﻿ Solution: Constant Acceleration Projectile/Target Problem | Primer on Geometric Algebra | David Hestenes

## Solution: Constant Acceleration Projectile/Target Problem

Problem: Determine

(a)  the range r of a target sighted in a direction  r/r  that has been hit by a projectile,
launched with velocity
v0,

(b)  the launching angle for maximum range

(c)  the time of flight.

(We'll work out the formulas first, then show an interactive diagram at the bottom of this page.)

We're given the sighting direction, call it ϕ  above the horizontal (it can be negative if the target is in a valley),
the initial muzzle speed v0, and the firing angle, call it θ above the horizontal. Common sense restrictions:

- π/2  <  ϕ  ≤  θ  <  π/2

We'll take
i to be the unit CCW bivector in the vertical plane. Then, multiplying any vector in the plane on the right
by
i will rotate that vector by  π/2  CCW. Here's a diagram summarizing our notation and the equations of motion:

Knowing the angles in the
red/blue/green triangle and the magnitude v0, we can solve the triangles for r, v, and t
We can summarize the given angles in the problem by writing the following useful rotor equations:

$\Large{\begin{array}{l}{{{\bf{\hat r}}}}{{{\bf{\hat v}}}_0}\,\,\, = \,\,\,{e^{{\bf{i}}\left( {\theta - \phi } \right)}}\,\,\,\,\,\,\,~~~~~~~~~~~~~~\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,{{{\bf{\hat r}}}}\, \wedge {{{\bf{\hat v}}}_0}\,\,\, = \,\,\,{\bf{i}}\,\sin \left( {\theta - \phi } \right)\\\\{\bf{\hat g}}\,{{{\bf{\hat v}}}_0}\,\,\, = \,\,\,{e^{{\bf{i}}\left( {\pi /2\,\, + \,\,\theta } \right)}}\,\,\, = \,\,\,{\bf{i}}\,{e^{{\bf{i}}\theta }}\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,{\bf{\hat g}}\, \wedge {{{\bf{\hat v}}}_0}\,\,\, = \,\,\,{\bf{i}}\,\cos \theta \\\\{\bf{\hat g}}\,{{{\bf{\hat r}}}}\,\,\, = \,\,\,{e^{{\bf{i}}\left( {\pi /2\,\, + \,\,\phi } \right)}}\,\,\, = \,\,\,{\bf{i}}\,{e^{{\bf{i}}\phi }}\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{\bf{\hat g}}\, \wedge {{{\bf{\hat r}}}}\,\,\, = \,\,\,{\bf{i}}\,\cos \phi \end{array}}$

We could use the law of sines to solve the triangle. As we've seen before when discussing vector identities, the law
of sines can be derived from equating three different expressions to calculate the area of the triangle, using wedge
products. Using those wedge products directly:

$\Large{\begin{array}{l}\frac{{\bf{r}}}{t} \wedge {{\bf{v}}_0}\,\,\, = \,\,\,\frac{1}{2}\,{\bf{g}}\,t\, \wedge \frac{{\bf{r}}}{t}\,\,\, = \,\,\,\frac{1}{2}\,{\bf{g}}\,t\, \wedge {{\bf{v}}_0}\\\\\\\frac{{{v_0}r}}{t}\,\,{{{\bf{\hat r}}}} \wedge {{{\bf{\hat v}}}_0}\,\,\, = \,\,\,\frac{{gr}}{2}\,{\bf{\hat g}} \wedge {{{\bf{\hat r}}}}\,\,\, = \,\,\,\frac{{{v_0}gt}}{2}\,{\bf{\hat g}} \wedge {{{\bf{\hat v}}}_0}\end{array}}$

The first equality allows us to solve for time of flight:

$\Large{\bbox[8px,border:2px solid red]{t\,\,\, = \,\,\,\frac{{2{v_0}}}{g}\,\,\frac{{{{{\bf{\hat r}}}} \wedge {{{\bf{\hat v}}}_0}}}{{{\bf{\hat g}} \wedge {{{\bf{\hat r}}}}}}\,\,\, = \,\,\,\frac{{2{v_0}}}{g}\,\,\frac{{\sin \left( {\theta - \phi } \right)}}{{\cos \phi }}}}$

The second equality of the triangle wedge products above gives the
total range (now that we know t):

$\Large{\bbox[8px,border:2px solid red]{r\,\,\, = \,\,\,{v_0}t\,\,\frac{{{\bf{\hat g}} \wedge {{{\bf{\hat v}}}_0}}}{{{\bf{\hat g}} \wedge {{{\bf{\hat r}}}}}}\,\,\, = \,\,\,\frac{{2{v^2}_0}}{g}\,\,\frac{{{{{\bf{\hat r}}}} \wedge {{{\bf{\hat v}}}_0}}}{{{\bf{\hat g}} \wedge {{{\bf{\hat r}}}}}}\,\,\frac{{{\bf{\hat g}} \wedge {{{\bf{\hat v}}}_0}}}{{{\bf{\hat g}} \wedge {{{\bf{\hat r}}}}}}\,\,\, = \,\,\,\frac{{2{v^2}_0}}{g}\,\,\frac{{\sin \left( {\theta - \phi } \right)\,\,\cos \theta }}{{{{\cos }^2}\phi }}}}$

Now we've answered parts (a) and (c) of the problem. For part (b), we are asked, given a fixed muzzle speed,
what firing angle θ should be use to get maximum range along that sighting direction? There are various
ways to calculate this from the range formula.

We can see from the numerators (bivectors or trig functions) that one term gets larger when
θ gets closer to ϕ+π/2,
while the other gets larger when
θ appoaches zero. One reasonable guess (at this point it's only a guess, but it turns out
to be correct) is that the answer is halfway between:
θr_maxϕ/2 + π/4. Of course, we could take the derivative of
the trig numerator with respect to
θ and set that derivative to zero to find the desired firing angle (which confirms
our guess). Or, we could use trig sum-of-two-angles identities to get:

$\Large{\begin{array}{l}\sin \left( {\theta - \phi } \right)\,\,\cos \theta \,\,\, = \,\,\,\frac{1}{2}\left[ {\sin \left( {\theta - \phi + \theta } \right)\,\, + \,\,\sin \left( {\theta - \phi - \theta } \right)} \right]\,\,\, = \,\,\,\frac{1}{2}\left[ {\sin \left( {2\theta - \phi } \right)\,\, - \,\,\sin \left( \phi \right)} \right]\\\\~~~~~2{\theta _{{\rm{r}}\_\max }}\,\, - \,\,\phi \,\,\, = \,\,\,\pi /2\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{\theta _{{\rm{r}}\_\max }}\,\,\, = \,\,\,\phi /2\,\, + \,\,\pi /4\end{array}}$

As an extra check (and practice with GA), we can get the same result another way:

$\Large{\begin{array}{l}{{{\bf{\hat v}}}_0}\,\,\, = \,\,\,{\bf{\hat g}}\,{\bf{i}}\,{e^{{\bf{i}}\theta }}\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\frac{d}{{d\theta }}\,\,{{{\bf{\hat v}}}_0}\,\,\, = \,\,\,{\bf{\hat g}}\,{\bf{i}}\,{e^{{\bf{i}}\theta }}\,{\bf{i}}\,\,\, = \,\,\,{{{\bf{\hat v}}}_0}\,{\bf{i}}\\\\0\,\,\, \leftarrow \,\,\,\frac{d}{{d\theta }}\,\,\left( {{\bf{\hat r}} \wedge {{{\bf{\hat v}}}_0}\,\,{\bf{\hat g}} \wedge {{{\bf{\hat v}}}_0}} \right)\,\,\, = \,\,\,{\bf{\hat r}} \wedge \left( {{{{\bf{\hat v}}}_0}\,{\bf{i}}} \right)\,\,{\bf{\hat g}} \wedge {{{\bf{\hat v}}}_0}\,\, + \,\,{\bf{\hat r}} \wedge {{{\bf{\hat v}}}_0}\,\,{\bf{\hat g}} \wedge \left( {{{{\bf{\hat v}}}_0}\,{\bf{i}}} \right)\\\\\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,{\bf{i}}\,\,{\bf{\hat r}} \cdot {{{\bf{\hat v}}}_0}\,\,{\bf{\hat g}} \wedge {{{\bf{\hat v}}}_0}\,\, + \,\,{\bf{\hat r}} \wedge {{{\bf{\hat v}}}_0}\,\,{\bf{\hat g}} \cdot {{{\bf{\hat v}}}_0}\,\,{\bf{i}}\\\\\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,{\bf{i}}\,\,{\left\langle {\left( {{\bf{\hat r}} \cdot {{{\bf{\hat v}}}_0} + \,\,{\bf{\hat r}} \wedge {{{\bf{\hat v}}}_0}} \right)\,\,\left( {{\bf{\hat g}} \cdot {{{\bf{\hat v}}}_0}\,\, + \,\,{\bf{\hat g}} \wedge {{{\bf{\hat v}}}_0}} \right)} \right\rangle _2}\,\,\, = \,\,\,{\bf{i}}\,\,{\left\langle {{\bf{\hat r}}\,{{{\bf{\hat v}}}_0}\,{\bf{\hat g}}\,\,{{{\bf{\hat v}}}_0}} \right\rangle _2}\\\\\\\,\,\,0\,\,\, = \,\,\,{\bf{i}}\,\,{\left\langle {{e^{{\bf{i}}\left( {\theta - \phi } \right)}}\,{\bf{i}}\,{e^{{\bf{i}}\theta }}} \right\rangle _2}\,\,\, = \,\,\, - \,\,\cos \left( {2\theta - \phi } \right)\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{\theta _{{\rm{r}}{\bf{\_}}\max }}\,\,\, = \,\,\,\phi /2\,\, + \,\,\pi /4\end{array}}$

Let's take another look at what this angle means. When looking at the trig form of the numerator in the range
formula, we thought of it as the average between 0 and
ϕ+π/2, an interpretation based on the formula without
much geometric insight. However, we might also think of it as the average of (halfway between)
ϕ and π/2 (the
vertical direction, the direction of
−g). This interpretation comes about naturally from the derivation of this
angle found in NFCM. This interpretation makes the result more intuitive. If we want maximum range,
choose the firing angle to be halfway between the sighting angle and the vertical.

We've now solved the three parts of the problem, but there are still some interesting results to mention.

We should also find that maximum range and perhaps time of flight for maximum range. We could insert
the angle found for max range into the formulas for range and time of flight, and then manipulate to a clean
result with trig identities. Instead, we'll use GA. It will be easier to find the max range first. This last diagram
aids in simplification. For one thing, it shows that the two bivector ratios in the range formula are the same.

$\Large{\begin{array}{l}{r_{\max }}\,\,\, = \,\,\,\frac{{2{v^2}_0}}{g}\,\,\frac{{{\bf{\hat r}} \wedge {{{\bf{\hat v}}}_{0{\mathop{\rm m}\nolimits} }}}}{{{\bf{\hat g}} \wedge {\bf{\hat r}}}}\,\,\frac{{{\bf{\hat g}} \wedge {{{\bf{\hat v}}}_{0{\mathop{\rm m}\nolimits} }}}}{{{\bf{\hat g}} \wedge {\bf{\hat r}}}}\,\,\, = \,\,\,\frac{{2{v^2}_0}}{g}\,\,\frac{{{\bf{\hat r}} \wedge {{{\bf{\hat v}}}_{0{\mathop{\rm m}\nolimits} }}}}{{{\bf{\hat g}} \wedge {\bf{\hat r}}}}\,\,\frac{{{{{\bf{\hat v}}}_{0{\mathop{\rm m}\nolimits} }} \wedge \left( { - {\bf{\hat g}}} \right)}}{{{\bf{\hat g}} \wedge {\bf{\hat r}}}}\,\,\, = \,\,\,\frac{{2{v^2}_0}}{g}\,\,{\left( {\frac{{{\bf{\hat r}} \wedge {{{\bf{\hat v}}}_{0{\mathop{\rm m}\nolimits} }}}}{{{\bf{\hat g}} \wedge {\bf{\hat r}}}}} \right)^2}\\\\\\{{{\bf{\hat v}}}_{0{\mathop{\rm m}\nolimits} }}\,\,\, = \,\,\,{{{\bf{\hat v}}}_{0\_r\_\max }}\,\,\, = \,\,\,\frac{{{\bf{\hat r}} - {\bf{\hat g}}}}{{\left| {{\bf{\hat r}} - {\bf{\hat g}}} \right|}}\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{\bf{\hat r}} \wedge {{{\bf{\hat v}}}_{0{\mathop{\rm m}\nolimits} }}\,\,\, = \,\,\,\frac{{0 - {\bf{\hat r}} \wedge \left( { - {\bf{\hat g}}} \right)}}{{\left| {{\bf{\hat r}} - {\bf{\hat g}}} \right|}}\,\,\, = \,\,\,\frac{{{\bf{\hat g}} \wedge {\bf{\hat r}}}}{{\left| {{\bf{\hat r}} - {\bf{\hat g}}} \right|}}\\\\\\\\ \bbox[8px,border:2px solid red]{{r_{\max }}\,\,\, = \,\,\,\frac{{2{v^2}_0}}{g}\,\,{\left( {\frac{1}{{{\bf{\hat r}} - {\bf{\hat g}}}}} \right)^2}\,\,\, = \,\,\,\frac{{{v^2}_0}}{g}\,\,\frac{1}{{1 - {\bf{\hat r}} \cdot {\bf{\hat g}}}}\,\,\, = \,\,\,\frac{{{v^2}_0}}{g}\,\,\frac{1}{{1 + \sin \phi }}}\\\\\\\\ \bbox[8px,border:2px solid red]{{t_{r\_\max }}\,\,\, = \,\,\,\sqrt {\frac{{2{r_{\max }}}}{g}\,} \,\,\, = \,\,\,\frac{{2{v_0}}}{g}\,\,\frac{1}{{\left| {{\bf{\hat r}} - {\bf{\hat g}}} \right|}}\,\,\, = \,\,\,\frac{{{v_0}}}{g}\,\,\sqrt {\frac{2}{{1 + \sin \phi }}}} \end{array}}$

Below is an interactive diagram for this problem.
We can try various experiments and draw quite a lot of conclusions
from this diagram.
Don't forget the handy reset button in the top right corner of the diagram (it's always there even though
not always visible). We're using the fixed values
g = 9.8 m/s2 and v0 = 7 m/s. Our problem is for a fixed muzzle speed and
we've chosen this one as convenient for the dimensions of the diagram. We have a slider for sighting angle
− 89° ≤ ϕ ≤ 89°,
one for firing angle
ϕ < θ ≤ 90°, and one for time 0 < t ≤ time of flight. The parabolic flight path modifies to its correct
shape r(t) = v0t + ½gt2 as the angles change. The time slider moves the projectile along its path. Turning on animation
shows the projectile slow down at higher altitudes and speed up at lower ones. Turning on the trace makes the projectile
leave a track in its wake. The values of the current projectile position (Pos), the firing angle for maximum range and
the value of that
max range are show dynamically on the left side of the diagram. Varying the firing angle slider should
allow you to verify these values.

Ideas to investigate using the interactive diagram:

1. Make the sighting angle to the target horizontal (
ϕ = 0). Vary the firing angle and verify that maximum range
occurs at 45
° and that the max range is v02/g = 72 / 9.8 = 5m.

2. Recreate the
trajectory envelope as shown in Figure 2.5, Chapter 3 of NFCM. Make the sighting angle
horizontal. Turn on the trace. As you vary through the range of firing angles, use the time slider to keep the
projectile "as far out as possible." You'll notice the shape of the enclosing envelope develop. Does the envelope
seem to be 5 units wide (from origin to max range) and 2.5 units high as stated in
Figure 2.5

You can reset and check out the shape of the envelope for different sighting angles.

3. Recreate the
locus of vertex ellipse in the same Figure 2.5 mentioned in our last paragraph. Reset the diagram,
set the sighting angle to zero degrees, set the time slider for the vertex (peak) of the parabolic path, then
turn on the trace. Now vary through all possible firing angles. Notice the trace draws what looks like an elliptical
path showing the locus of vertices for the family of projectile paths.

What happens to that apparent elliptical locus as the sighting angle is changed to larger angles?

4. Reset the diagram. Set the sighting angle to 0 degrees. Move the time slider all the way to the right (
time of
flight
for this path). Note the time. Move the time slider until the particle is at max height (vertex). Is the time
exactly half the time of flight? Now try different sighting angles. It should be obvious (since we hit the target
at a different elevation than we start out with) that the time at the vertex is then not half the total time of flight.

5. Reset. Set the sighting angle to 0 degrees. Move the time slider all the way to the right (time of flight).

Notice there is only one firing angle which will get you to
max range (5m), and it takes about 1.0s. Now,
suppose our target is only at a range of 4m.
You can get there with two different firing angles, one direct
and low (takes about 0.63s), the other a lob (takes twice as long in this case for the flight).

Is it a coincidence that the lob seems to take exactly twice as long as the more direct shot? What are the two
times if the target is only at 3m?

Notice
this is different from (but easily confused with) the standard freshman physics problem of solving
for time in the path equation
r(t) = v0t + ½gt2 for when we achieve a particular altitude. We usually break
the equation into horizontal and vertical parts and solve the vertical quadratic equation for
then correspond to the time to reach that altitude on the way up and the longer time to drop past that
altitude again on the way down.
That's not what we're looking at here. Yes, it's the same path equation,
and it's clearly quadratic in t
, but we're not asking for the times to pass a certain altitude along the path,
total time of flight to hit our target

If the target r and the velocity v0 is given (that is, both speed and firing angle are given), there is only one time
of flight
possible (given in our formulas above).

And, if we are going for maximum range, there is only one time of flight possible. But, if the muzzle speed is
larger than needed for max range (that is, our target is closer than max range) and the firing angle is allowed to
vary, we see in the interactive diagram that there are
two different firing angles which reach our target,
naturally in
two different times.

Let's gather data for two targets and see if anything interesting pops out.

Case 1:  Reset, trace off, ϕ = 0°, target at 4m. Two hits: (θ = 26.6°, t = 0.64s) & (θ = 63.4°, t = 1.28s).
Remember our max range firing angle for this case is 45°. Notice our two hits correspond to

θ = 45° ± 18.4°. Is this symmetry a coincidence (the ratio of times equal to 2 was a coincidence)?

Case 2:  Reset, trace off, ϕ = 30°, target at 1.81m. Two hits: (θ = 40.5°, t = 0.294s) & (θ = 79.8°, t = 1.26s).
O
ur max range firing angle can be seen to be 60°. Our two hits correspond to θ = 60° ± 19.6°.
This symmetry has been seen now at different sighting angles.
Perhaps there's a theorem here?

We've calculated the max range firing angle in the equations above, so these experimental results are not too
surprising. When the sighting angle is
ϕ = 0°, halfway between that and the vertical is 45°. When the sighting
angle is
ϕ = 30°, halfway between that and the vertical is 60°.

But, it is interesting that the two firing angles to hit a closer target are spaced symmetrically around the max
range firing angle. This turns out not to be a coincidence. While we explain this, go back to the interactive
diagram, reset the diagram, and click on the
Show Velocity Space checkbox. You'll see a blue parallelogram
v0 sides as blue vectors and the v sides as pink
vectors. These vectors (and the parallelogram) always follow the projectile dot, showing the initial firing
velocity (whose magnitude is fixed, but angle can vary) and the final velocity at the current dot position.
You'll see the parallelogram follow along as you move the three sliders.

not to scale with the rest of the diagram, and it can't
be, because its sides represent velocity (m/s), while the rest of the diagram is showing positions (m). It
shouldn't be too difficult to remember we are looking at two different kinds of graphs on the same diagram.

As shown in
NFCM, we can glance at the diagram at the top of this page and get two pieces of very useful
information:

$\Large{\begin{array}{l}\left( {{\bf{v}} - {{\bf{v}}_0}} \right)\left( {{\bf{v}} + {{\bf{v}}_0}} \right)\,\,\, = \,\,\,\left( {{\bf{g}}t} \right)\left( {\frac{{2{\bf{r}}}}{t}} \right)\,\,\, = \,\,\,2{\bf{gr}}\\\\{v^2}\,\, - \,\,{v_0}^2\,\,\, = \,\,\,2{\bf{g}} \cdot {\bf{r}}\,\,\,\,\,\,\,\,\,\,\,\,{\rm{and}}\,\,\,\,\,\,\,\,\,\,\,\,{\bf{v}} \wedge {{\bf{v}}_0}\,\,\, = \,\,\,{\bf{g}} \wedge {\bf{r}}\end{array}}$

These relationships are true for any of the particle paths in our interactive diagram. We are considering the
case where the target vector is given and we want to compare two paths to get there, the direct shot, which
we'll call path 1, and the lob, which we'll call path 2. Our initial speed is constant, only the firing angle varies.
The scalar part of our last equation is the well-known
conservation of energy equation for our projectile model.
Since our two paths end up at the common target, this equation tells us the landing speeds of both paths are
equal to each other. The right hand side of the bivector part of our last equation tells us for the two paths:

$\Large{{{\bf{v}}_1} \wedge {{\bf{v}}_{01}}\,\,\, = \,\,\,{{\bf{v}}_2} \wedge {{\bf{v}}_{02}}}$

That is, the areas of our blue parallelogram in the interactive diagram have the same value for either
the direct shot or the lob.
With the parallelogram now visible, revisit Case 1 and Case 2 above. The only
extra thing to do is to set the time slider to max (time of flight) so that our final speeds are the same for the
two paths as described in the last paragraph. Then,
keep your eye on the area value on the bottom left of
the window. Notice how it varies as the firing angle (and therefore the shape of the parallelogram) varies. But,
also notice that
both paths leading to the same target always have the same parallelogram area, as
predicted by our last equation. For the two associated paths, since the sides of the parallelogram have the
same values (speeds) and the areas are the same,
the two parallelograms for these paths are similar,
and therefore the ratios of, for example, long diagonal to short diagonal must be the same for both:

$\Large{\frac{{g{t_2}}}{{2r/{t_2}}}\,\,\, = \,\,\,\frac{{2r/{t_1}}}{{g{t_1}}}\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{t_1}{t_2}\,\,\, = \,\,\,\frac{{2r}}{g}}$

So, the product of the times of flight for these two paths is a known quantity. We can use this in the path
equations to learn something about the firing angles of these two paths:

$\Large{\begin{array}{l}{\bf{r}}\,\,\, = \,\,\,{{\bf{v}}_0}t\,\, + \,\,\frac{1}{2}{\bf{g}}{t^2}\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{{\bf{\hat v}}}_0}\,\,\, = \,\,\,\frac{{\bf{r}}}{{{v_0}t}}\,\, - \,\,\frac{{{\bf{g}}t}}{{2{v_0}}}\\\\\\{{{\bf{\hat v}}}_{01}}\,\,\, = \,\,\,\frac{{\bf{r}}}{{{v_0}{t_1}}}\,\, - \,\,\frac{{{\bf{g}}{t_1}}}{{2{v_0}}}\,\,\, = \,\,\,\frac{{{\bf{\hat r}}g{t_2}}}{{2{v_0}}}\,\, - \,\,\frac{{{\bf{g}}{t_1}}}{{2{v_0}}}\,\,\, = \,\,\,\frac{g}{{2{v_0}}}\,\left( {{\bf{\hat r}}{t_2}\,\, - \,\,{\bf{\hat g}}{t_1}} \right)\\\\{{{\bf{\hat v}}}_{02}}\,\,\, = \,\,\,\frac{{\bf{r}}}{{{v_0}{t_2}}}\,\, - \,\,\frac{{{\bf{g}}{t_2}}}{{2{v_0}}}\,\,\, = \,\,\,\frac{{{\bf{\hat r}}g{t_1}}}{{2{v_0}}}\,\, - \,\,\frac{{{\bf{g}}{t_2}}}{{2{v_0}}}\,\,\, = \,\,\,\frac{g}{{2{v_0}}}\,\left( {{\bf{\hat r}}{t_1}\,\, - \,\,{\bf{\hat g}}{t_2}} \right)\\\\\\\\{{{\bf{\hat v}}}_{01}}\,\, + \,\,{{{\bf{\hat v}}}_{02}}\,\,\, = \,\,\,\frac{{g\left( {{t_1}\,\, + \,\,{t_2}} \right)}}{{2{v_0}}}\,\,\left( {{\bf{\hat r}}\,\, - \,\,{\bf{\hat g}}} \right)\end{array}}$

Remember from above that the last vector in this equation points midway between the sighting direction and
the vertical, and we saw that was the direction of the firing angle for maximum range. But now we see two
unit vectors add to this same direction. That is,
we've proven our suspected theorem that the two firing
angles (for direct and lob paths) are spaced symmetrically around the max range firing angle.

The last question to answer here is, given the target position, what are the two times of flight and firing angles
we could use? It might be nice to have formulas to check against our interactive diagram. Actually, we have
pretty simple formulas above for the firing directions if we can first find the times of flight for the two paths.

The problem we started with at the top of this page was, given the initial speed and firing angle, find the time
of flight and range along the target line of sight. So, although we found a formula for time of flight, it involved
the given firing angle. But now, we're given the range (less than max range) and wish to find times of flight
and firing angles. We can eliminate the firing angle by squaring it out of the path equation, leaving
t as the
only unknown. It's a bit of an ugly quadratic, but we can clean it up some with an algebra theorem:

$\Large{\begin{array}{l}{\left( {2{{\bf{v}}_0}t\,\,\, = \,\,\,2{\bf{r}}\,\, - \,\,{\bf{g}}{t^2}} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,4{v_0}^2{t^2}\,\,\, = \,\,\,4{r^2}\,\, - \,\,4{\bf{r}} \cdot {\bf{g}}{t^2} + \,\,{\left( {{\bf{g}}{t^2}} \right)^2}\\\\{\left( {g{t^2}} \right)^2} - \left( {\frac{{4{v_0}^2\,\, + \,\,4{\bf{r}} \cdot {\bf{g}}}}{g}} \right)\,\left( {g{t^2}} \right)\,\, + \,\,4{r^2}\,\,\, = \,\,\,0\\\\g{t^2}\,\,\, = \,\,\,\frac{{4{v_0}^2\,\, + \,\,4{\bf{r}} \cdot {\bf{g}}}}{{2g}}\,\,\, \pm \,\,\,\frac{1}{2}\,\,\sqrt {{{\left( {\frac{{4{v_0}^2\,\, + \,\,4{\bf{r}} \cdot {\bf{g}}}}{g}} \right)}^2}\,\, - \,\,16{r^2}} \\\\~~~~t\,\,\, = \,\,\,\sqrt {\frac{{2{v_0}^2\,\, + \,\,2{\bf{r}} \cdot {\bf{g}}}}{{{g^2}}}\,\,\, \pm \,\,\,\sqrt {{{\left( {\frac{{2{v_0}^2\,\, + \,\,2{\bf{r}} \cdot {\bf{g}}}}{{{g^2}}}} \right)}^2}\,\, - \,\,\frac{{4{r^2}}}{{{g^2}}}} } \\\\\,~~~~~~\,\, = \,\,\,\frac{{\sqrt 2 \,{v_0}}}{g}\,\,\sqrt {1\,\, + \,\,\frac{{{\bf{r}} \cdot {\bf{g}}}}{{{v_0}^2}}\,\,\, \pm \,\,\,\sqrt {{{\left( {1\,\, + \,\,\frac{{{\bf{r}} \cdot {\bf{g}}}}{{{v_0}^2}}} \right)}^2}\,\, - \,\,{{\left( {\frac{{rg}}{{{v_0}^2}}} \right)}^2}} } \\\\\\{\rm{Theorem}}\,\,{\rm{(check}}\,\,{\rm{it}}\,\,{\rm{by}}\,\,{\rm{squaring)}}:\,\,\,\,\,\,\\\\\\\,\,\,\,\,\,\,\,\,\,\,\,\sqrt {a\,\, \pm \,\,\sqrt {{a^2}\,\, - \,\,{b^2}} } \,\,\, = \,\,\,\sqrt {\frac{{a + b}}{2}} \,\, \pm \,\,\sqrt {\frac{{a - b}}{2}} \\\\\\\\\,\,\,\,\,\,\,\,\,\,\,\,\alpha \,\,\, \equiv \,\,\,\sqrt {1\,\, + \,\,\frac{{{\bf{r}} \cdot {\bf{g}}}}{{{v_0}^2}}\,\, + \,\,\frac{{rg}}{{{v_0}^2}}} \,,\,\,\,\,\,\,\,\,\,\,\,\,\beta \,\,\, \equiv \,\,\,\sqrt {1\,\, + \,\,\frac{{{\bf{r}} \cdot {\bf{g}}}}{{{v_0}^2}}\,\, - \,\,\frac{{rg}}{{{v_0}^2}}} \\\\\\\\{t_1}\,\,\, = \,\,\,\frac{{{v_0}}}{g}\,\,\left( {\alpha \,\, - \,\,\beta } \right)\,,\,\,\,\,\,\,\,\,\,\,\,\,{t_2}\,\,\, = \,\,\,\frac{{{v_0}}}{g}\,\,\left( {\alpha \,\, + \,\,\beta } \right)\\\\{\rm{Check}}:\,\,\,\,\,\,{t_1}\,{t_2}\,\,\, = \,\,\,\frac{{{v_0}^2}}{{{g^2}}}\,\,\left( {{\alpha ^2}\,\, - \,\,{\beta ^2}} \right)\,\,\, = \,\,\,\frac{{{v_0}^2}}{{{g^2}}}\,\,\frac{{2rg}}{{{v_0}^2}}\,\,\, = \,\,\,\frac{{2r}}{g}\end{array}}$

As a final check, let's work out Case 2 above and compare with the results found with the interactive
diagram:

$\Large{\begin{array}{l}\frac{{rg}}{{{v_0}^2}}\,\,\, = \,\,\,\frac{{(1.81)(9.8)}}{{{{(7)}^2}}}\,\,\, = \,\,\,0.362\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\1\,\, + \,\,\frac{{{\bf{r}} \cdot {\bf{g}}}}{{{v_0}^2}}\,\,\, = \,\,\,1\,\, - \,\,\frac{{rg}}{{{v_0}^2}}\,\sin \phi \,\,\, = \,\,\,1\,\, - \,\,0.362\,\,\sin 30^\circ \,\,\, = \,\,\,0.819\\\\\\\alpha \,\,\, = \,\,\,\sqrt {0.819 + 0.362} \,\,\, = \,\,\,1.087\\\\\beta \,\,\, = \,\,\,\sqrt {0.819 - 0.362} \,\,\, = \,\,\,0.676\\\\\\{t_1}\,\,\, = \,\,\,\frac{{{v_0}}}{g}\,\,\left( {\alpha \,\, - \,\,\beta } \right)\,\,\, = \,\,\,\frac{7}{{9.8}}\,\,\left( {1.087\,\, - \,\,0.676} \right)\,\,\, = \,\,\,0.294s\\\\{t_2}\,\,\, = \,\,\,\frac{{{v_0}}}{g}\,\,\left( {\alpha \,\, + \,\,\beta } \right)\,\,\, = \,\,\,\frac{7}{{9.8}}\,\,\left( {1.087\,\, + \,\,0.676} \right)\,\,\, = \,\,\,\,1.26s\\\\\\{\bf{\hat r}}\,\,\, = \,\,\,\left\langle {\cos 30^\circ \,,\,\,\sin 30^\circ } \right\rangle \,\,\, = \,\,\,\left\langle {0.866\,,\,\,0.5} \right\rangle \,,\,\,\,\,\,\,\,\,\,\,\,\,{\bf{\hat g}}\,\,\, = \,\,\,\left\langle {0\,,\,\, - 1} \right\rangle \\\\\\{{{\bf{\hat v}}}_{01}}\,\,\, = \,\,\,\frac{g}{{2{v_0}}}\,\left( {{t_2}\,{\bf{\hat r}}\,\, - \,\,{t_1}\,{\bf{\hat g}}} \right)\,\,\, = \,\,\,\frac{{9.8}}{{2(7)}}\,\left( {1.26\,\left\langle {0.866\,,\,\,0.5} \right\rangle \,\, - \,\,0.294\,\left\langle {0\,,\,\, - 1} \right\rangle } \right)\\\,\,\,\,~~~\,\,\,\,\,\, = \,\,\,\left\langle {0.764\,,\,\,0.647} \right\rangle \,\,\, = \,\,\,{e^{{\bf{i}}\left( {40.3^\circ } \right)}}\\\\\\{{{\bf{\hat v}}}_{02}}\,\,\, = \,\,\,\frac{g}{{2{v_0}}}\,\left( {{t_1}\,{\bf{\hat r}}\,\, - \,\,{t_2}\,{\bf{\hat g}}} \right)\,\,\, = \,\,\,\frac{{9.8}}{{2(7)}}\,\left( {0.294\,\left\langle {0.866\,,\,\,0.5} \right\rangle \,\, - \,\,1.26\,\left\langle {0\,,\,\, - 1} \right\rangle } \right)\\\,\,\,\,\,\,~~~\,\,\,\, = \,\,\,\left\langle {0.178\,,\,\,0.985} \right\rangle \,\,\, = \,\,\,{e^{{\bf{i}}\left( {79.8^\circ } \right)}}\end{array}}$