OT: Looking for a good physics site

[narf poit chez BOOM]

Ah. So I was in the general area.

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

[Suicide Junkie]

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)

[narf poit chez BOOM]

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.

[Suicide Junkie]

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)

[narf poit chez BOOM]

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.

[narf poit chez BOOM]

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?

[Suicide Junkie]

(Position difference) = (speed diff) * T + (Accel Diff) * T * T

After time T, you want the position change due to Speed and Accel to be equal to the current position difference.

I think I missed a negative on the left side:
X1=0; X2=4
V1=0; V2=0
A1=2; A2=0
-(0-4) = 0-0 *T + (2 - 0) * T^2
4 = 2 T^2
T = sqrt(2)

[narf poit chez BOOM]

(1/2) * A * T^2 + V * T + P = D

D - V * T + P = (1/2) * A * T^2

A = 5
T = 3
P = 50
V = 15
D = 117.5

117.5 - 15 * 3 + 50 = (1/2) * 5 * 3^2
117.5 + -45 + 50 = 2.5 * 9
72.5 + 50 = 22.5 // Ok, I know this isn't going to work...
122.5 != 22.5

D - V * T - P = (1/2) * A * T^2
72.5 - 50 = 22.5
22.5 = 22.5

Woot!

BBS

[Jack Simth]

narf poit chez BOOM said:
Why is it equal to 0, though?

Brain fried; can't answer post fully at the moment.

That aspect, though...

It's 0 because it's a manipulation of two separate position functions that I want to be in the same place at the same time (definition of interception). They are both position functions, and I want them equal, so I assigned them that way, and tossed the same t in there: p1(t) = p2(t). I then expanded the functions to their components, and merged them onto one side (leaving the other side as 0).

[Jack Simth]

narf poit chez BOOM said:
@Jack: That formula you gave me gives Time, right?

They're simultaneous equations - they give time (which can be later used for the x/y of the encounter), and acceleration on both axes when solved together.