I'm working on a js widget, and I've come across a positioning problem, which I can't seem to solve with my limited geometry knowledge or by help of Wikipedia/google.
I have a quadrilateral rectangle, which is positioned at an angle. I know its two opposite vertexes and width/height ratio. And there's a point on it, which coordinates I also know.
I need to find how far (in %s of width/height) is that point from rectangle's sides. Is it possible to do so?
Having two corners P1 = (x1,y1) and P2 = (x2,y2) and point Q, you can find diagonal length
dx = (x2 - x1)
dy = (y2 - y1)
dlen = sqrt(dx^2 + dy^2)
and unit direction vector
dx = dx / dlen
dy = dy / dlen
and center of rectangle
cx = x1 + dx/2
cy = y1 + dy/2
Width and height (with known r = w/h ratio)
w = dlen / sqrt(1 + r^2)
h = w / r
Now we need direction of side of length w. Note that given information does not allow to choose exact rectangle orientation from two possible cases.
Angle between diagonal and side
sina = r / sqrt(1 + r^2)
cosa = 1 / sqrt(1 + r^2)
Side direction vector
wx = dx * cosa - dy * sina
wy = dx * sina + dy * cosa
and for the second orientation
wx' = dx * cosa + dy * sina
wy' = -dx * sina + dy * cosa
The second side vector
hx = -wy
hy = wx
Now we can find length of projection of point p onto sides W and H using dot product
qx = q.x - x1
qy = q.y - y1
qw = qx * wx + qy * wy
qh = qx * hx + qy * hy
The last values are coordinates in W-H basis, so value qw varies from 0 for points at the "left" to w for points at the "right" side. You can divide these values by w and h to get percent values.
Note again - there are two possible rectangles and correspondingly two positions of point Q
I come accross a math problem about Interactive Computer Graphics.
I summarize and abstract this problem as follows:
I'm going to rotation a 3d coordinate P(x1,y1,z1) around a point O(x0,y0,z0)
and there are 2 vectors u and v which we already know.
u is the direction to O before transformation.
v is the direction to O after transformation.
I want to know how to conduct the calculation and get the coordinate of Q
Thanks a lot.
Solution:
Rotation About an Arbitrary Axis in 3 Dimensions using the following matrix:
rotation axis vector (normalized): (u,v,w)
position coordinate of the rotation center: (a,b,c)
rotation angel: theta
Reference:
https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxnbGVubm11cnJheXxneDoyMTJiZTZlNzVlMjFiZTFi
for just single point no rotations is needed ... so knowns are:
u,v,O,P
so we now the distance is not changing:
|P-O| = |Q-O|
and directions are parallel to u,v so:
Q = O + v*(|P-O|/|v|)
But I suspect you want to construct rotation (transform matrix) such that more points (mesh perhaps) are transformed. If that is true then you need at least one known to get this right. Because there is infinite possible rotations transforming P -> Q but the rest of the mesh will be different for each ... so you need to know at least 2 non trivial points pair P0,P1 -> Q0,Q1 or axis of rotation or plane parallel to rotation or any other data known ...
Anyway in current state you can use as rotation axis vector perpendicular to u,v and angle obtained from dot product:
axis = cross (u,v)
ang = +/-acos(dot(u,v))
You just need to find out the sign of angle so try both and use the one for which the resultinq Q is where it should be so dot(Q-O,v) is max. To rotate around arbitrary axis and point use:
Rodrigues_rotation_formula
Also this might be helpfull:
Understanding 4x4 homogenous transform matrices
By computing dot product between v and u get the angle l between the vectors. Do a cross product of v and u (normalized) to produce axis of rotation vector a. Let w be a vector along vector u from O to P. To rotate point P into Q apply the following actions (in pseudo code) having axis a and angle l computed above:
float4 Rotate(float4 w, float l, float4 a)
{
float4x4 Mr = IDENTITY;
quat_t quat = IDENTITY;
float4 t = ZERO;
float xx, yy, zz, xy, xz, yz, wx, wy, wz;
quat[X] = a[X] * sin((-l / 2.0f));
quat[Y] = a[Y] * sin((-l / 2.0f));
quat[Z] = a[Z] * sin((-l / 2.0f));
quat[W] = cos((-l / 2.0f));
xx = quat[X] * quat[X];
yy = quat[Y] * quat[Y];
zz = quat[Z] * quat[Z];
xy = quat[X] * quat[Y];
xz = quat[X] * quat[Z];
yz = quat[Y] * quat[Z];
wx = quat[W] * quat[X];
wy = quat[W] * quat[Y];
wz = quat[W] * quat[Z];
Mr[0][0] = 1.0f - 2.0f * (yy + zz);
Mr[0][1] = 2.0f * (xy + wz);
Mr[0][2] = 2.0f * (xz - wy);
Mr[0][3] = 0.0f;
Mr[1][0] = 2.0f * (xy - wz);
Mr[1][1] = 1.0f - 2.0f * (xx + zz);
Mr[1][2] = 2.0f * (yz + wx);
Mr[1][3] = 0.0f;
Mr[2][0] = 2.0f * (xz + wy);
Mr[2][1] = 2.0f * (yz - wx);
Mr[2][2] = 1.0f - 2.0f * (xx + yy);
Mr[2][3] = 0.0f;
Mr[3][0] = 0.0f;
Mr[3][1] = 0.0f;
Mr[3][2] = 0.0f;
Mr[3][3] = 1.0f;
w = Mr * w;
return w;
}
Point Q is at the end of the rotated vector w. Algorithm used in the pseudo code is quaternion rotation.
If you know u, v, P, and O then I would suggest that you compute |OP| which should be preserved under rotations. Then multiply this length by the unit vector -v (I assumed u, v are unit vectors: if not - normalize them) and translate the origin by this -|OP|v vector. The negative sign in front of v comes from the description given in your question:"v is the direction to O after transformation".
P and Q are at the same distance R to O
R = sqrt( (x1-x0)^2 + (y1-y0)^2 + (z1-z0)^2 )
and OQ is collinear to v, so OQ = v * R / ||v|| where ||v|| is the norm of v
||v|| = sqrt( xv^2 + yv^2 + zv^2 )
So the coordinates of Q(xq,yq,zq) are:
xq= xo + xv * R / ||v||
yq= yo + yv * R / ||v||
zq= zo + zv * R / ||v||
Assuming I have a spaceship (source); And an asteroid (target) is somewhere near it.
I know, in 3D space (XYZ vectors):
My ship's position (sourcePos) and velocity (sourceVel).
The asteroid's position (targetPos) and velocity (targetVel).
(eg. sourcePos = [30, 20, 10]; sourceVel = [30, 20, 10]; targetPos = [600, 400, 200]; targetVel = [300, 200, 100]`)
I also know that:
The ship's velocity is constant.
The asteroid's velocity is constant.
My ship's projectile speed (projSpd) is constant.
My ship's projectile trajectory, after being shot, is linear (/straight).
(eg. projSpd = 2000.00)
How can I calculate the interception coordinates I need to shoot at in order to hit the asteroid?
Notes:
This question is based on this Yahoo - Answers page.
I also searched for similar problems on Google and here on SO, but most of the answers are for 2D-space, and, of the few for 3D, neither the explanation nor the pseudo-codes explain what is doing what and/or why, so I couldn't really understand enough to apply them on my code successfully. Here are some of the pages I visited:
Danik Games Devlog, Blitz3D Forums thread, UnityAnswers, StackOverflow #1, StackOverflow #2
I really can't figure out the maths / execution-flow on the linked pages as they are, unless someone dissects it (further) into what is doing what, and why;
Provides a properly-commented pseudo-code for me to follow;
Or at least points me to links that actually explain how the equations work instead of just throwing even more random numbers and unfollowable equations in my already-confused psyche.
I find the easiest approach to these kind of problems to make sense of them first, and have a basic high school level of maths will help too.
Solving this problem is essentially solving 2 equations with 2 variables which are unknown to you:
The vector you want to find for your projectile (V)
The time of impact (t)
The variables you know are:
The target's position (P0)
The target's vector (V0)
The target's speed (s0)
The projectile's origin (P1)
The projectile's speed (s1)
Okay, so the 1st equation is basic. The impact point is the same for both the target and the projectile. It is equal to the starting point of both objects + a certain length along the line of both their vectors. This length is denoted by their respective speeds, and the time of impact. Here's the equation:
P0 + (t * s0 * V0) = P1 + (t * s0 * V)
Notice that there are two missing variables here - V & t, and so we won't be able to solve this equation right now. On to the 2nd equation.
The 2nd equation is also quite intuitive. The point of impact's distance from the origin of the projectile is equal to the speed of the projectile multiplied by the time passed:
We'll take a mathematical expression of the point of impact from the 1st equation:
P0 + (t * s0 * V0) <-- point of impact
The point of origin is P1
The distance between these two must be equal to the speed of the projectile multiplied by the time passed (distance = speed * time).
The formula for distance is: (x0 - x1)^2 + (y0 - y1)^2 = distance^2, and so the equation will look like this:
((P0.x + s0 * t * V0.x) - P1.x)^2 + ((P0.y + s0 * t * V0.y) - P1.y)^2 = (s1 * t)^2
(You can easily expand this for 3 dimensions)
Notice that here, you have an equation with only ONE unknown variable: t!. We can discover here what t is, then place it in the previous equation and find the vector V.
Let me solve you some pain by opening up this formula for you (if you really want to, you can do this yourself).
a = (V0.x * V0.x) + (V0.y * V0.y) - (s1 * s1)
b = 2 * ((P0.x * V0.x) + (P0.y * V0.y) - (P1.x * V0.x) - (P1.y * V0.y))
c = (P0.x * P0.x) + (P0.y * P0.y) + (P1.x * P1.x) + (P1.y * P1.y) - (2 * P1.x * P0.x) - (2 * P1.y * P0.y)
t1 = (-b + sqrt((b * b) - (4 * a * c))) / (2 * a)
t2 = (-b - sqrt((b * b) - (4 * a * c))) / (2 * a)
Now, notice - we will get 2 values for t here.
One or both may be negative or an invalid number. Obviously, since t denotes time, and time can't be invalid or negative, you'll need to discard these values of t.
It could very well be that both t's are bad (in which case, the projectile cannot hit the target since it's faster and out of range). It could also be that both t's are valid and positive, in which case you'll want to choose the smaller of the two (since it's preferable to hit the target sooner rather than later).
t = smallestWhichIsntNegativeOrNan(t1, t2)
Now that we've found the time of impact, let's find out what the direction the projectile should fly is. Back to our 1st equation:
P0 + (t * s0 * V0) = P1 + (t * s0 * V)
Now, t is no longer a missing variable, so we can solve this quite easily. Just tidy up the equation to isolate V:
V = (P0 - P1 + (t * s0 * V0)) / (t * s1)
V.x = (P0.x - P1.x + (t * s0 * V0.x)) / (t * s1)
V.y = (P0.y - P1.y + (t * s0 * V0.y)) / (t * s1)
And that's it, you're done!
Assign the vector V to the projectile and it will go to where the target will be rather than where it is now.
I really like this problem since it takes math equations we learnt in high school where everyone said "why are learning this?? we'll never use it in our lives!!", and gives them a pretty awesome and practical application.
I hope this helps you, or anyone else who's trying to solve this.
If you want a projectile to hit asteroid, it should be shoot at the point interceptionPos that satisfy the equation:
|interceptionPos - sourcePos| / |interceptionPos - targetPos| = projSpd / targetVel
where |x| is a length of vector x.
In other words, it would take equal amount of time for the target and the projectile to reach this point.
This problem would be solved by means of geometry and trigonometry, so let's draw it.
A will be asteroid position, S - ship, I - interception point.
Here we have:
AI = targetVel * t
SI = projSpd * t
AS = |targetPos - sourcePos|
vector AS and AI direction is defined, so you can easily calculate cosine of the SAI angle by means of simple vector math (take definitions from here and here). Then you should use the Law of cosines with the SAI angle. It will yield a quadratic equation with variable t that is easy to solve (no solutions = your projectile is slower than asteroid). Just pick the positive solution t, your point-to-shoot will be
targetPos + t * targetVel
I hope you can write a code to solve it by yourself. If you cannot get something please ask in comments.
I got a solution. Notice that the ship position, and the asteroid line (position and velocity) define a 3D plane where the intercept point lies. In my notation below | [x,y,z] | denotes the magnitude of the vector or Sqrt(x^2+y^2+z^2).
Notice that if the asteroid travels with targetSpd = |[300,200,100]| = 374.17 then to reach the intercept point (still unknown, called hitPos) will require time equal to t = |hitPos-targetPos|/targetSpd. This is the same time the projectile needs to reach the intercept point, or t = |hitPos - sourcePos|/projSpd. The two equations are used to solve for the time to intercept
t = |targetPos-sourcePos|/(projSpd - targetSpd)
= |[600,400,200]-[30,20,10]|/(2000 - |[300,200,100]|)
= 710.81 / ( 2000-374.17 ) = 0.4372
Now the location of the intetception point is found by
hitPos = targetPos + targetVel * t
= [600,400,200] + [300,200,100] * 0.4372
= [731.18, 487.45, 243.73 ]
Now that I know the hit position, I can calculate the direction of the projectile as
projDir = (hitPos-sourcePos)/|hitPos-sourcePos|
= [701.17, 467.45, 233.73]/874.52 = [0.8018, 0.5345, 0.2673]
Together the projDir and projSpd define the projectile velocity vector.
Credit to Gil Moshayof's answer, as it really was what I worked off of to build this. But they did two dimensions, and I did three, so I'll share my Unity code in case it helps anyone along. A little long winded and redundant. It helps me to read it and know what's going on.
Vector3 CalculateIntercept(Vector3 targetLocation, Vector3 targetVelocity, Vector3 interceptorLocation, float interceptorSpeed)
{
Vector3 A = targetLocation;
float Ax = targetLocation.x;
float Ay = targetLocation.y;
float Az = targetLocation.z;
float As = targetVelocity.magnitude;
Vector3 Av = Vector3.Normalize(targetVelocity);
float Avx = Av.x;
float Avy = Av.y;
float Avz = Av.z;
Vector3 B = interceptorLocation;
float Bx = interceptorLocation.x;
float By = interceptorLocation.y;
float Bz = interceptorLocation.z;
float Bs = interceptorSpeed;
float t = 0;
float a = (
Mathf.Pow(As, 2) * Mathf.Pow(Avx, 2) +
Mathf.Pow(As, 2) * Mathf.Pow(Avy, 2) +
Mathf.Pow(As, 2) * Mathf.Pow(Avz, 2) -
Mathf.Pow(Bs, 2)
);
if (a == 0)
{
Debug.Log("Quadratic formula not applicable");
return targetLocation;
}
float b = (
As * Avx * Ax +
As * Avy * Ay +
As * Avz * Az +
As * Avx * Bx +
As * Avy * By +
As * Avz * Bz
);
float c = (
Mathf.Pow(Ax, 2) +
Mathf.Pow(Ay, 2) +
Mathf.Pow(Az, 2) -
Ax * Bx -
Ay * By -
Az * Bz +
Mathf.Pow(Bx, 2) +
Mathf.Pow(By, 2) +
Mathf.Pow(Bz, 2)
);
float t1 = (-b + Mathf.Pow((Mathf.Pow(b, 2) - (4 * a * c)), (1 / 2))) / (2 * a);
float t2 = (-b - Mathf.Pow((Mathf.Pow(b, 2) - (4 * a * c)), (1 / 2))) / (2 * a);
Debug.Log("t1 = " + t1 + "; t2 = " + t2);
if (t1 <= 0 || t1 == Mathf.Infinity || float.IsNaN(t1))
if (t2 <= 0 || t2 == Mathf.Infinity || float.IsNaN(t2))
return targetLocation;
else
t = t2;
else if (t2 <= 0 || t2 == Mathf.Infinity || float.IsNaN(t2) || t2 > t1)
t = t1;
else
t = t2;
Debug.Log("t = " + t);
Debug.Log("Bs = " + Bs);
float Bvx = (Ax - Bx + (t * As + Avx)) / (t * Mathf.Pow(Bs, 2));
float Bvy = (Ay - By + (t * As + Avy)) / (t * Mathf.Pow(Bs, 2));
float Bvz = (Az - Bz + (t * As + Avz)) / (t * Mathf.Pow(Bs, 2));
Vector3 Bv = new Vector3(Bvx, Bvy, Bvz);
Debug.Log("||Bv|| = (Should be 1) " + Bv.magnitude);
return Bv * Bs;
}
I followed the problem formulation as described by Gil Moshayof's answer, but found that there was an error in the simplification of the quadratic formula. When I did the derivation by hand I got a different solution.
The following is what worked for me when finding the intersect in 2D:
std::pair<double, double> find_2D_intersect(Vector3 sourcePos, double projSpd, Vector3 targetPos, double targetSpd, double targetHeading)
{
double P0x = targetPos.x;
double P0y = targetPos.y;
double s0 = targetSpd;
double V0x = std::cos(targetHeading);
double V0y = std::sin(targetHeading);
double P1x = sourcePos.x;
double P1y = sourcePos.y;
double s1 = projSpd;
// quadratic formula
double a = (s0 * s0)*((V0x * V0x) + (V0y * V0y)) - (s1 * s1);
double b = 2 * s0 * ((P0x * V0x) + (P0y * V0y) - (P1x * V0x) - (P1y * V0y));
double c = (P0x * P0x) + (P0y * P0y) + (P1x * P1x) + (P1y * P1y) - (2 * P1x * P0x) - (2 * P1y * P0y);
double t1 = (-b + std::sqrt((b * b) - (4 * a * c))) / (2 * a);
double t2 = (-b - std::sqrt((b * b) - (4 * a * c))) / (2 * a);
double t = choose_best_time(t1, t2);
double intersect_x = P0x + t * s0 * V0x;
double intersect_y = P0y + t * s0 * V0y;
return std::make_pair(intersect_x, intersect_y);
}
I’ve written the below python script. The idea is to calculate the new location of point C after you rotate the globe from point A to point B. I first calculate point P, which is the rotation pole. With calculating point P already something goes wrong. With the following input f.e. I would assume point P to be having latitude 90 or –90.
I asked this question before here: Rotate a sphere from coord1 to coord2, where will coord3 be?
But I figured it's better to ask again with the script included ;)
# GreatCircle can be downloaded from: http://www.koders.com/python/fid0A930D7924AE856342437CA1F5A9A3EC0CAEACE2.aspx?s=coastline
from GreatCircle import *
from math import *
# Points A and B defining the rotation:
LonA = radians(0)
LatA = radians(1)
LonB = radians(45)
LatB = radians(1)
# Point C which will be translated:
LonC = radians(90)
LatC = radians(1)
# The following equation is described here: http://articles.adsabs.harvard.edu//full/1953Metic...1...39L/0000040.000.html
# It calculates the rotation pole at point P of the Great Circle defined by point A and B.
# According to http://www.tutorialspoint.com/python/number_atan2.htm
# atan2(x, y) = atan(y / x)
LonP = atan2(((sin(LonB) * tan(LatA)) - (sin(LonA) * tan(LatB))), ((cos(LonA) * tan(LatB)) - (cos(LonB) * tan(LatA))))
LatP = atan2(-tan(LatA),(cos(LonP - LonA)))
print degrees(LonP), degrees(LatP)
# The equations to calculate the translated point C location were found here: http://www.uwgb.edu/dutchs/mathalgo/sphere0.htm
# The Rotation Angle in radians:
gcAP = GreatCircle(1,1,degrees(LonA),degrees(LatA),degrees(LonP),degrees(LatP))
gcBP = GreatCircle(1,1,degrees(LonB),degrees(LatB),degrees(LonP),degrees(LatP))
RotAngle = abs(gcAP.azimuth12 - gcBP.azimuth12)
# The rotation pole P in Cartesian coordinates:
Px = cos(LatP) * cos(LonP)
Py = cos(LatP) * sin(LonP)
Pz = sin(LatP)
# Point C in Cartesian coordinates:
Cx = cos(radians(LatC)) * cos(radians(LonC))
Cy = cos(radians(LatC)) * sin(radians(LonC))
Cz = sin(radians(LatC))
# The translated point P in Cartesian coordinates:
NewCx = (Cx * cos(RotAngle)) + (1 - cos(RotAngle)) * (Px * Px * Cx + Px * Py * Cy + Px * Pz * Cz) + (Py * Cz - Pz * Cy) * sin(RotAngle)
NewCy = (Cy * cos(RotAngle)) + (1 - cos(RotAngle)) * (Py * Px * Cx + Py * Py * Cy + Py * Pz * Cz) + (Pz * Cx - Px * Cz) * sin(RotAngle)
NewCz = (Cz * cos(RotAngle)) + (1 - cos(RotAngle)) * (Pz * Px * Cx + Pz * Py * Cy + Pz * Pz * Cz) + (Px * Cy - Py * Cx) * sin(RotAngle)
# The following equation I got from http://rbrundritt.wordpress.com/2008/10/14/conversion-between-spherical-and-cartesian-coordinates-systems/
# The translated point P in lat/long:
Cr = sqrt((NewCx*NewCx) + (NewCy*NewCy) + (NewCz*NewCz))
NewCLat = degrees(asin(NewCz/Cr))
NewCLon = degrees(atan2(NewCy, NewCx))
# Output:
print str(NewCLon) + "," + str(NewCLat)
I have two points in space, L1 and L2 that defines two points on a line.
I have three points in space, P1, P2 and P3 that 3 points on a plane.
So given these inputs, at what point does the line intersect the plane?
Fx. the plane equation A*x+B*y+C*z+D=0 is:
A = p1.Y * (p2.Z - p3.Z) + p2.Y * (p3.Z - p1.Z) + p3.Y * (p1.Z - p2.Z)
B = p1.Z * (p2.X - p3.X) + p2.Z * (p3.X - p1.X) + p3.Z * (p1.X - p2.X)
C = p1.X * (p2.Y - p3.Y) + p2.X * (p3.Y - p1.Y) + p3.X * (p1.Y - p2.Y)
D = -(p1.X * (p2.Y * p3.Z - p3.Y * p2.Z) + p2.X * (p3.Y * p1.Z - p1.Y * p3.Z) + p3.X * (p1.Y * p2.Z - p2.Y * p1.Z))
But what about the rest?
The simplest (and very generalizable) way to solve this is to say that
L1 + x*(L2 - L1) = (P1 + y*(P2 - P1)) + (P1 + z*(P3 - P1))
which gives you 3 equations in 3 variables. Solve for x, y and z, and then substitute back into either of the original equations to get your answer. This can be generalized to do complex things like find the point that is the intersection of two planes in 4 dimensions.
For an alternate approach, the cross product N of (P2-P1) and (P3-P1) is a vector that is at right angles to the plane. This means that the plane can be defined as the set of points P such that the dot product of P and N is the dot product of P1 and N. Solving for x such that (L1 + x*(L2 - L1)) dot N is this constant gives you one equation in one variable that is easy to solve. If you're going to be intersecting a lot of lines with this plane, this approach is definitely worthwhile.
Written out explicitly this gives:
N = cross(P2-P1, P3 - P1)
Answer = L1 + (dot(N, P1 - L1) / dot(N, L2 - L1)) * (L2 - L1)
where
cross([x, y, z], [u, v, w]) = x*u + y*w + z*u - x*w - y*u - z*v
dot([x, y, z], [u, v, w]) = x*u + y*v + z*w
Note that that cross product trick only works in 3 dimensions, and only for your specific problem of a plane and a line.
This is how I ended up doing it in come code. Luckily one code library (XNA) had half of what I needed, and the rest was easy.
var lv = L2-L1;
var ray = new Microsoft.Xna.Framework.Ray(L1,lv);
var plane = new Microsoft.Xna.Framework.Plane(P1, P2, P3);
var t = ray.Intersects(plane); //Distance along line from L1
///Result:
var x = L1.X + t * lv.X;
var y = L1.Y + t * lv.Y;
var z = L1.Z + t * lv.Z;
Of course I would prefer having just the simple equations that takes place under the covers of XNA.