gilrfiend stepped from the dock to the deck. He swung in a fair arc, but didn't come near a knockdown. <g>


Regards, Neil
s/v LIQUIDITY
Cape Dory 28 #167



neil@nrgordon.com

[John D]

Mitch is right..it's 'righting moment' that will determine the effect on your boat. Fortunately, the math is not complicated.

Lets set up an example, and see if this makes sense.
Using my Cd30 for example. Keel weight is 4000 lb, and the center of this mass is lets say, 4 ft below the waterline. Since the waterline is the fulcrum in this equation, the weight on one side of the fulcrum must exceed the weight on the other side before the fulcrum will pivot.

So on the boats lower portion, we have a 4000 lb weight at 4 ft. from the waterline, or 16000 lb of torque when the boat is lying at +90 degrees to the vertical.

The mast is say 40 ft. tall, and I weigh 200 lbs (I wish, heh)..so my lever arm if at the top of the mast would be 40 * 200 or 8000 lbs. when the boat is heeled over at 90 degrees to the vertical.

The reason I mention being over at 90 degrees to the vertical is that is the strongest effect either the keel or mast weight will exert on the boat. Less than 90 degrees over, and the vector addition will reveal that the effect grows considerably less quite fast.

In our example then, we have an 8000 lb. torque or force on the mast, counteracted by a 16000 lb torque on the keel. Ergo, the keel would win, and will continue to win as long as it is greater than the mast torque. Lets say I weighed 400 lbs (jeesh, better get the diet going there bud..) and then climbed (or was lifted) to the mast top. At 40 ft. long and positioned at +90 deg. from the vertical, that would be 40 * 400 or 16,000 lbs torque, the same as the keel exerts on the boat. Now if the boat were heeled to 90 degrees, it would be in balance and would not right itself automatically. This balance between the weight at the top of the mast and the keel would also be mostly true from the perfectly vertical position all the way to the the horizontal position, as each weight moves the same amount off of vertical. However, it is true that hull shape, beam of the boat, and other secondary factors confuse the equation somewhat, making an exact number something less straightfoward. I do believe that the basic idea presented here is however, workable and true to a first order approximation.

Hope this helps..

Cheers!

Larry DeMers
s/v DeLaMer
Cape Dory 30

You need to plug your own numbers in here to see what your boat's torque is. Now this exercise is not out of a book or reference source..just from common science. So, critiques are welcome.

When is a boat too light or a mast too narrow to climb? Is there a rule of thumb or a ballast/climber weight ratio? What if you find out at the top of the mast that your boat is too light? A bosun's chair may not offer quick release.

OK Mr Wizard Since you are hereafter known as the math guru, how would the math work for the heeling moment on a boat? Being relatively new to sailing, my mind still tells me that in Puget Sound it would be a rare wind to push my boat over enough to be dangerous, but my stomach tells me otherwise! Given the ballast of 3500 on my boat (Cape Dory Intrepid 9M (30 ft)) and the sail area, depending on which sails are flown, what are the mathmatics determining pressure on the sails and ballast vectors? I know the force of the wind on the sails increases geometrically (?) or in twice the wind there is four times the pressure per square foot of sail. Is there a good book or some place to research this. Really, I do want to know. I need to hook up with someone here to help me get over that "falling over" feeling that I'm sure you all had to get over at one time. Thanks.

John
S/V Mariah



john_dupras@hotmail.com

[Larry DeMers]

Hi,

Well, I don't claim to be a math wizard..just a reader of sailing books for 30 years or more, and a similar length of time sailing. I think that Catherine has made available the righting moments for your boat as well as others, and she she can direct you to that source.
But really, it isn't hard to picture..but it is hard for someone new to sailing to counteract their gut feeling of avoidance to heeling. When we bring non-sailors aboard, I whip out a model sailboat that I modified for this purpose. Maybe a textual version of this demo would help you out. Try this on for size: This may get a bit wordy..sorry.

Picture a sailboat with sails up, a wood dowl drilled through it's exact center, from stem to stern (usually coinciding with the water line by the way). Holding the ends of the wood dowel, you should be able to heel the boat port and stbd. and it should return to an upright and neutral position. Why? Gravity working on the keels weight is counteracting your fickle finger that is pushing on the sails (wind pressure, right?), causing the good ship to return to upright. Now heel that puppy over quite strongly..to 90 degrees lets say. Feel that keel pulling the boat back to upright? That is gravity working for you, counteracting the wind pressure in the sails.

Now if you climb to the top of the mast, and the boat is already heeling abit while underway, it will of course heel more with you up there. It the boat is sitting still when you get to the top, it will continue to remain still as before. Your weight is multiplied directly by the distance from the center of balance..the water line in this case. Lets say you are 40 ft. up and weigh 200 lbs. 40 x 200 = 8000lbs. That is the torque that your weight represents should it be at right angles to the vertical postiion of the mast. Between fully vertical and 90 degrees from vertical, the torque will go from basicly 00.0 to 8000 lbs, as you approach that 90 deg. from vertical point.

Ok, the same holds true for the keel. Upright, the keel has a righting moment or torque of 00.0. Now as we heel the boat over, the righting moment increases by the same math as the mast did. It is distance from the fulcrum or center of balance times the weight. In this example, the keel weighs 4000lbs. and is 4 ft. from the center of balance. 4000 x 4= 16,000 lbs. This means that at it's strongest point (again, with the boat at 90 deg. heel)the keel has the ability to counteract 16,000lbs of force trying to heel the boat. If the force heeling the boat is only 8000 lbs..as it is in this case, then the mast will not reach the 90 degree heeled point, but something less than half way..quite a bit less in fact.
Try this if this is still mud. Picture a 40 ft.long see-saw. One side has 4,000 lb. sitting on the end. The other side has 200 lbs. on it's end. Obviously, the heavier side will be down..lighter side will be up. In boat language..the mast will be upright.

To carry this example even further towards the boat example, lets move the fulcrum (pivot) point closer to the heavy end, so that it is 4 ft. from the heavy end. Now the effective weights working agianst each other are aided and hurt by the distance from the fulcrum. The heavy weight is 4 ft. from the fulcrum..so 4ft. x 4000 lbs. is 16,000 ft./lbs. of torque. The light weight is 40 ft. from the fulcrum, so that would be 40 ft. x 200 lbs, or 8000 ft./lbs. of torque. Comparing the two, 8000 ft./lbs. vs. 16,000 ft.lbs, it is obvious that the heavier side wins again. In order for you to cause the mast to heel over 90 deg. from the vertical and stay there, you will need to generate 16,000 ft. lb of torque. You would need to weigh 400 lbs (400lbs x 40ft = 16,000 ft. lbs).

Hope this is easier to understand..

Larry

Mitch is right..it's 'righting moment' that will determine the effect on your boat. Fortunately, the math is not complicated.

Lets set up an example, and see if this makes sense.
Using my Cd30 for example. Keel weight is 4000 lb, and the center of this mass is lets say, 4 ft below the waterline. Since the waterline is the fulcrum in this equation, the weight on one side of the fulcrum must exceed the weight on the other side before the fulcrum will pivot.

So on the boats lower portion, we have a 4000 lb weight at 4 ft. from the waterline, or 16000 lb of torque when the boat is lying at +90 degrees to the vertical.

The mast is say 40 ft. tall, and I weigh 200 lbs (I wish, heh)..so my lever arm if at the top of the mast would be 40 * 200 or 8000 lbs. when the boat is heeled over at 90 degrees to the vertical.

The reason I mention being over at 90 degrees to the vertical is that is the strongest effect either the keel or mast weight will exert on the boat. Less than 90 degrees over, and the vector addition will reveal that the effect grows considerably less quite fast.

In our example then, we have an 8000 lb. torque or force on the mast, counteracted by a 16000 lb torque on the keel. Ergo, the keel would win, and will continue to win as long as it is greater than the mast torque. Lets say I weighed 400 lbs (jeesh, better get the diet going there bud..) and then climbed (or was lifted) to the mast top. At 40 ft. long and positioned at +90 deg. from the vertical, that would be 40 * 400 or 16,000 lbs torque, the same as the keel exerts on the boat. Now if the boat were heeled to 90 degrees, it would be in balance and would not right itself automatically. This balance between the weight at the top of the mast and the keel would also be mostly true from the perfectly vertical position all the way to the the horizontal position, as each weight moves the same amount off of vertical. However, it is true that hull shape, beam of the boat, and other secondary factors confuse the equation somewhat, making an exact number something less straightfoward. I do believe that the basic idea presented here is however, workable and true to a first order approximation.

Hope this helps..

Cheers!

Larry DeMers
s/v DeLaMer
Cape Dory 30

You need to plug your own numbers in here to see what your boat's torque is. Now this exercise is not out of a book or reference source..just from common science. So, critiques are welcome.

When is a boat too light or a mast too narrow to climb? Is there a rule of thumb or a ballast/climber weight ratio? What if you find out at the top of the mast that your boat is too light? A bosun's chair may not offer quick release.

OK Mr Wizard Since you are hereafter known as the math guru, how would the math work for the heeling moment on a boat? Being relatively new to sailing, my mind still tells me that in Puget Sound it would be a rare wind to push my boat over enough to be dangerous, but my stomach tells me otherwise! Given the ballast of 3500 on my boat (Cape Dory Intrepid 9M (30 ft)) and the sail area, depending on which sails are flown, what are the mathmatics determining pressure on the sails and ballast vectors? I know the force of the wind on the sails increases geometrically (?) or in twice the wind there is four times the pressure per square foot of sail. Is there a good book or some place to research this. Really, I do want to know. I need to hook up with someone here to help me get over that "falling over" feeling that I'm sure you all had to get over at one time. Thanks.

John
S/V Mariah

demers@sgi.com

[Ken Coit]

Larry,

Since the mast is likely to weigh a bit and its center of gravity is about 20 ft in the air in your example, we probably shouldn't forget it in any mathematical analysis. 100 lbs at 20 ft. adds 2000 ft. lbs. to the capsize side of the equation at 90 degrees.

However, as several have pointed out, the key is your comfort level. I like being on deck.

Ken

Mitch is right..it's 'righting moment' that will determine the effect on your boat. Fortunately, the math is not complicated.

Lets set up an example, and see if this makes sense.
Using my Cd30 for example. Keel weight is 4000 lb, and the center of this mass is lets say, 4 ft below the waterline. Since the waterline is the fulcrum in this equation, the weight on one side of the fulcrum must exceed the weight on the other side before the fulcrum will pivot.

So on the boats lower portion, we have a 4000 lb weight at 4 ft. from the waterline, or 16000 lb of torque when the boat is lying at +90 degrees to the vertical.

The mast is say 40 ft. tall, and I weigh 200 lbs (I wish, heh)..so my lever arm if at the top of the mast would be 40 * 200 or 8000 lbs. when the boat is heeled over at 90 degrees to the vertical.

The reason I mention being over at 90 degrees to the vertical is that is the strongest effect either the keel or mast weight will exert on the boat. Less than 90 degrees over, and the vector addition will reveal that the effect grows considerably less quite fast.

In our example then, we have an 8000 lb. torque or force on the mast, counteracted by a 16000 lb torque on the keel. Ergo, the keel would win, and will continue to win as long as it is greater than the mast torque. Lets say I weighed 400 lbs (jeesh, better get the diet going there bud..) and then climbed (or was lifted) to the mast top. At 40 ft. long and positioned at +90 deg. from the vertical, that would be 40 * 400 or 16,000 lbs torque, the same as the keel exerts on the boat. Now if the boat were heeled to 90 degrees, it would be in balance and would not right itself automatically. This balance between the weight at the top of the mast and the keel would also be mostly true from the perfectly vertical position all the way to the the horizontal position, as each weight moves the same amount off of vertical. However, it is true that hull shape, beam of the boat, and other secondary factors confuse the equation somewhat, making an exact number something less straightfoward. I do believe that the basic idea presented here is however, workable and true to a first order approximation.

Hope this helps..

Cheers!

Larry DeMers
s/v DeLaMer
Cape Dory 30

You need to plug your own numbers in here to see what your boat's torque is. Now this exercise is not out of a book or reference source..just from common science. So, critiques are welcome.

When is a boat too light or a mast too narrow to climb? Is there a rule of thumb or a ballast/climber weight ratio? What if you find out at the top of the mast that your boat is too light? A bosun's chair may not offer quick release.

parfait@nc.rr.com

[John D]

Hi,

Well, I don't claim to be a math wizard..just a reader of sailing books for 30 years or more, and a similar length of time sailing. I think that Catherine has made available the righting moments for your boat as well as others, and she she can direct you to that source.
But really, it isn't hard to picture..but it is hard for someone new to sailing to counteract their gut feeling of avoidance to heeling. When we bring non-sailors aboard, I whip out a model sailboat that I modified for this purpose. Maybe a textual version of this demo would help you out. Try this on for size: This may get a bit wordy..sorry.

Picture a sailboat with sails up, a wood dowl drilled through it's exact center, from stem to stern (usually coinciding with the water line by the way). Holding the ends of the wood dowel, you should be able to heel the boat port and stbd. and it should return to an upright and neutral position. Why? Gravity working on the keels weight is counteracting your fickle finger that is pushing on the sails (wind pressure, right?), causing the good ship to return to upright. Now heel that puppy over quite strongly..to 90 degrees lets say. Feel that keel pulling the boat back to upright? That is gravity working for you, counteracting the wind pressure in the sails.

Now if you climb to the top of the mast, and the boat is already heeling abit while underway, it will of course heel more with you up there. It the boat is sitting still when you get to the top, it will continue to remain still as before. Your weight is multiplied directly by the distance from the center of balance..the water line in this case. Lets say you are 40 ft. up and weigh 200 lbs. 40 x 200 = 8000lbs. That is the torque that your weight represents should it be at right angles to the vertical postiion of the mast. Between fully vertical and 90 degrees from vertical, the torque will go from basicly 00.0 to 8000 lbs, as you approach that 90 deg. from vertical point.

Ok, the same holds true for the keel. Upright, the keel has a righting moment or torque of 00.0. Now as we heel the boat over, the righting moment increases by the same math as the mast did. It is distance from the fulcrum or center of balance times the weight. In this example, the keel weighs 4000lbs. and is 4 ft. from the center of balance. 4000 x 4= 16,000 lbs. This means that at it's strongest point (again, with the boat at 90 deg. heel)the keel has the ability to counteract 16,000lbs of force trying to heel the boat. If the force heeling the boat is only 8000 lbs..as it is in this case, then the mast will not reach the 90 degree heeled point, but something less than half way..quite a bit less in fact.
Try this if this is still mud. Picture a 40 ft.long see-saw. One side has 4,000 lb. sitting on the end. The other side has 200 lbs. on it's end. Obviously, the heavier side will be down..lighter side will be up. In boat language..the mast will be upright.

To carry this example even further towards the boat example, lets move the fulcrum (pivot) point closer to the heavy end, so that it is 4 ft. from the heavy end. Now the effective weights working agianst each other are aided and hurt by the distance from the fulcrum. The heavy weight is 4 ft. from the fulcrum..so 4ft. x 4000 lbs. is 16,000 ft./lbs. of torque. The light weight is 40 ft. from the fulcrum, so that would be 40 ft. x 200 lbs, or 8000 ft./lbs. of torque. Comparing the two, 8000 ft./lbs. vs. 16,000 ft.lbs, it is obvious that the heavier side wins again. In order for you to cause the mast to heel over 90 deg. from the vertical and stay there, you will need to generate 16,000 ft. lb of torque. You would need to weigh 400 lbs (400lbs x 40ft = 16,000 ft. lbs).

Hope this is easier to understand..

Larry

Mitch is right..it's 'righting moment' that will determine the effect on your boat. Fortunately, the math is not complicated.

Lets set up an example, and see if this makes sense.
Using my Cd30 for example. Keel weight is 4000 lb, and the center of this mass is lets say, 4 ft below the waterline. Since the waterline is the fulcrum in this equation, the weight on one side of the fulcrum must exceed the weight on the other side before the fulcrum will pivot.

So on the boats lower portion, we have a 4000 lb weight at 4 ft. from the waterline, or 16000 lb of torque when the boat is lying at +90 degrees to the vertical.

The mast is say 40 ft. tall, and I weigh 200 lbs (I wish, heh)..so my lever arm if at the top of the mast would be 40 * 200 or 8000 lbs. when the boat is heeled over at 90 degrees to the vertical.

The reason I mention being over at 90 degrees to the vertical is that is the strongest effect either the keel or mast weight will exert on the boat. Less than 90 degrees over, and the vector addition will reveal that the effect grows considerably less quite fast.

In our example then, we have an 8000 lb. torque or force on the mast, counteracted by a 16000 lb torque on the keel. Ergo, the keel would win, and will continue to win as long as it is greater than the mast torque. Lets say I weighed 400 lbs (jeesh, better get the diet going there bud..) and then climbed (or was lifted) to the mast top. At 40 ft. long and positioned at +90 deg. from the vertical, that would be 40 * 400 or 16,000 lbs torque, the same as the keel exerts on the boat. Now if the boat were heeled to 90 degrees, it would be in balance and would not right itself automatically. This balance between the weight at the top of the mast and the keel would also be mostly true from the perfectly vertical position all the way to the the horizontal position, as each weight moves the same amount off of vertical. However, it is true that hull shape, beam of the boat, and other secondary factors confuse the equation somewhat, making an exact number something less straightfoward. I do believe that the basic idea presented here is however, workable and true to a first order approximation.

Hope this helps..

Cheers!

Larry DeMers
s/v DeLaMer
Cape Dory 30

You need to plug your own numbers in here to see what your boat's torque is. Now this exercise is not out of a book or reference source..just from common science. So, critiques are welcome.

OK Mr Wizard Since you are hereafter known as the math guru, how would the math work for the heeling moment on a boat? Being relatively new to sailing, my mind still tells me that in Puget Sound it would be a rare wind to push my boat over enough to be dangerous, but my stomach tells me otherwise! Given the ballast of 3500 on my boat (Cape Dory Intrepid 9M (30 ft)) and the sail area, depending on which sails are flown, what are the mathmatics determining pressure on the sails and ballast vectors? I know the force of the wind on the sails increases geometrically (?) or in twice the wind there is four times the pressure per square foot of sail. Is there a good book or some place to research this. Really, I do want to know. I need to hook up with someone here to help me get over that "falling over" feeling that I'm sure you all had to get over at one time. Thanks.

John
S/V Mariah

Thanks Larry - that actually does help. I guess it's the initial tenderness in heeling easily to 10 degrees or so that makes it feel like it won't stop. Too many years in canoes and kayaks I guess. I can picture what your saying. Just the right amount of words if you ask me. Thanks again.

John
S/V Mariah



john_dupras@hotmail.com

[Chris Johnson]

I used to do it all the time, althought i only weigh 145..I've done it underway with a good 15 degree heel and it didn't seem to affect the balance that much; I think the force you exert is relatively small compared to what the sails exert. just my 2 cents



chrisj@resnet.gatech.edu

[Neil Gordon]

>>I know the force of the wind on the sails increases geometrically (?) or in twice the wind there is four times the pressure per square foot of sail.<<

True, except that as the boat heels, the effect of the wind on the sails decreases. With the boat over 90 degrees, the sails have no effect at all.

>>I need to hook up with someone here to help me get over that falling over" feeling that I'm sure you all had to get over at one time.<<

It's because you forget that the keel is down there, keeping the boat upright. Experience makes it look and feel like you are capsizing, but you're not.

If you don't have a clinometer on the boat, get one, so you know what the angle of heel is. Then, button up the cabin, close the hatch, ports, etc. ('cause in your head you know the boat's gonna fill up and sink!), and go out on a breezy day. See how far you can get the boat to heel... try to get the rail in the water. It's not easy! If you start to panic, just let go of the tiller. You'll head up and the boat will stop. After a while, you figure out that the tilt is both normal and something you control.


Regards, Neil
s/v LIQUIDITY
Cape Dory 28 #167



neil@nrgordon.com

[Bob Luby]

I'm not a naval architect, but the basic lever-arm type analysis is somewhat simplistic in the case of our boats, because as the boat heels, the shape of the hull in the water changes. This in turn changes buoyancy and thus the righting moment. This goes by the name of "form stability"

I think that the answers lie in the righting moment curves for our hulls. Have any such curves been specifically computed for our boats?

Perhaps the simple righting arm analysis is good enough for practical use, with form stability providing a safety factor?



Rluby@aol.com

[M. R. Bober]

I think you are correct. Given two vessels with the same draft, ballast, and mast height, a wider beam or hard chines would offer greater (initial) resistance to heeling (all other things being equal). I suspect with "beautiful full keel sailboats" (if you know what I mean), that the simple righting arm analysis is probably adequate. Still the case for sending the lightest crew member aloft is strong.

I thought that I had seen a posting of righting arm curves for CDs somewhere, but cannot find a valid URL.
Every besy wish.
Mitchell Bober
RESPITE CD330

[Larry DeMers]

Hi,
Well, my example first of all is totally simplisitic (as has been pointed out) -on purpose. It was meant as a tool to get the original thread writter to visualize what his boat is doing when heeling. Confusing the theory with beam, additional waterline due to heeling etc. is necessary if you are a N.A. lofting lines for your next product, but not for a new sailor trying to understand the heeling effect which terrorizes him.

Larry DeMers
s/v DeLaMer
Cape Dory 30

Larry,

Since the mast is likely to weigh a bit and its center of gravity is about 20 ft in the air in your example, we probably shouldn't forget it in any mathematical analysis. 100 lbs at 20 ft. adds 2000 ft. lbs. to the capsize side of the equation at 90 degrees.

However, as several have pointed out, the key is your comfort level. I like being on deck.

Ken

Mitch is right..it's 'righting moment' that will determine the effect on your boat. Fortunately, the math is not complicated.

Lets set up an example, and see if this makes sense.
Using my Cd30 for example. Keel weight is 4000 lb, and the center of this mass is lets say, 4 ft below the waterline. Since the waterline is the fulcrum in this equation, the weight on one side of the fulcrum must exceed the weight on the other side before the fulcrum will pivot.

So on the boats lower portion, we have a 4000 lb weight at 4 ft. from the waterline, or 16000 lb of torque when the boat is lying at +90 degrees to the vertical.

The mast is say 40 ft. tall, and I weigh 200 lbs (I wish, heh)..so my lever arm if at the top of the mast would be 40 * 200 or 8000 lbs. when the boat is heeled over at 90 degrees to the vertical.

The reason I mention being over at 90 degrees to the vertical is that is the strongest effect either the keel or mast weight will exert on the boat. Less than 90 degrees over, and the vector addition will reveal that the effect grows considerably less quite fast.

In our example then, we have an 8000 lb. torque or force on the mast, counteracted by a 16000 lb torque on the keel. Ergo, the keel would win, and will continue to win as long as it is greater than the mast torque. Lets say I weighed 400 lbs (jeesh, better get the diet going there bud..) and then climbed (or was lifted) to the mast top. At 40 ft. long and positioned at +90 deg. from the vertical, that would be 40 * 400 or 16,000 lbs torque, the same as the keel exerts on the boat. Now if the boat were heeled to 90 degrees, it would be in balance and would not right itself automatically. This balance between the weight at the top of the mast and the keel would also be mostly true from the perfectly vertical position all the way to the the horizontal position, as each weight moves the same amount off of vertical. However, it is true that hull shape, beam of the boat, and other secondary factors confuse the equation somewhat, making an exact number something less straightfoward. I do believe that the basic idea presented here is however, workable and true to a first order approximation.

Hope this helps..

Cheers!

Larry DeMers
s/v DeLaMer
Cape Dory 30

You need to plug your own numbers in here to see what your boat's torque is. Now this exercise is not out of a book or reference source..just from common science. So, critiques are welcome.

When is a boat too light or a mast too narrow to climb? Is there a rule of thumb or a ballast/climber weight ratio? What if you find out at the top of the mast that your boat is too light? A bosun's chair may not offer quick release.

demers@sgi.com