Development of the Mathematical Ballistics Model (4)

 We left off with equation (E) from Part 3:

r =  k ₁ ² /c -  k _{2 cos (}q  + k ₃)

Where k ₃ is a new constant based on:

sin ( q  + K ₃)  =   cos ( q  + K ₃ + p/2) = _{cos (}q  + k ₃)

Equation (E) is the geometric solution of the system of differential equations (A) and gives r as a function of q. This equation is the polar form of a conic with one focus at the origin.  There are three arbitrary constants that have been identified in this equation: k ₁ ,  k ₂ and k ₃. And which we will now evaluate.

We begin by differentiating (E) with respect to t, then evaluating the derivative at   q =  0:

r' _o  = k ₁ ² k ₂ sin ( q)/[ c -  k ₂ cos ( q)] ²

^{Since then: V}_o   =  Ö r ²_o   + r² _o q' _o²

^V_o   =  r _o q' _o

_{We have from eqn. (B) in Pt. 3:}

  _{r^-2  q'  =  k 1 = const.}

_So:  k ₁ =  r _o (r _o q')  =  r _{o V o}   

The value for k ₂ can now be found by letting q =  0 in equation (E) from Part 3, so:

r _o =  k ₁ ² /c -  k ₂  =  r² _o V² _o /c -  k ₂ 

_k 2 = c -  r_o V² _o

Before substituting the constants into equation (E) it is useful to define a dimensionless quantity which simplified the expression.  This is:

 p = 1   -  r_o V² _o  /c

The solution of (E) then becomes:

r =   (1 -  p) r_o  / 1 - p cos q

Recall from analytic geometry, e.g.

• Analytic Geometry Conclusion: Generalizing 2nd Degree Curves...

that the eccentricity of the conic is the coefficient  (- p) of the cos q term and it determines the character of the conic. We are now in position to compare the geometric characteristics of the trajectories, i.e.

ELLIPSE { Apogee at q =  0,  0 < p < 1

               { Circular path:  p = 0  or  ^V_o   =   Ö  c/ r _o

                      {Perigee at q = 0,  -1 < p <  0

PARABOLA  { p=  -1  or:   ^V_o   =   Ö  2c/ r _o

HYPERBOLA  { p   < -1  or:   ^V_o   >   Ö  2c/ r _o                      

_{Some cases of interest are shown in the diagram below:}

_{Incorporating solutions involving t - the time - leads to more detailed results.  For example, the magnitude of the speed can be obtained from:}

 _{^Vo   =  Ö r' ²   + r²  q' ²}

_{The magnitude of the velocity vector in terms of} q  can be obtained from:

        x'/ y'      =  sin q / p  - cos q

_{direction of the velocity vector in terms of}   q will be given by:

 tan  f  =  p sin q/  1 - p cos q

The path angle can be obtained from the appropriate velocity diagram, starting with the one for our basic conventions (See Oct. 7 post. Fig. 1).  Then we have from the applicable geometry: