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<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=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 2:44
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>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
style="PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: blue 2px solid"><FONT
style="BACKGROUND-COLOR: transparent" face=Arial color=#000000 size=2>
<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>
<BLOCKQUOTE
cite=mid:mailman.112.1397640356.840.personal_submersibles@psubs.org
type="cite">
<META content="MSHTML 6.00.6000.21264" name=GENERATOR>
<STYLE></STYLE>
<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>
<DIV><BR> </DIV></BLOCKQUOTE><BR><BR><BR>_______________________________________________<BR>Personal_Submersibles
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