Rods from God: Part Deux

[Mr.Creak]

Another question: How do you compute volume from mass?

If you know the density then it's straightforward.
Density = mass/ volume.

[KJ_Lesnick]

The problem is that I have missing variables when I did it.  I was basing it on a projectile of 6.1 x 0.3 meter conical projectiles and the drawings often show 12 of them.  So when I did the mathematical calculations I got a volume and from that a mass which I did two calculations WC (tungsten carbide) or just pure W.  Regardless, the problem is when I was doing calculations for different sizes I multiplied by 12, then divided by say 8 or 6.  I got a number and I found myself unable to determine length and diameter from them.

I think I missed a variable or two

[PR19_Kit]

Unless you know the density of the material you can't.............

[pyro-manic]

The problem is that I have missing variables when I did it.  I was basing it on a projectile of 6.1 x 0.3 meter conical projectiles and the drawings often show 12 of them.  So when I did the mathematical calculations I got a volume and from that a mass which I did two calculations WC (tungsten carbide) or just pure W.  Regardless, the problem is when I was doing calculations for different sizes I multiplied by 12, then divided by say 8 or 6.  I got a number and I found myself unable to determine length and diameter from them.

I think I missed a variable or two

What exactly are you trying to calculate?

[KJ_Lesnick]

PR-19

Unless you know the density of the material you can't.............

Well WC is 15.63 times water if I recall, tungsten is 19.3


Pyro-Manic

Well the basic rod would be 6.1 x 0.3 m in which 12 would be carried; the idea would be larger rods that would weigh the same but instead of being 12 small rods (which are actually quite substantial in weight), you'd have 8, 6, 4, 3, or 1.

[pyro-manic]

So the same total weight, but fewer, larger rods? If you have calculated your total mass, you can get the mass of each rod, and therefore the volume. So decide your diameter or your length, and with your given volume you can work out the other dimension.

[KJ_Lesnick]

So the same total weight, but fewer, larger rods?

Yup...

If you have calculated your total mass, you can get the mass of each rod, and therefore the volume. So decide your diameter or your length, and with your given volume you can work out the other dimension.

Understood... I'll put in what I got here.

For the sake of simplicity, I'm going to have the "rods" be "cones" because it's simpler to do the computation.

Volume of one cone

• V = 1/3(B)*h
  
• V = 1/3[(πr²)](h)
  
• V = 1/3[(3.1416)(0.3m)²]*(6.1m)
  
• V = 1/3[(3.1416)(0.09m²)]*(6.1m)
  
• V = 1/3(0.282744 m²)*(6.1m)
  
• V = (0.094248m²)(6.1m)
  
• V = 0.5749128m³
  

Volume of 12 cones

• 12*(0.5749129m³)
  
• 6.8989536m³
  

• Eight cones would have a cone that takes up 0.8623692 cubic meters
  
• Six cones would take up 1.1498256 cubic meters
  
• Four cones would take up 1.7247384 cubic meters
  
• Three cones would take up 2.2296512 cubic meters
  
• One would take up 6.8989536 cubic meters
  

The problem is that I can't seem to work out the base when you consider the cone has a fineness ratio of 20 1/3

[pyro-manic]

http://www.cleavebooks.co.uk/scol/calcone.htm

http://www.calculatorsoup.com/calculators/geometry-solids/cone.php

If your fineness ratio is (roughly) 20:1, then the length is 20 times the radius.

So V = (1/3)πr^2h

V = (1/3)πr^2(20r)

This gives you one variable, rather than two. You can then re-arrange for r, as you know the volume.

[KJ_Lesnick]

So V = 42.5861333....r^3