TRANSPORTATION ON LAND BY VEHICLES:-THE ROADS 



43 



3. Construct curve, starting- at E as follows :- 



Prolong E E, over E by length of the pole P. At the end of the pole, make 

 a perpendicular offset equalling half the length of the hammer, or 3-9 feet, the 

 end -point of which is at B. Mark B by a stake; join E with B, and prolong EB 

 over B by P ; at the end of P, make another perpendicular offset, this time the full 

 length of h, ending at C. Join B to C and prolong BC over C by P ; make perpen- 

 dicular offset at end of prolongation, and so on, and so on, till F F, is reached. 



In the logger's practice, the curves of a road can be constructed 

 readily with the help of an ordinary tapeline 21' 8',," long. Here, 

 for P equal to 21' 8'V', a "hammer" or offset of 1 foot (12 inches) 

 corresponds with a road curve of 12 degrees, a hammer of 2 feet 

 (24 inches) with a road curve of 24 degrees, &c. In other words: -every inch of "hammer" is the 

 equivalent (approximately) of one degree of curvature. 



To be e.xact. 



the first prolongation should be 

 and the subsequent prolongations 

 the first perpendicular should be 

 and the subsequent perpendiculars 



6. The engineer working with transit or compass obtains a given curvature of A" bv a deflection 



A" 

 of ^ from the line of the tangent toward the desired side for a distance of 100 feet, and thereafter 



by a similar deflection of A" for every 100 feet. Intermediate points are established by halving 

 or by quartering the angles and the distances. More properly, the lengths of shorter intermediate 

 chords should be prorated to the length of a 100-foot chord at the ratio of the sines of the 

 angles formed respectively between the tangent and the chords. 



sin B: sin A = short chord: 100 



^fo'^ 



PROBLEM:-Connect, using compass and chain, course AC and course C B by a curve starting at A. 



Given are:- 



Bearing A C is N 10° E, and CB is N90"E 



Distance A C is 194-4 feet = T. 



