I was unfortunate enough to attend a high school that did not care about preparing students for college, but I'm studying some game programming, and everything programming I get, but I'm taking a Game AI course and we're learning about FSM's and AI movements, and I came across "normalize", looked into it and it kind of makes sense, but how would I use this in game programming? Maybe I just don't understand the concept that well (never really took an algebra class, although I use way higher level mathematics everyday in other programming, however I learn as I go), but I just don't get why I would need to do that.
Vectors represent both length and direction. Sometimes it's handy to use a "unit vector" (with length one) for multiplication, because it's easy to control the result's length.
Multiply a one-pound weight by five, and what do you have? A five-pound weight.
Multiply a unit vector by five, and what do you have? A vector pointing five units in that direction.
As a practical example, let's say I'm writing a camera system for a simple game: two cars race against each other on a straight track. The camera should orbit around the player's car while looking in the direction of the opponent.
Since you tagged the question with unity3d, I'll offer C# code that would compile in Unity3D. The basic concepts are the same for any use, though.
Let's assume we're given two points:
Vector3 playerPos, enemyPos;
Some operations we might find useful:
//find separation between two points
Vector3 playerToEnemy = enemyPos - playerPos;
//find distance between them
float distance = playerToEnemy.magnitude;
//find direction of separation
Vector3 enemyDirection = playerToEnemy.normalized;
Finally, we position the camera:
Vector3 camPos = playerPos; //start at player's position
camPos -= enemyDirection * 10f; //move ten units away from enemy
camPos += Vector3.up * 5f; //move five units up
Transform tr = Camera.main.transform;
tr.position = camPos;
tr.LookAt(enemyPos);
You may notice that I used one of Unity3D's built-in unit vectors (including Vector3.up, Vector3.right, and Vector3.forward). Every transform object in Unity also has local-space equivalents.
Another use: with vector dot products, you can check if two normalized vectors point in similar directions because the dot product's output will range from -1 (opposite directions) to 0 (perpendicular) to 1 (parallel).
Related
I've written a simple A* path finding algorithm to quickly find a way through a tile based dungeon in which the tiles contain the information of walls.
An example of a dungeon (only 1 path for simplicity):
However now I'd like to add a variable amount of "Bombs" to the algorithm which would allow the path-finding to ignore 1 wall. However now it doesn't find the best paths anymore,
for example with use of only 1 bomb the generated path looks like the first image here:
Edit: actually it would look like this: https://i.stack.imgur.com/kPoAA.png
While the correct path would be the second image
The problem is that "Closed Nodes" now interfere with possible paths. Any ideas of how to tackle this problem would be greatly appreciated!
Your "game state" will no longer only be defined by your location, but also by an integer representing the number of bombs you have left. If you're following the pseudocode of A* on wikipedia, this means you cannot simply implement the closedSet as a grid of booleans. It should probably be implemented as, for example, a hash map / hash set, where every entry holds the following data:
x coordinate
y coordinate
number of bombs left
By visiting a certain position in the search process, you'll no longer mark just that position as closed. You'll mark the combination of position + number of bombs left as closed. That way, if later on in the same search process you run into a position where you're at the same location, but have more bombs left, you will not ignore it as closed but will actually continue searching that possibility.
Note that, if the maximum possible number of bombs is relatively low, you could also implement the closedSet as an array of boolean grids, where you first index by number of bombs, then by x and y coordinates to find out if a specific position is closed or not.
Doesn't this mean that you just pretend to not have any walls at all?
Use A* to find the shortest path from start to end and then check how many walls you'd have to go through. If you have enough bombs, you can use the path. Otherwise, try the next longest path and so on.
By the way: you might want to check http://gamedev.stackexchange.com for questions like this one.
You need to tweak the cost function to cost something for a bomb, then run the algorithm normally with an infinite cost for a second bomb. To get the bomb approximately halfway, play about with cost function, it should probably cost about the heuristic A-B distance times the cost for an empty tile. If you have two bombs, half the cost and of course then use of three bombs costs infinite.
But don't expect very good results. A* isn't designed for that kind of optimisation.
First of all a disclaimer: I'm posting this question here even if I realize it is quite maths heavy, because I have trouble figuring out on what other site it could belong.
I'm writing a 2d spaceships game where the player will have to select the ship's destination and a course will be automatically plotted.
Along with this, I'm offering various options to control the ship's acceleration while it gets there. All these options have to do with the target velocity at the destination.
One option is to select the desired destination and velocity vector, in which case the program will use cubic interpolation, since starting and target coordinates and velocity are available.
Another option is to just select the destination point, but let the game calculate the final velocity vector. This is done through quadratic interpolation (ie. acceleration is constant).
I would like to introduce another option: let the player select the destination and the maximum absolute value of the velocity vector, as in
sqrt( vf_x^2 + vf_y^2 ) <= Vf_max
I take I shall use a 3rd order polynomial to model the course in this case, but I'm having a pretty hard time figuring out the coefficients, since I miss one equation for each coordinate (the one given by velocity at destination). Furthermore, I'm confused as to how I should use the Vf_max constraint to help me figure out the missing coefficients.
I suspect it could be an optimization problem, but I'm quite ignorant about this topic.
Can anybody help me find a solution or point me in the right direction, please?
I have balls bouncing around and each time they collide their speed vector is reduced by the Coefficient of Restitution.
Right now my balls CoR for my balls is .80 . So after many bounces my balls have "stopped" rolling because their speed has becoming some ridiculously small number.
In what stage is it appropriate to check if a speed value is small enough to simply call it zero (so I don't have the crazy jittering of the balls reacting to their micro-velocities). I've read on some forums before that people will sometimes use an epsilon constant, some small number and check against that.
Should I define an epsilon constant and do something like:
if Math.abs(velocity.x) < epsilon then velocity.x = 0
Each time I update the balls velocity and position? Is this what is generally done? Would it be reasonable to place that in my Vector classes setters for x and y? Or should I do it outside of my vector class when I'm calculating the velocities.
Also, what would be a reasonable epsilon value if I was using floats for my speed vector?
A reasonable value for epsilon is going to depend on the constraints of your system. If you are representing the ball graphically, then your epsilon might correspond to, say, a velocity of .1 pixels a second (ensuring that your notion of stopping matches the user's experience of the screen objects stopping). If you're doing a physics simulation, you'll want to tune it to the accuracy to which you're trying to measure your system.
As for how often you check - that depends as well. If you're simulating something in real time, the extra check might be costly, and you'll want to check every 10 updates or once per second or something. Or performance might not be an issue, and you can check with every update.
Instead of an epsilon for an IsStillMoving function, maybe you could use an UpdatePosition function, scheduled on an object-by-object basis based on its velocity.
I'd do something like this (in my own make-it-up-as-you-go pseudocode):
void UpdatePosition(Ball b) {
TimeStamp now = Clock.GetTime();
float secondsSinceLastUpdate = now.TimeSince(b.LastUpdate).InSeconds;
Point3D oldPosition = b.Position;
Point3D newPosition = CalculatePosition(b.Position, b.Velocity, interval);
b.MoveTo(newPosition);
float epsilonOfAccuracy = 0.5; // Accurate to one half-pixel
float pixelDistance = Camera.PixelDistance(oldPosition, newPosition);
float fps = System.CurrentFramesPerSecond;
float secondsToMoveOnePixel = (pixelDistance * secondsSinceLastUpdate) / fps;
float nextUpdateInterval = secondsToMoveOnePixel / epsilonOfAccuracy;
b.SetNextUpdateAt(now + nextUpdateInterval);
}
Balls moving very quickly would get updated on every frame. Balls moving more slowly might update every five or ten frames. And balls that have stopped (or nearly stopped) would update only very very rarely.
IMO your epsilon approach is fine. I would just experiment to see what looks or feels natural to the animation in the game.
Epsilon by nature is the smallest possible increment. Unfortunately, computers have different "minimal" increments of their own depending on the floating point representation. I would be very careful (and might even go higher than what I would calculate just for safety) playing around with that, especially if I want a code to be portable.
You may want to write a function that figures out the minimal increment on your floats rather than use a magic value.
As a programmer I think it is my job to be good at math but I am having trouble getting my head round imaginary numbers. I have tried google and wikipedia with no luck so I am hoping a programmer can explain in to me, give me an example of a number squared that is <= 0, some example usage etc...
I guess this blog entry is one good explanation:
The key word is rotation (as opposed to direction for negative numbers, which are as stranger as imaginary number when you think of them: less than nothing ?)
Like negative numbers modeling flipping, imaginary numbers can model anything that rotates between two dimensions “X” and “Y”. Or anything with a cyclic, circular relationship
Problem: not only am I a programmer, I am a mathematician.
Solution: plow ahead anyway.
There's nothing really magical to complex numbers. The idea behind their inception is that there's something wrong with real numbers. If you've got an equation x^2 + 4, this is never zero, whereas x^2 - 2 is zero twice. So mathematicians got really angry and wanted there to always be zeroes with polynomials of degree at least one (wanted an "algebraically closed" field), and created some arbitrary number j such that j = sqrt(-1). All the rules sort of fall into place from there (though they are more accurately reorganized differently-- specifically, you formally can't actually say "hey this number is the square root of negative one"). If there's that number j, you can get multiples of j. And you can add real numbers to j, so then you've got complex numbers. The operations with complex numbers are similar to operations with binomials (deliberately so).
The real problem with complexes isn't in all this, but in the fact that you can't define a system whereby you can get the ordinary rules for less-than and greater-than. So really, you get to where you don't define it at all. It doesn't make sense in a two-dimensional space. So in all honesty, I can't actually answer "give me an exaple of a number squared that is <= 0", though "j" makes sense if you treat its square as a real number instead of a complex number.
As for uses, well, I personally used them most when working with fractals. The idea behind the mandelbrot fractal is that it's a way of graphing z = z^2 + c and its divergence along the real-imaginary axes.
You might also ask why do negative numbers exist? They exist because you want to represent solutions to certain equations like: x + 5 = 0. The same thing applies for imaginary numbers, you want to compactly represent solutions to equations of the form: x^2 + 1 = 0.
Here's one way I've seen them being used in practice. In EE you are often dealing with functions that are sine waves, or that can be decomposed into sine waves. (See for example Fourier Series).
Therefore, you will often see solutions to equations of the form:
f(t) = A*cos(wt)
Furthermore, often you want to represent functions that are shifted by some phase from this function. A 90 degree phase shift will give you a sin function.
g(t) = B*sin(wt)
You can get any arbitrary phase shift by combining these two functions (called inphase and quadrature components).
h(t) = Acos(wt) + iB*sin(wt)
The key here is that in a linear system: if f(t) and g(t) solve an equation, h(t) will also solve the same equation. So, now we have a generic solution to the equation h(t).
The nice thing about h(t) is that it can be written compactly as
h(t) = Cexp(wt+theta)
Using the fact that exp(iw) = cos(w)+i*sin(w).
There is really nothing extraordinarily deep about any of this. It is merely exploiting a mathematical identity to compactly represent a common solution to a wide variety of equations.
Well, for the programmer:
class complex {
public:
double real;
double imaginary;
complex(double a_real) : real(a_real), imaginary(0.0) { }
complex(double a_real, double a_imaginary) : real(a_real), imaginary(a_imaginary) { }
complex operator+(const complex &other) {
return complex(
real + other.real,
imaginary + other.imaginary);
}
complex operator*(const complex &other) {
return complex(
real*other.real - imaginary*other.imaginary,
real*other.imaginary + imaginary*other.real);
}
bool operator==(const complex &other) {
return (real == other.real) && (imaginary == other.imaginary);
}
};
That's basically all there is. Complex numbers are just pairs of real numbers, for which special overloads of +, * and == get defined. And these operations really just get defined like this. Then it turns out that these pairs of numbers with these operations fit in nicely with the rest of mathematics, so they get a special name.
They are not so much numbers like in "counting", but more like in "can be manipulated with +, -, *, ... and don't cause problems when mixed with 'conventional' numbers". They are important because they fill the holes left by real numbers, like that there's no number that has a square of -1. Now you have complex(0, 1) * complex(0, 1) == -1.0 which is a helpful notation, since you don't have to treat negative numbers specially anymore in these cases. (And, as it turns out, basically all other special cases are not needed anymore, when you use complex numbers)
If the question is "Do imaginary numbers exist?" or "How do imaginary numbers exist?" then it is not a question for a programmer. It might not even be a question for a mathematician, but rather a metaphysician or philosopher of mathematics, although a mathematician may feel the need to justify their existence in the field. It's useful to begin with a discussion of how numbers exist at all (quite a few mathematicians who have approached this question are Platonists, fyi). Some insist that imaginary numbers (as the early Whitehead did) are a practical convenience. But then, if imaginary numbers are merely a practical convenience, what does that say about mathematics? You can't just explain away imaginary numbers as a mere practical tool or a pair of real numbers without having to account for both pairs and the general consequences of them being "practical". Others insist in the existence of imaginary numbers, arguing that their non-existence would undermine physical theories that make heavy use of them (QM is knee-deep in complex Hilbert spaces). The problem is beyond the scope of this website, I believe.
If your question is much more down to earth e.g. how does one express imaginary numbers in software, then the answer above (a pair of reals, along with defined operations of them) is it.
I don't want to turn this site into math overflow, but for those who are interested: Check out "An Imaginary Tale: The Story of sqrt(-1)" by Paul J. Nahin. It talks about all the history and various applications of imaginary numbers in a fun and exciting way. That book is what made me decide to pursue a degree in mathematics when I read it 7 years ago (and I was thinking art). Great read!!
The main point is that you add numbers which you define to be solutions to quadratic equations like x2= -1. Name one solution to that equation i, the computation rules for i then follow from that equation.
This is similar to defining negative numbers as the solution of equations like 2 + x = 1 when you only knew positive numbers, or fractions as solutions to equations like 2x = 1 when you only knew integers.
It might be easiest to stop trying to understand how a number can be a square root of a negative number, and just carry on with the assumption that it is.
So (using the i as the square root of -1):
(3+5i)*(2-i)
= (3+5i)*2 + (3+5i)*(-i)
= 6 + 10i -3i - 5i * i
= 6 + (10 -3)*i - 5 * (-1)
= 6 + 7i + 5
= 11 + 7i
works according to the standard rules of maths (remembering that i squared equals -1 on line four).
An imaginary number is a real number multiplied by the imaginary unit i. i is defined as:
i == sqrt(-1)
So:
i * i == -1
Using this definition you can obtain the square root of a negative number like this:
sqrt(-3)
== sqrt(3 * -1)
== sqrt(3 * i * i) // Replace '-1' with 'i squared'
== sqrt(3) * i // Square root of 'i squared' is 'i' so move it out of sqrt()
And your final answer is the real number sqrt(3) multiplied by the imaginary unit i.
A short answer: Real numbers are one-dimensional, imaginary numbers add a second dimension to the equation and some weird stuff happens if you multiply...
If you're interested in finding a simple application and if you're familiar with matrices,
it's sometimes useful to use complex numbers to transform a perfectly real matrice into a triangular one in the complex space, and it makes computation on it a bit easier.
The result is of course perfectly real.
Great answers so far (really like Devin's!)
One more point:
One of the first uses of complex numbers (although they were not called that way at the time) was as an intermediate step in solving equations of the 3rd degree.
link
Again, this is purely an instrument that is used to answer real problems with real numbers having physical meaning.
In electrical engineering, the impedance Z of an inductor is jwL, where w = 2*pi*f (frequency) and j (sqrt(-1))means it leads by 90 degrees, while for a capacitor Z = 1/jwc = -j/wc which is -90deg/wc so that it lags a simple resistor by 90 deg.
I'm working on a 2D game where I'm trying to accelerate an object to a top speed using some basic physics code.
Here's the pseudocode for it:
const float acceleration = 0.02f;
const float friction = 0.8f; // value is always 0.0..1.0
float velocity = 0;
float position = 0;
move()
{
velocity += acceleration;
velocity *= friction;
position += velocity;
}
This is a very simplified approach that doesn't rely on mass or actual friction (the in-code friction is just a generic force acting against movement). It works well as the "velocity *= friction;" part keeps the velocity from going past a certain point. However, it's this top speed and its relationship to the acceleration and friction where I'm a bit lost.
What I'd like to do is set a top speed, and the amount of time it takes to reach it, then use them to derive the acceleration and friction values.
i.e.,
const float max_velocity = 2.0;
const int ticks; = 120; // If my game runs at 60 FPS, I'd like a
// moving object to reach max_velocity in
// exactly 2 seconds.
const float acceleration = ?
const float friction = ?
I found this question very interesting since I had recently done some work on modeling projectile motion with drag.
Point 1: You are essentially updating the position and velocity using an explicit/forward Euler iteration where each new value for the states should be a function of the old values. In such a case, you should be updating the position first, then updating the velocity.
Point 2: There are more realistic physics models for the effect of drag friction. One model (suggested by Adam Liss) involves a drag force that is proportional to the velocity (known as Stokes' drag, which generally applies to low velocity situations). The one I previously suggested involves a drag force that is proportional to the square of the velocity (known as quadratic drag, which generally applies to high velocity situations). I'll address each one with regard to how you would deduce formulas for the maximum velocity and the time required to effectively reach the maximum velocity. I'll forego the complete derivations since they are rather involved.
Stokes' drag:
The equation for updating the velocity would be:
velocity += acceleration - friction*velocity
which represents the following differential equation:
dv/dt = a - f*v
Using the first entry in this integral table, we can find the solution (assuming v = 0 at t = 0):
v = (a/f) - (a/f)*exp(-f*t)
The maximum (i.e. terminal) velocity occurs when t >> 0, so that the second term in the equation is very close to zero and:
v_max = a/f
Regarding the time needed to reach the maximum velocity, note that the equation never truly reaches it, but instead asymptotes towards it. However, when the argument of the exponential equals -5, the velocity is around 98% of the maximum velocity, probably close enough to consider it equal. You can then approximate the time to maximum velocity as:
t_max = 5/f
You can then use these two equations to solve for f and a given a desired vmax and tmax.
Quadratic drag:
The equation for updating the velocity would be:
velocity += acceleration - friction*velocity*velocity
which represents the following differential equation:
dv/dt = a - f*v^2
Using the first entry in this integral table, we can find the solution (assuming v = 0 at t = 0):
v = sqrt(a/f)*(exp(2*sqrt(a*f)*t) - 1)/(exp(2*sqrt(a*f)*t) + 1)
The maximum (i.e. terminal) velocity occurs when t >> 0, so that the exponential terms are much greater than 1 and the equation approaches:
v_max = sqrt(a/f)
Regarding the time needed to reach the maximum velocity, note that the equation never truly reaches it, but instead asymptotes towards it. However, when the argument of the exponential equals 5, the velocity is around 99% of the maximum velocity, probably close enough to consider it equal. You can then approximate the time to maximum velocity as:
t_max = 2.5/sqrt(a*f)
which is also equivalent to:
t_max = 2.5/(f*v_max)
For a desired vmax and tmax, the second equation for tmax will tell you what f should be, and then you can plug that in to the equation for vmax to get the value for a.
This seems like a bit of overkill, but these are actually some of the simplest ways to model drag! Anyone who really wants to see the integration steps can shoot me an email and I'll send them to you. They are a bit too involved to type here.
Another Point: I didn't immediately realize this, but the updating of the velocity is not necessary anymore if you instead use the formulas I derived for v(t). If you are simply modeling acceleration from rest, and you are keeping track of the time since the acceleration began, the code would look something like:
position += velocity_function(timeSinceStart)
where "velocity_function" is one of the two formulas for v(t) and you would no longer need a velocity variable. In general, there is a trade-off here: calculating v(t) may be more computationally expensive than simply updating velocity with an iterative scheme (due to the exponential terms), but it is guaranteed to remain stable and bounded. Under certain conditions (like trying to get a very short tmax), the iteration can become unstable and blow-up, a common problem with the forward Euler method. However, maintaining limits on the variables (like 0 < f < 1), should prevent these instabilities.
In addition, if you're feeling somewhat masochistic, you may be able to integrate the formula for v(t) to get a closed form solution for p(t), thus foregoing the need for a Newton iteration altogether. I'll leave this for others to attempt. =)
Warning: Partial Solution
If we follow the physics as stated, there is no maximum velocity. From a purely physical viewpoint, you've fixed the acceleration at a constant value, which means the velocity is always increasing.
As an alternative, consider the two forces acting on your object:
The constant external force, F, that tends to accelerate it, and
The force of drag, d, which is proportional to the velocity and tends to slow it down.
So the velocity at iteration n becomes: vn = v0 + n F - dvn-1
You've asked to choose the maximum velocity, vnmax, that occurs at iteration nmax.
Note that the problem is under-constrained; that is, F and d are related, so you can arbitrarily choose a value for one of them, then calculate the other.
Now that the ball's rolling, is anyone willing to pick up the math?
Warning: it's ugly and involves power series!
Edit: Why doe the sequence n**F** in the first equation appear literally unless there's a space after the n?
velocity *= friction;
This doesn't prevent the velocity from going about a certain point...
Friction increases exponentially (don't quote me on that) as the velocity increases, and will be 0 at rest. Eventually, you will reach a point where friction = acceleration.
So you want something like this:
velocity += (acceleration - friction);
position += velocity;
friction = a*exp(b*velocity);
Where you pick values for a and b. b will control how long it takes to reach top speed, and a will control how abruptly the friction increases. (Again, don't do your own research on this- I'm going from what I remember from grade 12 physics.)
This isn't answering your question, but one thing you shouldn't do in simulations like this is depend on a fixed frame rate. Calculate the time since the last update, and use the delta-T in your equations. Something like:
static double lastUpdate=0;
if (lastUpdate!=0) {
deltaT = time() - lastUpdate;
velocity += acceleration * deltaT;
position += velocity * deltaT;
}
lastUpdate = time();
It's also good to check if you lose focus and stop updating, and when you gain focus set lastUpdate to 0. That way you don't get a huge deltaT to process when you get back.
If you want to see what can be done with very simple physics models using very simple maths, take a look at some of the Scratch projects at http://scratch.mit.edu/ - you may get some useful ideas & you'll certainly have fun.
This is probably not what you are looking for but depending on what engine you are working on, it might be better to use a engine built by some one else, like farseer(for C#).
Note Codeplex is down for maintenance.