Downward pressure

novelman17128 · Jan 20, 2007

Mast pressure

Each of the stays exert a pressure, both downward and outward. That's what keeps the mast up. The weight of the mast also exerts a downward pressure. THE DOWNWARD PRESSURE IS MUCH GREATER THAN THE OUTWARD/SIDEWARD/FORWARD/REARWARD and can be calculated. Almost all sailboats, both large and small have a solid wooden or metal post just below where the mast attaches to the deck, on deck mounted masts and on the floor/keel on other boats. There is also downward and outward pressure exerted when the mast is under pressure from the sails (when you are sailing.) All of these pressures are calculated when the boat is designed and built, so don't make any major modifications on your mast/sails or other things that can add pressure to the mast plate, either on the deck or the keel without checking with the designer for safety. You can get all of the above info from the designer/manufacturer of your boat or roughly calculate it using gages that give you the tension on your stays. If you are concerned about the mast possibly coming down just understand that under virtually all conditions you might sail in, the mast will be secure as long as you keep all of the stays adequately tightened, locked and taped (to prevent them from "unscrewing.")Novelman

2 Hulls Dave · Jan 20, 2007

You bet!

"When a mast is rigged and properly tuned, does the rigging exert a lot of downward pressure?"Yes, there's a hell of a lot of compressive mast loads in a wind loaded rig - even if it's improperly tuned. The mast is trying to drive itself downward right through whatever it's stepped on. These compressive loads are "equal" to the tension loads in the standing rigging. This is a statics problem in freshman engineering. All the forces summed up (vectorly) have to equal zero since the "structure" is static (not accelerating).However, an unstayed rig is an entirely different animal - theoretically no compressive loads (other than its own weight and whatever running rigging adds).DaveS/V Pas de DeuxCatana 471-44

Ross · Jan 20, 2007

Manny, The shrouds and stays pull up

on the boat and down on the mast. The mast is a long lever, the mast step is the fulcrum. Imagine that the distance from the base of the mast to any one chain plate is the other portion of the lever. Now the mast step serves as a hinge in the lever. Not unlike a wrecking bar for pulling nails but hinged where it meets the surface. The shroud now serves to connect the ends of the lever together. The mast and the deck carry the compression load component and the shroud carries the tension.. If your mast is 40 feet tall and the distance from the mast base to the chain plate is 4 feet, then you have a 10:1 lever. One pound of force at right angles to the top of the mast and in line with the chainplate will be felt as 10 pounds of pull on the shroud. Since the assembly doesn't fly off into space there must be 10 pounds of force down at the base of the mast.Therefore as you assemble the shrouds and stays you seek to balance the loads port and starboard, fore and aft. Because the angles are large for the forestay and back stay the tension on the stay will be more effective in keeping the mast vertical. (the lever ratio again)

David11054 · Jan 21, 2007

Around 2 tons on our '77 h27

Because the shrouds form small angles with the mast when they attach to the hull, you can just sum the tension on each shroud, and then add the weight of mast, boom, mainsail and any mast-mounted equipment to find the load on the mast step in port.It's about 4,000 pounds, or 2 tons on our '77 h27.There is no idea of minimizing this load. The stays should be tensioned so that there is never any slack on the leeward stays, so that the forestay holds the jib luff reasonably straight, and (on some rigs) the mast has the proper bend. Don't worry about the load (which goes up substantially on larger rigs), the hull was designed to carry it.DavidLady Lillie

NewportVic · Jan 21, 2007

couple of years ago a Tartan 41 had to pull out

of the race to mexico when it developed a leak at the base of the mast where it sits on the hull. i think the guys had tightened it up a hair too much. Required pull out to repair. as I heard. So there are serious problems in over tuning.Vic "Seven"

Ross · Jan 21, 2007

If you allow that the mast is 5 inches wide and 8

inches long then it has a foot print of about 40 square inches. That translates to about 100 pounds per square inch. I absolutely don't believe that the hull of a 41 foot boat will fail under that load. There was undoubtedly another factor.

Robert Gainer · Jan 21, 2007

18,000 not 4,000

Ross,I have not been following this thread but if you are talking about a Tartan 41 it might have 18,000 pounds of compression in the mast just from dynamic loads not including preload from improper tuning of the rig. That is much greater then the 4,000 pounds you are discussing. And I think the mast is 11” X 7” with a .186 wall thickness and 7/16” upper shrouds and 3/8” lowers which are in line with that sort of strain in the rig. The compression in the mast can’t exceed the tension in the wires and 3/8” stainless is good for about 17,000 pounds I think. I don’t have any reference books handy but that number sticks in my mind.All the best,Robert Gainer

Ross · Jan 21, 2007

Robert, you are probably correct! I would hope

that the builder had done he numbers right and added a large safety factor. Sinking you boat with your own mast is embarassing at best and deadly at worst.

Moody Buccaneer · Jan 21, 2007

Right you are, Robert

3/8" wire is 17,500# in 302 Alloy, or 14,800# in 316 Alloy.The tension in the shrouds is governed by the righting moment and the width of the shroud base.If the boat has 25,000 pounds of righting moment and the shroud base is 10 feet (5 from centre) the tension in the shroud system is 5,000 pounds. This 5,000 pounds of load is shared by all the shrouds on the boat.The wire should remain elastic up to 55-60% of rated load, so it is very hard to damage the rigging with over tension. Distort the boat perhaps, but not damage the rigging or mast. It is not uncommon to hear of boats that have doors that stick when the rig is loaded, the boat is flexing. It so happens that 1x19 rigging wire stretches about .01% for each 1% of rated strength. At 55% of rated strength, the wire will stretch about .55% of it's length. A 55 foot shroud will stretch about 3-5/8", greater than the adjustment range of most turnbuckles.If the static load + the sailing load is greater than 55-60% of the wire's rated strength, either the rig is too tight (static tune) or the wire is too small.In any case, a good design has a safety factor of 2-3 built in. There is no way that anything but a gross tuning error should cause a problem with a mast step or compression post. Rot under mast steps can cause problems and damage boats but that is a maintenance problem not a design problem. Well designed boats just don't have failures from mast compression loads.

Fred · Jan 21, 2007

Don't forget shock load.

When the boat comes off a wave and slams the bow down hard, the mast whips forward and aft, or side to side or all of the above, while the sails go from slack to full as the boat rolls. This can cause a momentary load that exceeds "normal" forces of heel and righting moment.Add an accidental jibe to the above, and most production boats are still strong enough to take it, repeatedly. You can call it engineering or you can call it magic. As far as I'm concerned it's both.

Moody Buccaneer · Jan 21, 2007

Righting moment

The righting moment can be measured, or it can come from calculations.It is only important if you are designing a rig, racing, or a techno-file.In the range of heel that moat boats sail well (0-25 deg) the RM curve is linear. If you figure out how much weight it takes to heel the boat 2 or 3 degrees you can plot the line.If you want to measure your boat, you can do it by yourself but a couple of friends helps. Take a bucket and half fill it with water. Make a plumb bob and suspend it from the cabin roof with the weight in the water bucket (to keep the weight from swinging wildly). Tape a ruler to the top of the bucket (athwartship) as a scale, the string is your pointer. Measure from the top of the string to the ruler and make a note of the distance. Have one person sit in the middle of the boat where they can read the ruler without moving. get everyone off the boat and let the boat settle down. The recorder centres the string and notes or marks a "0" point. Have a friend stand on the rail and let the boat settle down again, make another mark on the ruler. Keep adding friends until the boat is heeled enough so the weight is up against the side of the bucket or you run out of friends.

Now to make enemies. As each person steps of the boat, have them stand on a scale and record their weight (if you ask their weight, they will lie. If you ask to weigh them before the test, they might not want to play).Measure the distance from the centre of the boat to the rail. Multiply that distance by the weight and you know how many foot/pounds of force each of your data points represents. Some simple math will tell you how many degrees you heeled the boat from the amount the string moved on the ruler.Plot these points on a piece of graph paper and you have the RM curve for your boat.If you want to do it without making enemies, you could use water jugs and weigh them, but 5-6 guys standing on the rail then bitching about their weight is *much* more fun.

Say you had 5 guys at 200# each standing on a rail that is 5 feet off centre (half the beam). That would equal 5,000 foot pounds. If the boat heeled 5 degrees, the RM at 10 degrees would be 10,000 foot/pounds, 20 deg = 20,000 foot pounds etc.What use is this? None, unless you are changing the rigging and need to know what number to use to select wire sizes.BTW, compression load is (RM30 x 1.5 x 1.85) / 1/2 beamFor a 12 foot beam and 30,000 ft/lb that would be about 14,000 pounds compression.It is also of some use in tuning at the dock. If you know RM20 (20deg) and measure out to the upper shroud you can figure what tension the wire will be at 20deg heel. If you have 20,000 ft/lb RM20 and the upper is 5 feet off centre you get a total of 4,000 pounds tension in *all* the shrouds. Single spreader rigs have about 45% of the load in the upper. The upper will have about 1800 pounds of load at 20deg. If static tune is set to 1800 pounds, the leeward upper should be just slack when the boat heels 20 degrees under sail. The shroud in this case could be either 9/32" 302 alloy or 5/16" 316 alloy to have static tune between 15% and 20% of rated strength.Brion Toss' book, Rigger's Apprentice is a good read if you are into this stuff.Randy

Robert Gainer · Jan 21, 2007

That was a great post, very informative

Randy,That was a great post, very informative and I especially enjoyed your comments on wire.You say “In the range of heel that moat boats sail well (0-25 deg) the RM curve is linear.” and I would quibble just a little bit with that and say that with modern boats it becomes non-linear much sooner then with the older boats. Boats today have a much harder bilge then the older ones when that rule of thumb started so form stability has more effect then before. You also say “compression load is (RM30 x 1.5 x 1.85) / 1/2 beam” and that’s very true. I was just wondering why you stated it that way. I usually see it expressed as (RM30 X 2.78) /1/2 Beam. Aside from the small difference in the factor 2.78 and your factor of 2.775 (that being 1.5 X 1.85) they are the same just with two factors simplified into one rounded. When you are discussing this with someone do you get into the derivation of the equation and that’s why you keep the two factors? I tie my presentations in with Skene’s Elements of Yacht Design and he uses the simplified version so that’s what I describe. I also make it a point to say that this rule of thumb is for older boats and needs to taken with a grain of salt with today’s boats. Because I use Skene’s as a source I also need to point out that his chart of righting moment at 30 degrees is for older CCA style boats and doesn’t apply to today’s boats at all. Of course if you want the maximum compression the inclining experiment is awkward at 30 degrees and having the lines to calculate GM is the best way. And at that point you might just resolve all the forces in the rig and calculate an accurate number for compression.When something like this discussion starts I usually say something about the angle of the shrouds and how that affects the compression. The rough rule of thumb doesn’t lend itself to extremes of beam so a word about shroud angle seems appropriate. I usually say something like if you had the shrouds at 90 degrees or parallel to the water there is no compression in the mast at all and as the angle decreases the compression goes up and approaches infinity as the angle approaches zero. Something like that gives you the opportunity to segue into resolving all the forces in the rig on the drawing board and knowing how to resolve a force is useful for so many things besides designing the rig.I also enjoyed Brion Toss's book, Rigger's Apprentice but would add The Nature of Boats by Dave Gerr. A great book that covers so many of the reasons why a boat looks like she does.All the best,Robert Gainer

NewportVic · Jan 22, 2007

re the Tartan ... their biggest problem was to get

into a port that was deep enough to haul them out fast ... on the SW FL gulf side. Heavy wind, big seas ... Race obsession. one of my racing buddies was on it geared up to party in Mexico and didn't make it. Was upset.Vic "Seven"

Moody Buccaneer · Jan 22, 2007

Thanks Robert

The formula should be: RMmax / 1/2 shroud base for the shrouds.I certainly hope no one takes this to the bank and changes the rig on their boat without a professional evaluation! (Don't try this at home, kids)

RM30 x 1.5 is an approximation of RMmax. The 1.85 is a fudge factor for fore and aft stays to handle pitching and sail loads. If you have the actual value for RMmax, I would use RMmax x 1.85 / 1/2 shroud base to size the wire for your rig. The 1/2 beam assumes the chainplates are at maximum beam as was the case for many years.You are quite correct that modern hulls may not follow the chart in Skene's very well. Skene's suggests a range of just over 10,000 lb/ft to just under 20,000 lb/ft for a 25 ft waterline. My tubby C30 (D/L 300) has a RM25 of just over 15,000 and assuming a linear slope, it should have RM30 just over 18,000.Max righting arm is about 1.8 between 55 and 60 deg for a RMmax of 21,600.RM30 x 1.5 = 27,000, so a rig built to the best guesstimates will be on the heavy side.If I didn't have the stability curve and used Skene's middle figure of 15,000 as RM30, RM30 x 1.5 = 22,600. That is darn close to the real number.If I had calculated RM2 correctly at 1400, assuming a linear slope, I would have estimated RM30 to be 21,000, RMmax as 31,500 and I'd be concerned that the rig was woefully undersized. The fact that my calculations recommended wire that was bigger than any other 30 footer on the planet *should* be a tip off that I need to have someone check my numbers. *yksIf the boat had a D/L of 80, I would not be so quick to use the old rules of thumb. I'd do the math at both extremes from Skene's, to see if a bad guess would put me too close to the limits. If so, I'd want to see a stability curve or do some real in the water measurements to see which end of the range the boat tends to. The thought of having a rig come down because it was under spec'd haunts me."Nothing too strong ever broke" is the safe approach to building a rig that might not get the best of care and not always be tuned properly.

Downward pressure

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2 Hulls Dave

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Fred

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