West and east highway segments are separated by 1000 ft horizontally. The west segment has a 0% constant grade and the east segment has a –1% grade. The east segment has a higher elevation than the west segment, and the two segments are connected by a joining sag and crest curve combination (so PVTs = PVCc). If the road is designed for 60 mi/h, what is the elevation difference between the west and east highway segments?

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

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The intent of this problem is to reinforce the concepts, and the set up for this problem. In looking at Example 3.11, the current problem will be similar except it will be

reversed with the lower highway segment on the left of that figure (instead of on the right) and the higher roadway segment on the right instead of the left. Also, in the current problem the ycon = 0 (since the problem states that sag and crest curves are joined) and ys = 0 since the initial grade of the sag curve (the grade of the west segment) is 0%.

To begin this problem, it is important to first get the K-values for both the sag and crest curves. With the 60-mi/h design speed given in the problem, from Table 3.2 we find the K-value for the crest curve is Kc =151, and from Table 3.3 we find the sag curve’s K-value is Ks = 136. let Gcon be the constant grade connecting the two curves so that G2s = G1c = Gcon. To determine the absolute value of the algebraic difference in the grades (in percent) for the sag and crest curves, we have As = Gcon (since G1s = 0 as given in the problem) and Ac = Gcon + 1 (since G1c = Gcon and G2c = −1, note that the sign of G2c becomes +1 since A’s are based on the absolute value in the differences of the grades which means you add the absolute values of two grades if they are opposite in sign, as they are here, and you subtract the lower absolute value grade from the higher absolute value grade if the two grades are the same sign).

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