Re: OT: Looking for a good physics site
 

That would happen when you are just barely able to intercept it.

One thing to note is that if you try to burn at 100% power up to halfway, and then decelerate at 100% power the rest of the way... then you have NO ACCEL left over to correct for rounding errors.

You should probably have the ship try to use 90% of its maximum accel, and try more thrust if that isn't enough...
If both ships are of equal accel, you've got to just head for it and hope the target isn't actively trying to avoid you.

Re: OT: Looking for a good physics site
 

That's a good suggestion. And it would fit in with how engine sare generally operated - Few are run at 100% power.

Re: OT: Looking for a good physics site
 

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

No, I don't mean that. I mean every quadratic function has 2 zeros/roots/x-intercepts. They are only a function of where the graph of the quadratic equation in question crosses the x axis. There will usually be two such locations, due to quadratic polynomial equations being parabolas and all (a parabola that has its vertex on the x axis will have both root solutions equal to each other).

The concept of zeros/roots/x-intercepts is unrelated to the problem of calculating where two ships should intercept each other in space. There is a lot of overlapping terminology, which is probably the root (hehehe) of the confusion. The zeros/roots/x-intercepts of the particular function we derived happen to be the time values (T) when the two objects will be in the same location at once, which is why they are useful for the solution.

Re: OT: Looking for a good physics site
 

Ah. So I was in the general area.

So...Can I get a hint on where I'm implementing it wrong?

Re: OT: Looking for a good physics site
 

You're getting out of the intercept routine a time and a location right?

The ship should simply accelerate at the rate passed into the routine, and in such a direction as to end up at the returned location at the returned time.

That part is just another (Once for each dimension) application of the base formula S(t) = S(0) + Vt +A(t^2).

Within rounding error, the total acceleration will be equal to the acceleration passed into the routine originally.
(Don't forget Pythagoras)

Re: OT: Looking for a good physics site
 

Uh, I think the formula Jack gave me gives me a time, if used properly.

I have said this before, but to repeat: What rate of acceleration do I pass into the routine and/or where do I get it?

To wit: I have Formula X and Formula Z which require an acceleration number. The acceleration in the X and Z directions will determine which direction the ship will accelerate in. I'm using Formula X and Formula Z (Formula being the one Jack posted) to try to determine in which direction to accelerate. Am I using it wrong or something? Becuase there seems to be some sort of basic miscommunication.

Oh, and I'm only getting one result per axis from that formula. I have no idea how I would get two results out of that formula.

Re: OT: Looking for a good physics site
 

The acceleration for X will be cos(theta) times the ship's engine power, while the acceleration for Y will be sin(theta) times the engine power.

And theta will be the direction to accelerate in.

So, lets see. (x,y are coords, * is multiplication)
For X:
S1x-S2x = (V1x-V2x) * T + ((A1*cos(Theta)) -A2x) * T * T
For Y:
S1y-S2y = (V1y-V2y) * T + ((A1*sin(Theta)) - A2y) * T * T

Unknowns are Theta and T, common to both.
You can rearrange the first formula to get Theta = F(T)... there will be inverse cos or inverse sin in it.
Replace the Theta in the second equation with F(T).

Now you've got a big horrid, fearsome beasty with only T as an unknown. Solve for T. You'll probably need a big sheet of paper, or maybe maple to simplify it down to a one-line equation to code in.

Plug T into the first formula again, and Solve for Theta.

Your ship/missile can then accelerate in the direction of Theta for time T. (Theta = 0 means the +X direction, 90 degrees = +Y)

Re: OT: Looking for a good physics site
 

I'm not so much solving for T as making a connection between dimensions out of it. It gets around having to play with sine and cosine.

Ax and Ay are related by the equation: Amax = sqrt(Ax^2+Ay^2) (for two dimensions, to get soonest interception for a "ram target" scenario). Amax is a constant. You now have three equations:

(1/2)(<font color="red">Ax</font> - <font color="green">Ax</font>)(T^2)+(<font color="red">Vx</font> - <font color="green">Vx</font>)T+(<font color="red">Sx</font>- <font color="green">Sx</font>) = 0
(1/2)(<font color="red">Ay</font> - <font color="green">Ay</font>)(T^2)+(<font color="red">Vy</font> - <font color="green">Vy</font>)T+(<font color="red">Sy</font>- <font color="green">Sy</font>) = 0
<font color="red">Amax</font> = sqrt(<font color="red">Ax</font>^2+<font color="red">Ay</font>^2)

We have many symbols:
T: Unknown, time of interception.
<font color="red">Ax</font>: Unknown, acceleration in X for the chaser.
<font color="green">Ax</font>: Known, acceleration in X for the chased.
<font color="red">Vx</font>: Known, starting velocity in X for the chaser.
<font color="green">Vx</font>: known, starting velocity in X for the chased.
<font color="red">Sx</font>: known, starting position in X for the chaser.
<font color="green">Sx</font>: known, starting position in X for the chased.
<font color="red">Ay</font>: unknown, acceleration in y for the chaser.
<font color="green">Ay</font>: known, acceleration in y for chased.
<font color="red">Vy</font>: known, starting velocity in y for chaser.
<font color="green">Vy</font>: known, starting velocity in y for chased.
<font color="red">Sy</font>: known, starting position in y for chaser.
<font color="green">Sy</font>: known, starting position in y for chased.
<font color="red">Amax</font>: known, based on the pursuer - maximum acceleration used(which is the acceleration you'll normally want to use for the chaser, and will likely be some arbitrary % of "true max").

However, we've only got three unknown values: T, <font color="red">Ax</font>, and <font color="red">Ay</font>. We have three equations that relate them (if we had three dimensions, we'd have a fourth unknown, but a fourth equation to go with it, so they don't matter).

I'm a little too tired to do the brute-force logic manipulations on this at the moment (overtime at work) but at this point you should be looking at nothing more than time-consuming symbol manipulation and replacement set to solve for all of them. You do them all at essentially the same time. Manipulate the equations until you have each in terms of the other three, then replace. Eventually, you'll get one unknown on one side and a (somewhat big) equation of knowns on the other - at which point you _finally_ apply actual math to it (or rather, tell the computer to do so). Then that value is "known". Once that value is known, you plug it into the spots on the other equations, and repeat the process.

Re: OT: Looking for a good physics site
 

The acceleration for X will be cos(theta) times the ship's engine power, while the acceleration for Y will be sin(theta) times the engine power.

And theta will be the direction to accelerate in.

Got that far. It's the whole 'How do I calculate theta' that's the problem.

S1x-S2x = (V1x-V2x) * T + ((A1*cos(Theta)) -A2x) * T * T

V1x would be Enemy Velocity X, A1 would be Acceleration Vector Length, Theta would be the angle to accelerate in, A2x would be Target Acceleration X and T would be Time?
What are S1 and S2?

Unknowns are Theta and T, common to both.
You can rearrange the first formula to get Theta = F(T)... there will be inverse cos or inverse sin in it.
Replace the Theta in the second equation with F(T).

What is 'F'?

Now you've got a big horrid, fearsome beasty with only T as an unknown. Solve for T. You'll probably need a big sheet of paper, or maybe maple to simplify it down to a one-line equation to code in.
Now I'm wondering which variation of that formula you gave I should use to figure out T or Theta or whatever it's supposed to give me.
Plus, I think that 'Big sheet of paper' part may put it somewhat outside my mathematical ability.

Plug T into the first formula again, and Solve for Theta.

Your ship/missile can then accelerate in the direction of Theta for time T. (Theta = 0 means the +X direction, 90 degrees = +Y)
Yeah, I may need a bit of help to figure out exactly how to do that.

Going to have to fiddle with the angle, then. C# uses radians for 3d object angles and 0 is +Y (Although I'm using X and Z for horizontal and vertical).

@Jack: That formula you gave me gives Time, right?

I'll go let my brain collapse for a while, then come back and see if I can figure this out. If brute force can figure it out, then I may be able to turtle my way through before my brain starts screaming at me.

Re: OT: Looking for a good physics site
 

Let's see if I can de-rust my brain...

Ok, we have: (1/2)(Ax - Ax)(T^2)+(Vx - Vx)T+(Sx- Sx) = 0

Which can be shortened down to (1/2) * A * T^2 + V * T + P = 0
A being Relative Acceleration,
T being Time,
V being Relative Velocity,
P being Position. Because using an S is not intuitive for me.

Why is it equal to 0, though? Pulling numbers out of a hat,
1/2 * 5 * 3^2 + 15 * 3 + 50
2.5 * 9 + 45 + 50
22.5 + 45 + 50
117.5

Um, not equal to zero. That would be total distance covered.
So: (1/2) * A * T^2 + V * T + P = D
D being Total Distance

(1/2) * A * T^2 - D + P = V * T?
1/2 * 5 * 3^2 - 117.5 + 50 = 45?
2.5 * 9 - 117.5 + 50
22.5 + -117.5 + 50
-45

Er, well, sorta. Except that should probably be positive. Also, the reply form should be larger.

(1/2) * A * T^2 + D - P = V * T?
1/2 * 5 * 3^2 + 117.5 - 50 = 45?
2.5 * 9 + 117.5 - 50
22.5 + 117.5 + -50
90

Nope...However, it would work for (1/2) * A * T^2 - D + P = -(V * T)

How am I doing so far?