176 ELEMENTARY TREATISE ON STOCK FEEDS AND FEEDING 



Per cent. 



butter 



fat 



Protein 

 pounds 



Carbohy- 

 drates 

 pounds 



Fat 

 pounds 



Nutritive 

 ratio 



Maintenance per 100 lbs. 



live weight 



10 pounds of milk 



10 " " " 



10 



15 ' 



15 " •' " 



15 " " " 



20 " " " 



22 " " " 



20 " " " 



25 " " " 



25 " " " 



35 " " " 



30 " " " 



30 " " ■' 



30 " " " 



35 



35 



40 " " " 



40 " " " 



40 " " " 



50 ' 



For each additional 10 lbs, 



0.07 



1. 10 

 1. 17 

 1.24 

 1.30 

 1.405 



I-5I 

 1.50 

 1.64 

 1.78 

 1.70 



1-875 



2.05 



1.90 



2. 11 

 2.32 

 2.10 

 2.345 

 2.59 

 2.30 



2.58 

 2.86 

 0.40 

 0.47 

 0.54 



0.7 

 8.81 

 9.14 

 9.48 



9-715 

 10.21 

 10.72 

 10 62 

 11.28 

 11.96 

 11-525 

 12.35 

 13.20 



12-43 



13-42 



14-44 



13-335 



14-49 



15-69 



14.24 



15-56 



16.92 



1. 81 



2.14 



2.48 



o.oi 

 0.24 

 0.26 

 0.28 

 0.31 

 0.34 

 037 

 0.38 

 0.42 

 0.46 



045 

 0.50 



0-55 

 0.52 

 0.58 

 0.64 



0-59 

 0.66 



0-73 

 0.66 

 0.74 

 0.82 

 0.14 

 0.16 

 0.18 



1:8.5 

 1:8.3 

 1:8.2 

 1:8.0 

 1:7.8 

 1:7.6 

 1:7.6 

 1:7-5 

 1:7-3 

 1:7.4 

 1:7.2 

 1:7.0 

 1:7 2 

 1:7.0 

 1:6.8 

 1:7.0 

 1:6.8 

 1:6.7 

 1:6.8 

 1:6.7 

 1:6.6 



Use of the Table. — Let us make this clearer by computing the 

 standard for a cow weighing 850 lbs. producing 23 lbs. of milk 

 daily, testing 5 per cent, butter fat. 



Since 850 is 8.5 X 100, we must multiply our maintenance 



(0.07 lb. protein; 0.7 lb. carbohydrates and o.oi fat) by 8.5. 



0.07 X 8.5 = 0,595 lb. of protein I Maintenance require- 



0.7 X 8.5 ^ 5.950 lbs. of carbohydrates >• tnent for a cow weigh- 

 o.oi X 8.5 ^= 0.085 lb. of fat j ing 850 lbs. 



Since our cow is producing 23 lbs. of milk, we divide 23 by 10 

 which gives us 2.3. Multiply the standard for each additional 

 ID lbs. of 5 per cent, butter fat milk (0.54 lb. protein, 2.48 lbs., 

 carbohydrates and 0.18 lb. fat) by 2.3 which gives us the milk 

 production requirement. 



0.54 X 2.3 -1.182 lbs. of protein ] Milk production requiren.ent 



2.48 X 2.3 = 5.704 lbs. of carbohydrates ^°r,^ cowproducing 23 lbs. of 

 0.18 X 2.1 = 0.44 lb. of fat I ^^^J}^y *^^*'"g 5 per cent. 



