Twenty male Friesian crossbred calves of 180 kg average live body weight
(LSW) were divided into four groups. Animals were fed according to NRC (1984)
allowances (T,), animal heart girth (HG)xO.028 (T2) , animal (HG)xO.032 (T3)and
animal (HG)x 0.034 (T4). The average daily gain in T4 was significantly higher than T2
and non significantly higher than T, and T3 The fourth treatment was economically
more efficient as well as in feed conversion than the other treatments with no
significant differences among them. Every cm of HG in T4 (The best treatment in
average daily gain, feed conversion and economic efficiency) required 0.055, 0.048,
0.006, 0.025 and 0.033 kg from OM, OM, CP, SE and TON, respectively. Also, the
results showed that every kg of OM, OM, CP, SE and TON in T4 was consumed by the
following numbers of cm of animal HG: 18.0, 21.0, 162.9, 39.6 and 30.2, respectively.
Therefore, the nutrient requirements of male Friesian crossbred calves can be
predictied from the following equations:- OM requirement (kg) = Animal HG x 0.055
OM requirement (kg) = Animal HG x 0.048
CP requirement (kg) = Animal HGxO.006
SE requirement (kg) = Animal HG x 0.025
TON requirement (kg)= Animal HG x 0.033 Or: OM requirement (kg) = Animal HG 118.0
OM requirement (kg) = Animal HG 121.0
CP requirement (kg) = Animal HG 1162.9
SE requirement (kg) = Animal HG 1 39.6
TON requirement (kg)= Animal HG 1 30.2 The results of this study showed that the small holder can use the following
equations to obtain the feed quantity for male Friesian crossbred calves which fed
concentrate feed mixture (CFM), berseem hay and rice straw:-                         The requirement of CFM (kg)                            = Animal HG x 0.042 or Animal HG/26.1, The requirement of berseem hay (kg)=Animal HG x 0.0094 or Animal
HG/106.7,         and      The requirement of rice straw (kg)      =Animal HG x 0.013 or Animal HG/76.6 Roughage:concentrate ratio will be 0.60:0.40 by using the previous equations.