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<DIV><FONT size=3>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.</FONT></DIV>
<DIV><FONT size=3></FONT> </DIV>
<DIV><FONT size=3>Jim. </FONT></DIV>
<DIV> </DIV>
<DIV>
<DIV>In a message dated 4/16/2014 11:41:59 A.M. Central Daylight Time,
personal_submersibles@psubs.org writes:</DIV>
<BLOCKQUOTE
style="PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: blue 2px solid"><FONT
style="BACKGROUND-COLOR: transparent" color=#000000 size=2
face=Arial><BR>Jim,<BR>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.<BR>Hank --------------------------------------------<BR>On Wed,
4/16/14, Personal Submersibles General Discussion
<personal_submersibles@psubs.org> wrote:<BR><BR>Subject: Re:
[PSUBS-MAILIST] K3000 spherical shell calculations<BR>To:
personal_submersibles@psubs.org<BR>Received: Wednesday, April 16, 2014, 11:12
AM<BR><BR><BR> <BR><BR><BR>Hi <BR>Les, <BR>The <BR>basic formula for the
volume of a sphere is . Don't<BR>accidentally <BR>plug in the diameter
instead of the radius (I've done<BR>that). To simplify the
formula,<BR>convert the 4/3 <BR>to a decimal carried to as many places as you
wish for<BR>accuracy: <BR>1.333333. So it now reads V=1.3333
<BR>π r3. Since π = 3.14159<BR>(rounded), you can go ahead and
multiply <BR>it by your 1.333333 to get 4.1888. <BR>Your simplified
formula now reads V =<BR>4.1888 x <BR>r3 <BR>or V = 4.1888 x r <BR>x r <BR>x
<BR>r. You can use that simplified formula<BR>for <BR>calculating the
volume of any sphere by plugging in the<BR>r3. The 4.1888 is a
<BR>constant. <BR> <BR>In <BR>your case since the diameter of the
sphere is 2 meters, your<BR>radius is 1 meter <BR>and the volume of your
sphere is 4.1888 cubic meters. Having the simplified formula saves
a<BR><BR>lot number crunching when you are calculating
different<BR>sizes. If you can set up a<BR>spreadsheet <BR>containing
that formula it will be even easier. You can also use that formula to
<BR>calculate the volume of a hemispherical tank head on a<BR>cylinder by
dividing it <BR>by 2. <BR> <BR>To <BR>calculate the volume of a
cylinder, first calculate the area<BR>of a circle of that <BR>radius and
multiply it by the length. A <BR>= π r2<BR>. For your radius
of 0.6 meters,<BR>A = 1.13 <BR>m2 or 4.524 m3 for a 4 meter
long<BR>cylinder. <BR> <BR>Add a<BR>hemispherical tank head on the
other <BR>end: V = 4.1888 x <BR>.63 and you get a volume of
<BR>.905 m3. <BR> <BR>Add <BR>the three figures together:
<BR>Sphere <BR>4.189
<BR>Cylinder <BR>4.524
<BR>Head <BR><BR>0.905
<BR> <BR><BR>9.618
m3 Total volume <BR> <BR>As <BR>you can see, these figures pretty
well match up with<BR>Sean’s. Your sub would have to<BR>weigh at least
<BR>9858 kg (21,688 lb) in air in <BR>order to submerge in sea water.
<BR>Adding external ballast tanks will not reduce that<BR>figure. Adding
internal ballast<BR>tanks will <BR>reduce it by the weight of the water in
those internal <BR>tanks. <BR> <BR>Don’t <BR>worry about dumb
questions. <BR>I’ve <BR>had a few. If anything I’ve<BR>written
<BR>above is inaccurate, someone will correct it for the benefit<BR>of
all. I wanted to keep it<BR>simple instead of <BR>adding too much
detail. That can<BR>be <BR>done later. <BR> <BR>Best
<BR>regards, <BR>Jim <BR>T. <BR> <BR><BR>In a message dated 4/16/2014
12:58:11 A.M. Central<BR>Daylight Time, <BR>personal_submersibles@psubs.org
writes:<BR><BR> Les, the total<BR>mass of the trimmed-out craft
will <BR> be exactly the displacement volume of your proposed
craft<BR>multiplied by the <BR> density of seawater, if you expect
to be neutrally<BR>buoyant. Back of <BR> envelope
calcs: a 2m sphere is 4.189 m^3, a cylinder<BR>1.2m OD x 4m is
<BR> 4.524 m^3, for a total of 8.713 m^3. Multiplying by
1025<BR>kg/m^3 (seawater <BR> density) gives 8930.825 kg.
Subtract some for the<BR>common volume, add <BR> some for
superstructure, conning tower etc., but<BR>that's the ballpark. Or
<BR> are your worried about the dry weight of the steel used
in<BR><BR> construction?<BR><BR>Sean<BR><BR><BR>On 2014-04-15
23:25, Personal <BR> Submersibles General Discussion
wrote:<BR><BR> <BR> <BR>
<BR><BR> Hello everybody<BR>,anybody, Les here , <BR>
<BR> Attatched myself to<BR>this email for
convenience <BR> (similar subject) been away from psubs for
quite some<BR>time wanting to start <BR> again.<BR>
Now it might sound<BR>dumb, but I tried to follow <BR>
the calc sheet for material and depth etc with
ring<BR>stiffeners but <BR> ufortunately had a few
problems, perhaps a sample calc<BR>attached to it would <BR>
assist me and maybe others on how to use it
correctly?<BR><BR> In between time I do<BR>need to get a
rough <BR> indication of the thickness
of steel<BR> and approx size <BR>
of ring stiffener size and<BR>quantity, to roughly
<BR> calculate the weight of what I wish to build,
to<BR>see if what I <BR> want to do is feasible or
not...WEIGHT IS CRITICAL for<BR>my project <BR> <BR>
Can anyone help me<BR>please my reqirements are; <BR>
<BR> A Sphere 2 meters<BR>diameter<BR>
A Cylinder attached to<BR>that 1.2m diameter x <BR>
4meters long<BR> ( I understand<BR>there
will be a flaring <BR> attatchment to the sphere, however
at this point for the<BR>exercise, just to <BR> calc the
min weight that would be possible on these two<BR>items would be an <BR>
indicator for me andd give me a mental appreciation of<BR>my
limitations <BR> )<BR> The desired depth
is<BR>300m, ( 984ft ) <BR> ( 452 psi ) or I could
settle for 250 meters( 820ft<BR>) ( 379 psi ) <BR>
both maximum dive depth not crush depth.<BR> Sorry to
be pain<BR>but can any-one help me <BR> <BR>
Thank you <BR> Les<BR>
<BR> P.S. In for a penny in<BR>for a pound, guess I
<BR> will make myself look completely dumb ....just
as<BR>an indication, with <BR> something like the
above how would I calculate<BR>the <BR>
<BR> <BR>
volume hence the size required for soft tanks for<BR>maximum
submergance <BR> <BR>
<BR>
<BR> <BR><BR><BR>_______________________________________________<BR>Personal_Submersibles
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