Hi.
When I was younger, I wrote a very simple flight combat game on the Commodore 64.
The plane could only fly in 8 directions UP (-40), Down (+40), Left (-1), Right (+1), and the four diagonals (-39, -41, +39, +41)
Those numbers represent the flight path since the system had a 40x25 text screen.
Anyway, when your plane would fire at an enemy aircraft, instead of showing the artillery on screen and checking to see if it collided with the enemy,
I made an immediate calculation on if it would would be a direct hit or not.
I would take the text location of your plane, subtract that from the text location of the enemy plane, and then divided this by your plane's current
direction.
Since I no longer have the game, I'll use the follow screen to show this example.
Imagine the "R" being your plane, and it's currently heading UP+RIGHT (-39) on the text screen.
The enemy plane is "*" that is also circled in red.
The "R" is at text location 1224.
The "*" is at text location 1068.
If I subtract the difference between these two and divide that with the plane's current direction (-39), the result would be (an absolute) 4 in this
scenario.
4 text locations from where the plane is heading at the time you fired.
Whenever I did that calculation and the result was a perfect integer, this would be a direct hit at the time you fired.
Left (-1) and Right (+1) directions would always yield such a result so I put a limit on distance, something like the number couldn't be above 12 or
so.
Anyway, all of that was done in a virtual 2D world.
My question is could such a formula also somehow apply to real life, 3D environments.
Say you have the exact known geo location of System A, whether by land or air, and also it's exact angle or mathematical direction.
Say you also have Target B's known geo location.
If you subtract these locations between the two of them and then divide that with System A's direction or aim, and the result is a positive integer,
could that mean System A's site is directly looking at Target B?
I really wanted to post this in "Ask a Physicist" thread, but it seems that has been locked so no more further replies are available anymore.
Sorry for the very long winded question.
I am, however, very excited for any possible nerdy responses on this.
Thanks ahead.