[PSUBS-MAILIST] K3000 spherical shell calculations

Personal Submersibles General Discussion personal_submersibles at psubs.org
Thu Apr 17 04:19:02 EDT 2014


Les and Jim,


Sorry, saw this late. I gave a similar answer on another email. You figure the displacement (which would include hard variable ballast tanks if mounted externally). That way you add water, which changes the sub's weight without changing its displacement). Main ballast or fairwater tanks are an entirely different issue, as they do in fact change the sub's displacement for surface work.


Keep in mind, however, that syntactic foam (as a for instance) is AIR--hollow glass micro balloons in an epoxy matrix, so it is in fact a method to use air for overall buoyancy. It is not a controllable quantity beyond the design and fabrication phase. Syntactic imposes weight and displacement issues of its own (roughly 50% lift per cubic whatever--A cubic foot of syntactic as an example, weighs about 30 pounds and the other 34 pounds of the displaced volume is positive. Without it, Alvin would be dragging its shiny new butt in the dirt all the time, and we certainly couldn't have that.


My apologies if this has become patently obvious by now. It's late, or early depending on whether you are on night shift or not (I am). In either case, my reading glasses are fogging up down here at the gateway to the Florida Keys.


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 3:56 am
Subject: Re: [PSUBS-MAILIST] K3000 spherical shell calculations



Hi again Jim 
Yes we are on different page my initial calc was purly to find out how much lighter the shell material would be if I used stiffeners for the same depth. I had not gotten to your stage......and I would still like to know this?
However now the embarrasing part...... I was however considering using the water as ballast weight ......which you have now reminded me that it is okay after neutral bouyancy for submerging  but not possible to use water tanks to it to creat neutral bouancy ...........dont know where my thinking was.
Thanks for your patience 
Cheers 
>From dumb Aussie 
Les 
  
----- Original Message ----- 
  
From:   Personal Submersibles General   Discussion 
  
To: personal_submersibles at psubs.org   
  
Sent: Thursday, April 17, 2014 3:58   PM
  
Subject: Re: [PSUBS-MAILIST] K3000   spherical shell calculations
  


  
Les,
  
 
  
I'm talking about how much the whole sub would have to   weigh while sitting on a dock or trailer regardless of how much any   individual component such as the sphere weighs.  You have to start with   displacement to determine the weight you'll need to design to.    For example, if the dimensions calculate to a volume equal to 22 tons of   displacement, then you can begin to add up the weight of the interior items   and the net-weight-in-water of the exterior items to see where you   stand.  If you have an exterior item that weighs 10 lb in air but would   displace 1 lb in water, it has a net-weight-in-water (NWW) of 9   lb.
  
 
  
So:
  
Hull + radios + electrical system + ballast tanks + all   other item + payload (people) = 20 tons on land.  You're still 2 tons   shy.  This means you're going to have to add 2 tons of something (lead   weights or whatever) in order to make the sub submerge.  No matter how   you slice it, it all has to add up to a weight equal to the   displacement.
  
 
  
For these purposes I'm ignoring the role of motors in   powering down to the desired depth when you have a slight positive   buoyancy.
  
 
  
Jim
  
 
  
 
  
  
In a message dated 4/17/2014 12:41:39 A.M. Central Daylight Time,   personal_submersibles at psubs.org writes:
  
    
Hi again Jim , please have patience with me, either I am completely not     thinking straight or we are talking apples and oranges ?
    
I am talking about dry air surface land weight , you make a     cylinder out of 3/4" steel plate 1.2m diam 4m long with same material     end caps 
    
What weight are you going to have to lift it with a crane?
    
cheers 
    
Les
    
      
----- Original Message ----- 
      
From:       Personal Submersibles       General Discussion 
      
To: personal_submersibles at psubs.org       
      
Sent: Thursday, April 17, 2014 2:44       PM
      
Subject: Re: [PSUBS-MAILIST] K3000       spherical shell calculations
      


      
Hi Les,
      
 
      
The figure of 9858 kg (21,688 lb.) is based       solely on the combined volume of the sphere, cylinder,       and endcap of the sub.  That is, it will displace that many       kg or lb of seawater when submerged.  Therefore the total       vessel must weigh that much in order to submerge       (nuetral) whether your hull is 1/4" thick or 2" thick.  That       includes the sphere, cylinder, radios, lead weights, occupants, lunch, and       everything else.  Those things that are exterior will increase the       displacement some and therefore the total weight requirement as       well.  Since you don't know the weight and volume (displacement) of       those yet, you can't calculate them.  However based on your       dimensions, the sub would have to weigh in the neighborhood of 11       tons.  See the first sentence of Sean's post.  If you're       comfortable with that, then you can proceed to the other steps in       evaluating the feasibility of the project.  It's way beyond anything       I would desire to tackle.
      
 
      
The weight of the water that enters your exterior       ballast tanks will not contribute to meeting the required       weight of the sub since that water only offsets the displacement       of the tanks themselves.  Adding more or bigger exterior ballast       tanks does not increase the sub's ability to submerge (other than by the       weight of the materials composing the tanks).  The tanks are for       adding buoyancy to the extent of the amount of air within       them.
      
 
      
Now if your hull and other components total more than 11       tons, you'll need to add static flotation such as syntactic       foam to compensate for the excess weight.  Yes, you could       accomplish the same thing with your ballast tanks, but it's not as       safe.
      
 
      
Best regards,
      
Jim
      
 
      
      
In a message dated 4/16/2014 8:44:20 P.M. Central Daylight Time,       personal_submersibles at psubs.org writes:
      
        
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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<clip_image002.png>
          
            
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