[PSUBS-MAILIST] K3000 spherical shell calculations

Wed Apr 16 12:41:10 EDT 2014

Jim,
You missed a step, If the sphere is welded to the cylinder like a giant deep worker.  Then your calculation needs to remove part of the volume of the sphere.  The portion of sphere volume  that intrudes into the cylinder needs to be removed from the total volume.
Hank --------------------------------------------
On Wed, 4/16/14, Personal Submersibles General Discussion <personal_submersibles at psubs.org> wrote:

 Subject: Re: [PSUBS-MAILIST] K3000 spherical shell calculations
 To: personal_submersibles at psubs.org
 Received: Wednesday, April 16, 2014, 11:12 AM

 Hi 
 Les, 
 The 
 basic formula for the volume of a sphere is .  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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