of the telescoping tube determines the
overall length accommodates the firing
effective area, and from this calculation
mechanism and mounting arrangement).
the diameter becomes 2.0 inches.
d. Propellant Charge Weight.
determined that a catapult with a
separation velocity of 80 fps minimum
(1) The propellant charge weight is estimated
and an acceleration of 20 g maximum
would require a stroke of 72 inches.
may be seen that the approximate ratio of
Since it is not practical to design to the
propellant charge to propelled weight for a
minimum velocity nor the maximum
catapult with a terminal velocity of 85 fps
acceleration, it is determined, using figure
is 0.44 gm/lb. Since the propelled weight,
i terminal velocity
of 85 fps can
W, is 350 pounds, the propellant charge,
be obtained with a stroke of 90 inches
with an acceleration of 18 g. With these
approximate. values established. the
design is continued.
b. Stroke Time.
Using equation (2) from
(2) It is possible to refine this approximation
using equation (31) from the refinements
found in the previous approximation, and the rate of
(3) In this equation, m' is the modified
propelled mass, modified to correct for
friction effects and direction of ejection. It
may be calculated from equation (32):
c. Peak Pressure.
either the peak pressure or the diameter
(4) The following values are assumed for use
of the catapult must be known. Since no
in equation (32):
m = propelled mass 350/g slugs
requirements, it is advantageous to
S = stroke to separation 90 in.
choose a peak pressure which experience
g = acceleration due to gravity
has shown will satisfy the design
requirements without requiring unusually
v = velocity at separation 85 fps
large diameter tubes. The selection of the
a = angle between direction of
preliminary peak pressure is also
thrust and horizontal +
determined, in part., by the characteristics
of available propellants. Such a peak
pressure is 2,000 psi. Using this figure in
equation (3), the effective area (a function
of diameter) of the catapult is found as
This corresponds very closely to a
(5) The charge weight may then be
diameter of 2 inches.
calculated using equation (31), assuming
when the telescoping tube (intermediate
tube) and the inner tube are traveling
together. Therefore, the outside diameter