338 TABLE XIII. CART-LOADS OF MANURE IN HEAPS PER ACRE. 



the other distance is 6 yards, and the number of heaps to be 

 niadt from a cart-load is 6 ? 



Find 13 J feet, or 4J yards, in the left-hand column, opposite 

 which, under 6 heaps per cart-load, stands . 40 



Which multiply by the same distance, or 4 J 



And divide by the other distance, or . 6)180 



Which gives the cart-loads required = 30 

 Table VII. may also be used as a manure Table, by consid- 

 ering the plants as so many heaps of manure laid down at the 

 various distances in the Table. 



RULE. To find the number of cart-loads or cubic yards re- 

 quired for any given quantity of land by Table VII., divide the 

 number opposite the given quantity of land, under the given 

 distance, by the number of heaps in a cart-load or cubic yard ; 

 the quotient is the quantity required. 



EXAMPLE 4. How many cart-loads of manure will be required 

 for 5 acres, 3 roods, 36 poles of land, laid down in heaps 5 J yards 

 distant from each other, dividing the cart-load into 4 heaps? 

 In Table VII., under 16 feet, or 5J yards, opposite 



5 acres stands . . . .800 



Opposite 3 roods stands . . .120 



And opposite 36 poles stands ... 36 



Which, divided by the heaps in a cart-load = 4)956 



Gives the cart-loads required *= 239 



To solve the same question by the present Table 

 Opposite 16 J feet, and under 4 heaps, stands the 



quantity for 1 acre . . . 40 



This multiply by 5 for the 5 acres , 5 



200 



For 3 roods take f of an acre . . = 30 



And for 36 poles take T % of a rood . = 9 



Cart-loads required as above . 239 



. A cubic yard of each of the following 

 manures weighs cwt - V- Its< 



New dung . . . 9 3 18 



(Dung, when fermented, a cwt. or two more) 

 Leaves and sea- weed . . 903 



Compost of dung with weeds and lime 

 which had been once turned over in 9 

 months . . . 14 5 



Garden- mould . . 19 3 25 



Water . . , . 15 3 



