Listen up, you pointy-headed nerds.

http://www.theatlantic.com/technology/archive/2011/12/the-physics-of-great-white-sharks-leaping-out-of-the-water-to-catch-seals/249799/

   Don't forget to factor in the absorption of the impact due to the  flexation of the hull.     Better contact  Dr. Sheldon Cooper for his hypothesis..

Thanks, Trianlinguini!

Here is the published paper mentioned in that Atlantic article.

http://www.tandfonline.com/doi/full/10.1080/17451000.2011.614255

It looks like 6 or 7 m/s is a better velocity estimate.

Based on your parameters, its 8.8 m/ . Prove me otherwise.

Based on your parameters, its 8.8 m/ . Prove me otherwise.

I was talking about the shark's velocity... based on the literature

My brain is far to valuable to waste on math, spelling, or punctuation. For I can recall almost any Simpson's scene I wish on YouTube at will, a far more important skill.

I hate these theoretical math questions as they seldom relate to the real world. For example, we all know that there are swells in the ocean. It seems to me that it would make a difference, that at the time of impact, the felt energy would be different if you were rising or dropping with a swell. At least it might make a difference as to how airborne the person on the kayak becomes. 

sometimes they approach nice and slow though

LOL  I hereby withdraw my variable 

Funny thing is i raised the surface tension question in our conversations prior to this post going public.  Also the mythbusters angle.

Both appear to have been summarily rejected.

We also shortly discussed the flex of the hull but I believe that made it into the first post.

The whole reason this idea came about is that Fisherman X provided that bit of research about "Polaris" attacks, and the idea is that a larger shark would come up for the exploratory bite instead of just headbutting the target. In that post, it also discussed that Carcharodon carcharias do not become mammalivorous until approximately 3 metres in length (10ft) but may give up on the Polaris attacks after 4 metres in length (13 feet) when they start using more teeth and less headbutt.

So with that being factored in, the expectation of this being a mammalivorous Carcharodon carcharias attack would posit that the fish would be of the following:
1. 3-4 meters in length
2. 544kg-680kg (1200-1500lbs)
3. Top speed of approximately 40km/h (25mph)

While the kayak is:
1. Hobie outback - 12'5"- ~100lbs with gear
2. Tasty Morsel - ~200lbs with gear

Questions are:
1. How much force would be required to pitchpole such kayak with expectation of a mid-ship hit?
2. Would the mammalivorous (BAN) fish have enough power to make Q1 happen?
3. Would any other non-mammalivorous species have enough power to do so?  We would expect a large whale would be able to do it but that would have been a little more obvious I think to our Tasty Morsel.

Questions are:
1. How much force would be required to pitchpole such kayak with expectation of a mid-ship hit?
2. Would the mammalivorous (BAN) fish have enough power to make Q1 happen?
3. Would any other non-mammalivorous species have enough power to do so?  We would expect a large whale would be able to do it but that would have been a little more obvious I think to our Tasty Morsel.

I think a hit on the posterior one fourth of the boat is needed to produce a pitchpole.  I hate to think what would happen with a mid-ship hit where there is less vertical structural integreity. 
In that instance you might need some steel undies fo sho.

Questions are:
1. How much force would be required to pitchpole such kayak with expectation of a mid-ship hit?
2. Would the mammalivorous (BAN) fish have enough power to make Q1 happen?
3. Would any other non-mammalivorous species have enough power to do so?  We would expect a large whale would be able to do it but that would have been a little more obvious I think to our Tasty Morsel.

I think a hit on the posterior one fourth of the boat is needed to produce a pitchpole.  I hate to think what would happen with a mid-ship hit where there is less vertical structural integreity. 
In that instance you might need some steel undies fo sho.

I'd agree that the rear quarter would be the most likely spot for a pitchpole.  but it would also be the most narrow area, so the hit would have to be almost perfect for that to happen and not be a glancing blow.  I'd say right behind the seat and the front of the tankwell area would be the most likely attack spot for this to happen and for your particular circumstance.  That area is also in the prop wash for the mirage drive, so it would make some sense as well.  Mr. Dr. Marx is the primary proponent of the mid-ship, but I think the amount of force would not be lessened significantly. 

Feel like submitting something to mythbusters?  I do recall them having a giant mechanical shark available

I’m not a math or physics dude but there are a lot of variables in the pitchpole problem, not only where the hit occurs but also the weight distribution in the kayak.  In addition to the angle of attack there is the momentum and continued propulsion of the shark and the flexibility of the kayak.  The pivot point (bow) of the kayak isn’t fixed so that would have to be accounted for and as mentioned above there might be forces or surface tension.  I imagine there are other variables I am not thinking of.

I had to go back and look up my old Paramedic Kinetic energy formula:

Kinetic energy : KE = 1/2 (M * (V * V))   (M= mass and V=velocity)

1/2(544 KG*(11ms *11ms)=32912 Joules

I am not going to say this is correct, because it doesn't give you the other half of the puzzle.

I had to go back and look up my old Paramedic Kinetic energy formula:

Kinetic energy : KE = 1/2 (M * (V * V))   (M= mass and V=velocity)

1/2(544 KG*(11ms *11ms)=32912 Joules

I am not going to say this is correct, because it doesn't give you the other half of the puzzle.

Is this the next equation?  W = F*d*cos(Theta)
Given the distance is unknown, can you then solve for that?

Of course the entire weight of the kayak is not lifted if the lift occurs at one end and the other end is still at least partially supported by the water............... and the boyance may impart more resistance to movement along the transverse axis through the frontal plane than would be required to lift from the center.