[PSUBS-MAILIST] K3000 spherical shell calculations
Personal Submersibles General Discussion
personal_submersibles at psubs.org
Thu Apr 17 10:41:41 EDT 2014
Vance,
Counterclockwise and typhoon.
Hmm, "the Spanish fleet was sunk by a typhoon"....just doesn't sound quite right does it?
Joe
Sent from my iPhone
On Apr 17, 2014, at 4:43 AM, Personal Submersibles General Discussion <personal_submersibles at psubs.org> wrote:
> I see that. And the hurricane bathtub thing, too.
> 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 4:33 am
> Subject: Re: [PSUBS-MAILIST] K3000 spherical shell calculations
>
> 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 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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