# Tank geometries

## Intro

As mentioned in RocketDesignSpreadsheet#Section_2_Oxidizer_Tank_Structure_Mass, there are some important dimensionless quantities which determine how tanks weight scales with volume:

## (Maxium Circumfrance)/(Volume^(1/3))

The point of the tank with maximum circumference will be near the weakest cross section of the tank with regards to pressure containment (see Pressure_Vessel). Strictly speaking for reasons of ease of computation, we take a cross section, near the weakest section, and account for any possible error by including a factor of safety. For instance, in the case of a cylinder, the plane of the cross section is taken to be parallel to the axis of the cylinder. If this is not the weakest cross section with regards to pressure containment it will be close.

#### Sphere

In the case of a sphere all cross sections, though the center of the sphere have the same circumference.

$C=2*\pi*r$

The volume of a sphere is given by:

$V=(4/3)\pi*r^3$

Therefore:

$\frac{C}{V^(1/3)}=\frac{2*\pi}{((4/3)\pi)^{1/3}} \approx 3.9$

#### Cylinder

As mentioned above the maximum circumference of a cylinder will occur near the cross section, which is parallel to the axis of the cylinder.

If h is the height of the cylinder and r is the radius then, the circumfrance of this cross section, will go around half a sphere twice (at the ends) and along the tank height twice. Therefore:

$C=2h+2*\pi*r$

The volume of this cylinder, will be equal to the cross sectional area multiplied by the hieght plus the area at the ends (two half spheres), which gives.

$V=\pi*r^2*h+4/3*pi*r^3$

Now if the hight is some multiple (n) of the radius we get:

$C=2(n+\pi)r$

$V=\pi*r^2*n*r+4/3*pi*r^3=\pi(n+(4/3))r^3(/math> Therefore: [itex](C/V^(1/3))=\frac{2(n+\pi)}{(\pi(n+(4/3)))^(1/3)}$

## (Maximum Cross Sectional Area)/(Volume^(2/3))

The point of the tank with maximum circumference will be near the weakest cross section of the tank with regards to pressure containment (see Pressure_Vessel). Strictly speaking for reasons of ease of computation, we take a cross section, near the weakest section, and account for any possible error by including a factor of safety. For instance, in the case of a cylinder, the plane of the cross section is taken to be parallel to the axis of the cylinder. If this is not the weakest cross section with regards to pressure containment it will be close.

#### Sphere

The cross sectional area of any cross section though the center of the sphere is the same regardless of the cross section and is given by:

$C=\pi r^2$

The volume of a sphere is:

$V=(4/3)\pi*r^3$

Therefore

$(C/V^(2/3))=\frac{\pi}{((4/3)\pi)^{2/3}}$

#### Cylinder

As mentioned above the maximum cross sectional area of a cylinder will occur near the cross section, which is parallel to the axis of the cylinder. The cross sectional area will consist of two parts, the main part of the cylinder which will have an area of 2*r*h, and the two half spheres on the end will have an area of pi*r^2 This gives:

$A_c=2 \ r \ h+\pi r^2$

and if h=nr then we get:

$A_c=2 \ r \ h+\pi r^2$

$A_c=2 \ r \ n \ r+\pi r^2=(2n+pi)r^2$

And, therefore

$(C/V^(2/3))=\frac{(2n+pi)}{((4/3)\pi)^{2/3)}$

## (Surface Area)/(Volume^(2/3))

The total mass of the tank, is the surface area multiplied by the wall thickness, multiplied by the density of the material.

#### Sphere m

The surface area of a sphere is given by:

$A_s=4 \pi r^2$

Therefore:

$(A_s)/V^(2/3)=\frac{4 \pi}{((4/3)\pi)^{2/3)}$

#### Cylinder

The surface area for the main part of the cylinder will be the circumference (2 pi r) multiplied by the height (h). The ends are two half spheres and will have a surface area of 4 pi r^2. Therefore

$A_s=2 \pi r \ h + 4 \pi r^2$

and if h=n r then:

$A_s=2 \pi r \ n \ r + 4 \pi r^2=\pi(n +r)r^2$

which gives:

$(A_s)/(V^2/3)=\frac{\pi(n +r)}{((4/3)\pi)^{2/3)}$