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Thanksgiving turkey 2

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SAIL3

Structural
Oct 7, 2010
751
The thanksgiving holiday is fast approaching and in that spirit, here is a problem to ponder over the holidays.
A submarine is submerged and stationary 500ft below the surface of the ocean.
In a cabin right next to the chow hall, a 100lb turkey was resting on a table when it happened to overhear two cooks discussing plans for the upcoming thanksgiving dinner.
What it heard threw it into a panic and it flew into the air and began to fly around the cabin in great distress.

Question: has the equilibrium/buoyancy of the submarine changed?
 
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I recommend that they get that baby in the oven soon if they plan on eating by thanksgiving.
We once raised a 60lb turkey. That sucker took 8 hours to cook. It kept springing the oven rack so we supported it with 2x4 blocking.

As for the buoyancy. No it will be the same. The mass hasn't moved outside of the sub.
Myth busters covered this.(except with pigeons and then a toy helicopter) The air from the wings is pushing down with a force equal to the mass lifted.
 
Must have been on Captain Nemo's Nautilus--one of those birds they hunted on Mysterious Island.
 
Back to the OP, by what mechanism do you postulate the turkey was flying? There is an equal and opposite reaction - do a free body diagram of the wing in air. So as the way-too-fat-to-fly turkey takes wing, the downforce generated within the air mass by Tom's wings pushed down with the same net force as if he were still standing on the floor.
 
Only 100 lb turkey I ever met was my brother when he was about 6!!

BTW - Port is right!!
 
TX..forgot to mention that while flying the turkey glides for a period then remembers the gravity of his situation and starts flapping his wings frantically again....
I don't fully understand how a local pressure differential at the wing of the turkey can translate into a gravity event for the submarine as awhole...
IR..if the turkey stops generating lift for a second or two(begins to freefall), is the equilibrium condition still vaild?
 
The centre of buoyancy will remain unchanged, but the centre of gravity will have changed. So there will have been 'some' movement to your sub. But the changes will be so miniscule.

So what happens once all the guys catch, and eat said turkey?
 
I've never quite bought into the notion that it's a steady state result. I'm not convinced that Mythbusters carried out a valid experiment, as they often do.

However, one can possibly look at it as a conservation of momentum problem, so there are impulsive events that are directed at the submarine, as the turkey flies or changes direction.

I completely agree that such impulses are absurdly tiny, given that something like the SSN John Warner has a gross weight of 7800 tons.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize
 
SAIL3,

It does not matter if the turkey is flapping or gliding. The wings vector air downwards to generate lift. If the turkey were to switch from flying to ballistically hurtling for a bit, there would be an increase in the sub's bouyancy followed by a reduction in the bouyancy when the turkey either crashed or generated extra lift to climb back to his original altitude.

Critter.gif
JHG
 
NO - it doesn't matter whether the turkey is flying or setting on his butt. The buoyancy will remain the same - but as noted the CG may change if he walks or flies the length of the sub.

FedEx flies baby chicks all over the US and they still charge by the pound - because they KNOW!
 
Oh boy, a dynamic question on a structural engineering blog!

Using equivalent static analysis, the Turkey Load, TL = W x Ct
W = 100 lbs
Ct = St / R, where St = Design turkey response acceleration parameter (we'll use 1.5 for a fat yet agile turkey) and R = turkey table response modification factor(not given but we'll assume 3.5 for ordinary moment resisting steel table frame design).

So TL = 43 lbs over the fundamental buoyancy period of the submarine, T.

Since T requires an incoherent model we will use the approximate period, Ta = Cs h^x where Cs = approximate period structural parameter (0.01 for submersed steel cylinders), h = length of the submarine (560 ft for a typical Ohio class) and x = approximate period exponent parameter (0.95 for submersed steel cylinders)

So T = Ta = 4.08 seconds

In conclusion, a Turkey Load of 43 lbs will unbalance the submarine for a period of 4.08 seconds until equilibrium is returned.

 
SAIL3,
Please advise what engineering school you graduated from (if you did). My son needs to make his choice soon ...
 
ok..let me continue with this saga...
the turkey is flying around the cabin and this went on for 5 days and nights, no rest and no food, as a result the turkey began to lose weight and strength until finally it collapses back on the table, after having lost 50lbs and too exhausted to care anymore, come what may(kinda like how some design projects wind up)...

same question: has the equilibrium/buoyancy of the submarine changed?
 
Of course it changed after such a long flight. The sub used up 5 days worth of fuel.

Probably one of those old diesel subs where they had to bring in livestock for food since they had no modern refrigeration. Serious odor problems on board.
 
It takes 2.5lb of food to raise 1 lb of turkey

This gives the equation -->
Total weight = 2.5*Delta Turkey

Also this particular turkey lost 50 lbs (Delta turkey = -50)

Because we know all equations are reversable this gives us
Toal wt = 2.5 (-50)=-125

Therefore the sub lost 125 lbs thus changing the boyancey.
 
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