[PSUBS-MAILIST] K3000 spherical shell calculations
Personal Submersibles General Discussion
personal_submersibles at psubs.org
Wed Apr 16 13:09:19 EDT 2014
True, Hank, and Sean alluded to that consideration. Same applies to the
CT/hatch. I was limiting my comments to the basics to avoid getting into
that as well as several others not mentioned which would have to be addressed
later. Good observation though, and it is big "detail" such as I was
referencing in the last paragraph of my post. For now I only addressed how to
calculate volume of two basic shapes in order to get a general idea of
displacement and resultant weight. Also the length of the cylinder as specified
by Les might already include the tank head whether hemispherical or other.
I didn't ask because I just wanted to give him the tools and so he could
run his own basic calcs. I have some questions of my own to ask later.
Jim.
In a message dated 4/16/2014 11:41:59 A.M. Central Daylight Time,
personal_submersibles at psubs.org writes:
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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