Bernd
11/17/2017 (Fri) 20:57:41
[Preview]
No.11981
del

So. How do we calculate structural tension in the ring? We need to understand what generates this tension.

So, we have a centrifugal effect in the ring. There needs to be some structural integrity in the ring that holds it together, so that it doesn't fall apart. To do that we need to employ differential calculus, but to make it clearer what's going on I'll use finite sections of ring and just linearise everything (which gives the same result).

So, if you take a section of the ring of angle Δφ, and balance out the forces, we have three forces: the structural tension on both sides, and the centrifugal effect that we need to balance out. The section has mass of ΔM = Δφ * μ, where by μ I have introduced angular mass density, which has units of mass per angle (in radians, or in degrees, doesn't matter, it just needs to be in same units as Δφ). By F = M * a and plugging everything in, we can get the equation for centripetal force F_c = Δφ * μ * g₀ (standard gravitational acceleration).

On the other hand, the tension generates, from vector-summing tension force F_t on both sides, a counteracting force towards the centre. That force can be calculated by using chord function, crd(θ) = 2 sin (θ/2). I'm providing a quick sketch of how this summing is done; but the end result is F_c = crd(Δφ) * F_t.

Now we need to linearise the expressions which will get us to the differential equations. As Δφ goes to zero, we can expand it to get that crd(dφ) is simply dφ (just like sine) — provided that we use radians as units of angle. So, crd(dφ) * F_t = dφ * F_t, which is in turn equal to dφ * μ * g₀ from previous equation. We can just divide by dφ on both sides (we sent dφ to zero, but it's just *almost* zero, so we can still divide by it). This gives us final equation for tension force: F_t = μ * g₀. Here we will need to plug in the value for angular mass density μ, or rather since mass of whole ring is M = 2π * μ (we're using radians), we can also write this in terms of mass of the ring F_t = M * g₀.

We'd like to compare this with tensile strength. We can write mass as product of ordinary density times volume, and volume as cross-section times circumference, where circumference is of course 2πR; M = ρ * S * 2πR. Tensile strength is given as tension force times cross-section; σ = F_t/S. Dividing both sides by S thus gives the equation: σ = ρ * 2πR * g₀. To compare with real materials let's just calculate the ratio between tensile strength and density σ/ρ from given dimensions; we get value of 1.3*10¹³ m²/s², which in usual units of tensile strength and density is 1.3*10¹⁰ MPa per g/cm³. For most metals, fibres, etc.. this is in range 100-1000; for graphene it's in range of 10⁵, and for carbon nanotubes it's theoretically up to 10⁷. So, the strongest structural material known to man is still 1000 times weaker than what one would require to hold up such a structure — and it's a nanomaterial, not something one would construct a huge-ass space structure out of it. To give another perspective of the scale of tension, if one assumed density of the ring close to that of rock or metal, several g/cm³, the tension force would be comparable to the pressure at the centre of Sun.