OT: Looking for a good physics site

[narf poit chez BOOM]

@Dogscoff: That would generate a pretty good intercept, but it wouldn't be perfect. To get a perfect intercept, you also have to calculate how far the target moves while the enemy is moving to where it will be then. Thanks for the idea, though.

There was distance. In fact, I'm pretty sure it was the position equation, now that I think about it. We never plugged it into the zeros-solving quadratic equation, hence no 4ac.

By 'distance' I meant we didn't take into account position. That wasn't worded well.

+/- is from the quadratic equation. There are two zeros (values of x where the curve intercepts the x axis and y is zero) in the solution to every quadratic equation. The handy cheat sheet formula compacts them into the same function, which varies only by making one version a + and the other a - where that +/- symbol goes.

So I should make two formulas, one with a '+' there and the other with a '-' there and if one returns a zero, I use the other?

The thing about the position function is that it assumes constant acceleration of both parties. It is calculating how much time passes until the objects intercept. From this, you can plot out where they intercept and get the necessary angle of motion. Remember that you had to do this every so often to find a new intercept heading, based on changes to the acceleration vectors of both objects.
I did get that.

This is what I have currently.

...Erm, does that formula give the distance that will be covered or the time to cover it?

[Fyron]

By 'distance' I meant we didn't take into account position. That wasn't worded well.

Yes we did; the very definition of intercepting a target object is when the positions are equal. We have two position equations, with two unknowns (time and position). Since we know the positions must be equal, we can simply substitute one of the equations for the S value (position) in the other equation, thus eliminating it.

So I should make two formulas, one with a '+' there and the other with a '-' there and if one returns a zero, I use the other?

No... both will return zeros of the quadratic function (aka x-intercepts). I wasn't talking about the integer value 0. A negative zero/intercept is thrown out, as per previous post.

[narf poit chez BOOM]

...I'm missing something here. You're saying the positions must be equal, but at the same time they start unequal.

No... both will return zeros of the quadratic function (aka x-intercepts). I wasn't talking about the integer value 0. A negative zero/intercept is thrown out, as per previous post.
...I have no idea what you just said. I can't even decypher the grammer. For one thing, the only zero I've ever known is an integer, except for the fact that zero isn't actually all that numerical.

[Fyron]

...I'm missing something here. You're saying the positions must be equal, but at the same time they start unequal.

Huh? Unknown and unequal are not the same thing.

...I have no idea what you just said.

http://en.wikipedia.org/wiki/Zero_of_a_function

[narf poit chez BOOM]

I'm missing something. Probably something obvious. Are you saying the function backtracks from where the positions are equal to where they are at start?

Ok, so occasionally the formula will return zero.

Perhaps it would help if you looked at my code? It could help us figure out the difference between what I think I should be doing versus what I should actually be doing.

[Suicide Junkie]

The two final positions are variables, and unknown. But they are equal to each other by definition, since equal positions is the definition of interception.
Knowing that they are both equal reduces the number of unknowns by one (Instead of positionA and positionB, you have two copies of positionIntercept). This allows you to solve for Time. If the interception is impossible, Time will end up being an imaginary number.

[narf poit chez BOOM]

Ok, thanks, that makes sense.

One of the problems is that it tends to flicker between thinking it is unsolveable and having an intercept angle. (I have a bit of code that gives it an angle of '0' if it's unsolveable). That's one reason I posted the code.

You guys are great, you know that, right?

[Fyron]

Ok, so occasionally the formula will return zero.

If one of the zeros/x-intercepts of the equation occurs at 0,0, then sure... The formula always returns two zeros/x-intercepts; that is the entire purpose of it.

[narf poit chez BOOM]

Oh, you mean it returns an intercept that is zero relative to the target.

Ok, so it should work. Now I just have to figure out why my implementation doesn't.