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.

Ummm.. k and V are normalized, right? The velocity terms are V/c and the “deceleration constant” is an acceleration also divided by light speed.

Otherwise the dimensions look very weird.

Sorry to quibble.

Hi Mike

The constants aren’t normalised, though they can be, if you like. I haven’t given k any dimensions, though you can probably work that out for yourself to make everything come out right. I’ll comment further in my next post.

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