Launch Position Effect



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 force-displacement curve. This position is refered to as the velocity-peak 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 efficiency-peak delta (EPD) and VPD as the capacitance gets smaller. This might be a reflection of the force-displacement 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 force-displacement 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 I2 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-0.5t (from basic motion equations). The work done on the projectile is now:

EP = g k s I

and the copper losses are:

EL = (g I)2 R (g-0.5 t)

= g1.5 I2 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 g0.5 over the projectile energy. The efficiency is defined as:

Efficiency = EP / (EP + EL)

but if Ep is small compared to El then the efficiency can be approximated to:

Efficiency = EP / EL

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.


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 current-time graphs, the displacement during this rise is a small fraction of the overall displacement, therefore the current-displacement curve has a faster rate of rise and looks more 'rectangular'.