[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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