Quantcast Attainable Vehicle Speed on a Straight Path

Share on Google+Share on FacebookShare on LinkedInShare on TwitterShare on DiggShare on Stumble Upon
Custom Search
 
  
 


MIL-HDBK-1013/14
4) Locate point D on the horizontal axis so that the distance between points C and D is
the accelerating distance [300 feet (91.5 m) in this example].
5) Draw a vertical line up from point D until it intersects the curve (at point E) for a =
10 feet per second squared (3.05 meters per second squared).
6) Draw a horizontal line from point E until it intersects the vertical axis (point F).
7) The value of the speed, "v," at point F, 58.4 mph (36.29 kph), is the answer.
Note: If "v0" = 0, the graph can be used to determine velocity from a dead start.
b) On a Slope. Due to gravitational effect, to achieve the same final speed as that on a
horizontal path, the required distance for acceleration on a slope will be shorter (longer) if the
vehicle is traveling downhill (uphill). Let, "s," be the acceleration distance needed to also attain
final speed, "v," on a horizontal path, and let, "s'," be the acceleration distance needed to attain, "v,"
on a sloped path. The following relationship shown in Equation (2) applies:
s'/s = 1/[1 + (g/a)sinθ]
EQUATION:
(2)
where:
s' = acceleration distance needed to attain final speed on a sloped path
s = acceleration distance needed to attain final speed on a horizontal path
g = gravitational constant = 32.2 feet per second squared (9.82 meters per second
squared)
a = acceleration of the vehicle, feet per second squared
θ = angle between the slope and the horizontal in degrees
This correction factor relationship is plotted as Figure 6.
17





 


Privacy Statement - Copyright Information. - Contact Us

Integrated Publishing, Inc.