Ballistic Trajectory for Static Target

Intro

Abstract

How to derive the ballistic trajectory for a projectile aimed at a static target, including gravity, using vector notation to find time of flight and initial direction.

Projectile position as a function of time

• Using the position formula from the Kinematic Equations:

$$p_t=p_{p0}+u_{p0}{s}_p{t}_c+1/2g_p{t}_c^2$$

• Where $u_{p0}$ is a normalized 3D direction vector (unit length)

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Components in 3D

$$x(t_c)=x_0+s_p{u}_x{t}_c+\frac{1}{2}{g}_x{t}_c^2$$ $$y(t_c)=y_0+s_p{u}_y{t}_c+\frac{1}{2}{g}_y{t}_c^2$$ $$z(t_c)=z_0+s_p{u}_z{t}_c+\frac{1}{2}{g}_z{t}_c^2$$

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Unit vector check

• A fourth equation can be added taking the Vector Magnitude of the Unit Vector $||u_{p0}||$

$$sqrt(u_x^2+u_y^2+u_z^2)=1$$

• And then squaring both sides

$$||u_{p0}||=(sqrt(u_x^2+u_y^2+u_z^2){)}^2=1^2$$

• Therefore

$$||u_{p0}||=1$$

Isolate $u_{p0}$

$$u_{p0}=(p_t-p_{p0}-1/2g_p{t}_c^2)/(s_p{t}_c)$$

• Simplify using $\Delta =p_t-p_{p0}$

$$u_{p0}=(\Delta -1/2g_p{t}_c^2)/(s_p{t}_c)$$

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Apply the Unit Vector Constraint to each dimension of $u_{p0}$

• Each component of the initial direction vector $u_{p0}$ can be written as:

$$u_x=({\Delta }_x-1/2g_x{t}_c^2)/(s_p{t}_c)$$ $$u_y=({\Delta }_y-1/2g_y{t}_c^2)/(s_p{t}_c)$$ $$u_z=({\Delta }_z-1/2g_z{t}_c^2)/(s_p{t}_c)$$

• Since $u_{p0}$ is a unit vector:

$$||u_{p0}|{|}^2=u_x^2+u_y^2+u_z^2=1$$

• Substituting the components:

$${\left(\frac{{\Delta }_x-1/2g_x{t}_c^2}{s_p{t}_c}\right)}^2+{\left(\frac{{\Delta }_y-1/2g_y{t}_c^2}{s_p{t}_c}\right)}^2+{\left(\frac{{\Delta }_z-1/2g_z{t}_c^2}{s_p{t}_c}\right)}^2=1$$

• Multiply through by $s_p^2{t}_c^2$ to eliminate the denominator:

$$({\Delta }_x-1/2g_x{t}_c^2{)}^2+({\Delta }_y-1/2g_y{t}_c^2{)}^2+({\Delta }_z-1/2g_z{t}_c^2{)}^2=s_p^2{t}_c^2$$

• Expand the squares (slide-style quartic)

$${\Delta }_x^2+{\Delta }_y^2+{\Delta }_z^2-({\Delta }_x{g}_x+{\Delta }_y{g}_y+{\Delta }_z{g}_z)t_c^2+\frac{1}{4}(g_x^2+g_y^2+g_z^2)t_c^4=s_p^2{t}_c^2$$

• Vector operations like Euclidean Vector Multiplication and Self Vector Multiplication let us multiply and square vectors directly in the trajectory equations.

Simplify

• Simplify $s_p^2{t}_c^2$ to

$$\Vert \Delta {\Vert }^2-({\mathbf{g}}_p\cdot \Delta )t_c^2+\frac{1}{4}\Vert {\mathbf{g}}_p{\Vert }^2{t}_c^4=s_p^2{t}_c^2$$

• Rearrange to standard quartic form

$$\frac{1}{4}\Vert {\mathbf{g}}_p{\Vert }^2{t}_c^4-({\mathbf{g}}_p\cdot \Delta +s_p^2)t_c^2+\Vert \Delta {\Vert }^2=0$$

Quadratic Formula

• Let (y = t_c^2). Then the quartic becomes a quadratic in (y):

$$\frac{1}{4}\Vert {\mathbf{g}}_p{\Vert }^2{y}^2-({\mathbf{g}}_p\cdot \Delta +s_p^2)y+\Vert \Delta {\Vert }^2=0$$

• Apply the Quadratic Formula:

$$y=\frac{({\mathbf{g}}_p\cdot \Delta +s_p^2)\pm \sqrt{({\mathbf{g}}_p\cdot \Delta +s_p^2{)}^2-\Vert {\mathbf{g}}_p{\Vert }^2\Vert \Delta {\Vert }^2}}{\frac{1}{2}\Vert {\mathbf{g}}_p{\Vert }^2}$$

• Simplify the denominator:

$$y=t_c^2=2\frac{({\mathbf{g}}_p\cdot \Delta +s_p^2)\pm \sqrt{({\mathbf{g}}_p\cdot \Delta +s_p^2{)}^2-\Vert {\mathbf{g}}_p{\Vert }^2\Vert \Delta {\Vert }^2}}{\Vert {\mathbf{g}}_p{\Vert }^2}$$

• Take the positive square root to find the physical solution for time of flight:

$$t_c=\sqrt{2\frac{({\mathbf{g}}_p\cdot \Delta +s_p^2)\pm \sqrt{({\mathbf{g}}_p\cdot \Delta +s_p^2{)}^2-\Vert {\mathbf{g}}_p{\Vert }^2\Vert \Delta {\Vert }^2}}{\Vert {\mathbf{g}}_p{\Vert }^2}}$$

• Usually, the smaller positive root corresponds to the direct trajectory (discard negative candidates).

Compute the initial direction vector

• Once $t_c$ is known, compute ${\mathbf{u}}_{p0}$ directly:

$${\mathbf{u}}_{p0}=(\Delta -\frac{1}{2}{\mathbf{g}}_p{t}_c^2)/(s_p{t}_c)$$

• This gives the unit vector pointing from the launch point to the target, accounting for gravity.