I have a line and a few points and I need to determine which points are under and which are beyond the line. I tried to find a line that is in 90degrees angle with my line and crosses the points but i couldnt figure out whether the vectors orientation is up or down. Can you help? Thank you
You can find line equation and substitute points in thin equation.
Easy case: Let's line is not vertical, so it might be described by equation
y = a * x + b
for every query point (px, py) calculate value
S = py - a * px - b
When S is positive, point is above the line, when negative - below.
If your line is defined by base point B and direction vector D, you can determine - what semi-plane (against the line) query point P belongs to - using cross product sign
Sign (D x (P-B))
Note that in this case term "below" depends also on sign of X-component of vector D
Related
I get the positions of 5 enemies in the game in vectors. Depending on the distance I choose, the number of enemies can vary from 0 to 5. I need to know their vectors each time to check whether it is possible to draw a straight line through a certain number of heroes (vectors).
After that, my hero will have to use his ability called wall. It consists of 2 start and end vectors. Thus, check whether my hero can put a wall on the enemies in the line to catch them
Let's say there are 3 enemy heroes whose positions I can get. I need to find out if I can pass through them directly, in order to use the ability on them.
Here's what using the ability looks like in the game
Here is getting the vector of one of the heroes
The ability itself can be twisted at a certain point. But anyway, it is necessary that the wall would touch several heroes
Wherever I move the mouse, I can put it in the desired position. But unfortunately it takes a lot of time, so I would like to automate
The coordinates of the wall itself, or rather its two edges, I can also get, but only after the ability has been used
If one prefers geometry to linear algebra...
Then One can compute the dot product of (unit-vector1. Unit-Vector2). That is equal to the SIN of the angle between them.
So if unit vector is the shooter position to target1, unit vector2 is the shooter to target2, etc... then when DOTPRODUCT(Vector1,vector2) = 1 and DOTPRODUCT(Vector1,vector3) = 1, then the three points are in syzygy.
And repeat from shooter to as many targets as you have to determine whether some or all of the points are in syzygy.
From your statement that there is a start and an endpoint I take that you select two enemys and want to trap anything in between.
So you're actually not looking for a straight line that can be drawn through your enemy positions but if they are withn a rectangle. It would be very unlikely and for more points nearly impossible that they are all collinear anyway.
So it becomes quite trivial. You draw a line through start and end enemy. Then you check the remaining enemies distance to that line vs the width of your AoE. Maye you want to also handle some body width in that calculation.
https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line
You can describe all points belonging to line (x0, y0) + (dx, dy)t = (x1, y1). Chose any two points and t as 1 and you will get (dx, dy) for line connecting two dots. Now you will need to find distance between this line and (x2, y2). it is distance between (x2, y2) and (xd, yd), where on one hand (xd, yd) = (x0, y0) + t1(dx, dy) and on other hand (xd, yd) = (x2, y2) + t2*(-dy, dx). Solving this two equations you will find t1, t2, (xd, yd) and distance between (x2, y2) and (xd, yd), which is distance between (x2, y2) and line, connecting (x0, y0) and (x1, y1).
Knowing this, you select dots with min_x and max_x and calculate ditance between line, connecting said dots and rest of the dots. If distance is lesser then some threshold of your choice, then you can assume that you can have line passing through all dots.
Any line in the plane can be described by an equation a*x + b*y + c = 0 with (a, b) ≠ (0, 0). Note that if you have an equation of this form, then multiplying each coefficient a, b, c with the same number yields an equation describing the same line. That's the reason (a, b, c) is called a homogeneous coordinate vector for that line.
How do you find a, b, c? One simple approach would be treating this as three linear equations in three unknowns. You plug in the x and y coordinates for all your three points, and get tree equations for a thorough c. However, there is a catch. Since the right hand side of each equation is zero, a = b = c = 0 is always a solution. In those cases where there is only one solution, that will be it. So in order for there to be a line, you need more than one solution. The mathematical tool to determine whether a set of equations had more than one solution is the determinant. It is zero if the system has no single unique solution.
Long story short: three points are collinear (on a line) if
⎛x1 y1 1⎞
det ⎜x2 y2 1⎟ = 0
⎝x3 y3 1⎠
The homogeneous coordinate vector describing the line world correspond to the kernel of that matrix.
Of course, if your input coordinates are floating point numbers, exact zero is unlikely. Presumably that wall does allow for some error in some way, and you'd need to tell us about that in order to get an answer that models this aspect correctly. In the mean time, know that the absolute value of the determinant above is proportional to the area of the triangle created by these three points. So if your were to pick a constant threshold value, the farther your enemies are apart along the direction of the wall, the less they could deviate from the straight line without violating that threshold.
I have:
axis-aligned rectangle R;
non-vertical line A (specified by two points);
vertical line B (specified by x-coordinate).
All objects above is specified by integer coordinates.
I need to check, if result of crossing A and B is in the R. But I can't just calculate result of crossing A and B because it may be unpresentable in integer coordinates. Also I can't convert it to doubles because I need to get absolutely accurate result without any inaccuracy.
So, how can I check it?
If lines are specified by integer coordinates, then their crossing point has rational coordinates, which may be compared with absolutely precision.
Let's A goes through points (ax0, ay0) and (ax1, ay1), and B is at X0 coordinate.
Then crossing point Y-coordinate is (ay0*(ax1-ax0)+(X0-ax0)*(ay1-ay0))/(ax1-ax0) = p/q, where p and q are integer, and q is positive (negate nominator if needed).
Then p may be compared with R.Top*q and R.Bottom*q
[let's ignore not interesting case when B does not intersect R]
R cuts a segment from B, let's call it S. Intersection of B and A lies inside of R if and only if A crosses S.
Finding points of S is trivial.
To check whether a line crosses a segment, you should check that ends of S lie on different sides from A - this can be checked by signed angles.
Let's say that the 2 points on A are (x0, y0) and (x1, y1), with 'x0 < x1'.
The rectangle is determined by a x-coordinate xR and a y-coordinate yR
The B line is determined by the x-coordinate xB
The point you're looking for is (xB, yB), where yB is to be determined somehow, by manipulating integers only :
So first, you need to check
if xB * xR >= 0 (same side of the x-coordinate)
if abs(xB) <= abs(xR) (the B line cuts the rectangle)
If it's ok, then you need to check that this integer (which is equal to (x1-x0)yB)
Y = (y1-y0)(xB-x0)+(x1-x0)y0
verifies
Y * yR >=0 (same side of the y-coord)
abs(Y) <= (x1-x0) * abs(yR) (meaning that your intersection point is in the R area regarding its y-coord)
Your point is inside R if and only if the 4 conditions are true.
Hope it helps.
I am trying to find circumcenter of Given Three point of Triangle……..
NOTE: all these three points are with X,Y and Z Co-Ordinate Means points are in 3D
I know that the circumcenter is the point where the right bisectors intersect….
But for that I have to find middle point of each side then the right bisectors and then intersection point of that …..this is long and error some process……
Is there not any formula which just takes as input these three points of triangle and giving us the Circumcenter of Triangle ……?
Thanks………
The wiki page on Circumscribed circle has it in terms of dot and cross products of the three vertex vectors. It also has a formula for the radius of the circle, if you are so interested.
First of all, you need to make sure that points are not collinear. i.e. do not lie in the same line. For that you need to find the direction cosines of the lines made by three points, and if they have same direction cosines, halt, you can't get circle out of it.
For direction cosine please check this article on wikipedia.
(A way of finding coordinate-geometry and geometry -- based on the theorem that, a perpendicular line from center of circle bisects a chord)
Find the equation of the plane.
This equation must reduce to the form
and the direction cosines (of the line perpendicular to plane determines the plane), so direction cosines of the line perpendicular to this line is
given by this link equations -- 8,9,10 (except replace it for l, m, n).
Find the equation of the lines (all three) in 3-d
(x-x1)/l=(y-y1)/m=(z-z1)/n (in terms of direction cosines) or
(x-x1)/(x2-x1)=(y-y1)/(y2-y1)=(z-z1)/(z2-z1)
Now we need to find the equation of line
a) this perpendicular to the line, from 2 (let l1, m1, n1 be direction cosines of this line)
b) must be contained in place from 1 (let l2, m2, n2 be direction cosines of this line perpendicular to plane)
Find and solve (at least two lines) from 3, sure you will be able to find the center of the circle.
How to find out equation ??? as we are finding the circum-center, we will get our points (i.e. it is the midpoint of the two points) and for a) we have
l1*l+m1*m+n1*n = 0 and l2*l+m2*m+n2*n = 0 where l, m, n are direction cosines of our, line, now solving this two equation, we can get l, m interms of n. And we use this found out (x1,y1,z1) and the value of l, m, 1 and we will have out equation.
The other process is to solve the equation given in this equation
https://stackoverflow.com/questions/5725871/solving-the-multiple-math-equations
Which is the deadliest way.
The other method is using the advantage of computer(by iteration) - as I call it (but for this you need to know the range of the coordinates and it consumes lot of memory)
it's like this (You can make it more precise by incrementing at 1/10) but certainly bad way.
for(i=minXrange, i>=maxXrange; i++){
for(j=minYrange, j>=maxYrange; j++){
for(i=minZrange, i>=maxZrange; k++){
if(((x1-i)^2 + (y1-j)^2 + (z1-k)^2) == (x2-i)^2 + (y2-j)^2 + (z2-k)^2) == for z)){
return [i, j, k];
}
}
}
}
Okay so I'm trying to get a separating axis theorem algorithm to work (for collision detection) and I need to find the minimal perpendicular vector between a point and a line. I'm not asking for the minimum perpendicular distance (which I know how to find) but rather the vector that would have the same magnitude as that distance and that goes from an arbitrary point and a point on the line. I know the location of the point, a point on the line, and a unit vector giving the direction of the line.
What I tried doing was first finding the minimal distance between the point and the line.
The next part is confusing but I:
1) Found the vector between the point and the point on the line I know
2) Found the vector between the point on the line and the point on the line plus the unit vector giving the direction of the line
3) Took the cross product of these two vectors (I'll call this cross product A)
4) Took the cross product of the unit vector giving the direction of the line and the vector from cross product A (I'll call this cross product B)
5) Normalized cross product B
6) Scaled cross product B by the minimal distance
Anyways that whole attempt failed miserably. Can anyone tell me how I am supposed to find this vector?
If I understood your question correctly, I believe this is what you're looking for:
P - point
D - direction of line (unit length)
A - point in line
X - base of the perpendicular line
P
/|
/ |
/ v
A---X----->D
(P-A).D == |X-A|
X == A + ((P-A).D)D
Desired perpendicular: X-P
where the period represents the dot product and |X-A| means magnitude.
From the above figure, you have:
q = p + s --> s = q - p = q - (p2-p1) = q + p1 - p2
==> s^ = |q - p2 - p1| / |s| (unitary vector)
Also: |s| = |q| sin c = |q|sin(b-a)
b = arcsin (qy / |q|); a = arcsin( p1y / |p1| )
where: |q| = (qx^2 + qy^2)^1/2
I have a line that I must do calculations on for each grid square the line passes through.
I have used the Superline algorithm to get all these grid squares. This gives me an array of X,Y coordinates to check.
Now, here is where I am stuck, I need to be able to calculate the distance traveled through each of the grid squares... As in, on a line not on either 90 degree or 45 degree angles, each grid square accommodates a different 'length' of the total line.
Image example here, need 10 reputation to post images
As you can see, some squares have much more 'line length' in them than others - this is what I need to find.
How do I work this out for each grid square? I've been at this for a while and request the help of the Stack Overflowers!
There may be some clever way to do this that is faster and easier, but you could always hack through it like this:
You know the distance formula: s=sqrt((x2-x1)^2+(y2-y1)^2). To apply this, you must find the x and y co-ordinates of the points where the line intersects the edges of each grid cell. You can do this by plugging the x and y co-ordinates of the boundaries of the cell into the equation of the line and solve for x or y as appropriate.
That is, each cell extends from some point (x0,y0) to (x0+1,y0+1). So we need to find y(x0), y(x0+1), x(y0), and x(y0+1). For each of these, the x or y value found may or may not be within the ranges for that co-ordinate for that cell. Specifically, two of them will be and two won't. The two that are correspond to the edges that the line passes through, and the two that aren't are edges that it doesn't pass through.
Okay, maybe this sounds pretty confusing, so let's work through an example.
Let's say your line has the equation x=2/3 * y. You want to know where it intersects the edges of the cell extending from (1,0) to (2,1).
Plug in x=1 and you get y=2/3. 2/3 is in the legal range for y -- 0 to 1 -- so (1,2/3) is a point on the edge where the line intersects this cell. Namely, the left edge.
Plug in x=2 and you get y=4/3. 4/3 is outside the range for y. So the line does not pass through the right edge.
Plug in y=0 and you get x=0. 0 is not in the range for x, so the line does not pass through the bottom edge.
Plug in y=1 and you get x=3/2. 3/2 is in the legal range for x, so (3/2,1) is another intersection point, on the top edge.
Thus, the two points where the line intersects the edges of the cell are (1,2/3) and (3/2,1). Plug these into the distance formula and you'll get the length of the line segement through this cell, namely sqrt((1-3/2)^2+(2/3-1)^2)=sqrt(1/4+1/9)=sqrt(13/36). You can approximate that to any desired level of precision.
To do this in a program you'd need something like: (I'll use pseudo code because I don't know what language you're using)
// Assuming y=mx+b
function y(x)
return mx+b
function x(y)
return (y-b)/m
// cellx, celly are co-ordinates of lower left corner of cell
// Upper right must therefore be cellx+1, celly+1
function segLength(cellx, celly)
// We'll create two arrays pointx and pointy to hold co-ordinates of intersect points
// n is index into these arrays
// In an object-oriented language, we'd create an array of point objects, but whatever
n=0
y1=y(cellx)
if y1>=celly and y1<=celly+1
pointx[n]=cellx
pointy[n]=y1
n=n+1
y2=y(cellx+1)
if y2>=celly and y2<=celly+1
pointx[n]=cellx+1
pointy[n]=y2
n=n+1
x1=x(celly)
if x1>=cellx and x1<=cellx+1
pointx[n]=x1
pointy[n]=celly
n=n+1
x2=x(celly+1)
if x2>=cellx and x2<=cellx+1
pointx[n]=x2
pointy[n]=celly+1
n=n+1
if n==0
return "Error: line does not intersect this cell"
else if n==2
return sqrt((pointx[0]-pointx[1])^2+(pointy[0]-pointy[1])^2)
else
return "Error: Impossible condition"
Well, I'm sure you could make the code a little cleaner, but that's the idea.
have a look at Siddon's algorithm: "Fast calculation of the exact radiological path for a three-dimensional CT array"
unfortunately you need a subscription to read the original paper, but it is fairly well described in this paper
Siddon's algorithm is an O(n) algorithm for finding the length of intersection of a line with each pixel/voxel in a regular 2d/3d grid.
Use the Euclidean Distance.
sqrt((x2-x1)^2 + (y2-y1)^2)
This gives the actual distance in units between points (x1,y1) and (x2,y2)
You can fairly simply find this for each square.
You have the slope of the line m = (y2-y1)/(x2-x1).
You have the starting point:
(x1,y2)
What is the y position at x1 + 1? (i.e. starting at the next square)
Assuming you set your starting point to 0 the equation of this line is simply:
y_n = mx_n
so y_n = (y2-y1)/(x2-x1) * x_n
Then the coordinates at the first square are (x1,y1) and at the nth point:
(1, ((y2-y1)/(x2-x1))*1)
(2, ((y2-y1)/(x2-x1))*2)
(3, ((y2-y1)/(x2-x1))*3)
...
(n, ((y2-y1)/(x2-x1))*n)
Then the distance through the nth square is:
sqrt((x_n+1 - x_n)^2 + (y_n+1 - y_n)^2)