mast load

[nemo]

Does anyone know parametric formulas for mainsail load on the mast?

[Dim]

Nemo,

SailDesign wrote at Dionisys's "About engineering forum":

"Dionysis - I just treat it as a beam with sail area centered at the mid-luff. Compression due to sail and boom is so small in comparison to bending loads that it just disappears.
Steve
PS - for pipe calcs, I use http://www.herne.com/nps.htm ".

Please, look.

Dim.

[nemo]

Thanks, but what's the wind load on the mainsail then?

[Dim]

Nemo,

In questions of sails and masts I don't understand at all. For this reason I don't accept participation in discussion of such questions. I have simply shown you a place at a forum where are discussed similar questions and people who to you can is qualified to answer. It's SailDesign and Òim B, at least.

Good luck, Dim.

[Tim B]

Um, yeah, sail forces, ask steve!

Basically, it is a function of area, wind speed and either Cd or Cl. Dependant on what orientation this thing is in. Obviously, downwind there is more drag than lift, and the reverse is true upwind. If one assumes a maximum Cl of say, 1.8, with Cd of .5 upwind, and a Cd of 1.5 with no Cl downwind, then stick it into the formula :

F=1/2*rho*V^2*area* Cl or Cd

where F=force , rho=density , V=apparent wind speed

Incidentally, I'm not taking any responsibility for the mast beaking if this is not correct. It is meant as a guide. ie. If you design for full sail at 30 kts, then sail it with full sail in 60kts, you can expect it to break.

Lars Larsson, explains this fully in his book, "principles of yacht design" It's well worth a look.

Anyway, you'd have to ask SailDesign for a full explanation, but that might help. I'd expect the answer to be big, but then, I presume you're trying to move a fair bit of boat.

Cheers,

Tim B.

[SailDesign]

Tim B says: "For sail forces ask Steve".....
Thanks......... ;-P

First of all, for practical use in calculating mast strengths, ignore the "lift" forces on the sail,and concentrate on the max force it is likely to see. In our case, it will be max righting moment. Usually, this is multiplied by a Factor of Safety (like 2) to allow for the boat having to be pushed to that angle. (Work done against rotational inertia as opposed to steady state forces)

So: Assuming a sail area of 600 sq.ft, a max RM of 30,000 ft.lbs at 45 degrees, and a luff length of 50 ft..... Boom is 10' off the waterline.
For equilibrium (i.e. to hold the boat steady at max RM, you have to apply the same moment as the Righting Moment. With the sail centred at (5 + 50/2) = 30 ft above DWL (let's assume VCG is at the WL to make life easier on ourselves), then we need a force of (30,000 / 30) = 1000 lbs on the sail. Since we have 600 sq.ft, then we need a wind pressure of (1000/600) = 1.6666667 lb/ft^2, which equates to about 20 knots of wind (pressure = 0.004 * V^2)

This is APPROXIMATE only! And this calculation is only really anywhere close to usable for a cat rig (main only). The minute you start adding headsails, spinnakers, etc. into the equation, you have to start paying for answers. ;-)

But feel free to ask - I get bored easily and need "displacement" activities.

Steve

[gonzo]

The load changes depending on the rig. It cna be a self standing mast, single or multiple spreader, fractional or masthead, etc. Also, the angle of the shrouds and stays will affect the compression.

[SailDesign]

Gonzo - you are totally and completely right.
Fortunately for me, the OP asked about "mainsail load on the mast", so I think I'm pretty well covered. Of course, what I writted earlier doesn't actually address the issue there, either (sorry...)
There are, as gonzo pointed out, so many variables (including sail cut, track/slot/battcar, etc) that there is no one formula that will cover it. My guesstimate above will allow you to calculate the required strength for a free-standing cat rig, however.
At last count, I have a collection of about 15 spreadsheets for rig calculations. ;-)
Steve

[gonzo]

I agree that they are very difficult calculations. Other variables include hull deformation and wether the rig is pretensioned or not. I think I'll take rocket science as a relaxing hobby

[Dim]

Gonzo,

You are the Best! It's very good answer.

Dim.

[grob]

just treat it as a beam

This one has been puzzling me, I tried to reverse engineer the free standing rig on a Windrider 16ft trimaran http://www.windride.com/wr16.shtml using simple beam theory and it comes out way too weak.

The mast is an aluminium tube which has the following dimensions OD=2,5" and thickness =.15". So the area moment of inertia =320000N/mm^2 and Distance of fiber from neutral axis is 30mm

If the beam is 3.7m the weight is 113kg and the max capacity is 227kg then the righting moment is RM=3.7*2270+1.85*1130=10512Nm or 10512000Nmm

Using simple beam theory stress = My/I

the beam stress is 10512000*30/320000=985.5 N/mm^2

the Yeild strength of Al is 250N/mm^2

This would indicate that the mast has a safety factor of 0.25 and so is 8 times too weak based on a safety factor of 2 using simple beam theory.

The boat has been in production since about 1995 and I have found no accounts of mast faliure.

Can anyone explain what is happening here, my guess is that the main sheets take some of the load. Does anyone know how to size a freestanding dinghy mast?

[SailDesign]

Grob - I think you are grossly over-guesstimating the righting moment here.
The max RM you can achieve on a tri is when the leeward ama is just submerged. On some pictures I found of the WindRider 16:
http://www.multihullsnorthwest.com/w...r_2145_002.jpg
http://www.multihullsnorthwest.com/w...r_2145_007.jpg

It looks as though the leeward ama submerges pretty quickly, and has maybe 250 lbs (1,112N) of buoyancy, since the main hull is still well immersed. This corresponds to a hull 12' x 1.25' x 9", with a block coefficient of 0.35, which looks about right from the pics.
Since the total beam is 3.7m, and the amas are about 9" in width,then the usable 1/2-beam would be 3.7/2-(0.114) = 1.736m
So RMmax would be 1,112N * 1.736m = 1930 Nm, or 1,930,000 Nmm.
The Section Modulus of the mast section is 320,000/30 = 10,667 mm^3, so the stress is 1,930,000 / 10,667 = 180 N/mm^2, which, while it is nowhere near being a safety factor of 2, is quite workable for a guess.
Steve

testing to see if I have the correct info entered

[brian eiland]

Isn't the 'parametric', or 'distributed' loading of the mainsail on the mast one of the factors that affects the mast staying in column, and thus is a real concern when figuring the strength of the mast to resist bending (getting out of column) in order to be able to accept even higher compression loads??

Or put another way, isn't the distributed loading of the traditional mainsail helpful in allowing the mast to stand up to the loading??

If this is the case, then shouldn't the distributed loading by the mainsail be included as a factor in determining the proper mast size rather than totally subrogated to 'the max force it is likely to see as a result of the righting moment'??

SailDesign had written,
"First of all, for practical use in calculating mast strengths, ignore the "lift" forces on the sail,and concentrate on the max force it is likely to see. In our case, it will be max righting moment."

I've been concerned with the lack of 'distributed loading' on my mast-aft design as I do not have a sail attached to the mast extrusion, but rather 'point loading' by the rigging strands that are transmitting my sail forces to the vessel...

....and that brings up another question of how exactly do the sail forces (the forward driving portions in particular) get transmitted to the vessel...primarily thru the mast or what piece(s) of rigging? and what share each assumes?

SailDesign had written:
At last count, I have a collection of about 15 spreadsheets for rig
calculations.

Brian ask:
Steve, would you mind sharing a few of these with me. Then I may have a few more questions of you. I am looking for a preferred format, or possibly a modified format I might use in a full structural anaysis of my rig that is to begin very soon.