Launch
Position
Effect
Conclusions
From
the various plots we can see that delta has significant effects
on the resulting velocity and efficiency. All of the previous coilguns
have utilised simple optical triggering which fixed the delta at
approximately 5 mm (the open loop triggering experiments also used
a delta of  5 mm). The efficiency at this delta is at best just
over 3 %. Contrast this with the best overall efficiency of 6 %
obtained during these experiments.
Points
of interest:
The
velocity peak occurs at a more negative delta when the capacitance
is increased. This is because the longer current pulse
makes it possible for the system to utilise more of the forcedisplacement
curve. This position is refered to as the velocitypeak delta (VPD).
Velocity
and efficiency become more sensitive to negative values of delta
as the capacitance is decreased. This is due to the shortening
of the current pulse combined with the low projectile force at negative
delta  the current dies away before the projectile reaches the
maximum force region of the coil, where it can gain energy more
efficiently.
Voltage
effect on the VPD. The voltage alters the maximum current
and hence projectile transit time. This is particulary evident with
the 13,600 uF capacitor. In this case we see the VPD changing from
6 mm for a charging voltage of 50 V, to 1.5 mm at a charging voltage
of 100 V.
Increasing
correlation between efficiencypeak delta (EPD) and VPD as the capacitance
gets smaller. This might be a reflection of the forcedisplacement
curve for the coil. A short current pulse (compared to the transit
time) localises the acceleration impulse such that it acts over
a limited region of the forcedisplacement curve. We see that the
EPD occurs between about 2_mm and 6
mm (8 % and 24 % of projectile length) over the whole parameter
space investigated.
The
peak efficiency reduces as the voltage increases. This
occurs because the copper losses increase more than the work done
on the projectile during the acceleration. To illustrate this point
as simply as possible let us assume that the current pulse is rectangular,
and that the acceleration is only a function of the current magnitude.
Now if the projectile takes a time t to accelerate over
the coil length with a current I, the copper losses will
be I^{2} R t. The work done on the projectile will
be k s I, where k is a proportionality constant
and s is the displacement over which the acceleration occurs.
If
the current is now increased by a factor of g, the time
taken to accelerate over s will be g^{0.5}t
(from basic motion equations). The work done on the projectile is
now:
E_{P} = g k s I
and
the copper losses are:
E_{L} = (g I)^{2} R (g^{0.5} t)
=
g^{1.5} I^{2} R t
Clearly
the energy converted to heat has increased by a greater proportion than the energy
delivered to the projectile. The copper losses increase by a factor of g^{0.5}
over the projectile energy. The efficiency is defined as:
Efficiency
= E_{P} / (E_{P} + E_{L})
but
if E_{p} is small compared to E_{l} then the efficiency can be
approximated to:
Efficiency
= E_{P} / E_{L}
Using
this approximation we can estimate the multiplication factor for the decrease
in efficiency as g^{0.5}
The
experimental configuration that most closely approaches this simplified analysis
is that involving the 150,000 uF capacitor (the current pulse stays fairly flat
over most of the acceleration displacement^{(1)}). Looking at the differences
in projectile energy and efficiency at two charging voltages we can see that the
actual and (expected) values are not too dissimilar, given the simplifications.
The charging voltages are 30 V and 50 V so the projectile energy multiplication
factor is 1.67 (g) and the efficiency multiplication factor is 0.775
(g^{0.5}). Table 1 shows the comparison.
Delta 
5 
0 
5 
Projectile
Energy Factor

1.82
(1.67) 
1.60
(1.67) 
1.55
(1.67) 
Efficiency
Factor 
0.870
(0.775) 
0.812
(0.775) 
0.787
(0.775) 
Table
1.
(1)
Although the rate of rise of current appears quite slow in the currenttime
graphs, the displacement during this rise is a small fraction of
the overall displacement, therefore the currentdisplacement curve
has a faster rate of rise and looks more 'rectangular'.
