Windsurfing Math

[Windnc]

A 200 pound sailor is ripping along at 25 knots when he suddenly hits a rough piece of chop. He loses control, is unable to unhook in time and is suddenly yanked forward off the board by the sail slamming down. Then almost instantaneoulsy he is violently yanked back again still hooked into the harness lines.

As he lays there entangled in the board and rig, trying to unhook and figure out if his body is still in once piece; he wonders how much Ibuprofen he will need tonight. He also wonders just how much force his being slammed put on the boom and his harness lines, let alone his back? He has a vague memory of F=MA from a long ago physics class but can't quite work it out.

How do you calculate the force on the harness lines and boom?

[isobars]

Without instrumentation, you can't. Getting catapulted (lifted off your feet) can involve anywhere from 1.0000001 g (plenty of time to pray, swear, snap a GoPro shot, unhook or let go, pull down even harder on the boom to go for distance. etc.) to many gs (who am I and why am I wet?) on that boom arm.
And that's just the vertical gs. There's also the horizontal component, which may run from 0 to multiple gs, whether we crash or not. There are just too many variables and unknowns to calculate much of anything in a crash.

At several gs, we're either going to get violently slammed or get the boom ripped out of our hands (think doing pullups with multiple friends hanging from your shoulders). Thus even without instrumentation we can guess that boom load is not likely to exceed something like 3-4 gs, if that much ... probably more like 2 in "normal" very aggressive sailing. So your 200# example might put a few hundred pounds of force on the boom arm in extreme conditions.

I've destroyed new booms by stepping on one boom arm and pulling up with one hand; that surely isn't hundreds of pounds ... and that boom disappeared from the market overnight. Some modern booms I can barely flex with both hands while standing on the other boom arm; that's hundreds of pounds with no significant effect. Those booms are still around.

I'm sure most major brands have tested their finished products' breaking strength on machines, but that's not as useful to the end user as weight, shape, grip diameter, head and tail hardware, price, etc. Some people value a great warranty, but unless it includes rescues west of the Golden Gate Bridge it's not of much use when it REALLY counts.

Mike \m/

[DanWeiss]

The force on the boom is measured at the boom head. clew and harness lines. I assume you ask about the force imparted by the lines during the yank as that might be the easiest to calculate using simple math and concept. No engineering here.

The force imparted to the boom by the harness lines is never greater than the total amount felt in the harness hook. Each harness line strap takes very roughly 1/2 the total force felt by the hook.

The magnitude of the force builds to a level necessary to rip the sailor from the straps then greatly reduces when the sailor releases from the straps and is launched forward. That force would steadily decrease until the mast hits either the board or the water and ceases its arc, thus no longer imparting force to the harness line straps through the lines but through compression of the boom head toward the clew. The harness line force may again spike depending on the continued trajectory of the sailor's body as it slams against the lower mast or is damped by impacting the board's nose.

I'm uncertain how to calculate the forces imparted to the clew and front end by converting the rotational forces into linear when the rigs suddenly stops. These forces must contribute with the harness line strap forces to the overall compression forces felt by the boom. We all know the effects of this: broken board, broken mast, torn harness or rarely, broken ribs. Just remember that different forces with different amounts and directions are nothing more exotic than different vectors. You do not simply add the vector numbers to get a huge sum. I recall there is long division involved.

Rough numbers (wild guess)

The force necessary to accelerate 200 lb. is anything over 200 lb. In this case, it's 200 lb. plus the resistance force of the foot straps (friction plus rotational resistance as the straps open to the moment of release). The sailor then accelerates to rig impact then smashes, more or less into something hard. All that force is split between the lower back/hips and hands, albeit momentarily. The rest is black art to me.

[bred2shred]

I agree it would be pretty difficult to calcuate the amount of force in a windsurfing crash. I think you'd need to use instrumentation to get accurate numbers. Unlike a situation such a car slamming into a brick wall (essentially a non-deformable object of infinite mass) or another car (object of known mass & velocity) the windsurfer is slamming into water. You hit the water and it displaces. The way in which your body hits the water directly effects how the water is displaced and therefore direcly effects the force on your body. This is totally apparant when you consider that under some catapult situations, you walk (swim) away with zero issues. In other cases, your body is violently torqued and your head is left ringing.

I recall one particularly violent catapult that I experienced several years ago. I'm about 200lbs I and was bump & jump sailing bayside on a fully wound 4.4 day. I was lit up on a broad reach back to the beach when I lost it and got slammed very hard. When I came up, my Reactor bar had been bent up at about a 45 degree angle. I don't know how much force it takes to do that, but I know it was a lot. Amazingly, there was no other equipment damage.

sm

[jsampiero]

[rigitrite]

"Without instrumentation, you can't. "

Completely untrue, we did similar problems in engineering dynamics many times.

However, we would need a couple of extra pieces of information:
1. Windspeed at the time of catapult
2. Air density
3. harness line length
4. Height of the sailor to approximate center of gravity and probable boom height.
5. Sail size and sail style (RAF, race sail, etc...)

That information, plus a couple of common sense assumptions (planing at max. board speed at time of wipeout) and it's a series of four or five straight forward (but not necessarily easy) engineering calculations to arrive at what you want.
F=ma is correct, but calculating "a" is the trick, it would go something like this:
1. calculate force on the sail and cg of the sail.
2. calculate angular accelleration of sail pivoting on mast base
3. calculate additional acceleration imparted by gravity on fixed pivot (not the same as freefall)
4. perform same calculations for the sailor on second set of pivots around boom via harness lines.
5. Calculate reduction to impact based on viscous properties of water
@70°F. This is proportional to force of impact, ie. if you hit water at 200 mph, it's equivalent to hitting any other solid.
A good guide would be a set of engineering calculations for a Trebucet (type of catapult), since it's the same principle. I'm sure such a thing is easy to find on the interwebz.

[coboardhead]

Remember that the actual stresses that are exerted on the body are also a function of elastic (bending) and frictional responses of the whole system. A fat (more elastic sailer) of similar weight might deform more as the load is applied and, therefore, experience a reduced magnitude of maximum stress that a leaner sailer might experience. Think crumple zones in automobile design.

I'm with Iso...this needs instrumentation to accurately evaluate the stresses. While, a relatively simple, calculation of F=ma may provide the total force available, how the reaction to that force...your body, harness line elasticity, boom deflection, mast deflection... is calculated would be subject to an almost infinite set of variables. Those are the actual stresses exerted on the harness line.

[dllee]

Not so bad, happenned to all of us multiple times if we sail enough.
Try clearing a triple, landing on the UP side, come instantly to a dead stop with NO rollover or slide. That's 45mph to ZERO in 1 second.

[isobars]

[pointster]

Suppose you weigh 160 pounds, and you are going 30 mph when you get catapulted, which takes one second. Using F=MA, where you mass in slugs is about 5, and your acceleration is 44 ft/sec squared, then the total force is 220 pounds, applied at the harness hook. This is horizontal component in direction of motion, but I think it is probably the upper bound.