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 MIL-HDBK-1013/14
To illustrate the use of this figure, consider the example used in 6.4.1a, except the vehicle
is traveling downhill on a 5-degree slope. The steps are:
1) Locate 5 degrees on the horizontal axis (point A).
2) Draw a vertical line up from point A until it intersects the curve (at point B) for a =
10 feet per second squared (3.05 meters per second squared).
3) Draw a horizontal line from point B toward the vertical axis and read off the "s'/s"
value at the intersecting point C.
4) The value of s'/s is 0.78. Because s' = s x (s'/s) and s = 300 feet (91.5 m), therefore
s' = 300 feet (91.5 m) x 0.78 = 234 feet (71.32 m).
This example shows that to accelerate the vehicle to the same 58.4 mph speed (36.29
kph), a 5-degree slope will help shorten the accelerating distance from 300 feet (91.5 m) to 234 feet
(71.32 m). It clearly demonstrates the increased vulnerability caused by local terrain sloping down
toward a protected area. Modifying the local terrain is an effective way to minimize vulnerability.
6.4.2
Attainable Vehicle Speed on a Curved Path. Centrifugal force makes it difficult to drive
fast on a curve unless the road surface is properly banked. The centrifugal force, "CF," of a vehicle
moving on a curved path depends on its weight, "w," the radius of the curvature, "r," and the speed,
"v," and g = gravitational constant = 32.2 feet per second squared (9.82 meters per second squared),
as shown in Equation (3).
CF = wv 2 /(gr)
EQUATION:
(3)
where:
CF = centrifugal force
W = vehicle weight
r = radius of curvature
v = vehicle speed
g = gravitational constant
When the "CF" is large enough, it will overcome the road friction and a vehicle will
skid. The vehicle could also topple if its center of gravity is too high. Because skidding usually
occurs first, only this condition will be considered here. Road friction force, "FF," equals the
product of the vehicle weight, "w," and the friction coefficient, "f," between the tires and the road
surface, as shown in Equation (4).
FF = fw
EQUATION:
(4)
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