Errata for SWCE 7th Edition
Last updated 12-Oct 2016
|
Page |
Location |
Existing |
Correct |
| xiv | 1st line | PE polyethylene | Delete line (duplicate) |
| 22 | Line for Lindane | catttle | cattle |
| 39 | Figure 3.5 caption | Theissen | Thiessen |
| 45 | Figure 3.7 caption | Herschfield | Hershfield |
| 52 | Example 3.7, solution, first equation | (36+8.44)0..76 | (36+8.44)0.76 |
| 55 | Paragraph 3, line 5 | guage | gauge |
| 55 | Last line | vapr | vapor |
| 76 | Paragraph 3, line 1 | Plants species | Plant species |
| 94 | Table 5.4, footnote [d] | (litter, grass, and brush overstory | (litter, grass, and brush overstory) |
| 94 | Table 5.4, footnote [d] | Good > 70% ground cover | Good: > 70% ground cover |
| 97 | Example 5.5, Solution, end of first paragraph | S = 43.4 mm | S = 44.8 mm |
| 97 | Example 5.5, Solution, first equation | … = 44.8 | … = 44.8 mm |
| 98 | Equation 5.11 | q = qu A Q Fp | q = qu A Q |
| 98 | Eqn 5.11 definitions | Fp = pond and swamp adjustment factor from Table 5.6 | Delete term from equation. |
| 99 | Table 5.6 | Delete entire table. NRCS no longer recommends the use of the Pond and Swamp adjustment factor. | |
| 100 | Paragraph 1, last line | is shown in Example 5.5. | is shown in Example 5.6. |
| 101 | 5th line from top | Find Fp = 0.97 for 0.2% pond … from Table 5.6. | Delete entire sentence. |
| 101 | Example 5.6, last line | q = 0.0012 × 100 × 10 × 0.97 = 1.16 m3/s | q = 0.0012 × 100 × 10 = 1.20 m3/s |
| 102 | Table 5.7, line for Railroad yard | 0.20 – 0.35 | 0.20-0.35 |
| 103 | Table 5.9, line 2 of 4 | 1.1. | 1.1 |
| 104 | Example 5.8, Solution, 1st equation | tc=0.0195L0.77s-0.385 = 0.0078(28000.77)… | tc=0.0195L0.77s-0.385 = 0.0195(28000.77)… |
| 113 | Problem 5.11, last sentence | point n | the point of inflection |
| 118 | After Equation 6.6 | q = discharge (L3/T), | q = discharge (L3/T), |
| 119 | Example 6.1, line 1 | b = 1.0 | b = 1.0 m |
| 123 | Line 4 | 6.6 | 6.4 |
| 125 | Table 6.1, Material column | Peak, muck and sand | Peat, muck and sand |
| 126 | Section 6.12, paragraph 3, line 1 | Figure 6.6 | Figure 6.7 |
| 128 | Part (b), table header, rightmost column | a | q |
| 130 | Line after Eqn 6.16 | into Equation 6.18 | into Equation 6.16 |
| 132 | Solution, line 1 | y = 0.596. | y = 0.596 m. |
| 150 | After Equation 7.4, second line | (dimensionless) | (kg m-1 s-1 Pa-1.5) |
| 151 | Second line | 0.1 kg Pa-1.5 | 0.1 kg m-1 s-1 Pa-1.5 |
| 257 | Paragraph 1, line 4 | than | then |
| 292 | Paragraph 3, line 4 | … or d uring extended … |
… or during extended … (Delete paragraph break.) |
| 312 | Example 13.2, second to last line | Total Volume 144.45 | Total Volume 144.05 |
| 335 | Example 14.3, Solution step (3) | Equation 8b | Equation 14.8b |
| 337 | Figure 14.8 | Spacing = 12.4 m | Spacing = 12.1 m |
| 339 | Figure 14.9, caption | Delete the second sentence. | |
| 354 | Example 15.1, second equation | Pe = 0.92 [1.25(100)0.824… | Pe = 0.92 [1.25(90)0.824… |
| 360 | Under Eqn 15.5 | salt tolerance threshold from Table 15.4 | salt tolerance threshold from Table 15.3 |
| 365 | Example 15.7, Solution, line 3 | Average depth = (0.95 + 0.98 … + 84 +0.88)/16 = 0.84 | Average depth = (0.95 + 0.98 … + 0.84 +0.88)/16 = 0.84 |
| 410 | Example 17.1, Solution, 2nd line | Table 15.4 | Table 15.3 |
| 410 | Example 17.1, Solution, 4th line | (0.9 m × 120 mm/m) 108 mm | 0.9 m × 120 mm/m = 108 mm |
| 410 | Example 17.1, Solution, 4th line | Table 15.5 | Table 15.4 |
| 438 | Figure 18.1 caption | Soil wetting patter | Soil wetting pattern |
| 450 | Step (9) | Ho = 13.83 + … | Ho = 13.88 + … |
| 451 | Step (16) | EU = … = 95.6% | EU = … = 95.3% |
| 454 | Step (10) | … flow in the manifold is 0.211 L/s. | .. flow in the manifold is 0.106 L/s. |