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)