<html><head><meta http-equiv="Content-Type" content="text/html; charset=UTF-8"></head><body><div>Ok. I figured as much. </div><div>Thanks,</div><div>Scott</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 <personal_submersibles@psubs.org> wrote:<br><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 href="mailto:personal_submersibles@psubs.org">personal_submersibles@psubs.org</a>> wrote:<br><br></div><blockquote type="cite"><div><meta http-equiv="Content-Type" content="text/html; charset=UTF-8"><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="font-family: 'Times New Roman', serif; color: black; "><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="font-family: 'Times New Roman', serif; color: black; ">x</span></i></span><span class="Apple-style-span" style="font-family: Arial; "><i style="mso-bidi-font-style: normal"><span style="font-family: 'Times New Roman', serif; color: black; "> r </span></i></span><span class="Apple-style-span" style="font-family: Arial; "><i style="mso-bidi-font-style: normal"><span style="font-family: 'Times New Roman', serif; color: black; ">x</span></i></span><span class="Apple-style-span" style="font-family: Arial; "><i style="mso-bidi-font-style: normal"><span style="font-family: 'Times New Roman', serif; color: black; "> r </span></i></span><span class="Apple-style-span" style="font-family: Arial; "><i style="mso-bidi-font-style: normal"><span style="font-family: 'Times New Roman', serif; color: black; ">x</span></i></span><span class="Apple-style-span" style="font-family: Arial; "><i><span style="font-family: 'Times New Roman', serif; color: black; "> length?</span></i></span></span></div><div><font class="Apple-style-span" face="'Times New Roman', serif" style="font-size: 10px;">Thanks,</font></div><div><font class="Apple-style-span" face="'Times New Roman', serif" style="font-size: 10px;">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 href="mailto:personal_submersibles@psubs.org">personal_submersibles@psubs.org</a>> wrote:<br><font id="role_document" color="#000000" size="2" face="Arial">
<div>
<p class="MsoNormal" style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black; 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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black; 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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black"><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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">x</span></i><i style="mso-bidi-font-style: normal"><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">
r<sup>3</sup></span></i><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">
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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">x</span></i><i style="mso-bidi-font-style: normal"><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black"> r
</span></i><i style="mso-bidi-font-style: normal"><span style="FONT-SIZE: 10pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">x</span></i><i style="mso-bidi-font-style: normal"><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black"> r
</span></i><i style="mso-bidi-font-style: normal"><span style="FONT-SIZE: 10pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">x</span></i><i style="mso-bidi-font-style: normal"><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">
r. </span></i><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black"><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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black"><o:p> </o:p></span></p>
<p class="MsoNormal" style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black"><o:p> </o:p></span></p>
<p class="MsoNormal" style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black"><span style="mso-spacerun: yes"></span></span> </p>
<p class="MsoNormal" style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black"><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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">x</span><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">
.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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black"><o:p> </o:p></span></p>
<p class="MsoNormal" style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black"><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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black"><o:p> </o:p></span></p>
<p class="MsoNormal" style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black"><o:p> </o:p></span></p>
<p class="MsoNormal" style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">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; FONT-FAMILY: "Times New Roman","serif"; COLOR: black"><o:p> </o:p></span></p>
<p class="MsoNormal" style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">Best
regards,<o:p></o:p></span></p>
<p class="MsoNormal" style="MARGIN: 0in 0in 0pt; LINE-HEIGHT: normal"><span style="FONT-SIZE: 14pt; FONT-FAMILY: "Times New Roman","serif"; COLOR: black">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 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" color="#000000" size="2" face="Arial">
<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 name="GENERATOR" content="MSHTML 6.00.6000.21264">
<style></style>
<div><font size="2" face="Arial">Hello everybody ,anybody, Les here ,
</font></div>
<div><font size="2" face="Arial">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 size="2" face="Arial">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 size="2" face="Arial">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 size="2" face="Arial">Can anyone help me please my reqirements are;
</font></div>
<div><font size="2" face="Arial">A Sphere 2 meters diameter</font></div>
<div><font size="2" face="Arial">A Cylinder attached to that 1.2m diameter x
4meters long</font></div>
<div><font size="2" face="Arial"> ( 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 size="2" face="Arial">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 size="2" face="Arial">Sorry to be pain but can any-one help me
</font></div>
<div><font size="2" face="Arial">Thank you </font></div>
<div><font size="2" face="Arial">Les</font></div>
<div> </div>
<div><font size="2" face="Arial">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 size="2" face="Arial">
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
mailing
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