HKJHWAY CONSTRUCTION 



diameter of the wheel, as the larger the wheel the less force required 

 to lift it over the obstruction or to roll it up the inclination due to the 

 indentation of the surface. 



The power required to draw a wheel over a stone or any ob- 

 stacle, such as S in Fig. 1, may be thus calculated. Let P represent 

 the power sought, or that which would just balance the weight on 



the point of the stone, and the 

 slightest increase of which 

 would draw it over. This 

 power acts in the direction 

 C P with the leverage of B C 

 or 1) K. Gravity, represented 

 by W, resists in the direction 

 C B with the leverage B D. 

 The equation of equilibrium 



S 

 Fit-. 1. 



will be P X C B == W X B 1), whence 



P_W B1) _ W I 

 V CB" 



CD -AD 



Let the radius of the wheel = C D = 20 inches, and the height 

 of the obstacle = A B == 4 inches. Let the weight W = 500 pounds, 

 of which 200 pounds may be the weight of the wheel and 300 pounds 

 the load on the axle. The formula then becomes 



i ; 



676-484 13.8 



9rt /I 0(J 99 



500 ' ^ = y - 500 13 2 f = 314.7 pounds. 



The pressure at the point I) is compounded of the weight and 

 the pow r er, and equals 



C D 9fi 



W~g- = 500X^ = 591 



and therefore acts with this great effect to destroy the road in its 

 collision with the stone, in addition there is to be considered the 

 effect of the blow given by the wheel in descending from it. For 

 minute accuracy the non-horizontal direction of the draught and 

 the thickness of the axle should be taken into account. The power 

 required is lessened by proper springs to vehicles, by enlarged wheels, 

 and by making the line of draught ascending. 



