Linear interpolation using actual SOI as weighting factor:
actual duration = ((actual SOI value [25,99 degrees] - duration #1 border value [22,01 degrees])/(duration #0 border value [30 degrees] - duration #1 border value [22,01 degrees]))*(duration #0 value) + ((duration #0 border value [30 degrees] - actual SOI value [25,99 degrees])/(duration #0 border value [30 degrees] - duration #1 border value [22,01 degrees]))*(duration #1 value)
actual duration = ((25,99-22,01)/(30-22,01))*25,01 + ((30-25,99)/(30-22,01))*22,87
actual duration = (3,98/7,99)*25,01 + (4,01/7,99)*22,87
actual duration ~= 0,4981*25,01 + 0,5019*22,87
actual duration ~= 12,4575 + 11,4785
actual duration ~= 23,9360 degrees of crankshaft
Taking into consideration SOI at 25,99 degrees of crankshaft before TDC EOI (end of injection) would be at 2,054 degrees of crankshaft before TDC.
Lets assume SOI corrections would advance SOI by 1 degree of crankshaft (all other values are same):
actual duration = ((26,99-22,01)/(30-22,01))*25,01 + ((30-26,99)/(30-22,01))*22,87
actual duration = (4,98/7,99)*25,01 + (3,01/7,99)*22,87
actual duration ~= 0,6233*25,01 + 0,3767*22,87
actual duration ~= 15,5887 + 8,6151
actual duration ~= 24,2038 degrees of crankshaft
Despite advancing SOI by 1 degree of crankshaft EOI would be at 2,7862 degrees of crankshaft before TDC, so only 0,7322 degree of crankshaft sooner than before because of longer duration!
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