14 



THE GAEDENER'S ASSISTANT. 



sight column be greater than that in the fore- 

 sight one, the ground is rising; if the sum in the 

 fore-sight column be the greater, the ground is 

 falling, as is the case in the present instance. 









Distance of surface 



Station. 



Back-sight. 



Fore-sight. 



below horizontal 

 line or datum, at 





Ft. Ins. 



Ft. Ins. 



Ft. Ins 

 A, 3 



1 



A, 3 



B, 5 



B, 5 



2 



B, 2 



C, 5 6 



C, 8 6 



3 



C, 3 6 



D, 3 6 



D, 8 6 



4 



D, 4 6 



E, 3 



E, 7 



5 



E, 4 



F, 4 6 



F, 7 6 



6 



F, 2 6 



G, 5 6 



G, 10 6 







19 6 



Difference of level. 



27 

 19 6 



2. To find the height of a horizontal line h I 

 (fig. 797), according to which the ground may be 

 made level with its own soil. — Find the area of 

 the perpendicular space between the surface of 

 the ground and the horizontal line AG. Divide 

 that area by the length of the horizontal line, and 

 the quotient will he the distance from the line 

 AG to a parallel horizontal line h /, according to 

 which the ground will be level; for there will be 

 as much of the solid ground above it as there 

 are vacant spaces below it; so that Avhen the 

 heights are taken down to that line and turned 

 into the hollows, the whole will be level. Sup- 

 posing the rods are placed at the distance of 

 50 feet from each other, the areas of the spaces 

 between each rod from the ground to the hori- 

 zontal line AG should be found, in order to 

 ascertain the area of the whole space. This can 

 be done as follows: — 



Perpend 

 culars, 



i-l A, 3 

 i B, 5 



B, 5 



C, 8-5 



C, 8-5 



D, 85 



D, 8-5 



E, 7-0 



E, 70 



F, 7 5 



F, 7 5 



G, 10-5 



Base, . . . 



8 

 50 



13 5 

 50 



17-0 

 50 



15.5 



50 



14 5 

 50 



18-0 

 50 





-4-2)400 



. ... 200 



675 

 337-5 



850 

 425 



775 

 387-5 



725 

 302-5 



900 



Areas, . . 



450 



The total of these areas is 2162-5 square feet. 



Now the object is, to find where a horizontal 

 line can be drawn that will leave as much space 

 or hollow below it as there will be of ground 

 above it, or that will form the end of a parallelo- 

 gram, the length of which is AG = 300 feet, and 

 which shall contain an area equal to that of the 

 whole space above ground, up to the line AG, or 

 21625 square feet. If it do the one it will cer- 

 tainly do the other; for a line that will include 

 all the clear space above it, must exactly leave 

 that which is, or ought to be made, solid ground 

 below it. We may therefore divide the area of 



the space b}^ the length, and it will give the 

 breadth of the parallelogram, or average per- 

 pendicular distance from the datum line to the 

 surface of the ground. The total area of the 

 irregular space is, as above found, 2162*5 square 

 feet, which, divided by 300, the length, gives 

 7-208 feet, or about 7 feet 2\ inches, and this 

 distance, measured clown from the datum hori- 

 zontal line A G, will determine the position of 

 the required horizontal ground line h I. 



In the above example, the rods are at equal 

 distances, and this being the case, the mean 

 distance from the datum line to the ground 

 could be found by merely adding to the half 

 sum of the first and last perpendiculars all the 

 others, and dividing this sum by the number 

 of perpendiculars, less one; thus — 



3 + 10-5 + 5 + 8-5 + 8-5 + 7 + 7-5zz 43-25, 

 2 



which, divided by 6, the number of perpendi- 

 culars, less 1, gives 7*208, or 7 feet 2 J inches, 

 as before. But cases may occur where the 

 rods, to give the correct area, must be placed 

 at unequal distances, so as to be at the ex- 

 tremities of the different slopes, and then the 

 mean perpendicular distance cannot be ob- 

 tained by averaging the respective heights, 

 but must be calculated by multiplying the 

 sum of each pair of perpendiculars by the 

 distance between them, and taking half the 

 product for the area; or the sum of any pair 

 of perpendiculars multiplied by half the dis- 

 tance, or half the sum of the perpendiculars 

 by the whole distance between them, will, in 

 either case, give the area of space included 

 between the two perpendiculars. The rods, as 

 before remarked, should be placed so as the 

 surface of the ground will deviate little from 

 a straight line between every two, or, in other 

 words, at the highest and lowest parts of the 

 surface. 



Thus, however irregular the surface may be, 

 levels can be taken on the rods, so that the 

 distance from the ground to a horizontal 

 datum line can be found at any point, and 

 consequently the area of the whole space or 

 section between that line and the ground can 

 be ascertained. This area, divided by the 

 whole base or distance between the rods placed 

 at the extremities, will give the distance of the 

 level ground line from the horizontal datum 

 line. Here it may be observed that the 

 centre of this ground level remains at the 

 same height, although the ground may be laid 

 to any slope. A border the length of the 



