Only a portion of the food ingredients wliicli are eaten is digested 

 and rendered soluble by the changes they undergo in the mouth, 

 stomach and intestines. This soluble portion is assimilated, and from 

 this alone is the animal nourished. The undigested part passes on 

 and is excreted as manure. This undigested part is of no use to the 

 animal. 



11. The value of feeding-stuff s vai'ies with the amount of diges- 

 tible food nutrlejits which they contain. — Chemical analysis shows 

 the total amount of nutritiv^e ingredients in food-stuffs, but the 

 digestible portion can be found only by carefully conducted feeding 

 experiments with farm animals. Since the amount of digestible 

 nutrients varies with different foods, it must be found for each one 

 by careful experimentation upon animals. Many such experiments 

 have been conducted with each of the common food-stuffs, so that 

 at the present time there are many tables of figures giving for each 

 feeding-stuff the digestible part of its protein, carbohydrate and fat, 

 the total amount of which has been shown by chemical analysis. 



In practical feeding experiments it has been found that one pound 

 of digestible fat will go as far as 2^ pounds of digestible carbohy- 

 drates ; or, in other words, that one pound of digestible fat will go 

 2J times as far as one pound of digestible carbohydrates. There- 

 fore, it is necessary to multiply the fat by 2 J in order to get its 

 equivalent in carbohydrates. 



12. The projportion hetween the digestible protein and digestible 

 carbohydrates -\- {fat x 2^) in a given food is called a nutritive 

 'ratio. — When we know the digestible nutrients in a food we can 

 easily finds its nutritive ratio. Thus a given food contains 2 parts 

 digestible protein, 10 parts digestible carbohydrates and 1 part fat : 

 the 1 part fat is equivalent to 2^ parts digestible carbohydrates ; 10 

 parts carbohydrates -\- 2^ is equivalent to 12^ parts of digestible 

 carbohydrates. That is : this particular food contains digestible 

 nutrients equivalent to '^ parts protein and \'^\ parts carbohydrates.^ 

 or for each part of digestible protein there are G-J parts of digestible 

 carbohydrates and fat. Therefore, this food has a nutritive ratio 

 of 1 : <o\. 



13. To find the nutritive ratio of any food.^ add the {faty.'^^ 



543 



