[PSUBS-MAILIST] K3000 spherical shell calculations

Wed Apr 16 21:43:15 EDT 2014

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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