CEMP-E
TI 850-02
AFMAN 32-1125(I)
1 MARCH 2000
TE =85,000 + 45,000 = 130,000 lbs.
WEng = 170 + 100 = 270 tons
WCar = 187 tons
NCar = 25
G = 130,000 / 20(270 + 25 x 187) - 0.15
= 1.16%
2 engines and 25 cars = 27 total
Grade length = 27 x 70 = 1890 Feet (or longer)
Thus, if this revised arrangement is acceptable, grades of 1890 feet or longer could have an effective
gradient as high as 1.16%. Table B-1 shows the curve compensation for curves on this grade.
Table B-1. Curve Compensation for a 1.16% Grade.
Uncompensated Grade
Compensated Grade
Degree of Curve
Actual
Effective
Actual
Effective
1
1.16
1.20
1.12
1.16
2
1.16
1.24
1.08
1.16
3
1.16
1.28
1.04
1.16
4
1.16
1.32
1.00
1.16
5
1.16
1.36
0.96
1.16
6
1.16
1.40
0.92
1.16
From the table, if a 5 degree curve was located within the ruling grade, the constructed (actual)
gradient through that curve must be limited to 0.96% to keep the effective gradient within 1.16%.
B-3
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