[PSUBS-MAILIST] K3000 spherical shell calculations

Personal Submersibles General Discussion personal_submersibles at psubs.org
Thu Apr 17 04:32:20 EDT 2014


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 
be3/4" 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 xr3or V = 4.1888 xr xr xr.  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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>>>
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