# Background

A McKendree cylinder is a rotating cylindrical space habitat comparable to the more well known O'Neill model. It was proposed by NASA engineer Thomas McKendree in 2000 as an update of O'Neill's, using carbon nanotubes instead of steel and aluminum to allow for much larger structures – up to 10,000km long/1,000km radius, compared to O'Neill's 32km length/8km radius.

A single McKendree cylinder therefore has about 66 million square kilometres of habitable space along the interior surface and potential for even more within the hull itself and interior structures.

# Problem

Internal stresses on the walls of the McKendree Cylinder due to massive objects inside the habitat.

I found in the paper written by Thomas McKendree some valuable information.

The maximum radius of such an O'Neill style colony is limited by the hoop stress of the spinning structure, and the tensile strength to density ratio of the material. The formula is

$$R < \frac{HoopStress}{gG}$$

Where R is the radius, g is the acceleration of pseudo-gravity at the rim, and G is the density. [Molecular nanotechnology (MNT)] offers a 5 x 10$$^{10}$$ Pa tensile strength. Using the design rule of 50% safety factors for O'Neill style colonies, a 3.3 x 10$$^{10}$$ Pa design tensile strength is reasonable. The associated material density is 3.51 x 10$$^3$$ kg/m$$^3$$. One goal of the architecture is for g to equal 9.8 m/s$$^2$$. This all gives a possible space station radius of 9.6 x 10$$^5$$ m, or nearly 1000 km. For comparison, the corresponding feasible radius for titanium is 14 km, and even at its ultimate tensile strength with no safety factor, the titanium limit would be 23 km.

At the 9.6 x 10$$^5$$ m radius, the entire available strength (at the safety factor) of the MNT-based material is being used to prevent the rotating structure from bursting, and there is no strength left over to hold the space station's contents, including an atmosphere. To do so, a lower radius must be set.

In the section immediately following, McKendree arrives at a radius of 461 km when atmosphere and fixtures are accounted for:

One can directly solve for the structure radius where the shell is 5000 kg/m$$^2$$. Using MNT materials, the structure will be 461 km in radius. For comparison, the equivalent number for titanium is 6.6 km, or for a titanium shell is at its ultimate tensile strength with no safety factor, 11 km.

You can now see that adding 5000 kg/m$$^2$$ made the radius drop from 1000 km to 461 km, which is something I can't understand since the stress of the 5000 kg/m$$^2$$ is 50 kPa which is so much less compared to the strength of habitat material (5 x 10$$^{10}$$ Pa), so why did it have such a big impact on the radius?

Please note that 5000 kg/m$$^2$$ used as either a shield, interior material or atmosphere is equal to two large stones of 1 cubic meter above each other, which is not something that high (or massive) compared to our modern structures.

# Partial Solution

I tried to calculate how massive an object can be inside the habitat at its maximum radius (1000 km) as best as I could disregarding the above calculations.

I used 3.3 x 10$$^{10}$$ Pa design tensile strength out of the available 5 x 10$$^{10}$$ Pa tensile strength as McKendree did for safety factors. I then proceeded to calculate how much mass needs to be added inside the habitat to the rotational stress made by the radius (3.3 x 10$$^{10}$$ Pa) to exceed the maximum strength of the material (5 x 10$$^{10}$$). This came out as 1.7 x 10$$^{10}$$ Pa or 1.7 x 10$$^{9}$$ kg/m$$^2$$, which is a far far larger number than the quoted 5000 kg/m$$^2$$.

# Questions

1- Apparently, McKendree's calculations are so much different from mine (Mckendree's calculations do not allow massive objects to exit inside while mine allow this much much better), so why is that?

2- Would increasing the thickness of the habitat walls allow more massive objects to be built inside? And wouldn't the additional mass of the wall count as stress as well?

## 1 Answer

Q1:

In the case of cylinders, pressures acting radially do not one-to one correspond to hoop stresses. Basically, you have to counteract a radial force with a force almost perpendicular to it (the cohesion force between adjecent wall elements). So there is a "penalty", which grows proportionally with the cylinder radius, since the bigger the radius is, the flatter it is locally.

More rigorous treatment is possible for example using the principle of virtual work, and the result is this:

$$\sigma = \frac{pR}{t}$$

Where $$\sigma$$ is the hoop stress, $$p$$ is internal pressure, $$R$$ is radius, and $$t$$ the tickness of the wall. (this works under the $$t$$ << $$R$$ assumption (thin walled cylinder))

Q2:

A thicker (but still thin) cylinder would help to balance greater internal pressure (due to buildings or atmosphere). But you can not increase the maximal unloaded radius, that is given by the equation you referenced, since the mass of the new material would exactly cancel out your gains.