Rocket Ascents... Can divers breach like a fish (split from Accident in Mich)

[MXGratefulDiver]

This is ridiculous.

But it's also the best thread ever (ever being defined as the short amount of time during which I've been a member).

"Best" ... in what sense?

That would be like saying that Jerry Springer is the best show on TV ... then again, that might be so (wouldn't know ... I killed my television years ago) ...

... Bob (Grateful Diver)

 

[pilot fish]

Even better

baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

 

[gangrel441]

I honestly think Rec Diver is sitting there, reading this thread, honestly thinking that we're all idiots for not believing him

You know, one of the reasons I joined this whole mess was because he didn't state this just as an aneicdote, but was asking bob why he didn't do that himself. As much as he has argued and insisted on the credibility of a diver breaching, he has not once addressed the my point that, even if plausible, is not advisable under any stressed or rescue circumstances. That seemed to be the basis of his post in the first place. :frown3:

 

[Blackwood]

"Best" ... in what sense?

That would be like saying that Jerry Springer is the best show on TV ... then again, that might be so (wouldn't know ... I killed my television years ago) ...

... Bob (Grateful Diver)

Most enjoyable read.

 

[bradshsi]

In order to prevent any further injury to the laws of physics I'll comment as follows...

The problem is one of two parts. Firstly what was the divers velocity underwater prior to breaching the surface and secondly upon breaching the surface how far would he/she travel above the surface ?

Taking the second partof the problem (above the water surface), the equations are quite simple

Vertical velocity Vy = Vo*Sin(phi) - g*t
Vertical position y = (Vo*Sin(phi))*t - 0.5*g*t^2

Where

Vo is the initial velocity of the diver at the surface
g is gravity (approximately 33 ft/s)
phi is the angle of the diver's trajectory relative to the horizontal (in degrees)
t is time

To solve the problem we need to know the divers velocity at the surface, solve for Vy becoming zero (highest point of his travel) to find t. Then substituting t in the second equation will yield the highest vertical position y.

Now as has been previously stated the claimed velocity of the diver was 10ft/sec (100 ft in 10 seconds). We will assume that the diver had no horizontal component to his velocity therefore phi is 90 degrees (and thus Sin(phi) = 1)).

Solving for these values yields:

t = 0.3 seconds
y = 1.5 feet

So for the ascent rate claimed, there is no way the diver could have cleared the water (unless he/she was very short...).

A more interesting question is what would the ascent velocity have to be in order to achieve what was described. If we assume a diver of average height with 18" long fins, the diver would have a total "height" of 88" (7.3 feet)

Solving the equations in reverse for this yields:

t = 0.67 seconds
Vo = 22 ft/second

This leads on nicely to the first part of the problem. Since we now know that the divers ascent rate would have to be 22 ft/second is this rate of ascent possible ?

There are two ways to answer this either by experiment or by calculation.

I suggested to my engineering group that I set them the task of solving the diver velocity problem by CFD (Computational Fluid Dynamics) instead of doing real work. They in turn suggested that doing real work was more important because otherwise the company would fail and their children would starve.

So unfortunately that leaves experimentation. Anyone willing to log such an ascent on their dive computers for posterity ?? Darwin Award candidates only should apply.

 

[D_B]

true , but do not have to be same material and we need to ignore air resistance

We cannot ignore resistance as it is greatly mangnified while assending through the water and is a very important part of the equation for breaching.

oh , I agree , but I was not ignoring water , just commenting on what you said ... "have an equal air resistance they will travel to the same height" but , no matter, as I do think "breaching" is hogwash

DB

 

[pants!]

In order to prevent any further injury to the laws of physics I'll comment as follows...

The problem is one of two parts. Firstly what was the divers velocity underwater prior to breaching the surface and secondly upon breaching the surface how far would he/she travel above the surface ?

Taking the second part, the equations are quite simple

Vertical velocity Vy = Vo*Sin(phi) - g*t
Vertical position y = (Vo*Sin(phi))*t - 0.5*g*t^2

Where

Vo is the initial velocity of the diver at the surface
g is gravity (approximately 33 ft/s)
phi is the angle of the diver's trajectory relative to the horizontal (in degrees)
t is time

To solve the problem we need to know the divers velocity at the surface, solve for Vy becoming zero (highest point of his travel) to find t. Then substituting t in the second equation will yield the highest vertical position y.

Now as has been previously stated the claimed velocity of the diver was 10ft/sec (100 ft in 10 seconds). We will assume that the diver had no horizontal component to his velocity therefore phi is 90 degrees (and thus Sin(phi) = 1)).

Solving for these values yields:

t = 0.3 seconds
y = 1.5 feet

So for the ascent rate claimed, there is no way the diver could have cleared the water (unless he/she was very short...).

A more interesting question is what would the ascent velocity have to be in order to achieve what was described. If we assume a diver of average height with 18" long fins, the diver would have a total "height" of 88" (7.3 feet)

Solving the equations in reverse for this yields:

t = 0.78 seconds
Vo = 25.7 ft/second

This leads on nicely to the first part of the problem. Since we now know that the divers ascent rate would have to be 25.7 ft/second is this rate of ascent possible ?

There are two ways to answer this either by experiment or by calculation.

I suggested to my engineering group that I set them the task of solving the diver velocity problem by CFD (Computational Fluid Dynamics) instead of doing real work. They in turn suggested that doing real work was more important because otherwise the company would fail and their children would starve.

So unfortunately that leaves experimentation. Anyone willing to log such an ascent on their dive computers for posterity ?? Darwin candidates only should apply.

What's all this noise?

The answer is 6 mph, because that's what I converted 83 ft/s to, and that's the limit of my mathematical skills.

6, see? Find clearing the water! That's all it takes! Bob says so!

 

[bradshsi]

What's all this noise?

The answer is 6 mph, because that's what I converted 83 ft/s to, and that's the limit of my mathematical skills.

6, see? Find clearing the water! That's all it takes! Bob says so!

Er

83 ft/s is about 56 MPH. I guess your calculations are as skewed as mine (since I got the Vo value wrong first time...)


Just to clarify :

at Vo = 25.7, the diver would breach the surface by 10 ft
at Vo = 22, the diver would breach by 7.3 ft

BTW the noise you hear is the sound of reality intruding into this thread...

 

[pants!]

Whoops, missed the decimal point. 83 feet in 10 seconds, or 8.3 ft/s, is the random number NWGratefulDiver came up with and Rec Diver used to get 6 mph.

 

[gangrel441]

Finally broke down and gave us a profile, eh pants!?