308 



CONCRETE DESIGN 



From e erect a perpendicular. Then, returning to the 

 formula, it is desired to find the vertical distance to the 



FIG. 10 



parabolic curve. This distance is denoted by Y. The fol- 

 lowing values are known: H = 9, X = ll, and 5 = 88. Sub- 

 stituting these values in the formula, 



= 8.4375 ft. 

 ' J 



Lay off e f equal to 8.4375 ft. Then / is a point on the 

 curve. 



Suppose it is desired to locate the curve at the first division 

 point from a, namely, g. Through g draw a perpendicular. 

 Here, H = 9, X = 33, and 5 = 88. ^Therefore, 



= 3.9375 ft. 



From g, there is plotted upwards to scale 3.9375 ft. to /t, 

 which is a point on the curve. Other points may be obtained 

 in the same manner and the parabolic curve drawn through 

 them. ' 



Instead of using the formula to determine the location of 

 points along the parabola, as just shown, it is customary 

 <50<r/r fo A? af/v/efeaf fa/o /0 eye/0/ ' farfs 



FIG. 11 



to divide an arch into ten equal horizontal parts and then 

 use the values given in Fig. 11. These values are derived 

 from the formula on page 307, and enable a parabola to 

 be plotted with ease when the span is divided into ten equal 

 parts. The rise at each one of these parts is given in terms 



