[PSUBS-MAILIST] K3000 spherical shell calculations
Personal Submersibles General Discussion
personal_submersibles at psubs.org
Thu Apr 17 02:28:05 EDT 2014
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_
(mailto:personal_submersibles at psubs.org)
To: _personal_submersibles at psubs.org_
(mailto: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_
(mailto:personal_submersibles at psubs.org)
To: _Personal Submersibles General Discussion_
(mailto:personal_submersibles at psubs.org)
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_ (mailto: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_ (mailto: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_ (mailto: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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