[PSUBS-MAILIST] K3000 spherical shell calculations

Thu Apr 17 04:43:59 EDT 2014

I see that. And the hurricane bathtub thing, too.
Vance

-----Original Message-----
From: Personal Submersibles General Discussion <personal_submersibles at psubs.org>
To: Personal Submersibles General Discussion <personal_submersibles at psubs.org>
Sent: Thu, Apr 17, 2014 4:33 am
Subject: Re: [PSUBS-MAILIST] K3000 spherical shell calculations

Vance,
.>>don't toilets flush counter clockwise in New Zealand.
  No that's bath tubs.
It's how this "American" program I'm using states it. It has readings for weight in air & weight in water.
If you weigh 200lb & jump in the water & you are neutrally buoyant then it would say you way nothing
in water & your air weight was 200lb. It is helpful to see at an instant what weight you would need to
add to achieve neutral buoyancy.
Alan

   From: Personal Submersibles General Discussion <personal_submersibles at psubs.org>
 To: personal_submersibles at psubs.org 
 Sent: Thursday, April 17, 2014 8:01 PM
 Subject: Re: [PSUBS-MAILIST] K3000 spherical shell calculations

Okay Alan, I give up. The weight of the hull in water? What does that mean? Displacement is calculated at the outer skin diameter of the pressure hull which, along with the displacement of externally added bits and pieces, give you a total payload number in pounds or kilos (depending on your preference).

So if we're talking about a 12,000 pound displacement in seawater for a finished sub, for instance, then you have a crane weight in air of the same amount less the variable payload (crew, vbt fluids, etc., call that 1,000 pounds for convenience).

If your pressure hull weighs 5500 pounds, for instance, then during the design phase you have essentially the other 5500 pounds of that 11,000 (12,000 minus the adjustable payload reserve of 1,000 pounds) to play with in permanent gear and so on (motors, decking, electrical distribution, batteries, drop weights, ballast tanks, guard rails, and so on). The 11,000 pound total is your deadweight, or crane weight. Add crew and ham sandwiches and pee bottles, plus some variable ballast water and, all things being equal, you vent the mains and reach (or at least strive for) neutral buoyancy.

I have this feeling that we are all talking about the same thing, only backwards. After all, don't toilets flush counter clockwise in New Zealand, or is that hurricanes? I get those two mixed up. In any case, was that what you just said, only different? I'm guessing the weight in water means the weight OF water displaced by the hull. Yes, no? Or am I missing something?

Vance

-----Original Message-----
From: Personal Submersibles General Discussion <personal_submersibles at psubs.org>
To: Personal Submersibles General Discussion <personal_submersibles at psubs.org>
Sent: Thu, Apr 17, 2014 3:11 am
Subject: Re: [PSUBS-MAILIST] K3000 spherical shell calculations

Les a correction,
looking at it again (after being emailed off list by Jim; thanks Jim)
I forgot to add in the weight of the cylinder to the calculation
should be the displacement = 6869 + the hull weight of 2380.7
= 9149.7 which is more in line with Sean & Jim.
Alan

   From: Personal Submersibles General Discussion <personal_submersibles at psubs.org>
 To: Personal Submersibles General Discussion <personal_submersibles at psubs.org> 
 Sent: Thursday, April 17, 2014 6:26 PM
 Subject: Re: [PSUBS-MAILIST] K3000 spherical shell calculations

Les,
in my calculations the dry weight of the sphere & the cylinder are
1056.3  kg + 1,324.4 kg = 2,380.7 kg.
The weight of the sphere & cylinder in water is -3378.9 liters + -3490.1 kg.
Which means that for these two items alone you are going to have to make your sub weigh
3378.9 kg + 3490.1 kg = 6869 kg to get it neutrally buoyant.
Alan

   From: Personal Submersibles General Discussion <personal_submersibles at psubs.org>
 To: Personal Submersibles General Discussion <personal_submersibles at psubs.org> 
 Sent: Thursday, April 17, 2014 3:02 PM
 Subject: Re: [PSUBS-MAILIST] K3000 spherical shell calculations

Hi Les (or Gidday)
The 2 meter inside diameter sphere made from 516 / 70 steel needs a crush depth of
561 meters to fit G.Ls perameters of the colapse dive pressure / nominal diving pressure 
> 1.87.
At 10.8mm thick you get a crush depth of 559.75 M.
It will weigh 1056.3 kg & be -3378.9 kg water weight salt. (need a lot of lead)
A tube 1.2M ID & 4 meters long of same material at 11.3mm thick crushes at 561 meters,
weighs 1,324.4 kg with a -3490.1 kg water weight.
In my program I am showing no advantage in having stiffeners at this thickness & diameter.
Someone might want to confirm this but I ran this tube as a 500mm, a 1 meter & 2 meter 
length & got the same crush depth for the diameter & thickness.
Cheers Alan

   From: Personal Submersibles General Discussion <personal_submersibles at psubs.org>
 To: Personal Submersibles General Discussion <personal_submersibles at psubs.org> 
 Sent: Thursday, April 17, 2014 1:43 PM
 Subject: Re: [PSUBS-MAILIST] K3000 spherical shell calculations

Thanks Guys for your response ...and my head goes around and around......good mental exercise??? let us start again
Psubs calcs for unstiffened cylinder 1.2m x 4 meters long  indicated that it need be 3/4" wall for 314psi at 706fsw (good I have an indicator)
I Needed to know the actual physical lifting weight of the two items, the 2m sphere and the 1.2d x 4m cylinder  okay so I calculated the surface area of the cylinder 
easy enough A = 2 (pi) r h + 2(pi) r squared  = 175.9ft squared .  Did the same for a sphere 4(pi)r squared = 113.097 ft squared
Total area of sphere and cylinder = 289 ft squared
Multiplied by 30.65 lbs (for 3/4 steel plate per ft.squared)  therefore   289ft squared x 30.65 lbs /foot squared =  8858 lbs (all soft conv)  = 3.95 ton approx.
This figure aligns with sean? I think,  not sure about Jim T 's 21,688 Lb unless he forgot already had it in lbs not kgs., as we all have done I am sure 
Anyway presuming I am right the original question I was wanting, was an indication of how  thinner steel plate I could use with what size stiffeners at what spaces to have the same depth capabilities and how much physical weight I might loose. This is all for an indication ...if weight feasability works then I can bother about details such as joining taper for sphere/cylinder etc.other equip weight etc.
The submersing tank question was again what volume of water required for this size craft so I again could calculate  physical  weight of additional fabricated external tanks
I hope I have not  confused everyone 
Cheers 
Les

----- Original Message ----- 

From:   Personal Submersibles General   Discussion 

To: Personal Submersibles General   Discussion 

Sent: Thursday, April 17, 2014 6:58   AM

Subject: Re: [PSUBS-MAILIST] K3000   spherical shell calculations

Uh...no.

Do a sphere calc and add it to a cylinder calc.

Vance

Sent from my iPhone

On Apr 16, 2014, at 5:16 PM, Personal Submersibles General Discussion   <personal_submersibles at psubs.org>   wrote:

This may be a dumb question, but is finding the volume of a cylinder     with two hemispherical heads

V =     4.1888 x r x r x length?

Thanks,

Scott Waters

Sent from my U.S. Cellular©     Smartphone

Personal Submersibles General Discussion <personal_submersibles at psubs.org>     wrote:

Hi     Les,

The     basic formula for the volume of a sphere is <clip_image002.png>.  Don't     accidentally plug in the diameter instead of the radius (I've done     that).  To simplify the formula,     convert the 4/3 to a decimal carried to as many places as you wish for     accuracy:  1.333333.  So it now reads V=1.3333     π r3.  Since π = 3.14159 (rounded), you can go ahead and     multiply it by your 1.333333 to get 4.1888.  Your simplified formula now reads V = 4.1888 x     r3     or V = 4.1888 x     r x     r x     r.  You can use that simplified formula     for calculating the volume of any sphere by plugging in the     r3.  The 4.1888 is a     constant.

In     your case since the diameter of the sphere is 2 meters, your radius is 1     meter and the volume of your sphere is 4.1888 cubic meters.  Having the simplified formula saves     a lot number crunching when you are calculating different sizes.  If you can set up a spreadsheet     containing that formula it will be even easier.  You can also use that formula to     calculate the volume of a hemispherical tank head on a cylinder by dividing     it by 2.

To     calculate the volume of a cylinder, first calculate the area of a circle of     that radius and multiply it by the length.  A = π r2 .  For your radius of 0.6 meters, A =     1.13 m2 or 4.524 m3 for a 4 meter long cylinder.  

Add a hemispherical tank head on the other     end:  V = 4.1888 x     .63 and you get a volume of      .905 m3.

Add     the three figures together:

Sphere             4.189

Cylinder           4.524

Head                0.905

                        9.618 m3 Total volume

As     you can see, these figures pretty well match up with Sean’s.  Your sub would have to weigh at     least 9858 kg (21,688 lb) in     air in order to submerge in sea water.  Adding external ballast tanks will     not reduce that figure.  Adding     internal ballast tanks will reduce it by the weight of the water in those     internal tanks.

Don’t     worry about dumb questions.      I’ve had a few.  If     anything I’ve written above is inaccurate, someone will correct it for the     benefit of all.  I wanted to     keep it simple instead of adding too much detail.  That can be done     later.

Best     regards,

Jim     T.

In a message dated 4/16/2014 12:58:11 A.M. Central Daylight Time, personal_submersibles at psubs.org     writes:

Les, the total mass of the trimmed-out craft       will be exactly the displacement volume of your proposed craft multiplied       by the density of seawater, if you expect to be neutrally buoyant.        Back of envelope calcs:  a 2m sphere is 4.189 m^3, a cylinder 1.2m OD       x 4m is 4.524 m^3, for a total of 8.713 m^3. Multiplying by 1025 kg/m^3       (seawater density) gives 8930.825 kg.  Subtract some for the common       volume, add some for superstructure, conning tower etc., but that's the       ballpark.  Or are your worried about the dry weight of the steel used       in construction?

Sean

On 2014-04-15 23:25, Personal       Submersibles General Discussion wrote:

Hello everybody ,anybody, Les here ,         

Attatched myself to this email for         convenience (similar subject) been away from psubs for quite some time         wanting to start again.

Now it might sound dumb, but I tried to         follow the calc sheet for material and depth etc with ring         stiffeners but ufortunately had a few problems, perhaps a sample         calc attached to it would assist me and maybe others on how to use it         correctly? 

In between time I do need to get a rough         indication of the thickness of steel  and  approx         size of  ring stiffener size and quantity, to         roughly calculate the weight of what I wish to build, to see if         what I want to do is feasible or not...WEIGHT IS CRITICAL for my         project 

Can anyone help me please my reqirements         are; 

A Sphere 2 meters diameter

A Cylinder attached to that 1.2m diameter x         4meters long

 ( I understand there will be a         flaring attatchment to the sphere, however at this point for the         exercise, just to calc the min weight that would be possible on these         two items would be an indicator for me andd give me a mental         appreciation of my limitations )

The desired depth is 300m, ( 984ft )         ( 452 psi ) or I could settle for 250 meters( 820ft ) ( 379         psi ) both maximum dive depth not crush depth.

Sorry to be  pain but can any-one help         me 

Thank you 

Les

P.S. In for a penny in for a pound, guess I         will make myself look completely dumb ....just as an indication,         with something like the above how would I calculate the          

                volume hence the size required for soft tanks for maximum submergance          

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