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The calculation of the corresponding temperatures is simple, since there is a linear temperature gradient between L1 and L6 (i.e. the temperature increment from one to the next line is always the same). Calculation: Divide range of gradient by five (L6 – L1), this is the temperature increment from one line to the adjacent lane. Example: calculation of temperature at line 2,5 in a temperature gradient from 40°C (L1) to 60°C (L6) • subtract temperature at L1 from temperature L6 (range of gradient: 60-40°C = 20°C) • divide temperature by 5 (increment per lane: 20°C/5 = 4°C) • multiply increment by 1,5 (1,5 increments from L1 to L2,5 = 6°C) • add this value to the temperature at L1 (40°C + 6°C) Result: temperature at L 2,5 is 46°C TGGE Manual October 2009 36
