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#1
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| How do I find out the force on a mast? Hello. I need to calculate the approximate rotational force on a mast and have no idea what formula I should use. Based on a mast height and boom length giving me an amount of sail square footage, and a given amount of wind in knots, how much (in foot pounds) of force will be applied to the mast trying to 'rotate' it around. So I'm looking for some kind of formula. Any help out there? Thank you in advance. |
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#2
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| Not entirely sure what you want to calculate, but the heeling moment from the rig is in equilibrium with the righting moment from the hull, usually taken at 30 degrees heel. If it's stayed then the mast is in compression and the stays are in tension. If not stayed then its the standard beam with bending moment and shear. You might want to read "Principles of Yacht design" or something similar for a primer. If you actually want to calculate torsional forces, they are the least of your worries, if the mast doesn't buckle it will be torsionally stiff enough.
__________________ Mike Johns. Last edited by MikeJohns : 01-28-2010 at 03:56 PM. |
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#3
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| Please provide a picture explaining the setup. Also provide any info, and then elaborate your question as detailed as possible. This may seem exaggerated but in this kind of setup one different info can mean lots of work. Actually, for each connection point a formula need to be created plus some more work depending on your questions. |
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#4
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| Quote:
If the mainsail and boom are free to rotate about the axis of the mainsail luff (which most are), there will be no rotational force on the mast. It's a hinged connection that has no moments. |
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#5
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| There is going to be a negligible force from the slides or boltrope.
__________________ Gonzo |
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#6
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| Good point Gonzo. It is the relatively small rotational force at the boltrope that allows a freely rotating mast to rotate. |
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#7
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| It is too small to worry about. I don't think anyone calculates masts taking that into consideration.
__________________ Gonzo |
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#8
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| Rotation,if any, depends on the positition of the shroud attachments. If the tangs are set rearward on the mast, then rotation moment is largely opposed. Whereas, if tangs are attached to the front of the mast, then rotation force is likely. In addition, the location of the gooseneck pin with respect to the mast center line is quite influential. I suspect that the calculation of rotational moment could become very involved. Sail area, sheeting force, trim angle, even the L/D ratio of the sail, along with the hardware positioning variables. |
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#9
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| To simplify: The sail calculations only need wind force and lift force on it. At the boom base there is always a torque force down due to weight. When turning centrifugal force is added. With wind an additional torque force is added radially. These same forces are transmitted to the connection point of the mast. Then there are torque forces with or with out wind along the sail edges or stays. This is also transmitted to the mast at the connection point. Then again there are torque forces along the inner sail edge and where it's connected. Then there is the wind resistance force on the mast itself. Add all up and you got the forces acting at the base of the mast. That's about it. Note these do not include forces introduce by a boat or waves etc. |
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#10
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| OK. Though I appreciate all the things everyone has talked about when it comes to booms and pins and goosenecks, etc. this isn't what I need. Imagine that the boom was hard mounted to the mast and cannot rotate separate from the mast. Now, I want to rotate the entire mast and boom as one object. I need to calculate the amount of pressure in foot pounds that will attempt to rotate the mast. The formula would need to have a variable for the length of the boom, height of the mast and force of the wind and it would output an approximate amount of force in foot pounds that was trying to make that rotating mast... rotate. Any ideas? |
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#11
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| Quote:
On thinking about it more, the boom wouldn't exert any rotational force but the sail in it's track would. It wouldn't have leverage like a solidly mounted boom without the main sheets, that is x amount of force applied to a lever several feet long, but x amount of force applied to a lever the length of the distance from the track to the center of the mast, a lever a few inches long. But then again, the boom would resist any rotation from the sail on the track. I know it doesn't answer your question on how to find the amount of force, but is just my guess on how the force might be applied. Does it make any sense? |
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#12
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| OK, I'll try again... Take away the sheet and let the wind move the mast freely. I still need to calculate this force. |
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#13
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| I understand that what I'm asking isn't normal for a typical rig. But go with me. If the full force of the wind on the sail was attempting to rotate the mast with nothing else to help, how much rotational force is applied to the mast? |
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#14
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| Sorry, I understand what you are asking but I'm no good with numbers and formulas. |
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#15
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| what you want is simply the wind pressure on the sail... you can assume its located at the geometric center of the sail which will take into account all the variables you mention, and that will give you the arm - the distance to the mast, which which will give you your pounds per foot, or what ever related measure you need. Someone else is going to have to tell you how to find that presure per sq foot (probably a search in the forum will find it) But I wonder why you want this? Are you going to try to rotate the mast via some gears or something at the base of the mast? If you are going to use the traditional main sheet, then the rotational force is not that important, but rather the vertical righting arm and the stresses it puts on your mast (downward) and rig (pull upward)... |
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