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.

(2) Referring again to figure 20, it may be

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

by using figure 21. From this figure it

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,

W, is 350 pounds, the propellant charge,

be obtained with a stroke of 90 inches

c, is:

with an acceleration of 18 g. With these

approximate. values established. the

design is continued.

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

to first order equations section (para. 51).

change of acceleration specified, the time is estimated

as follows:

(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):

(1) To use equation (3) from paragraph 22,

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):

diameters are specified in the design

requirements, it is advantageous to

choose a peak pressure which experience

has shown will satisfy the design

32.2 ft/sec2

requirements without requiring unusually

large diameter tubes. The selection of the

preliminary peak pressure is also

thrust and horizontal +

determined, in part., by the characteristics

90

(vertical-upward

of available propellants. Such a peak

ejection)

pressure is 2,000 psi. Using this figure in

equation (3), the effective area (a function

of diameter) of the catapult is found as

follows:

This corresponds very closely to a

(5) The charge weight may then be

diameter of 2 inches.

calculated using equation (31), assuming

(2) Peak pressure and acceleration occur

an

when the telescoping tube (intermediate

tube) and the inner tube are traveling

together. Therefore, the outside diameter

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