OT: Looking for a good physics site
 

I'm trying to code intercept calculations for two spaceships, both capable of velocity, acceleration and different positions.

I don't want hints. I have been working on this code off and on for months. I don't care about figuring it out myself. I just want to know how to code it.

Many explicitives were deleted in the making of this post.

Re: OT: Looking for a good physics site
 

Is the target ship going to have a constant acceleration, or will it actively evade?

Re: OT: Looking for a good physics site
 

I'm just looking for straight-line acceleration right now. More complex stuff can wait.

Thanks for any and all help.

Re: OT: Looking for a good physics site
 

Also, I just want the enemy ship to hit the target ship - Missile interception. I can probably manage to work out how to match velocities from that.

Re: OT: Looking for a good physics site
 

Wasn't that where you were having all the problems before?
Reversing thrust at half way causing all kinds of grief?

Re: OT: Looking for a good physics site
 

I've gone through a number of calculations. None of them worked completely. The closest I got was an angle calculation that works for velocity only.

I have never gotten any kind of interception that works perfectly for velocity, acceleration and position, either missile interception or matching interception.

And by 'perfectly', I mean, 'Intercepts, in a straight line, a target moving in a straight line'.

Re: OT: Looking for a good physics site
 

I'll see if I can work out the math for a least-time position only intercept, assuming the target has constant acceleration and the only variable is what direction the missile accelerates in.

Re: OT: Looking for a good physics site
 

Well, they also start at different positions.

Thanks, wether you manage it or not.

Re: OT: Looking for a good physics site
 

That's not physics, that's math (mind you, Physics IS math, but nowhere are you actually needing to deal with things like, oh, gravity - you're just looking at acceleration).

A position under constant acceleration vs. time for a given axis has an equation that can be formatted as:

P(T)=(1/2)A(T^2)+VT+S

P(T): Position at time T (the function)
T: Time (the only variable)
A: Acceleration (a constant for this exercise)
V: Initial Velocity (at T = 0; a constant)
S: Starting position (P(0); a constant)

You're looking to calculate an intercept - well, you've got a lot to worry about, but basically, you're looking for a way to get two different position functions at the same point at the same time:

So basically, you have P1(T) = P2(T), and want T and A (do I need to reword that?).

As each object has a different starting position (otherwise, T=0 always works; this is the definition of interception), Acceleration, and initial velocity are likely to be different; you're looking to solve:

(1/2)<font color="red">A</font>(T^2)+<font color="red">V</font> T+<font color="red">S</font> = (1/2)<font color="green">A</font>(T^2)+<font color="green">V</font> T+<font color="green">S</font>

Do note that your intercepting ship can potentially control it's acceleration. Well, let's start manipulating:

(1/2)<font color="red">A</font>(T^2)+<font color="red">V</font> T+<font color="red">S</font> = (1/2)<font color="green">A</font>(T^2)+<font color="green">V</font> T+<font color="green">S</font>

(1/2)(<font color="red">A</font> - <font color="green">A</font>)(T^2)+(<font color="red">V</font> - <font color="green">V</font>)T+(<font color="red">S</font>- <font color="green">S</font>) = 0

Oh, hey - we now have a quadratic equation; sub (1/2)(<font color="red">A</font> - <font color="green">A</font>) with a, (<font color="red">V</font> - <font color="green">V</font>) with b, and (<font color="red">S</font>- <font color="green">S</font>) with c, and T with x. The Quadratic formula will do the job:

x = (-b +/- sqrt(b^2 - 4ac))/(2a)

Note: a is a variable, as the interceptor will need to choose it. You've got a variable on both sides - but that's okay, as you'll want at least two dimensions, and you've got a separate equation for each dimension; you'll end up with a system of equations (one for each dimension) and some relationship between the a on the various axes (usually the Pythagorean therom).

Edit:
That is, of course, assuming you don't want to stop at the end, other than ramming the target....

Re: OT: Looking for a good physics site
 

Tankee! Tankee very much!

...But, um, if the enemy acceleration needs to be chosen to use the formula and the formula tells it in which direction to accelerate and thus, how much to accelerate...

...Isn't that a closed loop?

/me goes off to see what happens if /me uses that by inputing '0' for the enemy acceleration.

...Uh, the first one is the pursuer, right?

...'+/-'?

Re: OT: Looking for a good physics site
 

Aren't those pretty much the equations we came up with last time, Narf? I think we used a slightly different form of acceleration equation, but it amounts to the same thing.

Re: OT: Looking for a good physics site
 

There's a resemblance, but I'm pretty sure there was no distance, 4ac or 4a.

In any case, something about my implementation is off, possibly having something to do with the questions I asked.

Re: OT: Looking for a good physics site
 

Nope. You have a second, identical equation for the other dimension (or more - it's extendable to as many as you like; most games will use two (X, Y), some will use three (X, Y, Z) but you can have 30 if you feel like it), and you'll have a relationship between the two based on the overall max acceleration for the pursuer (based on the pythagorean therom; max acceleration = sqrt(a(x)^2 + a(y)^2 + a(z)^2 + ... ) and the persuer is picking an acceleration in the x, y, z, ... axises (axen? axes? - what's the plural of axis?); you'll end up with a system of equations, based on how much you're simulating.

It's simultaneous equations. If you've got four variables and three equations that relate them, you can usually solve for everything in terms of a single variable.
Whichever you'd prefer; the exact choice is completely immaterial. Math's like that, sometimes.
The Quadratic formula gives two results. One will usually give you negative time, which you discard. Occasionally it will come back with an imaginary number (by way of the square root of a negative) - in which case, interception is not possible.

Oh, and there's a REASON most games use flat velocities, rather than accelerations.

The simple way to do it, if you're going a little bit at a time anyway, is to simply accelerate in the direction of the target. The computer will have some overshooting issues, and it won't be the most efficient course, but it'll do the job eventually.

Re: OT: Looking for a good physics site
 

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.

+/- 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.

===

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.

Re: OT: Looking for a good physics site
 

"axes," pronounced like "ax'eees."

Re: OT: Looking for a good physics site
 

Nope. You have a second, identical equation for the other dimension (or more - it's extendable to as many as you like; most games will use two (X, Y), some will use three (X, Y, Z) but you can have 30 if you feel like it), and you'll have a relationship between the two based on the overall max acceleration for the pursuer (based on the pythagorean therom; max acceleration = sqrt(a(x)^2 + a(y)^2 + a(z)^2 + ... ) and the persuer is picking an acceleration in the x, y, z, ... axises (axen? axes? - what's the plural of axis?); you'll end up with a system of equations, based on how much you're simulating.

Er, yes, I'm currently using it for two axis. Three would probably result in a significantly smaller consumer base. Four would limit it to only the seriously dedicated, which does not include me, in both desire and ability.

I do know that equation. My problem is the whole 'Pursuer picking an acceleration' thing. See, the equation helps pick the pursuers' acceleration and the pursuers' acceleration is part of the equation...

Um, there must be something obvious I'm not seeing. Possibly it's in the communication.

It's simultaneous equations. If you've got four variables and three equations that relate them, you can usually solve for everything in terms of a single variable.

I get the general concept. However, my math skills are abysmal. I still carry a grudge with the school system here over that; they were near-genius level before I entered it. (I have a psychiatric assesment to prove it, too).

I am trying to raise my math skills. I have learnt a few things bashing my head against this.

Whichever you'd prefer; the exact choice is completely immaterial. Math's like that, sometimes.

Er, do you mean I can put '0' for the pursuers' acceleration or I can use either for the pursuer or both? Because I can see the second, but I'm not sure about the first.

The Quadratic formula gives two results. One will usually give you negative time, which you discard. Occasionally it will come back with an imaginary number (by way of the square root of a negative) - in which case, interception is not possible.

The negative time would explain the enemy ship accelerating south if it's headed south and sufficiently southward of the target.
If the result is not a number or infinity, it is discarded.

So, uh, how do I make sure the formula is giving me a good result?

Oh, and there's a REASON most games use flat velocities, rather than accelerations.

My left brain is perfectionist, precise and thorough. I believe you are all sufficiently aquanted with my right brain...

...Besides, one of the main features of this game will be multiple FTL and STL systems.

The simple way to do it, if you're going a little bit at a time anyway, is to simply accelerate in the direction of the target. The computer will have some overshooting issues, and it won't be the most efficient course, but it'll do the job eventually.

That would be a whole lot easier. But my brain has consistantly discarded that option in favour of the better (In the end) result.

Although it is getting tempting.

Ok, my brain is seriously fried and it's rather late. I'll look at this again tomorrow.

Re: OT: Looking for a good physics site
 

A variation on this which may (or may not) work better would be:

Missile looks at where target is now, and figures how much time it would take the missile to accellerate in a straight line and get to that point. Lets say it predicts it could get there at time U.

Missile now predicts where target will be at time U based on target's current course and accelleration. Lets call this Location K. (I like picking random letters to use a variables, it bugs reals mathematicians/ scientists=-).

Missile begins accelerating towards K.

Every time the target changes course or acceleration, go back to square one.

When missile gets within a certain "kill range" of the target, it startss accellerating directly toward the target instead of some predicted future position.

Again, it's not perfect. Overshooting may still occur (but less so than if always accellerating directly at the target), and it might do crazy things if the target takes violent evasive maneuvers, but it seems to me it ought to generate something akin to a decent intercept. As a bonus, it only has to recalculate when the target changes course, instead of all the time, thus saving you CPU cycles.

Just a thought.

Re: OT: Looking for a good physics site
 

@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?

Re: OT: Looking for a good physics site
 

The problem is in framing the question. So let me ask you about what you are trying to achieve: Does a dog chase a cat by predicting where the cat will be next frame? Does a Cheetah hunt a deer by moving towards a calculated destination? The answer of course is no.

What we are trying to implement here can be described by fighter pilots: The three modes of persuit are Pure Persuit, Lag Persuit, and Lead Persuit. Pure Persuit is probably most familiar. It involves pointing your nose directly at the target. Heat seaker missiles follow this course. Unfortunately, so do a lot of weak AI opponants. It is also fairly easy to outmaneuver; just turn so that the enemy is sleightly behind you and you will arc around him.Lag Persuit is what a dog does to a person. He points his nose behind the target, and this changes the way his target is facing fastest. It brings him around to the rear of his opponant.Lead Persuit is what is what is meant by an "intercept course", and involves pointing your nose in front of your opponant. It's what children playing a game of tag do.

On to an intercept algorithm. Instead of percieving the universe as a set of coordinates from "above", consider a 'ground-level' view. What you can do is turn to place him at any angle from you. You could turn to put him in your 12:00--dead ahead. What you can watch is what angle he is presenting to you. If you are in his 6:00 he is running straight away from you.

Let's start by saying you are both the same speed. We'll add speed difference in a moment. If he is facing directly at you, or directly away from you, then you should turn to face directly at him. What if he is facing so that you are in his 9:00 or 3:00 -- directly abreast of his run? Then you will have to run in the same direction on a parallel course until something changes. Do that by turning to put him in your 3:00/9:00.

So now we have straight-on, and parallel-course; 0~90 degrees, based on his presentation of 0~90 degrees. these are the two extremes, and the only thing left is to figure out what the middle values should be. It's easy enough for dumb animals, so we should be able to 'get it'. As the target turns more and more away from dead-on, so should you. Atan2 of his angle gives the fraction.

Now about that speed difference. If you are moving twice as fast as the target, then cut the angle change in half. Pretty cool, huh?

Re: OT: Looking for a good physics site
 

You just gave me a formula I already tried. It worked for velocity.

Also, you probably didn't mean to, but your post comes across as condescending.

Re: OT: Looking for a good physics site
 

Sorry, I cut/pasted from a post I did in another web site; the attitude probably comes from some hostility experienced there.

Re: OT: Looking for a good physics site
 

No problem.

Hey, if that thread has the solution, could you point me to it?

Re: OT: Looking for a good physics site
 

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.

Re: OT: Looking for a good physics site
 

...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.

Re: OT: Looking for a good physics site
 

...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

Re: OT: Looking for a good physics site
 

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.

Re: OT: Looking for a good physics site
 

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.

Re: OT: Looking for a good physics site
 

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?

Re: OT: Looking for a good physics site
 

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.

Re: OT: Looking for a good physics site
 

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.

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?

Re: OT: Looking for a good physics site
 

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.

Re: OT: Looking for a good physics site
 

(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)

Re: OT: Looking for a good physics site
 

(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

Re: OT: Looking for a good physics site
 

Could someone tell me why '1/2 * 5 * 3^2 - 117.5 + 50 = 45?' returned -45? Thanks. I can't see the error.

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

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

(1/2) * A * T^2 + V * T - D = P
(1/2) * 5 * 3^2 + 15 * 3 - 117.5 = 50?
2.5 * 9 + 45 - 117.5 = 50?
22.5 + 45 - 117.5 = 50? // Don't think so...
-50 != 50

Um...Where, exactly, did I make a mistake?

Re: OT: Looking for a good physics site
 

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).

Re: OT: Looking for a good physics site
 

Ok, thanks. That makes sense.

My sympathies on the fried brain.

Re: OT: Looking for a good physics site
 

It should probably be noted that constant acceleration is a basic premise of these formulae; to get a join-fleet, your acceleration will have to vary over time unless you are already in the same location (in which case, A1 = A2 for all time).

Re: OT: Looking for a good physics site
 

Yeah - they're the dumb-fire (accelerates in a straight line in whatever direction you pointed it - doesn't make decisions in-flight) missile targeting solution (goal: HIT THE TARGET HARD) for a target that isn't evading (e.g., another dumb-fire missile). If the target is evading, or you can track target mid-flight, you want something a little more organic (such as a simple "accelerate in direction of current target's location")

Re: OT: Looking for a good physics site
 

I was thinking I could probably get a matching intercpt by treating the targets' velocity and acceleration as 0, by adding them (Or perhaps subtracting them) from or to the enemies' velocity and acceleration.

Or something like that.

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

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

(1/2) * A / D^2 + V / D + P = T
1/2 * 5 / 117.5^2 + 15 / 117.5 + 50 = 3?
1/2 * 0.042 + 0.127 + 50 = 3?
0.021 + 0.127 + 50 = 3? // It was at this point that my suspicions became rather definite...
50.148 != 3

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

(1/2) * A * D^2 + V * D + P = T
1/2 * 5 * 117.5^2 + 15 * 117.5 + 50 = 3?
2.5 * 13,806.25 + 1,762.5 + 50 = 3? // Yeah, didn't think so.
34,515.625 + 1,812.5 != 3
36,328.125 != 3 // Curiousity.

(1/2) * A * D^0.5 + V / D + P = T?
1/2 * 5 * 117.5^0.5 + 15 / 117.5 + 50 = 3?
2.5 * 10.84 + 0.128 + 50 = 3? // Erm, no...
77.228 != 3.

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

(1/2) * A / D + V / D + P = T^2 + T
1/2 * 5 / 117.5 + 15 / 117.5 + 50 = 12?
1/2 * 0.043 + 0.128 + 50 = 12?
0.0215 + 0.128 + 50 = 12? // Er, no...
50.1495 != 12

Ok, my brain is frying...

Re: OT: Looking for a good physics site
 

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

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

T^2 + T = (1/2) * A + V + P?
3^2 + 3 = 1/2 * 5 + 15 + 50?
9 + 3 = 2.5 + 15 + 50? // Erm, no...
12 = 17.5 + 50?
12 != 67.5

T^2 * T = (1/2) * A + V + P?
3^2 * 3 = 1/2 * 5 + 15 + 50?
9 * 3 = 2.5 + 15 + 50? // Fiddle, fiddle...
27 = 17.5 + 50?
27 != 67.5

T^2 * T = (1/2) * A + V + P + D?
3^2 * 3 = 1/2 * 5 + 15 + 50 + 117.5?
9 * 3 = 2.5 + 15 + 50 + 117.5? // ...Lute?...
27 = 17.5 + 50 + 117.5?
27 != 67.5 + 117.5?

T^2 * T = ((1/2) * A + V) * 2 + P + D?
3^2 * 3 = (1/2 * 5 + 15) * 2 + 50 + 117.5?
9 * 3 = (2.5 + 15) * 2 + 50 + 117.5? // ...Mumble...
27 = 17.5 * 2 + 50 + 117.5?
27 = 35 + 50 + 117.5?
27 != 202.5

T = (((1/2) * A + V) * 2 + P + D) / ((1/2) * A + V + P)?
3 = ((1/2 * 5 + 15) * 2 + 50 + 117.5) / (1/2 * 5 + 15 + 50)?
3 = ((2.5 + 15) * 2 + 50 + 117.5?) / (2.5 + 15 + 50)?// ...Hmm...
3 = (17.5 * 2 + 50 + 117.5) / (17.5 + 50)?
3 = (35 + 50 + 117.5) / (17.5 + 50)?
3 = 202.5 / 67.5
3 = 3!

A = 8
T = 4
P = 35
V = 25
(1/2) * A * T^2 + V * T + P = D

1/2 * 8 * 4^2 + 25 * 4 + 35 = D
4 * 16 + 100 + 35 = D
64 + 135 = D
199 = D

A = 8
T = 4
P = 35
V = 25
D = 199

T = (((1/2) * A + V) * 2 + P + D) / ((1/2) * A + V + P)?
4 = ((1/2 * 8 + 25) * 2 + 35 + 199) / (1/2 * 8 + 25 + 35)?
4 = ((4 + 25) * 2 + 234) / (4 + 60)?
4 = (29 * 2 + 234) / 64?
4 = (58 + 234) / 64?
4 = 292 / 64?
4 != 4.5625 // Nope...Error check...Still nope. So close, yet so far.