document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=823904>Question 823904</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need to make a spear head equal to that of Goliath in the Bible. We know it weighed around 15 lbs and was made of iron. Well, the best we can tell. \r<BR>\n" );
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document.write( "So I know that 15 lbs of cast iron would equal 240 oz. \r<BR>\n" );
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document.write( "We know that cast iron weighs 4.166666666666667 oz per cubic inch? (450 lbs per cubic foot = 7,200 oz / 1728 inches per cubic foot = 4.166666666666667)\r<BR>\n" );
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document.write( "If my math is right, we are looking at 57.6 cubic inches of cast iron would be 15 lbs.\r<BR>\n" );
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document.write( "So to get an idea of how big I need to make the spear head, lets say it is almost a cone shape 4 inches at the wide end. How long would it need to be?\r<BR>\n" );
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document.write( "Cone volume = 57.6 cubic inches<BR>\n" );
document.write( "Cone base = 4 inches<BR>\n" );
document.write( "Cone height = ???\r<BR>\n" );
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document.write( "I will need to make a prototype. Use that to disperse water to the 57.6 volume. Then make a mold and cast the iron. I know the cone height will actually be short of what I need but at this point I don't have a clue how long it needs to be. At least this should get me into the ball park. From there I will keep working on it until I end up with the 57.6 or a bit more.\r<BR>\n" );
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document.write( "If you see any error in my math up to this point, please let me know.\r<BR>\n" );
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document.write( "Kind of a fun project you will have to admit. Thanks a bunch. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #495997 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=495997\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Cone volume = 57.6 cubic inches<BR>\n" );
document.write( "Cone base = 4 inches<BR>\n" );
document.write( "Cone height = ??? <BR>\n" );
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document.write( "The volume of a cone: V = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F3 ALIGN=MIDDLE  ALT=\"1%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" >*<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2Ar%5E2h ALIGN=MIDDLE  ALT=\"pi%2Ar%5E2h\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Assuming the diameter of the base of the cone = 4 inches, r=2 inches<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F3 ALIGN=MIDDLE  ALT=\"1%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" >*<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2A2%5E2%2Ah+=+57.6 ALIGN=MIDDLE  ALT=\"pi%2A2%5E2%2Ah+=+57.6\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "multiply both sides by 3<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2A4%2Ah+=+172.8 ALIGN=MIDDLE  ALT=\"pi%2A4%2Ah+=+172.8\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "divide both sides by <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=4%2Api ALIGN=MIDDLE  ALT=\"4%2Api\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "h = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=172.8%2F%284%2Api%29 ALIGN=MIDDLE  ALT=\"172.8%2F%284%2Api%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "h = 13.75 inches<BR>\n" );
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document.write( "A cone with a volume of 57.6, and a diameter of 4 will be 13.75 inches long<BR>\n" );
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