All should be taken care as long as right is side and pointing left to stage(I have pointed right w/ okay results) and left is mid and yes you decode when you mix.
FWIW, if I'm thinking this through properly, the outcome when having the (+) lobe of the figure-8 facing R instead of L is just a reversed stereo image.
<00> DecoderIn both cases below, we're using the same decoder, which calculates L and R as follows:
L = M + S
R = M - SThe following two cases walk through the impact on the decoding results with the S mic's
<01> (+) lobe facing L and (-) lobe facing R, and
<02> vice versa, (-) lobe facing L and (+) facing R
<01> Proper configuration: (+) lobe facing L and (-) lobe facing RGiven the proper configuration, the definition of our M and S channels is:
M = L + R
S = L - R (i.e. (+) lobe facing L, (-) lobe facing R)So the "math" - substituting the above values above for M and S - works out to:
L = M + S (per our <00> Decoder definition above)
= (L + R) + (L - R)
= L + R + L - R
= 2LR = M - S (per our <00> Decoder definition above)
R = (L + R) - (L - R)
= L + R - L + R
= 2R<02> "Improper" configuration, (-) lobe facing L and (+) lobe facing ROur S channel lobes are reversed now, so we have a different definition of our S channel (the mid remains the same):
M = L + R
S = -L + R
= R - L (just another way of stating -L + R)And since our decoder hasn't changed, the "math" - substituting the above values above for M and S - works out to:
L = M + S (per our <00> Decoder definition above)
= (L + R) + (R - L)
= L + R + R - L
= 2R R = M - S (per our <00> Decoder definition above)
= (L + R) - (R - L)
= L + R - R + L
= 2LSo...all this to say that if you have the (+) lobe of the S channel facing R instead of L, the result is just a reversed stereo field.