I'm planning some fiction involving a very long spaceship landing. Initially I wanted the ship to be the same length as the NEC railroad, at 750 km. However looking at the mathematics of the situation, it's going to end up with the ends way off of the ground and probably snap.

$$\text{overhang} = \sqrt{r^2+(x/2)^2} - r.$$

Where $r$ is the diameter of the planet (roughly earth-sized at ~6371 km), and $x$ is the length of the ship.

For a rough estimate of how long I can make the ship, I'm thinking of modelling it as a solid piece of steel (in terms of density and strength) with a length to width ratio of 1:3. And a length to thickness ratio of 1:30? I don't know how sensible that thickness is.

I'm also assuming it's heavy enough that it wraps the curvature of the planet. At some length I image that instead of bending elastically it plastically deforms and the ship begins to be damaged by fatigue. Further past that (due to fatigue) I imagine that it just snaps like a twig.

I've been poking around looking for formulas to work out what details I need to know to figure out a 'sensible' size for the spaceship. I can't figure out which one is more appropriate though, or how much the thickness affects matters, or even if modelling it as solid steel is going to be close to accurate.

Formula's I've looked at:

Euler-Bernoulli beam theory, static beam

$$\frac{\mathrm d^2}{\mathrm dx^2}\left(EI\frac{\mathrm d^2w}{\mathrm dx^2}\right) = q$$

Where $q$ is the 'distributed load, in other words a force per length', but I don't know how to figure that out. Nor do I know $w$ the 'deflection of the beam in the $z$ direction at some position $x$'. Is it going to be the exact fit of the ground beneath it, or will it 'flop' and still leave a gap?

Euler's critical load

$$P_\mathrm{cr} = \frac{\pi^2EI}{(KL)^2}$$

My issue here is that this is about compression, which is only half the story with the spaceship. Part of it is also under tension. Based on the diagrams it's not even the right sort of compression. But it does describe a moment of failure.

Johnson's parabolic formula - Critical slenderness

$$\left(\frac{l}{k}\right)_\mathrm{cr} = \sqrt{\frac{2\pi^2E}{\sigma_y}} $$

I don't fully grasp how this applies, but it seems relevant based on how slender the spaceship might be? Naively it looks like a rearrangement of the previous Euler's formula, but that doesn't seem right based on the Wikipedia graphs.

More equations


There're more equations here, especially to do with plates which seem even more complicated. The plate bending equations are detailed in that link, but they are going completely over my head. But maybe there is an equation here that models the spaceship better as a plate than a bar?


Similar to the plate bending equations, they are going over my head and seem to be overkill for a simple model. But again, it could be a better model?

  • $\begingroup$ Since you are talking about your ship snapping I'm assuming it's only supported in the middle, otherwise just add thrusters along the length and make it flexible enough to rest along the earths curvature. For a constant cross-section the length is normally irrelevant, what matters is the ratio between length and the second moment of area for your cross-section. In your case a longer ship is actually beneficial as the ends (assuming your ship is stiff enough) will be further from the earths center. $\endgroup$ Jun 11 '21 at 9:31
  • 1
    $\begingroup$ pt. 2 The ends would then be subjected to less gravity, and that gravity would be at a more beneficial angle (not perpendicular). I don't think a solid block of steel is a useful approximation, try a hollow square profile and add a distributed load representing cargo. $\endgroup$ Jun 11 '21 at 9:32
  • $\begingroup$ @Jollerprutt do you know how I would go about simply calculating that...? I looked into Second Moment of Area, and I got a little lost. $\endgroup$
    – Pureferret
    Jun 11 '21 at 10:34
  • $\begingroup$ In fact I found amesweb.info/section/second-moment-of-area-calculator.aspx into which I can plug numbers, but I now don't know what to do with them $\endgroup$
    – Pureferret
    Jun 11 '21 at 10:43
  • $\begingroup$ IMHO you have confused the definition of ratios. Length to thickness ratio 1:30 would mean (to my experience at least) that its very very thick indeed. $\endgroup$
    – NMech
    Jun 18 '21 at 5:51

From what I understand your goal is to design a spaceship with a known length made from common earth materials that will not snap in half (or more pieces) when it has landed on earth.

enter image description here

Figure 1: Spaceship: yellow, Earth:blue

And your question is:

how does thickness and width factor into the equation.

The answer to that question, is that a thin spaceship would have a better chance of flexing without breaking in half.


In order to understand this intuitively, I will give you the following example:

Take an cucumber and slice two slices.

  • A) a thick one (at least a cm or more)
  • B) a wafer thin one (less than a mm)

Both of them are from the same material, and you can cut them to the same length and width, but you can flex B almost infinitely without breaking, while A has a limited range.

Hopefully this will be sufficient to demonstrate the principle, before proceeding to the math. (If you are not convinced about this, the math will only confuse you).

First of all some simplification, because of the symmetry you can consider just half of that problem (the only catch is that the left side is fixed) i.e.

enter image description here

Figure 2: Symmetry and simplification


  • $q$ is the distributed load defined as $N/m$

distributed load

To calculate q you need to know the cross-section geometry at each point. To simplify things here lets assume that the spaceship is a solid rectangular bar (probably hollow is better but its easy to do it yourself).

  • L: length
  • B: width (breadth)
  • H: Height (thickness).

In that case the

  • volume of the spaceship would be V= LBH
  • mass of the spaceship would be $m =\rho*V= \rho* L*B*H$
  • weight of the spaceship would be $W =g\cdot\rho\cdot V= g\cdot\rho\cdot L\cdot B\cdot H$

The Q would be: $$q = \frac{W}{L} =g\cdot\rho\cdot B\cdot H$$

Maximum Bending Moment

The easiest approach then is to find the equations of the beam for a cantilever beam with a uniform load. From there you will find that the maximum bending moment will be at the support and its

$$M = \frac{qL^2}{2}$$

Bending stress (dominant quantity)

What will dominate whether the beam will break is if the normal bending stress exceeds the allowable tensile stress (there are also shear stress, but the have a smaller effect, so I won't factor them in for this simplistic calculation).

The bending stress is given by:

$$\sigma = \frac{M}{I}\cdot\frac{H}{2}$$


  • M is the Bending moment (already calculated it)
  • H is the height (thickness)
  • I is the second moment of area for the cross-section investigated. For a solid rectangular beam with breadth B and Height H, it is:

$$I = \frac{B \cdot H^3}{12}$$

So the equation for stress becomes (for the assumptions)

$$\sigma = \frac{M}{\frac{B \cdot H^3}{12}}\cdot\frac{H}{2}$$

$$\sigma = \frac{3 q\cdot L^2}{B \cdot H^2}$$

$$\sigma = \frac{3 (g\cdot\rho\cdot B\cdot H)\cdot L^2}{B \cdot H^2}$$

$$\sigma = \frac{3 g\cdot\rho \cdot L^2}{H}$$ Therefore, for a given allowable stress $\sigma_{all}$ the maximum thickness to avoid failure is:

$$H = \frac{3 g\cdot\rho \cdot L^2}{\sigma_{all}}$$

Now that you have the maximum thickness from the material you can estimate the actual value of I.

And then you can use the beam bending equation that defines the relationship between curvature $k$ (which is $\frac{1}{R_{earth}$), with the bending Moment M, the material property E, and the second moment of area.

$$k =\frac{1}{R_{earth}} =? \frac{M}{EI}$$

The relationship between the left and right hand side will decide whether its feasible.

  • 1
    $\begingroup$ This is the best kind of answer you get on stackexchange websites. Enough explanation, hand holding, and calculation to get me to the end without doing all of the work for me and leaving me wondering how we got here. $\endgroup$
    – Pureferret
    Jun 18 '21 at 8:58

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