MANURES. 353 



EXPLANATION. In the left hand column are placed the dis- 

 tances of the rows and the heaps in each row (i. e., the dis- 

 tances between the heaps in each direction), and at the top 

 of the columns will be noticed the number of heaps intended 

 to be made of each load; the point where the two meet 

 gives the number of loads per acre which will be required 

 for that purpose. 



EXAMPLE 1. Required the number of loads necessary to 

 manure an acre, dividing each load into six heaps, and pla- 

 cing them 4^ yards apart 2 



SOLUTION. In the left hand column find 4| (the distance 

 of the heaps apart), and opposite it to the right, under 6 

 (the number of heaps in each load), will be found 39f. 

 Ans. 



EXAMPLE 2. A farmer has a field containing 5J- acres, 

 over which he wishes to spread 82 loads of manure. Now, 

 82 divided by 5J- gives 15 loads per acre, and by referring 

 to the table it will be seen that the desired object can be at- 

 tained by making 4 heaps of each load, and placing them 9 

 yards apart, or by 9 heaps at 6 yards apart, as may be 

 thought most advisable. 



NOTES. A cubic foot of half- rotten stable manure will 

 weigh 56 Ibs. ; if coarse or dry, 48 Ibs. 



A load of manure is about 36 cubic feet, and of the first 

 quality will weigh 2016 Ibs. ; of the second, 1728 Ibs. 

 Eight loads of the first kind spread over an acre will give 



