Chapter Three
3.2. Turning in Space : Norem's Idea
In our phase of constant acceleration (Stage 1) we took a laser luminosity of L' = 43000 TW. Our ship
is being accelerated 1.6 years till it has a velocity of 1.62 . 108 ms-1.
The accelerating distance is 4.41 . 1015 m long. As we already mentioned in Sec.3.1
(Accelerating), our ship does not accelerate in direction to Epsilon Eridani.
Instead of this, the initial
direction of the starship and the direction to the target star form an angle
, see Fig.5. This angle - and the interstellar magnetic field
- enable us to brake the starcraft by the pressure of laser beams from the solar system:
In Chap.1 we have learned that we can use the
interstellar magnetic field for such a turning manoeuvre by charging up our starship. For instance, at the beginning of the charging, the
light sail will be separated into six very long and very narrow stripes which - after they are
charged up - are pushing away each other. Thus the light sail transforms into three mutually perpendicular axes, the ship itself being at the intersection of these axes.
In our estimation it should be possible, to maintain tremendous charges with such a construction. In Fig.6 we give an example,
how those six stripes could be joined to form the light sail. 
In the following we assume the interstellar field as orientated perpendicular to the direction Sun-Epsilon Eridani to simplify the scenario.
In Sec.3.3 (Braking) we shall see that the decelerating phase (Stage 4) takes as long as the accelerating phase (Stage 1). If we assume constant acceleration and deceleration, the distance d between Sun and Epsilon Eridani is as same length as the distance of the force-free flight (the green line).
In the last section we already determined the initial velocity at Stage 2 to (see above):
After having been accelerated, our ship overcomes the distance of ten light-years in
Here we see: It must be guaranteed that our brave space pioneers can overcome such a long
period in isolation - but this is another theme. Nevertheless, let us concentrate on Stage 3.
To describe the turning we remember
the definition of the angular velocity 
B , see also (2b.01)
...
(3b.01)
... and the relationship we discovered in Chap.2, Sec.2.1 (go
to):
(3b.02)
We realize: the angular velocity depends on the electric charge Q and the mass m of the body - as well as on
the magnetic induction B0. First of all we are interested in the Period T of a complete circulation. Please keep
in mind that our ship carries out only a part of this circulation. If we equalize both formulas (3b.01)
(3b.02), we have:
(3b.03)
How strong is the interstellar magnetic field?
According to Forward, it has a strength of B0 = 3 . 10-5 g = 3 . 10-9 T [Forward (1964)]. Perhaps this is a bit too optimistic; therefore we would like to refer to newer sources [Musser (2000)], in which a field strength of B0 = 5 . 10-10 T is given. The next interesting question concerning the ship is: how strong could we charge it up?
Here we have the problem, that the properties of the interstellar matter are not well known.
Ionized gas particles could - once the ship is charged - be attracted and captured by the ship's stripes. Thus, the generators on board must
compensate this neutralizing effects. We assume that the inert atomic nuclei of the (possible) ionized interstellar gas are
not as strong attracted by the ship as the lighter electrons. Therefore we want our ship be charged negatively, perhaps in future it is possible to generate a charge of Q
= - 1010 C.
By the way, we have an additional problem (but here we cannot go into the details): While traveling with relativistic speeds, crew and equipment
are being exposed to a high cosmic radiation.
In the previous section we took a ship mass of
m' = 7.58 . 107 Kg. According to formula (3b.02) we are now able to calculate
the angular velocity of the turning to
With this result, our ship needs for one circulation a time T of:
With the radius R of the circulation and the angle (see Fig.5)
we can calculate the duration 
of the turning (Stage 3). To find out the radius R we refer to equation (2b.02):
Therefore we have:
According to Fig.5, tan(/2) is given by
This means an angle of
= 2.86o.
With a few trigonometric calculus and the circulation period T
we can easily derive the duration of the turning phase. Here we intend you to calculate the relationship between those periods.
Our result:
(3b.04)
If we insert the
period T into this equation, we get a period of about = 1.52 a.
Finally, we
calculate the length of phase 3 (the blue line in Fig.5) to:
s = v.T
= 0.54 c . 1.52 a 
0.82 ly ,
referring
to a length of nearly s = 7.76 . 1015 m.
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