63 



Inclination of the Meridian. 



(13.) In projecting arcs of a great circle with the "solar transit," it is of the 

 utmost importance that the surveyor be able to tell the inclination of the meridians for 

 any latitude, and for any distance of eastings or westings. 



As this problem is not treated in elementary works on 

 surveying, perhaps the few following hints may be of use 

 to the young student. 



In the following figure, let the two arcs A G, and B G 

 be two arcs of a quadrant of the meridian, 1 of longitude 

 apart. Let A B = the arc of one degree of longitude on 

 the equator = 69.16 miles. 



Let D E be an arc of longitude on any parallel of 

 latitude. Also, let E H and D H be the tangents of those 

 meridians meeting in the earth's axis produced, and cor- 

 responding to the parallel of latitude D E. 



Then the line E F = D F = cos L = cos A D or B E. 

 Also, the angle D F E = l,and the angle D H E=the 

 inclination of the meridians, which is the angle we wish 

 to find, and which we will represent by X. And because 

 the two triangles F D E and D H E are on the same base 

 E D, and isosceles, their vertical angles vary inversely as 

 their sides ; and we have the equation, 



1XEF=XXEH, But 



E F = cos L, and E H = cot L, hence 

 X cot L = 1 cos L, or 

 X = cos L -f- cot L = sin L, . . 



(a) 



That is to say, 



The inclination of the meridians for any difference of longitude, varies as the sine 

 of the latitude. 



(14.) Since the sine of the latitude is the inclination in decimals of a degree, for 

 one degree of longitude, if we multiply by 3600" we shall have the inclination in 

 seconds of arc. Then, if we divide this by the number of miles in one degree of 

 longitude on that latitude, we shall have the inclination due to one mile on that 

 parallel. Thus, for 



Latitude 43 . 

 Multiply by 3600" 



log. sine = 9.833783 

 " = 3.556303 



Divide by 50^66, = 1 long, on that L, log. = 

 4S".46 = inclination for one mile of long. 



3.390086 

 1.704682 

 1.685404 



(15.) The use of the Inclination, as found by the preceding article, is to show 

 the surveyor how much he must deflect a line of survey from the due east or west, 

 to have it meet the parallel at a given distance from the initial point of the survey, 

 for it will be remembered that a parallel of latitude is a curve, having the cotangent 

 of the latitude for its radius. And the line due east or west is the tangent of the 

 curve. 



Thus, on latitude 43, I wish to project a six-mile line west, for the southerly 

 line of a township. 



Remembering that in an isosceles triangle, the angle at the base is less than a 

 right angle by half the angle at the vertex, I deflect my line towards the pole by the 

 inclination due to three miles, or in this case 48".46 X 3 = 2' .25", t. e., Deflection 

 y<z Inclination. 



(16.) Table No. Ill, which was computed from the formula (a) Art. 37, gives the 

 Inclination for one mile, and for six miles on any parallel, from 10 to 60 of latitude ; 

 also the Convergency for six miles, on any latitude. 



