I didn't, since 0.3c is mildly relativistic. Here's the set of formulas I've got:Jonathan_S wrote: Since RFC seems to ignore relativity for wedge acceleration, we can just use Newtonian formulas for all this.

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t, d, v, a are measured by the moving observer.

T, D, v, A are measured by a 'stationary' observer.

From a standing start, so at t_0 = T_0 = 0 s, d_0 = D_0 = 0 m and v_0 = 0 m/s;

a is a constant.

Newtonian rocket*: v(t) /c = at/c, while,

Einsteinian rocket: v(t) /c = tanh(at/c)

http://math.ucr.edu/home/baez/physics/R ... ocket.html

* Though a 'rocket' is an unlikely vehicle to be producing constant acceleration for long periods of time.

With x == at/c

gamma(x) = 1/sqrt{1 - [v(t)/c]^2} = cosh(x)

t(x) = c/a x

d(x) = c^2/a [1 - 1/gamma(x)] = c^2/a [1 - 1/cosh(x)]

v(x) /c = tanh(x)

T(x) = c/a sinh(x)

D(x) = c^2/a [gamma(x) - 1] = c^2/a [cosh(x) - 1]

A(x) = a /gamma(x)^3 = a/cosh^3(x)

With y == aT/c = sinh(x)

gamma(y) = sqrt{ y^2 + 1 }

t(y) = c/a arsinh(y)

d(y) = c^2/a [1 - 1/gamma(y)] = c^2/a [ 1 - 1/sqrt{ y^2 + 1 } ]

T(y) = c/a y

D(y) = c^2/a [gamma(y) - 1] = c^2/a [ sqrt{ y^2 + 1 } - 1 ]

v(y) /c = y /gamma(y) = 1/sqrt{ 1 + 1/y^2 }

A(y) = a /gamma(y)^3 = a { y^2 + 1 }^-3/2

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Setting a = 5000 m/s2, and plugging in v/c = 0.3, I got

T = 18,900 s = 5.24 hr,

D = 868 e9 m = 48.3 lt-min.

Setting a = 6000 m/s2,

T = 15,700 s = 4.36 hr,

D = 723 e9 m = 40.2 lt-min.

Just what I did! If you compare, you'll see the author's error isn't plot-critical; missiles run slower, but they run longer so they actually have longer ranges than advertised.Jonathan_S wrote: (Now in actuality I didn't work it out like this until now, I simply plugged the numbers into a spreadsheet I already had for missile accelerations; treating the ship like a very slow, very long endurance, missile)