.... I am actually able to sample the data at up to 20kHz (5kHz realistically) through a dedicated sensor box.

Therefore, each of the currents and voltages will be available as a discrete waveform (0.0002 secs between each data point). Does this make the calculation any easier?

The graphical solution seems fine for one-offs but I wanted to implement a system to calculate the phase currents automatically in software based on the measured line quantities - that is why I'm trying to do it mathematically.

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In terms of the Fermat solution, I think the idea is to calculate the centre point of the triangle made by the 'line' current vectors and use this with the original vector values to calculate the new 'phase' current values. This is what topgone has done in a previous post.

If this was done at each time instant (every 0.0002 seconds) with the instantaneous 'line' values, then the instantaneous 'phase' values could be calculated at each time instant. Then if this process was repeated which each set of instantatneous (sampled) values over an extended period of time (say 1 second) the waveform for each of the 3 'phase' currents would be produced.

Does anyone see a flaw in this approach?

Cheers,

W.

First let's make a distinction here. The graphical solution (including any math solution associated thereto) is for RMS data, not instantaneous data.

Values of an instant are processed with basic math. "Phase" voltage is the absolute difference of line-to-neutral values. For example, your earlier post gave the following values for a single instant:

v1 (phase-neutral) = 308V

v2 (phase-neutral) = -275V

v3 (phase-neutral) = -35V

Line-to-line voltages at that instant would be:

v12 = |v1 ? v2| = |308V ? (-275V)| = 583V

v23 = |v2 ? v3| = |(-35V) ? (-275V)| = 240V

v31 = |v3 ? v1| = |(-35V) ? 308V| = 343V

Calculating instant phase current also uses basic math, but gets slightly more complex because it requires the use all six discrete values to compute each phase current. With a delta connected 3? load (a wye connected 3? load would be a different calculation), you simply do a wye-delta transform substituting a correlated "vi" for each "R" in the formula. A wye-connected 3? load would be calculated differently because each line current goes through three resistances (except when one line current is zero; if two are zero, so would be the third, or there would be a ground fault), rather than two as in a delta configuration, and the neutral point is a floating voltage.