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<DIV>Hi again Jim </DIV>
<DIV>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?</DIV>
<DIV><STRONG>However now the embarrasing part</STRONG>...... 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.</DIV>
<DIV>Thanks for your patience </DIV>
<DIV>Cheers </DIV>
<DIV>From dumb Aussie </DIV>
<DIV>Les </DIV>
<BLOCKQUOTE
style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV style="FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: black"><B>From:</B>
<A title=personal_submersibles@psubs.org
href="mailto:personal_submersibles@psubs.org">Personal Submersibles General
Discussion</A> </DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A
title=personal_submersibles@psubs.org
href="mailto:personal_submersibles@psubs.org">personal_submersibles@psubs.org</A>
</DIV>
<DIV style="FONT: 10pt arial"><B>Sent:</B> Thursday, April 17, 2014 3:58
PM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> Re: [PSUBS-MAILIST] K3000
spherical shell calculations</DIV>
<DIV><BR></DIV><FONT id=role_document face=Arial color=#000000 size=2>
<DIV><FONT size=3>Les,</FONT></DIV>
<DIV><FONT size=3></FONT> </DIV>
<DIV><FONT size=3>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
<EM>displacement</EM> 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.</FONT></DIV>
<DIV><FONT size=3></FONT> </DIV>
<DIV><FONT size=3>So:</FONT></DIV>
<DIV><FONT size=3>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.</FONT></DIV>
<DIV><FONT size=3></FONT> </DIV>
<DIV><FONT size=3>For these purposes I'm ignoring the role of motors in
powering down to the desired depth when you have a slight positive
buoyancy.</FONT></DIV>
<DIV><FONT size=3></FONT> </DIV>
<DIV><FONT size=3>Jim</FONT></DIV>
<DIV><FONT size=3></FONT> </DIV>
<DIV> </DIV>
<DIV>
<DIV>In a message dated 4/17/2014 12:41:39 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" face=Arial color=#000000 size=2>
<DIV>Hi again Jim , please have patience with me, either I am completely not
thinking straight or we are talking apples and oranges ?</DIV>
<DIV>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 </DIV>
<DIV>What weight are you going to have to lift it with a crane?</DIV>
<DIV>cheers </DIV>
<DIV>Les</DIV>
<BLOCKQUOTE
style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV style="FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: black"><B>From:</B>
<A title=mailto:personal_submersibles@psubs.org
href="mailto:personal_submersibles@psubs.org">Personal Submersibles
General Discussion</A> </DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A
title=mailto:personal_submersibles@psubs.org
href="mailto:personal_submersibles@psubs.org">personal_submersibles@psubs.org</A>
</DIV>
<DIV style="FONT: 10pt arial"><B>Sent:</B> Thursday, April 17, 2014 2:44
PM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> Re: [PSUBS-MAILIST] K3000
spherical shell calculations</DIV>
<DIV><BR></DIV><FONT face=Arial color=#000000 size=2>
<DIV><FONT size=3>Hi Les,</FONT></DIV>
<DIV><FONT size=3></FONT> </DIV>
<DIV><FONT size=3>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.</FONT></DIV>
<DIV><FONT size=3></FONT> </DIV>
<DIV><FONT size=3>The weight of the water that enters your exterior
ballast tanks will <EM><U>not</U></EM> 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.</FONT></DIV>
<DIV><FONT size=3></FONT> </DIV>
<DIV><FONT size=3>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.</FONT></DIV>
<DIV><FONT size=3></FONT> </DIV>
<DIV><FONT size=3>Best regards,</FONT></DIV>
<DIV><FONT size=3>Jim</FONT></DIV>
<DIV> </DIV>
<DIV>
<DIV>In a message dated 4/16/2014 8:44:20 P.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" face=Arial color=#000000 size=2>
<DIV><FONT face=Arial size=2>Thanks Guys for your response ...and my
head goes around and around......good mental exercise??? let us start
again</FONT></DIV>
<DIV><FONT face=Arial size=2>Psubs calcs for <STRONG>unstiffened
cylinder</STRONG> 1.2m x 4 meters long indicated that it need
be<STRONG> 3/4" wall for 314psi at 706fsw</STRONG> (good I have an
indicator)</FONT></DIV>
<DIV><FONT face=Arial size=2>I Needed to know the actual physical
lifting weight of the two items, the 2m sphere and the 1.2d x 4m
cylinder </FONT><FONT face=Arial size=2>okay so I calculated the
surface area of the cylinder </FONT></DIV>
<DIV><FONT face=Arial size=2>easy enough A = 2 (pi) r h + 2(pi) r
squared = 175.9ft squared . </FONT><FONT
face=Arial size=2>Did the same for a sphere 4(pi)r squared = 113.097 ft
squared</FONT></DIV>
<DIV><FONT face=Arial size=2>Total area of sphere and cylinder = 289 ft
squared</FONT></DIV>
<DIV><FONT face=Arial size=2>Multiplied by 30.65 lbs (for 3/4 steel
plate per ft.squared) therefore 289ft squared x 30.65 lbs
/foot squared = <STRONG>8858 lbs (all soft conv) = 3.95
ton approx.</STRONG></FONT></DIV>
<DIV><FONT face=Arial size=2>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 </FONT></DIV>
<DIV><FONT face=Arial size=2>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 </FONT><FONT
face=Arial size=2>weight feasability works then I can bother
about details such as joining taper for sphere/cylinder etc.other equip
weight etc.</FONT></DIV>
<DIV><FONT face=Arial size=2>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</FONT></DIV>
<DIV><FONT face=Arial size=2>I hope I have not confused everyone
</FONT></DIV>
<DIV><FONT face=Arial size=2>Cheers </FONT></DIV>
<DIV><FONT face=Arial size=2>Les</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV>----- Original Message ----- </DIV>
<BLOCKQUOTE
style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: black"><B>From:</B>
<A title=mailto:personal_submersibles@psubs.org
href="mailto:personal_submersibles@psubs.org">Personal Submersibles
General Discussion</A> </DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A
title=mailto:personal_submersibles@psubs.org
href="mailto:personal_submersibles@psubs.org">Personal Submersibles
General Discussion</A> </DIV>
<DIV style="FONT: 10pt arial"><B>Sent:</B> Thursday, April 17, 2014
6:58 AM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> Re: [PSUBS-MAILIST]
K3000 spherical shell calculations</DIV>
<DIV><BR></DIV>
<DIV>Uh...no.</DIV>
<DIV><BR></DIV>
<DIV>Do a sphere calc and add it to a cylinder calc.</DIV>
<DIV><BR></DIV>
<DIV>Vance<BR><BR>Sent from my iPhone</DIV>
<DIV><BR>On Apr 16, 2014, at 5:16 PM, Personal Submersibles General
Discussion <<A title=mailto:personal_submersibles@psubs.org
href="mailto:personal_submersibles@psubs.org">personal_submersibles@psubs.org</A>>
wrote:<BR><BR></DIV>
<BLOCKQUOTE type="cite">
<DIV>
<DIV>This may be a dumb question, but is finding the volume of a
cylinder with two hemispherical <SPAN class=Apple-style-span
style="FONT-SIZE: 10px">heads</SPAN></DIV>
<DIV><SPAN class=Apple-style-span style="FONT-SIZE: 10px"><SPAN
class=Apple-style-span style="FONT-FAMILY: Arial"><SPAN
style="COLOR: black; FONT-FAMILY: 'Times New Roman', serif"><I>V =
4.1888 </I></SPAN></SPAN><SPAN class=Apple-style-span
style="FONT-FAMILY: Arial"><I
style="mso-bidi-font-style: normal"><SPAN
style="COLOR: black; FONT-FAMILY: 'Times New Roman', serif">x</SPAN></I></SPAN><SPAN
class=Apple-style-span style="FONT-FAMILY: Arial"><I
style="mso-bidi-font-style: normal"><SPAN
style="COLOR: black; FONT-FAMILY: 'Times New Roman', serif"> r </SPAN></I></SPAN><SPAN
class=Apple-style-span style="FONT-FAMILY: Arial"><I
style="mso-bidi-font-style: normal"><SPAN
style="COLOR: black; FONT-FAMILY: 'Times New Roman', serif">x</SPAN></I></SPAN><SPAN
class=Apple-style-span style="FONT-FAMILY: Arial"><I
style="mso-bidi-font-style: normal"><SPAN
style="COLOR: black; FONT-FAMILY: 'Times New Roman', serif"> r </SPAN></I></SPAN><SPAN
class=Apple-style-span style="FONT-FAMILY: Arial"><I
style="mso-bidi-font-style: normal"><SPAN
style="COLOR: black; FONT-FAMILY: 'Times New Roman', serif">x</SPAN></I></SPAN><SPAN
class=Apple-style-span style="FONT-FAMILY: Arial"><I><SPAN
style="COLOR: black; FONT-FAMILY: 'Times New Roman', serif"> length?</SPAN></I></SPAN></SPAN></DIV>
<DIV><FONT class=Apple-style-span style="FONT-SIZE: 10px"
face="'Times New Roman', serif">Thanks,</FONT></DIV>
<DIV><FONT class=Apple-style-span style="FONT-SIZE: 10px"
face="'Times New Roman', serif">Scott Waters</FONT></DIV>
<DIV><BR></DIV>
<DIV><BR></DIV>
<DIV><BR></DIV>
<DIV><BR></DIV>
<DIV>
<DIV style="FONT-SIZE: 75%; COLOR: #575757">Sent from my U.S.
Cellular© Smartphone</DIV></DIV><BR>Personal Submersibles General
Discussion <<A title=mailto:personal_submersibles@psubs.org
href="mailto:personal_submersibles@psubs.org">personal_submersibles@psubs.org</A>>
wrote:<BR><FONT face=Arial color=#000000 size=2>
<DIV>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'">Hi
Les,<O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'">The
basic formula for the volume of a sphere is <SPAN
style="mso-no-proof: yes"><clip_image002.png></SPAN>.
Don't accidentally plug in the diameter instead of the radius (I've
done that).<SPAN style="mso-spacerun: yes"> </SPAN>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
</SPAN><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'"><EM>V=1.3333
<SPAN class=texhtml>π</SPAN> r<SUP>3</SUP></EM>.<SPAN
style="mso-spacerun: yes"> </SPAN>Since <SPAN
class=texhtml>π</SPAN> = 3.14159 (rounded), you can go ahead
and multiply it by your 1.333333 to get 4.1888.<SPAN
style="mso-spacerun: yes"> </SPAN>Your simplified formula now
reads <I style="mso-bidi-font-style: normal">V = 4.1888
</I></SPAN><I style="mso-bidi-font-style: normal"><SPAN
style="FONT-SIZE: 10pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">x</SPAN></I><I
style="mso-bidi-font-style: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">
r<SUP>3</SUP></SPAN></I><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">
or <I style="mso-bidi-font-style: normal">V = 4.1888 </I></SPAN><I
style="mso-bidi-font-style: normal"><SPAN
style="FONT-SIZE: 10pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">x</SPAN></I><I
style="mso-bidi-font-style: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">
r </SPAN></I><I style="mso-bidi-font-style: normal"><SPAN
style="FONT-SIZE: 10pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">x</SPAN></I><I
style="mso-bidi-font-style: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">
r </SPAN></I><I style="mso-bidi-font-style: normal"><SPAN
style="FONT-SIZE: 10pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">x</SPAN></I><I
style="mso-bidi-font-style: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">
r. </SPAN></I><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'"><SPAN
style="mso-spacerun: yes"> </SPAN>You can use that simplified
formula for calculating the volume of any sphere by plugging in the
r<SUP>3</SUP>.<SPAN style="mso-spacerun: yes"> </SPAN>The <I
style="mso-bidi-font-style: normal">4.1888</I> is a
constant.<O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'"><O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">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.<SPAN style="mso-spacerun: yes"> </SPAN>Having the
simplified formula saves a lot number crunching when you are
calculating different sizes.<SPAN style="mso-spacerun: yes">
</SPAN>If you can set up a spreadsheet containing that formula it
will be even easier.<SPAN style="mso-spacerun: yes">
</SPAN>You can also use that formula to calculate the volume of a
hemispherical tank head on a cylinder by dividing it by
2.<O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'"><O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">To
calculate the volume of a cylinder, first calculate the area of a
circle of that radius and multiply it by the length. <SPAN
style="mso-spacerun: yes"> </SPAN><I
style="mso-bidi-font-style: normal">A = <SPAN
class=texhtml>π</SPAN> r<SUP>2</SUP></I> . <SPAN
style="mso-spacerun: yes"> </SPAN>For your radius of 0.6
meters, A = 1.13 m<SUP>2</SUP> or 4.524 m<SUP>3</SUP> for a 4 meter
long cylinder.<SPAN style="mso-spacerun: yes">
</SPAN></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'"><SPAN
style="mso-spacerun: yes"></SPAN></SPAN> </P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'"><SPAN
style="mso-spacerun: yes"></SPAN>Add a hemispherical tank head on
the other end:<SPAN style="mso-spacerun: yes"> </SPAN>V =
4.1888 </SPAN><SPAN
style="FONT-SIZE: 10pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">x</SPAN><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">
.6<SUP>3</SUP> and you get a volume of<SPAN
style="mso-spacerun: yes"> </SPAN>.905
m<SUP>3</SUP>.<O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'"><O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">Add
the three figures together:<O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">Sphere<SPAN
style="mso-tab-count: 1">
</SPAN>4.189<O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">Cylinder<SPAN
style="mso-tab-count: 1">
</SPAN>4.524<O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><U><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">Head<SPAN
style="mso-tab-count: 2">
</SPAN>0.905<O:P></O:P></SPAN></U></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'"><SPAN
style="mso-tab-count: 2">
</SPAN>9.618 m<SUP>3</SUP> Total volume<O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'"><O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">As
you can see, these figures pretty well match up with Sean’s.<SPAN
style="mso-spacerun: yes"> </SPAN>Your sub would have to weigh
at least 9858 kg (21,688 lb) <I
style="mso-bidi-font-style: normal"><U>in air</U></I> in order to
submerge in sea water.<SPAN style="mso-spacerun: yes">
</SPAN>Adding external ballast tanks will not reduce that
figure.<SPAN style="mso-spacerun: yes"> </SPAN>Adding internal
ballast tanks will reduce it by the weight of the water in those
internal tanks.<O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'"><O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">Don’t
worry about dumb questions.<SPAN style="mso-spacerun: yes">
</SPAN>I’ve had a few.<SPAN style="mso-spacerun: yes">
</SPAN>If anything I’ve written above is inaccurate, someone will
correct it for the benefit of all.<SPAN
style="mso-spacerun: yes"> </SPAN>I wanted to keep it simple
instead of adding too much detail.<SPAN
style="mso-spacerun: yes"> </SPAN>That can be done
later.<O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'"><O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">Best
regards,<O:P></O:P></SPAN></P>
<P class=MsoNormal
style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><SPAN
style="FONT-SIZE: 14pt; COLOR: black; FONT-FAMILY: 'Times New Roman','serif'">Jim
T.</SPAN><SPAN
style="FONT-SIZE: 14pt; FONT-FAMILY: 'Times New Roman','serif'"><O:P></O:P></SPAN></P></DIV>
<DIV> </DIV>
<DIV>
<DIV>In a message dated 4/16/2014 12:58:11 A.M. Central Daylight
Time, <A title=mailto:personal_submersibles@psubs.org
href="mailto:personal_submersibles@psubs.org">personal_submersibles@psubs.org</A>
writes:</DIV>
<BLOCKQUOTE
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<DIV class=moz-cite-prefix>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?<BR><BR>Sean<BR><BR><BR>On 2014-04-15
23:25, Personal Submersibles General Discussion wrote:<BR></DIV>
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<DIV><FONT face=Arial size=2>Hello everybody ,anybody, Les here
, </FONT></DIV>
<DIV><FONT face=Arial size=2>Attatched myself to this email for
convenience (similar subject) been away from psubs for quite
some time wanting to start again.</FONT></DIV>
<DIV><FONT face=Arial size=2>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? </FONT></DIV>
<DIV><FONT face=Arial size=2>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
</FONT></DIV>
<DIV><FONT face=Arial size=2>Can anyone help me please my
reqirements are; </FONT></DIV>
<DIV><FONT face=Arial size=2>A Sphere 2 meters
diameter</FONT></DIV>
<DIV><FONT face=Arial size=2>A Cylinder attached to that 1.2m
diameter x 4meters long</FONT></DIV>
<DIV><FONT face=Arial size=2> ( 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 )</FONT></DIV>
<DIV><FONT face=Arial size=2>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.</FONT></DIV>
<DIV><FONT face=Arial size=2>Sorry to be pain but can
any-one help me </FONT></DIV>
<DIV><FONT face=Arial size=2>Thank you </FONT></DIV>
<DIV><FONT face=Arial size=2>Les</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=Arial size=2>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 </FONT></DIV>
<DIV><FONT face=Arial
size=2> volume hence
the size required for soft tanks for maximum submergance
</FONT></DIV>
<DIV> </DIV>
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