Assemetric Bow Pole Calculation

[Paul B]

Assymetric Bow Pole Calculation

I have heard a couple of different opinions about how to calculate the load on an unsupported bow pole for an Assymetric spinnaker.

I would like to hear how others calculate this, using SA, wind speed, and unsupported tube length.

Assume carbon construction.

 

[PI Design]

Well, I've never done this calc but if I were to, I would treat it like this:

Estimate the force coefficient for the kite - values should be available from references such as Marchaj/Bethwaite, as a first guess I would imagine circa Cf = 2.5. Total force on sail is:

0.5*air density*sail area*max apparent wind speed^2*Cf

Although the kite is contsrained at the three corners, I would assume that 50% of the force is taken through the tack attachment (end of the pole). Although some of the force would actually act axially putting the pole in tension rather than pure bending, I would assume that the force acts normal to the pole (which is round) so all the force contributes to bending.

e.g. kite = 20sqm, max apparent wind = 20m/s, Cf = 2.5, density = 1.2kg/m^3

Force = 0.5*1.2*20^2*20*2.5 *0.5
= 6000N

Then calculate as a cantilever beam with end point load:

bending moment = force * unsupported length

stress = bending moment/section modulus

Compare stress with yield stress of the pole material, allowing for any safety factor deemed prudent (which will vary between an ocean going cruiser and a racing dinghy - but bare in mind that my assumption that he force is perpendicular to the pole is already pessimistic).

It may also be necessary to check pitch-poling loads.

 

[yades]

further check on results.....

Well, I've never done this calc but if I were to, I would treat it like this:

Estimate the force coefficient for the kite - values should be available from references such as Marchaj/Bethwaite, as a first guess I would imagine circa Cf = 2.5. Total force on sail is:

0.5*air density*sail area*max apparent wind speed^2*Cf

Although the kite is contsrained at the three corners, I would assume that 50% of the force is taken through the tack attachment (end of the pole). Although some of the force would actually act axially putting the pole in tension rather than pure bending, I would assume that the force acts normal to the pole (which is round) so all the force contributes to bending.

e.g. kite = 20sqm, max apparent wind = 20m/s, Cf = 2.5, density = 1.2kg/m^3

Force = 0.5*1.2*20^2*20*2.5 *0.5
= 6000N

Then calculate as a cantilever beam with end point load:

bending moment = force * unsupported length

stress = bending moment/section modulus

Compare stress with yield stress of the pole material, allowing for any safety factor deemed prudent (which will vary between an ocean going cruiser and a racing dinghy - but bare in mind that my assumption that he force is perpendicular to the pole is already pessimistic).

It may also be necessary to check pitch-poling loads.

Smart assumption...furthermore.... considering the force acting normally to the pole at the fore end and supposing the construction material being of carbon fiber, I would also deem rather important, besides the calc. for stress under tension forces on the .."outer" side of the gen pole, a further check for compression stress on the ..."inner" side (allegedly the top side) of the pole , which could be assumed being rather weak when working under compression

yades

 

[Steve B]

The tension in the luff of the sail is not just bending the pole, it may also
be putting it into compression.
Apart from calculating the bending stress you have to work out the axial compressive stress, combine the two and then ensure the factored yield stress is not exceeded. You also have to check the compression will not buckle the spar. However I imagine that pitch-poling is far more likely to cause a buckling failure.
Regards
Steve B

 

[raw]

meh?


How about take the load required to simply lift the forefoot of the boat clear of the water, apply a suitable factor of safety to the bending moment and calculate using normal engineering methods.

After that, take pick up the tube catalogue, and select the next size up to get that all important, dont mess withtha front of my boat look needed for start lines. Done.


Practical engineering 101........

 

[mark_m]

How about take the load required to simply lift the forefoot of the boat clear of the water, apply a suitable factor of safety to the bending moment and calculate using normal engineering methods.

Because it can be much more than that. The top of the asymmetric is pulling down on the mast and the main sail force is producing a forwards pitching moment, so potentially the forestay upward force could be more than the total weight of the boat.

I like your simple approach though.

How about finding the breaking strength of the rope attaching the spinnaker to the pole? The load cannot be greater than that, unless the rope is significantly stronger than its rated strength.

Assuming you're using 4mm, I would assume the breaking strength is about 1000kg. This sits quite well with PI Design's load calculation of 6000N. I would definately want the pole to be stronger than the rope. If you're using larger DIA rope for reasons other than strength, try looking at the strength of the fixings or blocks you're using to give other clues about the load's upper limit.

Also bear in mind that the pole is probably not perfectly built into the boat (unless its a fixed pole), so the support will be transferred to it at points, rather than distributed. This could cause local buckling unless there is sufficient reinforcement.

 

[Paul B]

Some good ideas so far. Let’s look at how the calculations square with the hands-on approaches.

If we take the example from PI and apply it to a well-known entity (Melges 24) we can calculate the following:

0.5*1.2*62*20^2*2.5*0.5 = 18,600 newtons, or 4,180 pounds force load on the tack line.

The block the Melges 24 uses for the Spinnaker Tack is a Harken Carbo 40, with a Safe Working Load of 485 pounds and a breaking strength of 1620. Considering the tack block is changing the sheet direction roughly 90 degrees that increases the block’s required load rating by 1.41, so the block should break in the neighborhood of 1150 pounds on the tack line. The block should actually not see more than 485/1.41, or 344 pounds force to remain in the SWL.

OK, so let’s re-consider factors. I know the boats will never see 20 m/s apparent wind speed. Maybe 8 m/s is a better factor. Any more than this and it is likely the boat is going to round up with the big ASO kite up.

0.5*1.2*62*8^2*2.5*0.5 = 2976 newtons, or 668 pounds force load on the tack line.

Multiply that by 1.41 and we’re looking at 942 pounds for the block. We’re under the Breaking strength of the Carbo 40 now, but still well beyond the SWL.

Next, let’s look at the assumption of ½ load on the tack. Le’ts assume 33% each to the tack, head, and clew.

0.5*1.2*62*8^2*2.5*0.33 = 1964 newtons, or 441 pounds force at the tack line. Hey, we’re under the SWL.

However, if we’re changing the direction of the sheet more than 60 degrees we could still have a problem. At our assumed 90 degrees the load on the tack block is 441 x 1.41 = 622 pounds. Beyond the SWL still.

If we changed the SWL to a safety factor of 2:1 instead of 3.3:1 we could use 810 as the SWL and now we’re in the ballpark.


Other considerations:

Now if we look at the ASO as a big genoa (it isn’t, but let’s say it is for the argument) then the Melges kite at 62 sq m in 15 knots of wind would have a sheet load of 650 pounds. Maybe we should be assuming the sheet is carrying more load than the tack and head, not equal?

I agree with Steve B, the pole is definitely in compression as well as in bending. If you let the “pole out” line off while the kite is loaded it will retract with a lot of force.

On most poles the tack line is running above the pole, so it is acting somewhat as a bowstring as well.


So far there is some good thinking, but we seem to be making a lot of assumptions. I think there are a lot of the boats out there that have had their poles sorted by "practical engineering". I’m offering a local spar supplier the use of my boat if they have access to a portable load cell, to get some actual loads to compare the theoretical calculations.


Finally, does anyone have a good number for yield stress MPa for a generic carbon/epoxy tube?

 

[water addict]

You gotta temper the sail load with righting moment. Going strictly by sail force may be off, as the full sail force must be reacted against something (ie. Newton).

Look in Principles of Yacht Design and adapt that approach assuming the sail will heel the boat. I would then add the additional load case assuming following winds with the reacting force of hull drag. As a practical engineering approach to this you could also size the backstay first and figure it's breaking strength. Then design the pole, spin tack to be somewhat weaker so that if you are in overload conditions, the tack fitting or pole breaks before the backstay does, as it would be better than the rig falling down.

 

[PI Design]

I agree with Steve B, the pole is definitely in compression as well as in bending. If you let the “pole out” line off while the kite is loaded it will retract with a lot of force.

Finally, does anyone have a good number for yield stress MPa for a generic carbon/epoxy tube?

You're right. I was having a brain dump when I said tension.

As CF is brittle, it does not yield. UTS is typically around 2500-3000MPa I think.

 

[Paul B]

You're right. I was having a brain dump when I said tension.

As CF is brittle, it does not yield. UTS is typically around 2500-3000MPa I think.

Thanks. I believed CF had no "yield stress". You mentioned it in your earlier post, so I thought maybe there was something I wasn't understanding.

I suppose in a deep running condition with the CE of the kite forward of the pole (probably a bit by the lee) there might be some tension on the pole, but in normal circumstances you are right, we're looking at the pole in bending and compression.

If I asked how to calculate the construction of a mast there would be 20 replies giving basically the same info, the basic calc once you have the mast height, number of spreaders, chainplate width, righting moment, etc. Ditto a rudder shaft once you have area and speed. Maybe the whole "prod" thing is too young yet for all the forces to be fully understood with a universally agreed method for calculation.

I know a rather large race boat around here broke their bow pole on about their second outing. I've sailed on a couple of boats with poles that bend to a frightening degree when tight reaching in chop.

I'm sure the big design offices have their own proprietary calcs.