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Does anyone know a very simple physics engine, or just a set of basic functions that could complete these tasks: Simple point, line, and rectangle collision detection? I looked at Box2D but it is way too advanced for what I am making. I just need some simple code. Thanks in advance!
Here's my shot at point/line collision detection. The important thing is to avoid trig functions, divisions, and other expensive operations so as not to slow things down too much.
As GMan's comment notes, you need to bear in mind that the point will be moving. So you'll have the current position of the point (let's call it A) and the possible new position of the point (B). You need to find out if, when the point moves from A to B, it would collide with a line.
Let's call the start and end points of the line C and D. A collision will occur only if the lines AB and CD intersect.
Let's describe the line AB using the standard equation Ux + Vy + W = 0. After some algebra, the equation comes out as:
(ay - by) x + (bx - ax) y + ax by - bx ay = 0
We can describe the line CD in terms of a parameter t and constants X Y U V:
x = X + Ut
y = Y + Vt
It's helpful if we set t=0 at point C, and t=1 at point D. By considering t=0 we can work out X and Y. And by considering t=1 we can work out U and V. This gives
x = cx + (dx - cx) t
y = cy + (dy - cy) t
To find the intersection point of the two lines, substitute these into the equation we found for AB, which gives
(ay - by) (cx + (dx - cx) t) + (bx - ax) (cy + (dy - cy) t) + ax by - bx ay = 0
This reduces to
t = (ax by - bx ay + bx cy - cx by + cx ay - ax cy) / q
where q = (ay - by)(cx - dx) - (ax - bx)(cy - dy)
If q is zero, the lines are parallel and do not meet.
If 0 < t < 1 then the line extrapolated from AB intersects CD.
But we still don't know that this intersection is actually between the points A and B. So we need to repeat all the previous steps, swapping AB and CD, and writing the line AB in terms of a parameter s. This gives:
s = (cx dy - dx cy + dx ay - ax dy + ax cy - cx ay) / q
If 0 < s < 1 then the line extrapolated from CD intersects AB.
That's it. So in your code you start by calculating q. If q is zero then the lines are parallel. Otherwise, go on to calculate t and s. If 0 < t < 1 and 0 < s < 1 then a collision is about to happen. To find the location of the collision, substitute t or s back into the original equations for CD.
For extra speed you can remove the divisions by q - it's possible to just check whether the top half of each fraction is in the correct range, and then each check should only need 10 multiplication operations.
Is there a reason you couldn't use basic geometry for this?
Two points collide when their coordinates are the same.
A point intersects with a line when its position is a solution to the equation for the line.
A point intersects a rectangle when the point is bounded by the rectangle.
More complex composite cases can be constructed by composing these cases. Were you specifically looking for a library for some reason?
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Currently I was trying to solve a problem with the condition below:
1. giving an elliptic curve y^2 = x^3 + ax + b
2. the user will input a, b and two points that exactly on the curve.
To put it simply, what I really need to do is make a secant line of the graph with the two points P and Q and try to check that if there is any point of intersection existing. If it exists, get the x and y of this point. I'm so struggled to solve this problem. Could someone give me some hints?
I'd homogenize this to x^3 + axz^2 + bz^3 - y^2z = 0 and points P = [Px : Py : Pz] and Q = [Qx : Qy : Qz]. Then any point R = λP + μQ with (λ, μ) ≠ (0, 0) lies on the line spanned by P and Q. If you wanted to avoid homogenizations, you'd require λ+μ=1 but that usually leads to divisions I'd rather avoid until the very end.
Plug the resulting coordinates of R into the homogenized equation of the elliptic curve, and you obtain a homogeneous cubic equation in λ and μ, i.e. something like
αλ³ + βλ²μ + γλμ² + δμ³ = 0
with the α, β, γ and δ depending on your a, b, P, Q. For μ=0 you get a coordinate vector which is a multiple of P, and as homogeneous coordinates identify multiples, you get point P itself, which lies on the curve. So μ=0 must satisfy the equation, hence you know α=0 even before you compute it. Likewise λ=0 represents Q so δ=0 if that point lies on the curve. You are left with
(βλ + γμ)λμ = 0
The trailing two factors encode the two known intersections I just mentioned. The parenthesis is the third intersection, the one you need. Now simply pick λ=γ and μ=−β to obtain a simple expression for the third point of intersection.
If you want to dehomogenize at the end, simply divide the first two coordinates of the resulting homogeneous coordinate vector by the third.
If I didn't mess up my sympy computation, you have
β = 3*Px^2*Qx + 2*Px*Pz*Qz*a - Py^2*Qz - 2*Py*Pz*Qy + Pz^2*Qx*a + 3*Pz^2*Qz*b
γ = 3*Px*Qx^2 + 2*Pz*Qx*Qz*a - Pz*Qy^2 - 2*Py*Qy*Qz + Px*Qz^2*a + 3*Pz*Qz^2*b
Which is expectedly very symmetric in P and Q. So essentially you just need a single function, and then you get β=f(P,Q) and γ=f(Q,P).
In C++ and with the whole homogenization / dehomogenization in place:
inline double f(double Px, double Py, double Qx, double Qy, double a, double b) {
return 3*Px*Px*Qx + 2*Px*a - Py*Py - 2*Py*Qy + Qx*a + 3*b;
}
std::pair<double, double> third_intersection(double Px, double Py, double Qx, double Qy, double a, double b) {
double beta = f(Px, Py, Qx, Qy, a, b);
double gamma = f(Qx, Qy, Px, Py, a, b);
double denominator = gamma - beta; // Might be zero if line PQ is an asymptote!
double x = (gamma*Px - beta*Qx) / denominator;
double y = (gamma*Py - beta*Qy) / denominator;
return std::make_pair(x, y);
}
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I am working with set of points, and my goal is to add 4 vectors together and calculate the resultant between start and end point,
Since I already prepared the code to do the above part which seems to be working fine but am not to sure about it.
Anyway the real reason I am posting this question is to do with the resultant direction and angle.
I find it hard to either understand the concept of finding resultant angle and direction as well as PROGRAMMING wise.
Consider this scenario....Image
1st add vectors "Head 2 Tail"
From what I have learned about vector addition is to subtract x2 - x1, y2 -y1 this will dive me the misplacement difference and do the same calculation for all the points from A-E
To get the resultant I square root all the points x to power of 2 and add all the y position to power of 2.
This ideology seem to work fine.....
But the QUESTION here is how do i get the angle and direction of that resultant....?
the code I use to calculate resultant:
double Pta;
double Ptb;
Point vect;
float R1, R2;
float resultant;
for(vector<Point>::iterator iter_a = Left_Arm_xy.begin()+1; iter_a != Left_Arm_xy.end(); ++iter_a)
{
if(center.y <= 240)
{
vect.x = iter_a->x - (iter_a -1)->x;
vect.y = iter_a->y - (iter_a -1)->y;
vect_add.push_back(Point(vect.x,vect.y));
for(vector<Point>::iterator iter_v = vect_add.begin(); iter_v - vect_add.begin() + 4 < vect_add.size(); iter_v+=4)
{
R1 = iter_v->x + (iter_v +1)->x + (iter_v +2)->x + (iter_v +3)->x;
R2 = iter_v->y + (iter_v +1)->y + (iter_v +2)->y + (iter_v +3)->y;
resultant = sqrt(pow(R1,2) + pow(R2,2));
}
}
Consider this..............
Ok lets consider Points A[2,4], B[4,8], C[10,12], To add this vectors i add vectors/points I subtract point B x4 - A x2 and point B y8 - A y4 and point C x10 - B x4 and point C y12 - B y8 this will give me the the displacements between points....Now to get the Resultant i add all the Points X's and Y's x's 2+4+10 = 16 y's 4+8+12 = 24, Next i would square root 16 ^2 + 24^2 = 28.84. So based on these calculations where resultant is a number not and x and y value how can i get the direction and angle....?
It is a simple summation of vectors.
(x, y) = (x1, y1) + (x2, y2) + ... = (x1+x2+..., y1+y2+...)
When you have the final vector, it's angle is found by using
tan(angle) = y/x
The angle between two vectors is generally defined as:
Angle = arccos( DotProduct(v1, v2) / ( Length(v1) * Length(v2) ) );
The direction is simply subtraction of the two vectors:
Direction = v2 - v1;
Usually, you have to normalize this to get a unit vector:
Len = SquareRoot( direction.x * direction.x + direction.y * direction.y );
Direction.x /= Len;
Direction.y /= Len;
Thus, you'll have a unit direction vector and the angle of the vector.
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I am doing cropping a object in real time using c++. Here I know 2 coordinates of points A and B. I want to find 3rd point Z which is perpendicular to AB line. Z(x3,y3). It mean ABZ angle is 90. I gained 2 huge equations when I have used BZ slope and BZ distance. I have simplified it and gain following equation.
y3 ( y3 (k1 + y3) + k) = k ;
here k1 , k2 , k3 are Constance which has given. But k1 , k2 , k3 are not equals to each other.
I want to find y3 here. Please help me.
This sounds overly complicated, at least if I've understood your
problem. If there are no other constraints on Z (other than it be on a
line perpendicular to AB at B), then:
Z.x = B.x - (A.y - B.y);
Z.y = B.y + (A.x - B.x);
solves the problem. If there are additional constraints on Z, the above
expression still gives a point that, with B, defines the line.
Depending on the constraints, calculating the actual Z may be more or
less complicated, but it still shouldn't involve cubic equations.
If you do need to take a cube root, of course, the simplest solution is to use the standard function cbrt.
Generally, if you want to solve a cubic equation
x^3 + ax^2 + bx + c = 0
and you already have two solutions x1, x2 and are looking for the third solution x3
then you can do this by observing that
x^3 + ax^2 + bx + c = (x-x1)(x-x2)(x-x3),
and hence that
-a = x1 + x2 + x3
or
x3 = -a -x1 -x2
Hence finding the third solution of a cubic given the other two solutions is trivial.
The cubic equation has an exact solution. It's ugly but there you go.
Bear in mind that there will be pathalogical cases where numeric instabilities show up. If accuracy is important you should use more than one method.
This guy has solution code, and Dr. Math has some explanation.
I'm creating a 2D game and want to test for collision between an OBB (Oriented Bounding Box) and a Circle. I'm unsure of the maths and code to do this. I'm creating the game in c++ and opengl.
Since both your shapes are convex, you can use the Separating Axis Theorem. Here's a tutorial on how to implement an algorithm to do this.
Essentially, you try to find if it's possible to put a line somewhere that's between the two shapes and if you can't find one, then you know they're colliding.
References and general answer taken from this question.
Here's what I would do, in pseudocode:
function does_line_go_through_circle (original_line, circle_centerpoint, radius):
original_slope = get_slope_of_line (original_line)
perpendicular_slope = 1/original_slope
perpendicular_line = create_line_with_slope_through_point (perpendicular_slope, circle_centerpoint)
intersect_point = intersection_of_infinite_lines (perpendicular_line, original_line)
if point_is_on_line (intersect_point, original_line):
finite_line_along_radius = create_finite_line_between_points (circle_centerpoint, intersect_point)
if length_of_line (finite_line_along_radius) < length_of_line (radius):
return true
end
end
return false
end
function does_box_intersect_with_circle (bounding_box, circle):
for each side in bounding_box:
if does_line_go_through_circle (side, circle.center, circle.radius):
return true
end
end
return false
end
Keep in mind, I'm a little rusty on this stuff, I might be wrong.
Anyway, it should be trivial to implement this in C++.
We will divide the rectangle into 4 finite lines.
We can construct the line equation ax + by + c = 0 connecting the points (x1, y1) and (x2, y2) as follows:
mx - y + c = 0
where m = (y2-y1)/(x2-x1)
Shortest (perpendicular) distance from a line ax + by + c = 0 to a point (xc, yc) is given by the expression:
d = (a*xc + b*yc + c) / sqrt(a*a + b*b)
or d = (m*xc - yc + c) / sqrt(m*m + 1) according the the above equation
For infinite lines, you can check if 'd' is less than the radius of the circle.
But for finite lines, you'd also have to make sure whether the point of contact is within the line.
Now m = tan(angle). You can have
cos = 1 / sqrt(m*m + 1); sin = m / sqrt(m*m + 1)
Then you can calculate the point of contact as
xp = xc + d*cos; yp = yc + d*sin
And to check whether (xp, yp) lies in between the points connecting line, you can do a simple check as
if(x1 < xp < x2 && y1 < yp < y2)
return true
depending upon which is greater among x1 x2 and y1 y2.
You can repeat the same algorithm for all the four lines of a rectangle by passing the points.
Please correct me if I'm wrong somewhere.
Sorry in advance, I'm struggling a bit with how to explain this... :)
Essentially, I've got a typical windows coordinate system (the Top, Left is 0,0). If anybody's familiar with the haversine query, like in SQL, it can get all points in a radius based on latitude and longitude coordinates.
I need something much simpler, but my math skills ain't all up to par! Basically, I've got random points scattered throughout about a 600x400 space. I have a need to, for any X,Y point on the map, run a query to determine how many other points are within a given radius of that one.
If that's not descriptive enough, just let me know!
Straightforward approach:
You can calculate the distance between to points using the Pythagorean theorem:
deltaX = x1 - x2
deltaY = y1 - y2
distance = square root of (deltaX * deltaX + deltaY * deltaY)
Given point x1,y1, do this for every other point (x2,y2) to see if the calculated distance is within (less than or equal to) your radius.
If you want to make it speedier, calculate and store the square of the radius and just compare against (deltaX * deltaX + deltaY * deltaY), avoiding the square root.
Before doing the Pythagoras, you could also quickly eliminate any point that falls outside of the square that can fully contain the target circle.
// Is (x1, y1) in the circle defined by center (x,y) and radius r
bool IsPointInCircle(x1, y1, x, y, r)
{
if (x1 < x-r || x1 > x+r)
return false;
if (y1 < y-r || y1 > y+r)
return false;
return (x1-x)*(x1-x) + (y1-y)*(y1-y) <= r*r
}
Use Pythagoras:
distance = sqrt(xDifference^2 + yDifference^2)
Note that '^' in this example means "to the power of" and not C's bitwise XOR operator. In other words the idea is to square both differences.
If you only care about relative distance you shouldn't use square root you can do something like:
rSquared = radius * radius #square the radius
foreach x, y in Points do
dX = (x - centerX) * (x - centerX) #delta X
dY = (y - centerY) * (y - centerY) #delta Y
if ( dX + dY <= rSquared ) then
#Point is within Circle
end
end
Using the equation for a circle:
radius ** 2 = (x - centerX) ** 2 + (y - centerY) ** 2
We want to find if a point (x, y) is inside of the circle. We perform the test using this equation:
radius ** 2 < (x - centerX) ** 2 + (y - centerY) ** 2
// (Or use <= if you want the circumference of the circle to be included as well)
Simply substitute your values into that equation. If it works (the inequality is true), the point is inside of the circle. Otherwise, it isn't.