Robert Zubrin and Dana Andrews invented the concept of the Magnetic Sail, or Mag-Sail, which is basically a large superconducting ring which generates a large magnetic field, which mimics a planet’s magnetosphere and interacts with the Sun’s plasma-wind. Within a solar system a Mag-Sail provides unlimited range, so long as coolant lasts for the superconductors. But the most interesting use for a Mag-Sail is providing a reaction-mass free way of slowing an interstellar space vehicle. According to a NIAC study by Zubrin et. al. a Mag-Sail can potentially slow a vehicle from 0.95c to just 0.0054c in just 820 days – with no reaction mass required.

To the mathematically inclined quoting a few figures doesn’t tell one much, and so I went a little further in my reading of Zubrin and developed his equations further. His various works state that the final velocity, V, for a deccelerating vehicle goes like so…

V = Vo / (1 + k.Vo^(1/3).t)^3, where Vo is the initial velocity, k the decceleration constant, and t the decceleration time.

…with a bit of rearrangement the time can be worked out from the velocities…

t = (1/k).[ V^(-1/3) – Vo^(-1/3) ]

…i.e. inverse of the decceleration constant times the difference of the inverse of the cube-root of the final velocity and the inverse of the cube-root of the initial velocity. Which is easier to understand in symbols, if you ask me.

Let’s take a step back to the first equation. Call (1+ k.Vo^(1/3).t) a separate name, B, thus we get…

V = Vo/B^3

…and the acceleration, dV/dt, is now…

a = -3k.Vo^(4/3)/B^4

…and the initial acceleration, a(0), is -3.k.Vo^(4/3). The minus sign indicates the vehicle is slowing down.

How far does the vehicle travel while deccelerating from Vo to V? It’s a straightforward integration – much to my surprise – and boils down to…

s = Vo/(2k.Vo^(1/3))*[(B^2 – 1)/B^2]

…which, after some algebra, becomes…

s = Vo.t/2*[(1 + B)/B^2]

…which isn’t so intimidating. Alternatively we can express the decceleration constant, k, as an expression…

k = 1/t * [V^(-1/3) – Vo^(-1/3)]

and substitute into our equation for the displacement to get…

s = Vo.t/2 * [(V/Vo)^(1/3).(1 + (V/Vo)^(1/3))]

which may seem a bit strange, but even the regular equation for displacement under constant acceleration can be rearranged into a similar form…

s = Vo.t/2 * [1 + (V/Vo)]

but what does that all mean? Part II coming soon.