Hallo Hiroshi

In the calculation of max yield you use 800(c*d)/25 c to calculate
time. You divide termal time  with mean temp, it should be termal
time/termal time. Thus 800/ (25-8)  [ termal time/ (tmean-tbase)]
give you 47 days  against the original value of 32. New yield should
be 13953.125 kg ha-1

Regards
Andre

<<<<<<Original message>>>>>>>
I determined the cultivar coeficients for NC+4616 (Values) as shown
above. Ranges represent ones for maize in the DSSAT cultivar file.

 I define (Maximum yield potential) = P5/25 * G2 * G3, assuming that
the mean daily temperature during the grain-filling period is 25C and
the plant density is 5.5 plants/m2. For NC+4616, the maximum yield
potential is 8 (mg/kernel/d) * 800 (C*d) / 25 (C) * 675 (kernel/plant)
* 5.5 (plants/m2)

= 950 g/m2 = 9500 kg/ha.

 This yield is too low as a potential for maize yield. Is the
definition for maximum yield potential inappropriate, or do I
misunderstand something?


I would be grateful if someone could kindly tell me.


Best wishes,



Hiroshi Hasegawa


Division of Upland Farming

Tohoku National Agricultural Experiment Station

  960-2156

  Fukushima-shi, Arai, Aza harajuku-minami 50

  JAPAN

  Phone; 81-24-593-5151

  FAX  ; 81-24-593-2155

  Email; [log in to unmask]

  http://ss.fk.affrc.go.jp/kenkyuu/sakutuke/hasegawa/
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