> I'd thought that the 'phase' of the lfo would be defined
> by the capacitor that was being shorted.
> So knowing this is not the case saves me from embarking on aa
> doomed project.

When the capacitor is discharged the LFO may start with the positive or
negative slope. It depends upon the situation when the electronic discharge
switch closes. As far as I remember the LFO starts from zero to the positive
if the triangle was in the rising slope (independend of the actual positive
or negative value) when the capacitor is discharged by the switch (and vice
versa). But I'm not 100% sure and have to take a look at the schematics. It
has to do with the Schmitt Trigger state of the circuit when the capacitor
is shortened. The state of Schmitt Trigger defines if the LFO starts with
rising or falling slope.

Best wishes
Dieter Doepfer