# Question about inertial reference frames

I am trying to develop some intuition regarding reference frames. Let's say I am trying to shoot a target moving at velocity $v$ parallel to my position such that the angle of impact with the target is 90 degrees to its direction of travel. If both the target and I were stationary, both of us would see the angle of impact as 90 degrees. If the target is moving, the angle I see the shell leave the gun should be different than the angle the target sees. Good. Now, lets say the target starts at some velocity v parallel to my position, I shoot at an angle based on my earlier calculations, but the bullet hits the target at an angle other than 90 degrees or misses it completely. The target at impact was still moving at velocity V. Assuming the target remained parallel to my position, I think I would be forced to conclude that the target accelerated at some point.

How in a military setting would you be able to improve target accuracy when there may be some unknown and variable degree of acceleration between firing the shell and the shell impacting the object?

• Are you just asking on how to calculate the lead angle with a moving target? Is the targeting system computer controlled or brain controlled? Commented Oct 2, 2017 at 19:54

Let us look at the geometry of the situation. The target path is a minimum distance $d$ from you. When the shot is fired the target is located a distance $b$ to the right of the "center" point and moving with speed $v$. The bullet is moving with speed $c \gg v$. At what angle $\theta$ do you need to shoot to hit the target, if the target is also accelerating with $v(t) = v+ a t$

The equations for the system are:

\begin{align} \sin \theta & = \frac{x}{c \,t} \\ \cos \theta & = \frac{d}{c \,t} \\ x & = -b + v t +\frac{1}{2} a t^2 \end{align}

Solving these when $a=0$ is doable, with $$\theta = \sin^{-1} \left( \frac{d \tfrac{v}{c} }{\sqrt{b^2+d^2}} \right) - \tan^{-1} \left( \frac{b}{d} \right)$$

and $$x = \frac{b c \sin \theta}{v - c \sin \theta}$$

But when $a \neq 0$ then math becomes too complex. Essentially there is 4-order polynomial in terms of time $t$ or distance $x$ that needs to be solved. Sorry.

The equation to be solved is

$$x = \frac{ a (d^2+x^2)}{2 c^2} + \frac{v}{c} \sqrt{d^2+x^2}-b$$