1 Introduction

In this section we want to know whether or not Rama can simulate the gravity that exists at the surface of the Earth, about $9.8\,m.s^{-2}$. In theory, by spinning around its axis at a sufficient angular speed, Rama can create an "artificial gravity" at its ground level that is as large as we want. However, this rotation (and the pressure from the Rama atmosphere) creates stress forces inside the shell, which could potentially destroy it. In this section we want to compute these forces in order to decide whether Rama could exist, at least in theory, or if it is way too big to withstand these stress forces.

Computing the exact stress forces inside the Rama shell would require to know its exact shape, it's internal structure, the materials used to build it, etc. However, in order to know whether a material could withstand these forces or not, we don't need a very precise value: an order of magnitude is sufficient. Therefore, in the following, we use two simplified models and compare their results in order to answer the question.

2 Basic model

We assume here that, in addition to being infinitely long and invariant by translation along its axis (see the Introduction), the shell is also infinitely thin (more precisely that the width $\Delta r\eqdef r_e-r_s$ of the shell is much smaller than its internal radius $r_s$ - i.e. $\Delta r/r_s\ll 1$).

We can now consider a small $\diff z,\diff\theta$ element of the shell (in cylindrical coordinates, with $z$ measured along the axis). The forces that apply on this element are (see Fig. 1):

the centrifugal force $\rho (\diff z\,\Delta r\,r_s \diff\theta) r_s\omega^2$, where $\rho$ is the volumetric mass of the shell and $\omega$ its angular speed,
the force exerted by the atmosphere $p_s \diff z\,r_s \diff\theta$, where $p_s$ is the atmospheric pressure at the ground level,
the tangential stress forces $\sigma_{\theta} \diff z\,\Delta r$ exerted on both ends of the $\diff\theta$ element.

Figure 1: The forces exerted on a small element of an infinitely long and thin cylinder.

The sum of these forces must be 0. This is the case by symmetry for the tangential forces. For the radial forces, we must first compute the radial component of the stress forces, which is simply $2 \sigma_{\theta} \diff z\,\Delta r \sin(\diff\theta/2) = \sigma_{\theta} \diff z\,\Delta r\,\diff\theta$. The equilibrium equation then gives: \begin{equation} \sigma_{\theta}=\rho r_s^2\omega^2+p_s\frac{r_s}{\Delta r}\label{eq:basic} \end{equation} Assuming that we have $r_s=8000\,m$, $\Delta r=1000\,m$, $\omega=2\pi/180\,s^{-1}$ (yielding an artificial gravity $r_s\omega^2=9.74\,m.s^{-2}$), $p_s=101325 Pa$ and a shell made of steel with a volumetric mass $\rho=8000\,kg.m^{-3}$, we get $\sigma_{\theta}=624.6\,MPa$. This is higher than the yield strength of ordinary steel, which is about $450\,MPa$ [1]. This is not good, because it means that the shell would deform in an inelastic, i.e. irreversible way, or would even break. However, this is lower than the yield strength of high strength steel or steel with prestressing strands (up to $1650\,MPa$), which have about the same volumetric mass.

Therefore, according to this very basic model, the Rama shell can withstand the stress forces created by its rotation, provided the appropriate materials are used (this is even more true if we consider a shell with a spongious or trabecular steel structure, instead of plain steel: the volumetric mass $\rho$, and thus the stress force $\sigma_{\theta}$, would then be much smaller, but the yield strength would probably remain about the same). But this model also shows that much bigger spining cylinders are not possible: if we impose a constant artificial gravity and a constant pressure at the ground level, i.e. $r_s\omega^2=cst$ and $p_s=cst$, then $\sigma_{\theta}$ increases linearly with $r_s$ and will thus eventually exceed the yield strength of any material.

3 Less basic model

Due to the simplifications made in the above model, the results otained with it may be far from the exact ones. In order to test this we want to compare these results with the ones obtained with a less simplified model: if the results are comparable, there is a good chance that they are not far from the exact ones. Therefore, we now consider the following models: an infinitely long but not infinitely thin cylinder, and a plain disc (to model the extremities), but not connected to each other. Both models use the same equations, but with different boundary conditions. Thus we first derive the common equations, and then solve them for the two sets of boundary conditions. The following is mostly based on [2].

3.1 Differential equation

Consider a small volume $\diff z,\diff r,\diff\theta$ in a spinning cylinder or disc. Due to the centrifugal force, this element will move away from its rest position $r$ to a new position $r+u(r)$. Its rest width $\diff r$ will thus become $r+\diff r+u(r+\diff r)-r-u(r)=(1+u'(r))\diff r$, and its rest length $r\diff\theta$ will become $(r+u(r))\diff\theta=(1+u(r)/r)r\diff\theta$. We thus get the radial and tangential strain for this element: \begin{align} \epsilon_r(r)&=u'(r)\\ \epsilon_\theta(r)&=\frac{u(r)}{r} \end{align}

Consider now the radial components of the forces that apply to this element (see Fig. 2). We have:

the centrifugal force $\rho (\diff z\,\diff r\,r \diff\theta) r\omega^2$,
the radial stress on the inner side $-\sigma_r(r)\,\diff z\,r\diff\theta$,
the radial stress on the outer side $(\sigma_r(r)+\diff\sigma_r(r))\,\diff z\,(r+\diff r)\diff\theta$,
the tangential stress $-\sigma_\theta\,\diff z\,\diff r\,\diff\theta$.

Figure 2: The forces exerted on a small element of a disc or thick cylinder. The sum of these forces must be 0, which gives: \begin{equation} \sigma_r(r)+r\sigma'_r(r)-\sigma_\theta(r)=-\rho r^2\omega^2 \end{equation}

We can finally assume that the material deforms linearly and use its Young's modulus $Y$ and its Poisson's ratio $\nu$ to relate the stress $\sigma_r,\sigma_\theta$ to the strain $\epsilon_r,\epsilon_\theta$ with [3]: \begin{align} \sigma_r(r)&=Y\frac{\epsilon_r(r)+\nu\epsilon_\theta(r)}{1-\nu^2}\\ \sigma_\theta(r)&=Y\frac{\epsilon_\theta(r)+\nu\epsilon_r(r)}{1-\nu^2} \end{align}

Combining the above equations gives a single differential equation for the displacement $u(r)$: \begin{equation} u''(r)-\frac{u(r)}{r}+u'(r)=-\frac{1-\nu^2}{Y}\rho r^2\omega^2 \end{equation} whose general solution is: \begin{equation} u(r)=-\frac{1-\nu^2}{8Y}\rho r^3\omega^2+c_1 r+\frac{c_2}{r}\label{eq:solution} \end{equation}

We can now substitute this general solution in the above equations to compute the quantities we are interested in, i.e. the stress forces: \begin{align} \sigma_r(r)&=-\frac{3+\nu}{8}\rho r^2\omega^2+\frac{Y}{1-\nu^2}\left(c_1(1+\nu)-\frac{c_2}{r^2}(1-\nu)\right)\label{eq:sigmar}\\ \sigma_\theta(r)&=-\frac{1+3\nu}{8}\rho r^2\omega^2+\frac{Y}{1-\nu^2}\left(c_1(1+\nu)+\frac{c_2}{r^2}(1-\nu)\right)\label{eq:sigmatheta} \end{align}

3.2 Plain disc

Consider now a solid rotating disc of radius $r_e=9\,km$, modeling the extremities of Rama. The boundary conditions for this case are 1) that no hole must form at the center and 2) that the radial stress at $r_e$ must be 0 (otherwise the disc would expand). This means $u(0)=0$ and $\sigma_r(r_e)=0$. Using Eq. \eqref{eq:solution} the first condition implies $c_2=0$. We can then use Eq. \eqref{eq:sigmar} to get the value of $c_1$, which in turn gives the maximum value of $\sigma_r$ and $\sigma_\theta$, reached at the center of the disc: \begin{equation} \sigma_r(0)=\sigma_\theta(0)=\frac{3+\nu}{8}\rho r_e^2\omega^2 \end{equation} We can also compute the tangential stress at the boundary of the disc, and we get: \begin{equation} \sigma_\theta(r_0)=\frac{1-\nu}{4}\rho r_e^2\omega^2 \end{equation}

This shows that the maximum stress is of the same order of magnitude as the result obtained with our first model in Eq. \eqref{eq:basic} for a infinitely thin and long cylinder. In fact it is about 2 times smaller (for steel $\nu=0.3$). This means that the plain discs at the north and south pole of Rama can also withstand the stress forces caused by their rotation, even if they were made of ordinary steel, and even if there were no hole in their structure to reduce their density.

We can also compute the deformation of the disc due to the rotation, $u(r_e)$. Substituting the value of $c_1$ obtained from the boundary conditions in Eq. \eqref{eq:solution}, we get \begin{equation} u(r_e)=\frac{1-\nu}{4}\frac{\rho r_e^3\omega^2}{Y} \end{equation} Assuming that $Y=200\,GPa$ (the Young's modulus for steel), we get $u(r_e)=6.2\,m$, which is a small deformation compared to the radius $r_e=9000\,m$.

3.2 Shell

Consider now an infinitely long cylinder with an internal surface of radius $r_s=8\,km$ and an external radius $r_e=9\,km$. The boundary conditions for this case are that the radial stress at $r_s$ is equal to the atmospheric pressure $p_s$ and that the radial stress at $r_e$ is 0. Using Eq. \eqref{eq:sigmar} this gives a linear system of 2 equations and 2 unknowns which give the values of $c_1$ and $c_2$. Substituting these values in Eq. \eqref{eq:sigmatheta} gives the radial and tangential stress at any $r$. In fact we can neglect the atmospheric pressure here ($p_s$ is 6000 times smaller than $\rho r_s^2\omega^2$ for Rama), which gives: \begin{align} \sigma_r(r)&\approx\frac{3+\nu}{8}\rho \omega^2\left[r_s^2+r_e^2-r^2-\frac{r_s^2r_e^2}{r^2}\right]\\ \sigma_\theta(r)&\approx\frac{3+\nu}{8}\rho \omega^2\left[r_s^2+r_e^2+\frac{r_s^2r_e^2}{r^2}\right]-\frac{1+3\nu}{8}\rho r^2\omega^2 \end{align} From this we find that $\sigma_r$ reaches its maximum value at $r^2=r_sr_e$, while $\sigma_\theta$ reaches its maximum at $r=r_s$. The corresponding maximum values are: \begin{align} \sigma_r^{max}&=\frac{3+\nu}{8}\rho(r_e-r_s)^2\omega^2\\ \sigma_\theta^{max}&=\frac{\rho\omega^2}{4}\left[(1-\nu)r_s^2+(3+\nu)r_e^2\right] \end{align}

Here again we find that the maximum stress is of the same order of magnitude as the result obtained with our first model in Eq. \eqref{eq:basic} for a infinitely thin and long cylinder. In fact the tangential stress has the same order of magnitude ($739\,MPa$), and the maximum radial stress is smaller ($16\,MPa$).

We can also compute the deformation of the shell due to the rotation, $u(r_s)$. Substituting the values of $c_1$ and $c_2$ obtained from the boundary conditions in Eq. \eqref{eq:solution}, we get \begin{equation} u(r_s)=\frac{\rho r_s\omega^2}{4Y}\left((1-\nu)r_s^2+(3+\nu)r_e^2\right) \end{equation} With $Y=200\,GPa$ as above we get $u(r_e)=25.7\,m$, which is also small deformation compared to the radius $r_s=8000\,m$ (0.3%), but more than 4 times larger than the one for the plain disc ($6.2\,m$).

4 Conclusion

Both models give about the same result, which is that Rama can withstand the stress forces caused by its rotation. According to the simplified model this requires high strength steel or steel with prestressing strands. According to the less simplified model, even ordinary steel would suffice, even if there were no hole in the shell's structure to reduce its density. However, both models also show that much bigger spining cylinders are not possible.

As a concluding remark, note that the mass of Rama can be estimated with $\rho\pi(r_e^2-r_s^2)l_s$. With $\rho=8000\,kg.m^{-3}$ this gives 17090 giga tons! Even if the shell was mostly hollow, with $\rho=8\,kg.m^{-3}$, we would still get 17 giga tons, i.e. 16 times the world production of steel of year 2004 [4]!