<html><body><div style="color:#000; background-color:#fff; font-family:HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, sans-serif;font-size:14pt"><div><span>Les,</span></div><div style="color: rgb(0, 0, 0); font-size: 18.88888931274414px; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;"><span>in my calculations the dry weight of the sphere & the cylinder are</span></div><div style="color: rgb(0, 0, 0); font-size: 18.88888931274414px; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;"><span>1056.3 kg + 1,324.4 kg = 2,380.7 kg.</span></div><div style="color: rgb(0, 0, 0); font-size: 18.88888931274414px; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style:
normal;"><span>The weight of the sphere & cylinder in water is -3378.9 liters + -3490.1 kg.</span></div><div style="color: rgb(0, 0, 0); font-size: 18.88888931274414px; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;"><span>Which means that for these two items alone you are going to have to make your sub weigh</span></div><div style="color: rgb(0, 0, 0); font-size: 18.88888931274414px; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;"><span>3378.9 kg + 3490.1 kg = 6869 kg to get it neutrally buoyant.</span></div><div style="color: rgb(0, 0, 0); font-size: 18.88888931274414px; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;"><span>Alan</span></div><div><br></div> <div
style="font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; font-size: 14pt;"> <div style="font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; font-size: 12pt;"> <div dir="ltr"> <hr size="1"> <font size="2" face="Arial"> <b><span style="font-weight:bold;">From:</span></b> Personal Submersibles General Discussion <personal_submersibles@psubs.org><br> <b><span style="font-weight: bold;">To:</span></b> Personal Submersibles General Discussion <personal_submersibles@psubs.org> <br> <b><span style="font-weight: bold;">Sent:</span></b> Thursday, April 17, 2014 3:02 PM<br> <b><span style="font-weight: bold;">Subject:</span></b> Re: [PSUBS-MAILIST] K3000 spherical shell calculations<br> </font> </div> <div class="y_msg_container"><br><div id="yiv2990392473"><div><div style="color: rgb(0, 0, 0); background-color: rgb(255, 255, 255); font-family: HelveticaNeue,
'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; font-size: 14pt;"><div><span>Hi Les (or Gidday)</span></div><div style="color: rgb(0, 0, 0); font-size: 18.88888931274414px; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;"><span>The 2 meter inside diameter sphere made from 516 / 70 steel needs a crush depth of</span></div><div style="color: rgb(0, 0, 0); font-size: 18.88888931274414px; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;"><span>561 meters to fit G.Ls perameters of the colapse dive pressure / nominal diving pressure </span></div><div style="color: rgb(0, 0, 0); font-size: 18.88888931274414px; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style:
normal;"><span>> 1.87.</span></div><div style="color: rgb(0, 0, 0); font-size: 18.88888931274414px; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;"><span>At 10.8mm thick you get a crush depth of 559.75 M.</span></div><div style="color: rgb(0, 0, 0); font-size: 18.88888931274414px; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;"><span>It will weigh 1056.3 kg & be -3378.9 kg water weight salt. (need a lot of lead)</span><img src="https://s.yimg.com/ok/u/assets/img/emoticons/emo1.gif" alt="*:) happy" style="font-size:14pt;"></div><div style="color: rgb(0, 0, 0); font-size: 14pt; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;">A tube 1.2M ID & 4 meters long of
same material at 11.3mm thick crushes at 561 meters,</div><div style="color: rgb(0, 0, 0); font-size: 14pt; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;">weighs 1,324.4 kg with a -3490.1 kg water weight.</div><div style="color: rgb(0, 0, 0); font-size: 14pt; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;">In my program I am showing no advantage in having stiffeners at this thickness & diameter.</div><div style="color: rgb(0, 0, 0); font-size: 14pt; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;">Someone might want to confirm this but I ran this tube as a 500mm, a 1 meter & 2
meter </div><div style="color: rgb(0, 0, 0); font-size: 14pt; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;">length & got the same crush depth for the diameter & thickness.</div><div style="color: rgb(0, 0, 0); font-size: 14pt; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;">Cheers Alan</div><div style="color: rgb(0, 0, 0); font-size: 14pt; font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; background-color: transparent; font-style: normal;"><br clear="none"></div><div><br clear="none"></div> <div style="font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande', sans-serif; font-size: 14pt;"> <div style="font-family: HelveticaNeue, 'Helvetica Neue', Helvetica, Arial, 'Lucida Grande',
sans-serif; font-size: 12pt;"> <div class="yiv2990392473yqt3593165016" id="yiv2990392473yqt88275"><div dir="ltr"> <hr size="1"> <font size="2" face="Arial"> <b><span style="font-weight:bold;">From:</span></b> Personal Submersibles General Discussion <personal_submersibles@psubs.org><br clear="none"> <b><span style="font-weight:bold;">To:</span></b> Personal Submersibles General Discussion <personal_submersibles@psubs.org> <br clear="none"> <b><span style="font-weight:bold;">Sent:</span></b> Thursday, April 17, 2014 1:43 PM<br clear="none"> <b><span style="font-weight:bold;">Subject:</span></b> Re: [PSUBS-MAILIST] K3000 spherical shell calculations<br clear="none"> </font> </div> <div class="yiv2990392473y_msg_container"><br clear="none"><div id="yiv2990392473">
<div>
<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 <b>unstiffened
cylinder</b> 1.2m x 4 meters long indicated that it need
be<b> 3/4" wall for 314psi at 706fsw</b> (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
= <b>8858 lbs (all soft conv) = 3.95 ton
approx.</b></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-color: rgb(228, 228, 228); font-style: normal; font-variant: normal; font-weight: normal; font-size: 10pt; line-height: normal; font-family: arial;"><b>From:</b>
<a rel="nofollow" shape="rect" title="personal_submersibles@psubs.org" ymailto="mailto:personal_submersibles@psubs.org" target="_blank" href="mailto:personal_submersibles@psubs.org">Personal Submersibles General
Discussion</a> </div>
<div style="font-style: normal; font-variant: normal; font-weight: normal; font-size: 10pt; line-height: normal; font-family: arial;"><b>To:</b> <a rel="nofollow" shape="rect" title="personal_submersibles@psubs.org" ymailto="mailto:personal_submersibles@psubs.org" target="_blank" href="mailto:personal_submersibles@psubs.org">Personal Submersibles General
Discussion</a> </div>
<div style="font-style: normal; font-variant: normal; font-weight: normal; font-size: 10pt; line-height: normal; font-family: arial;"><b>Sent:</b> Thursday, April 17, 2014 6:58
AM</div>
<div style="font-style: normal; font-variant: normal; font-weight: normal; font-size: 10pt; line-height: normal; font-family: arial;"><b>Subject:</b> Re: [PSUBS-MAILIST] K3000
spherical shell calculations</div>
<div><br clear="none"></div>
<div>Uh...no.</div>
<div><br clear="none"></div>
<div>Do a sphere calc and add it to a cylinder calc.</div>
<div><br clear="none"></div>
<div>Vance<br clear="none"><br clear="none">Sent from my iPhone</div>
<div><br clear="none">On Apr 16, 2014, at 5:16 PM, Personal Submersibles General Discussion
<<a rel="nofollow" shape="rect" ymailto="mailto:personal_submersibles@psubs.org" target="_blank" href="mailto:personal_submersibles@psubs.org">personal_submersibles@psubs.org</a>>
wrote:<br clear="none"><br clear="none"></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="yiv2990392473Apple-style-span" style="FONT-SIZE:10px;">heads</span></div>
<div><span class="yiv2990392473Apple-style-span" style="FONT-SIZE:10px;"><span class="yiv2990392473Apple-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="yiv2990392473Apple-style-span" style="font-family: Arial;"><i style=""><span style="color: black; font-family: 'Times New Roman', serif;">x</span></i></span><span class="yiv2990392473Apple-style-span" style="font-family: Arial;"><i style=""><span style="color: black; font-family: 'Times New Roman', serif;"> r </span></i></span><span class="yiv2990392473Apple-style-span" style="font-family: Arial;"><i style=""><span style="color: black; font-family: 'Times New Roman', serif;">x</span></i></span><span class="yiv2990392473Apple-style-span" style="font-family: Arial;"><i style=""><span style="color: black; font-family: 'Times New Roman', serif;"> r </span></i></span><span class="yiv2990392473Apple-style-span" style="font-family: Arial;"><i style=""><span style="color: black; font-family: 'Times New Roman', serif;">x</span></i></span><span class="yiv2990392473Apple-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="yiv2990392473Apple-style-span" style="FONT-SIZE:10px;" face="'Times New Roman', serif">Thanks,</font></div>
<div><font class="yiv2990392473Apple-style-span" style="FONT-SIZE:10px;" face="'Times New Roman', serif">Scott Waters</font></div>
<div><br clear="none"></div>
<div><br clear="none"></div>
<div><br clear="none"></div>
<div><br clear="none"></div>
<div>
<div style="FONT-SIZE:75%;COLOR:#575757;">Sent from my U.S. Cellular©
Smartphone</div></div><br clear="none">Personal Submersibles General Discussion <<a rel="nofollow" shape="rect" ymailto="mailto:personal_submersibles@psubs.org" target="_blank" href="mailto:personal_submersibles@psubs.org">personal_submersibles@psubs.org</a>>
wrote:<br clear="none"><font id="yiv2990392473role_document" face="Arial" color="#000000" size="2">
</font><div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;">Hi
Les,</span></div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;">The
basic formula for the volume of a sphere is <span style=""><clip_image002.png></span>. Don't
accidentally plug in the diameter instead of the radius (I've done
that).<span style=""> </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;"><i>V=1.3333
<span class="yiv2990392473texhtml">π</span> r<sup>3</sup></i>.<span style=""> </span>Since <span class="yiv2990392473texhtml">π</span> = 3.14159 (rounded), you can go ahead and
multiply it by your 1.333333 to get 4.1888.<span style=""> </span>Your simplified formula now reads <i style="">V = 4.1888 </i></span><i style=""><span style="font-size: 10pt; color: black; font-family: 'Times New Roman', serif;">x</span></i><i style=""><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="">V = 4.1888 </i></span><i style=""><span style="font-size: 10pt; color: black; font-family: 'Times New Roman', serif;">x</span></i><i style=""><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;">
r </span></i><i style=""><span style="font-size: 10pt; color: black; font-family: 'Times New Roman', serif;">x</span></i><i style=""><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;">
r </span></i><i style=""><span style="font-size: 10pt; color: black; font-family: 'Times New Roman', serif;">x</span></i><i style=""><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=""> </span>You can use that simplified formula
for calculating the volume of any sphere by plugging in the
r<sup>3</sup>.<span style=""> </span>The <i style="">4.1888</i> is a
constant.</span></div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;"></span></div>
<div class="yiv2990392473MsoNormal" 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=""> </span>Having the simplified formula saves
a lot number crunching when you are calculating different sizes.<span style=""> </span>If you can set up a spreadsheet
containing that formula it will be even easier.<span style=""> </span>You can also use that formula to
calculate the volume of a hemispherical tank head on a cylinder by dividing
it by 2.</span></div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;"></span></div>
<div class="yiv2990392473MsoNormal" 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=""> </span><i style="">A = <span class="yiv2990392473texhtml">π</span> r<sup>2</sup></i> . <span style=""> </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=""> </span></span></div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;"><span style=""></span></span> </div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;"><span style=""></span>Add a hemispherical tank head on the other
end:<span style=""> </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="">
</span>.905 m<sup>3</sup>.</span></div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;"></span></div>
<div class="yiv2990392473MsoNormal" 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:</span></div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;">Sphere<span style="">
</span>4.189</span></div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;">Cylinder<span style="">
</span>4.524</span></div>
<div class="yiv2990392473MsoNormal" 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="">
</span>0.905</span></u></div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;"><span style="">
</span>9.618 m<sup>3</sup> Total volume</span></div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;"></span></div>
<div class="yiv2990392473MsoNormal" 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=""> </span>Your sub would have to weigh at
least 9858 kg (21,688 lb) <i style=""><u>in
air</u></i> in order to submerge in sea water.<span style=""> </span>Adding external ballast tanks will
not reduce that figure.<span style=""> </span>Adding
internal ballast tanks will reduce it by the weight of the water in those
internal tanks.</span></div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;"></span></div>
<div class="yiv2990392473MsoNormal" 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="">
</span>I’ve had a few.<span style=""> </span>If
anything I’ve written above is inaccurate, someone will correct it for the
benefit of all.<span style=""> </span>I wanted to
keep it simple instead of adding too much detail.<span style=""> </span>That can be done
later.</span></div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;"></span></div>
<div class="yiv2990392473MsoNormal" style="MARGIN:0in 0in 0pt;LINE-HEIGHT:normal;"><span style="font-size: 14pt; color: black; font-family: 'Times New Roman', serif;">Best
regards,</span></div>
<div class="yiv2990392473MsoNormal" 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;"></span></div></div>
<div> </div>
<div>
<div>In a message dated 4/16/2014 12:58:11 A.M. Central Daylight Time, <a rel="nofollow" shape="rect" ymailto="mailto:personal_submersibles@psubs.org" target="_blank" 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">
</font><div class="yiv2990392473moz-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 clear="none"><br clear="none">Sean<br clear="none"><br clear="none"><br clear="none">On 2014-04-15 23:25, Personal
Submersibles General Discussion wrote:<br clear="none"></div>
<blockquote type="cite">
<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>
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