[PSUBS-MAILIST] K3000 spherical shell calculations
Personal Submersibles General Discussion
personal_submersibles at psubs.org
Thu Apr 17 01:28:11 EDT 2014
Calculating volume using surface area multiplied by thickness is not accurate unless you use calculus and integrate between the radius limits. Fortunately, this is how we derive the rule-of-thumb formulas for volume, which are simpler and easier to use. For a spherical shell,
V = (4/3) × π × (ro^3 - ri^3)
For a cylindrical shell,
V = π × (ro^2 - ri^2) × L
Once you have your material volume, multiply by its density to get the mass.
Sean
On April 16, 2014 7:43:15 PM MDT, Personal Submersibles General Discussion <personal_submersibles at psubs.org> wrote:
>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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