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I am trying to solve the following problem (I am using Matlab, though pseudo-code / solutions in other languages are welcome):
I have two circles on a Cartesian plane defined by their centroids (p1, p2) and their radii (r1, r2). circle 1 (c1 = [p1 r1]) is considered 'dynamic': it is being translated along the vector V = [0 -1]. circle 2 (c2 = [p2 r2]) is considered 'static': it lies in the path of c1 but the x component of its centroid is offset from the x component of c2 (otherwise the solution would be trivial: the distance between the circle centroids minus the sum of their radii).
I am trying to locate the distance (d) along V at which circle 1 will 'collide' with circle 2 (see the linked image). I am sure that I can solve this iteratively (i.e. translate c1 to the bounding box of c2 then converge / test for intersection). However, I would like to know if there is a closed form solution to this problem.
Shift coordinates to simplify expressions
px = p1.x - p2.x
py = p1.y - p2.y
And solve quadratic equation for d (zero, one, or two solutions)
px^2 + (py - d)^2 = (r1 + r2)^2
(py - d)^2 = (r1 + r2)^2 - px^2
d = py +/- Sqrt((r1 + r2)^2 - px^2)
That's all.
As the question title does not match the question and accepted answer which is dependent on a fixed vector {0, -1}, or {0, 1} rather than an arbitrary vector I have added another solution that works for any unit vector.
Where (See diagram 1)
dx, dy is the unit vector of travel for circle c1
p1, p2 the centers of the moving circle c1 and static circle c2
r1, r2 the radius of each circle
The following will set d to the distance c1 must travel along dx, dy to collide with c2 if no collision the d will be set to Infinity
There are three cases when there is no solution
The moving circle is moving away from the static circle. u < 0
The moving circle never gets close enough to collide. dSq > rSq
The two circles are already overlapping. u < 0 luckily the math makes
this the same condition as moving away.
Note that if you ignore the sign of u (1 and 3) then d will be the distance to first (causal) contact going backward in time
Thus the pseudo code to find d
d = Infinity
rSq = (r1 + r2) ^ 2
u = (p1.x - p2.x) * dx + (p1.x - p2.x) * dy
if u >= 0
dSq = ((p2.x + dx * u) - p1.x) ^ 2 + ((p2.y + dy * u) - p1.y) ^ 2
if dSq <= rSq
d = u - (rSq - dSq) ^ 0.5
The point of contact can be found with
cpx = p1.x + dx * d;
cpy = p1.x + dy * d;
Diagram 1
I will try to be very descriptive with this. I'm editing a game right now and the scenario is a 3D area.
I have an initial angle, writen as a direction vector, and another vector which haves different coordinates. As we know, the angle between 2 vectors is given by the formula: Theta = ACos( DotProduct( vec1, vec2 ) / ( VectorLength( vec1 ) * VectorLength( vec2 ) ) )
So let's describe the scenario: I'm currently programming some kind of stationary weapon, a sentry gun, this thing moves slowly his "head", shooting bullets to enemies. That angle rotation thing is my problem.
Let's imagine this: I have my sentry gun on a empty 3D area, and a "enemy" spawns over there. I can currently get the direction vector of my sentry's view angle, and the direction vector between my sentry and the player. Let's guess, using the formula described, his separation angle is 45 degrees. My sentry gun thinks (calls a function) at every 0.1 seconds, and I want to move his head 5 degrees at every thinking function until it reach the the player (ie, both vectors are nearly equal), and that means it will reach the player (if player keeps on its position...) in 0.9 seconds (5 degrees from 45)
How I can move sentry's view angle slowly until it reach a target? In 2D is easily but know I'm fighting with a 3D scenario, and I'm currently lost with this.
Any help would be appreciated, and about coding, I will be grateful with a pseudocode. Thanks! (and sorry for my english)
What you need is called SLERP - spherical linear interpolation
Your starting direction vector is p0 there, goal direction is p1, Omega is your Theta, and t parameter varies in range 0..1 with needed step
Delphi example for 2D case (it is easy to control)
var
p0, p1: TPoint;
i, xx, yy: Integer;
omega, InvSinOmega, t, a0, a1: Double;
begin
P0 := Point(0, 200);
P1 := Point(200, 0);
omega := -Pi / 2;
InvSinOmega := 1.0 / Sin(omega);
Canvas.Brush.Color := clRed;
Canvas.Ellipse(120 + P0.X, 120 + P0.Y, 120 + P0.X + 7, 120 + P0.Y + 7);
Canvas.Ellipse(120 + P1.X, 120 + P1.Y, 120 + P1.X + 7, 120 + P1.Y + 7);
for i := 1 to 9 do begin
t := i / 10;
a0 := sin((1 - t) * omega) * InvSinOmega;
a1 := sin(t * omega) * InvSinOmega;
xx := Round(P0.X * a0 + P1.X * a1);
yy := Round(P0.Y * a0 + P1.Y * a1);
Canvas.Brush.Color := RGB(25 * i, 25 * i, 25 * i);
Canvas.Ellipse(120 + xx, 120 + yy, 120 + xx + 9, 120 + yy + 9);
end;
For quite a while I've been performing rotations simply with:
setOrigin(screenWidth/2, screenHeight/2)
theta = approach(360) // cycles from 0 to 360
cP = vec2(0, 0) // center point
rp = vec2(0, 100) // rotated point
rP.x = cP.x + cos(theta) * 100 //cos accepts degrees
rP.y = cP.y + sin(theta) * 100 //sin accepts degrees
However I've found the following method to be quite widespread
setOrigin(screenWidth/2, screenHeight/2)
theta = approach(360) // cycles from 0 to 360
cP = vec2(0, 0) // center point
rp = vec2(0, 100) // rotated point
rPx = cP.x + (rP.x * cos(theta) - rP.y * sin(theta)) * 100 //cos accepts degrees
rPy = cP.y + (rP.y * sin(theta) + rP.x * cos(theta)) * 100 //sin accepts degrees
rP = {rPx, rPy}
I understand why the second version works mathematically i'm just not sure why on would prefer it over the simpler first method (apart from the obvious case of placing the values within a matrix and performing transformations)
Your first method is a simplified version of the second where you only allow the difference vector (from rotation center to rotation point) to be a horizontal line. If you plug in rp.y instead of 100, you will see that you get the formulas of the second variant, where the rp.x parts are removed.
Btw, the second formulas are not quite correct. The * 100 is wrong because you already multiplied with rp. In the second line, you have to swap rp.y and rp.x to:
rPy = cP.y + (rP.x * sin(theta) + rP.y * cos(theta))
Okay, this all takes place in a nice and simple 2D world... :)
Suppose I have a static object A at position Apos, and a linearly moving object B at Bpos with bVelocity, and an ammo round with velocity Avelocity...
How would I find out the angle that A has to shoot, to hit B, taking into account B's linear velocity and the speed of A's ammo ?
Right now the aim's at the current position of the object, which means that by the time my projectile gets there the unit has moved on to safer positions :)
I wrote an aiming subroutine for xtank a while back. I'll try to lay out how I did it.
Disclaimer: I may have made one or more silly mistakes anywhere in here; I'm just trying to reconstruct the reasoning with my rusty math skills. However, I'll cut to the chase first, since this is a programming Q&A instead of a math class :-)
How to do it
It boils down to solving a quadratic equation of the form:
a * sqr(x) + b * x + c == 0
Note that by sqr I mean square, as opposed to square root. Use the following values:
a := sqr(target.velocityX) + sqr(target.velocityY) - sqr(projectile_speed)
b := 2 * (target.velocityX * (target.startX - cannon.X)
+ target.velocityY * (target.startY - cannon.Y))
c := sqr(target.startX - cannon.X) + sqr(target.startY - cannon.Y)
Now we can look at the discriminant to determine if we have a possible solution.
disc := sqr(b) - 4 * a * c
If the discriminant is less than 0, forget about hitting your target -- your projectile can never get there in time. Otherwise, look at two candidate solutions:
t1 := (-b + sqrt(disc)) / (2 * a)
t2 := (-b - sqrt(disc)) / (2 * a)
Note that if disc == 0 then t1 and t2 are equal.
If there are no other considerations such as intervening obstacles, simply choose the smaller positive value. (Negative t values would require firing backward in time to use!)
Substitute the chosen t value back into the target's position equations to get the coordinates of the leading point you should be aiming at:
aim.X := t * target.velocityX + target.startX
aim.Y := t * target.velocityY + target.startY
Derivation
At time T, the projectile must be a (Euclidean) distance from the cannon equal to the elapsed time multiplied by the projectile speed. This gives an equation for a circle, parametric in elapsed time.
sqr(projectile.X - cannon.X) + sqr(projectile.Y - cannon.Y)
== sqr(t * projectile_speed)
Similarly, at time T, the target has moved along its vector by time multiplied by its velocity:
target.X == t * target.velocityX + target.startX
target.Y == t * target.velocityY + target.startY
The projectile can hit the target when its distance from the cannon matches the projectile's distance.
sqr(projectile.X - cannon.X) + sqr(projectile.Y - cannon.Y)
== sqr(target.X - cannon.X) + sqr(target.Y - cannon.Y)
Wonderful! Substituting the expressions for target.X and target.Y gives
sqr(projectile.X - cannon.X) + sqr(projectile.Y - cannon.Y)
== sqr((t * target.velocityX + target.startX) - cannon.X)
+ sqr((t * target.velocityY + target.startY) - cannon.Y)
Substituting the other side of the equation gives this:
sqr(t * projectile_speed)
== sqr((t * target.velocityX + target.startX) - cannon.X)
+ sqr((t * target.velocityY + target.startY) - cannon.Y)
... subtracting sqr(t * projectile_speed) from both sides and flipping it around:
sqr((t * target.velocityX) + (target.startX - cannon.X))
+ sqr((t * target.velocityY) + (target.startY - cannon.Y))
- sqr(t * projectile_speed)
== 0
... now resolve the results of squaring the subexpressions ...
sqr(target.velocityX) * sqr(t)
+ 2 * t * target.velocityX * (target.startX - cannon.X)
+ sqr(target.startX - cannon.X)
+ sqr(target.velocityY) * sqr(t)
+ 2 * t * target.velocityY * (target.startY - cannon.Y)
+ sqr(target.startY - cannon.Y)
- sqr(projectile_speed) * sqr(t)
== 0
... and group similar terms ...
sqr(target.velocityX) * sqr(t)
+ sqr(target.velocityY) * sqr(t)
- sqr(projectile_speed) * sqr(t)
+ 2 * t * target.velocityX * (target.startX - cannon.X)
+ 2 * t * target.velocityY * (target.startY - cannon.Y)
+ sqr(target.startX - cannon.X)
+ sqr(target.startY - cannon.Y)
== 0
... then combine them ...
(sqr(target.velocityX) + sqr(target.velocityY) - sqr(projectile_speed)) * sqr(t)
+ 2 * (target.velocityX * (target.startX - cannon.X)
+ target.velocityY * (target.startY - cannon.Y)) * t
+ sqr(target.startX - cannon.X) + sqr(target.startY - cannon.Y)
== 0
... giving a standard quadratic equation in t. Finding the positive real zeros of this equation gives the (zero, one, or two) possible hit locations, which can be done with the quadratic formula:
a * sqr(x) + b * x + c == 0
x == (-b ± sqrt(sqr(b) - 4 * a * c)) / (2 * a)
+1 on Jeffrey Hantin's excellent answer here. I googled around and found solutions that were either too complex or not specifically about the case I was interested in (simple constant velocity projectile in 2D space.) His was exactly what I needed to produce the self-contained JavaScript solution below.
The one point I would add is that there are a couple special cases you have to watch for in addition to the discriminant being negative:
"a == 0": occurs if target and projectile are traveling the same speed. (solution is linear, not quadratic)
"a == 0 and b == 0": if both target and projectile are stationary. (no solution unless c == 0, i.e. src & dst are same point.)
Code:
/**
* Return the firing solution for a projectile starting at 'src' with
* velocity 'v', to hit a target, 'dst'.
*
* #param ({x, y}) src position of shooter
* #param ({x, y, vx, vy}) dst position & velocity of target
* #param (Number) v speed of projectile
*
* #return ({x, y}) Coordinate at which to fire (and where intercept occurs). Or `null` if target cannot be hit.
*/
function intercept(src, dst, v) {
const tx = dst.x - src.x;
const ty = dst.y - src.y;
const tvx = dst.vx;
const tvy = dst.vy;
// Get quadratic equation components
const a = tvx * tvx + tvy * tvy - v * v;
const b = 2 * (tvx * tx + tvy * ty);
const c = tx * tx + ty * ty;
// Solve quadratic
const ts = quad(a, b, c); // See quad(), below
// Find smallest positive solution
let sol = null;
if (ts) {
const t0 = ts[0];
const t1 = ts[1];
let t = Math.min(t0, t1);
if (t < 0) t = Math.max(t0, t1);
if (t > 0) {
sol = {
x: dst.x + dst.vx * t,
y: dst.y + dst.vy * t
};
}
}
return sol;
}
/**
* Return solutions for quadratic
*/
function quad(a, b, c) {
let sol = null;
if (Math.abs(a) < 1e-6) {
if (Math.abs(b) < 1e-6) {
sol = Math.abs(c) < 1e-6 ? [0, 0] : null;
} else {
sol = [-c / b, -c / b];
}
} else {
let disc = b * b - 4 * a * c;
if (disc >= 0) {
disc = Math.sqrt(disc);
a = 2 * a;
sol = [(-b - disc) / a, (-b + disc) / a];
}
}
return sol;
}
// For example ...
const sol = intercept(
{x:2, y:4}, // Starting coord
{x:5, y:7, vx: 2, vy:1}, // Target coord and velocity
5 // Projectile velocity
)
console.log('Fire at', sol)
First rotate the axes so that AB is vertical (by doing a rotation)
Now, split the velocity vector of B into the x and y components (say Bx and By). You can use this to calculate the x and y components of the vector you need to shoot at.
B --> Bx
|
|
V
By
Vy
^
|
|
A ---> Vx
You need Vx = Bx and Sqrt(Vx*Vx + Vy*Vy) = Velocity of Ammo.
This should give you the vector you need in the new system. Transform back to old system and you are done (by doing a rotation in the other direction).
Jeffrey Hantin has a nice solution for this problem, though his derivation is overly complicated. Here's a cleaner way of deriving it with some of the resultant code at the bottom.
I'll be using x.y to represent vector dot product, and if a vector quantity is squared, it means I am dotting it with itself.
origpos = initial position of shooter
origvel = initial velocity of shooter
targpos = initial position of target
targvel = initial velocity of target
projvel = velocity of the projectile relative to the origin (cause ur shooting from there)
speed = the magnitude of projvel
t = time
We know that the position of the projectile and target with respect to t time can be described with some equations.
curprojpos(t) = origpos + t*origvel + t*projvel
curtargpos(t) = targpos + t*targvel
We want these to be equal to each other at some point (the point of intersection), so let's set them equal to each other and solve for the free variable, projvel.
origpos + t*origvel + t*projvel = targpos + t*targvel
turns into ->
projvel = (targpos - origpos)/t + targvel - origvel
Let's forget about the notion of origin and target position/velocity. Instead, let's work in relative terms since motion of one thing is relative to another. In this case, what we now have is relpos = targetpos - originpos and relvel = targetvel - originvel
projvel = relpos/t + relvel
We don't know what projvel is, but we do know that we want projvel.projvel to be equal to speed^2, so we'll square both sides and we get
projvel^2 = (relpos/t + relvel)^2
expands into ->
speed^2 = relvel.relvel + 2*relpos.relvel/t + relpos.relpos/t^2
We can now see that the only free variable is time, t, and then we'll use t to solve for projvel. We'll solve for t with the quadratic formula. First separate it out into a, b and c, then solve for the roots.
Before solving, though, remember that we want the best solution where t is smallest, but we need to make sure that t is not negative (you can't hit something in the past)
a = relvel.relvel - speed^2
b = 2*relpos.relvel
c = relpos.relpos
h = -b/(2*a)
k2 = h*h - c/a
if k2 < 0, then there are no roots and there is no solution
if k2 = 0, then there is one root at h
if 0 < h then t = h
else, no solution
if k2 > 0, then there are two roots at h - k and h + k, we also know r0 is less than r1.
k = sqrt(k2)
r0 = h - k
r1 = h + k
we have the roots, we must now solve for the smallest positive one
if 0<r0 then t = r0
elseif 0<r1 then t = r1
else, no solution
Now, if we have a t value, we can plug t back into the original equation and solve for the projvel
projvel = relpos/t + relvel
Now, to the shoot the projectile, the resultant global position and velocity for the projectile is
globalpos = origpos
globalvel = origvel + projvel
And you're done!
My implementation of my solution in Lua, where vec*vec represents vector dot product:
local function lineartrajectory(origpos,origvel,speed,targpos,targvel)
local relpos=targpos-origpos
local relvel=targvel-origvel
local a=relvel*relvel-speed*speed
local b=2*relpos*relvel
local c=relpos*relpos
if a*a<1e-32 then--code translation for a==0
if b*b<1e-32 then
return false,"no solution"
else
local h=-c/b
if 0<h then
return origpos,relpos/h+targvel,h
else
return false,"no solution"
end
end
else
local h=-b/(2*a)
local k2=h*h-c/a
if k2<-1e-16 then
return false,"no solution"
elseif k2<1e-16 then--code translation for k2==0
if 0<h then
return origpos,relpos/h+targvel,h
else
return false,"no solution"
end
else
local k=k2^0.5
if k<h then
return origpos,relpos/(h-k)+targvel,h-k
elseif -k<h then
return origpos,relpos/(h+k)+targvel,h+k
else
return false,"no solution"
end
end
end
end
Following is polar coordinate based aiming code in C++.
To use with rectangular coordinates you would need to first convert the targets relative coordinate to angle/distance, and the targets x/y velocity to angle/speed.
The "speed" input is the speed of the projectile. The units of the speed and targetSpeed are irrelevent, as only the ratio of the speeds are used in the calculation. The output is the angle the projectile should be fired at and the distance to the collision point.
The algorithm is from source code available at http://www.turtlewar.org/ .
// C++
static const double pi = 3.14159265358979323846;
inline double Sin(double a) { return sin(a*(pi/180)); }
inline double Asin(double y) { return asin(y)*(180/pi); }
bool/*ok*/ Rendezvous(double speed,double targetAngle,double targetRange,
double targetDirection,double targetSpeed,double* courseAngle,
double* courseRange)
{
// Use trig to calculate coordinate of future collision with target.
// c
//
// B A
//
// a C b
//
// Known:
// C = distance to target
// b = direction of target travel, relative to it's coordinate
// A/B = ratio of speed and target speed
//
// Use rule of sines to find unknowns.
// sin(a)/A = sin(b)/B = sin(c)/C
//
// a = asin((A/B)*sin(b))
// c = 180-a-b
// B = C*(sin(b)/sin(c))
bool ok = 0;
double b = 180-(targetDirection-targetAngle);
double A_div_B = targetSpeed/speed;
double C = targetRange;
double sin_b = Sin(b);
double sin_a = A_div_B*sin_b;
// If sin of a is greater than one it means a triangle cannot be
// constructed with the given angles that have sides with the given
// ratio.
if(fabs(sin_a) <= 1)
{
double a = Asin(sin_a);
double c = 180-a-b;
double sin_c = Sin(c);
double B;
if(fabs(sin_c) > .0001)
{
B = C*(sin_b/sin_c);
}
else
{
// Sin of small angles approach zero causing overflow in
// calculation. For nearly flat triangles just treat as
// flat.
B = C/(A_div_B+1);
}
// double A = C*(sin_a/sin_c);
ok = 1;
*courseAngle = targetAngle+a;
*courseRange = B;
}
return ok;
}
Here's an example where I devised and implemented a solution to the problem of predictive targeting using a recursive algorithm: http://www.newarteest.com/flash/targeting.html
I'll have to try out some of the other solutions presented because it seems more efficient to calculate it in one step, but the solution I came up with was to estimate the target position and feed that result back into the algorithm to make a new more accurate estimate, repeating several times.
For the first estimate I "fire" at the target's current position and then use trigonometry to determine where the target will be when the shot reaches the position fired at. Then in the next iteration I "fire" at that new position and determine where the target will be this time. After about 4 repeats I get within a pixel of accuracy.
I just hacked this version for aiming in 2d space, I didn't test it very thoroughly yet but it seems to work. The idea behind it is this:
Create a vector perpendicular to the vector pointing from the muzzle to the target.
For a collision to occur, the velocities of the target and the projectile along this vector (axis) should be the same!
Using fairly simple cosine stuff I arrived at this code:
private Vector3 CalculateProjectileDirection(Vector3 a_MuzzlePosition, float a_ProjectileSpeed, Vector3 a_TargetPosition, Vector3 a_TargetVelocity)
{
// make sure it's all in the horizontal plane:
a_TargetPosition.y = 0.0f;
a_MuzzlePosition.y = 0.0f;
a_TargetVelocity.y = 0.0f;
// create a normalized vector that is perpendicular to the vector pointing from the muzzle to the target's current position (a localized x-axis):
Vector3 perpendicularVector = Vector3.Cross(a_TargetPosition - a_MuzzlePosition, -Vector3.up).normalized;
// project the target's velocity vector onto that localized x-axis:
Vector3 projectedTargetVelocity = Vector3.Project(a_TargetVelocity, perpendicularVector);
// calculate the angle that the projectile velocity should make with the localized x-axis using the consine:
float angle = Mathf.Acos(projectedTargetVelocity.magnitude / a_ProjectileSpeed) / Mathf.PI * 180;
if (Vector3.Angle(perpendicularVector, a_TargetVelocity) > 90.0f)
{
angle = 180.0f - angle;
}
// rotate the x-axis so that is points in the desired velocity direction of the projectile:
Vector3 returnValue = Quaternion.AngleAxis(angle, -Vector3.up) * perpendicularVector;
// give the projectile the correct speed:
returnValue *= a_ProjectileSpeed;
return returnValue;
}
I made a public domain Unity C# function here:
http://ringofblades.com/Blades/Code/PredictiveAim.cs
It is for 3D, but you can easily modify this for 2D by replacing the Vector3s with Vector2s and using your down axis of choice for gravity if there is gravity.
In case the theory interests you, I walk through the derivation of the math here:
http://www.gamasutra.com/blogs/KainShin/20090515/83954/Predictive_Aim_Mathematics_for_AI_Targeting.php
I've seen many ways to solve this problem mathematically, but this was a component relevant to a project my class was required to do in high school, and not everyone in this programming class had a background with calculus, or even vectors for that matter, so I created a way to solve this problem with more of a programming approach. The point of intersection will be accurate, although it may hit 1 frame later than in the mathematical computations.
Consider:
S = shooterPos, E = enemyPos, T = targetPos, Sr = shooter range, D = enemyDir
V = distance from E to T, P = projectile speed, Es = enemy speed
In the standard implementation of this problem [S,E,P,Es,D] are all givens and you are solving either to find T or the angle at which to shoot so that you hit T at the proper timing.
The main aspect of this method of solving the problem is to consider the range of the shooter as a circle encompassing all possible points that can be shot at any given time. The radius of this circle is equal to:
Sr = P*time
Where time is calculated as an iteration of a loop.
Thus to find the distance an enemy travels given the time iteration we create the vector:
V = D*Es*time
Now, to actually solve the problem we want to find a point at which the distance from the target (T) to our shooter (S) is less than the range of our shooter (Sr). Here is somewhat of a pseudocode implementation of this equation.
iteration = 0;
while(TargetPoint.hasNotPassedShooter)
{
TargetPoint = EnemyPos + (EnemyMovementVector)
if(distanceFrom(TargetPoint,ShooterPos) < (ShooterRange))
return TargetPoint;
iteration++
}
Basically , intersection concept is not really needed here, As far as you are using projectile motion, you just need to hit at a particular angle and instantiate at the time of shooting so that you get the exact distance of your target from the Source and then once you have the distance, you can calculate the appropriate velocity with which it should shot in order to hit the Target.
The following link makes teh concept clear and is considered helpful, might help:
Projectile motion to always hit a moving target
I grabbed one of the solutions from here, but none of them take into account movement of the shooter. If your shooter is moving, you might want to take that into account (as the shooter's velocity should be added to your bullet's velocity when you fire). Really all you need to do is subtract your shooter's velocity from the target's velocity. So if you're using broofa's code above (which I would recommend), change the lines
tvx = dst.vx;
tvy = dst.vy;
to
tvx = dst.vx - shooter.vx;
tvy = dst.vy - shooter.vy;
and you should be all set.
While working on SVG implementation for Internet Explorer to be based on its own VML format I came to a problem of translation of an SVG elliptical arc to an VML elliptical arc.
In VML an arc is given by: two angles for two points on ellipse and lengths of radiuses,
In SVG an arc is given by: two pairs of coordinates for two points on ellipse and sizes of ellipse boundary box
So, the question is: How to express angles of two points on ellipse to two pairs of their coordinates.
An intermediate question could be: How to find the center of an ellipse by coordinates of a pair of points on its curve.
Update: Let's have a precondition saying that an ellipse is normally placed (its radiuses are parallel to linear coordinate system axis), thus no rotation is applied.
Update: This question is not related to svg:ellipse element, rather to "a" elliptical arc command in svg:path element (SVG Paths: The elliptical arc curve commands)
So the solution is here:
The parametrized formula of an ellipse:
x = x0 + a * cos(t)
y = y0 + b * sin(t)
Let's put known coordinates of two points to it:
x1 = x0 + a * cos(t1)
x2 = x0 + a * cos(t2)
y1 = y0 + b * sin(t1)
y2 = y0 + b * sin(t2)
Now we have a system of equations with 4 variables: center of ellipse (x0/y0) and two angles t1, t2
Let's subtract equations in order to get rid of center coordinates:
x1 - x2 = a * (cos(t1) - cos(t2))
y1 - y2 = b * (sin(t1) - sin(t2))
This can be rewritten (with product-to-sum identities formulas) as:
(x1 - x2) / (2 * a) = sin((t1 + t2) / 2) * sin((t1 - t2) / 2)
(y2 - y1) / (2 * b) = cos((t1 + t2) / 2) * sin((t1 - t2) / 2)
Let's replace some of the equations:
r1: (x1 - x2) / (2 * a)
r2: (y2 - y1) / (2 * b)
a1: (t1 + t2) / 2
a2: (t1 - t2) / 2
Then we get simple equations system:
r1 = sin(a1) * sin(a2)
r2 = cos(a1) * sin(a2)
Dividing first equation by second produces:
a1 = arctan(r1/r2)
Adding this result to the first equation gives:
a2 = arcsin(r2 / cos(arctan(r1/r2)))
Or, simple (using compositions of trig and inverse trig functions):
a2 = arcsin(r2 / (1 / sqrt(1 + (r1/r2)^2)))
or even more simple:
a2 = arcsin(sqrt(r1^2 + r2^2))
Now the initial four-equations system can be resolved with easy and all angles as well as eclipse center coordinates can be found.
The elliptical curve arc link you posted includes a link to elliptical arc implementation notes.
In there, you will find the equations for conversion from endpoint to centre parameterisation.
Here is my JavaScript implementation of those equations, taken from an interactive demo of elliptical arc paths, using Sylvester.js to perform the matrix and vector calculations.
// Calculate the centre of the ellipse
// Based on http://www.w3.org/TR/SVG/implnote.html#ArcConversionEndpointToCenter
var x1 = 150; // Starting x-point of the arc
var y1 = 150; // Starting y-point of the arc
var x2 = 400; // End x-point of the arc
var y2 = 300; // End y-point of the arc
var fA = 1; // Large arc flag
var fS = 1; // Sweep flag
var rx = 100; // Horizontal radius of ellipse
var ry = 50; // Vertical radius of ellipse
var phi = 0; // Angle between co-ord system and ellipse x-axes
var Cx, Cy;
// Step 1: Compute (x1′, y1′)
var M = $M([
[ Math.cos(phi), Math.sin(phi)],
[-Math.sin(phi), Math.cos(phi)]
]);
var V = $V( [ (x1-x2)/2, (y1-y2)/2 ] );
var P = M.multiply(V);
var x1p = P.e(1); // x1 prime
var y1p = P.e(2); // y1 prime
// Ensure radii are large enough
// Based on http://www.w3.org/TR/SVG/implnote.html#ArcOutOfRangeParameters
// Step (a): Ensure radii are non-zero
// Step (b): Ensure radii are positive
rx = Math.abs(rx);
ry = Math.abs(ry);
// Step (c): Ensure radii are large enough
var lambda = ( (x1p * x1p) / (rx * rx) ) + ( (y1p * y1p) / (ry * ry) );
if(lambda > 1)
{
rx = Math.sqrt(lambda) * rx;
ry = Math.sqrt(lambda) * ry;
}
// Step 2: Compute (cx′, cy′)
var sign = (fA == fS)? -1 : 1;
// Bit of a hack, as presumably rounding errors were making his negative inside the square root!
if((( (rx*rx*ry*ry) - (rx*rx*y1p*y1p) - (ry*ry*x1p*x1p) ) / ( (rx*rx*y1p*y1p) + (ry*ry*x1p*x1p) )) < 1e-7)
var co = 0;
else
var co = sign * Math.sqrt( ( (rx*rx*ry*ry) - (rx*rx*y1p*y1p) - (ry*ry*x1p*x1p) ) / ( (rx*rx*y1p*y1p) + (ry*ry*x1p*x1p) ) );
var V = $V( [rx*y1p/ry, -ry*x1p/rx] );
var Cp = V.multiply(co);
// Step 3: Compute (cx, cy) from (cx′, cy′)
var M = $M([
[ Math.cos(phi), -Math.sin(phi)],
[ Math.sin(phi), Math.cos(phi)]
]);
var V = $V( [ (x1+x2)/2, (y1+y2)/2 ] );
var C = M.multiply(Cp).add(V);
Cx = C.e(1);
Cy = C.e(2);
An ellipse cannot be defined by only two points. Even a circle (a special cased ellipse) is defined by three points.
Even with three points, you would have infinite ellipses passing through these three points (think: rotation).
Note that a bounding box suggests a center for the ellipse, and most probably assumes that its major and minor axes are parallel to the x,y (or y,x) axes.
The intermediate question is fairly easy... you don't. You work out the centre of an ellipse from the bounding box (namely, the centre of the box is the centre of the ellipse, as long as the ellipse is centred in the box).
For your first question, I'd look at the polar form of the ellipse equation, which is available on Wikipedia. You would need to work out the eccentricity of the ellipse as well.
Or you could brute force the values from the bounding box... work out if a point lies on the ellipse and matches the angle, and iterate through every point in the bounding box.
TypeScript implementation based on the answer from Rikki.
Default DOMMatrix and DOMPoint are used for the calculations (Tested in the latest Chrome v.80) instead of the external library.
ellipseCenter(
x1: number,
y1: number,
rx: number,
ry: number,
rotateDeg: number,
fa: number,
fs: number,
x2: number,
y2: number
): DOMPoint {
const phi = ((rotateDeg % 360) * Math.PI) / 180;
const m = new DOMMatrix([
Math.cos(phi),
-Math.sin(phi),
Math.sin(phi),
Math.cos(phi),
0,
0,
]);
let v = new DOMPoint((x1 - x2) / 2, (y1 - y2) / 2).matrixTransform(m);
const x1p = v.x;
const y1p = v.y;
rx = Math.abs(rx);
ry = Math.abs(ry);
const lambda = (x1p * x1p) / (rx * rx) + (y1p * y1p) / (ry * ry);
if (lambda > 1) {
rx = Math.sqrt(lambda) * rx;
ry = Math.sqrt(lambda) * ry;
}
const sign = fa === fs ? -1 : 1;
const div =
(rx * rx * ry * ry - rx * rx * y1p * y1p - ry * ry * x1p * x1p) /
(rx * rx * y1p * y1p + ry * ry * x1p * x1p);
const co = sign * Math.sqrt(Math.abs(div));
// inverse matrix b and c
m.b *= -1;
m.c *= -1;
v = new DOMPoint(
((rx * y1p) / ry) * co,
((-ry * x1p) / rx) * co
).matrixTransform(m);
v.x += (x1 + x2) / 2;
v.y += (y1 + y2) / 2;
return v;
}
Answering part of the question with code
How to find the center of an ellipse by coordinates of a pair of points on its curve.
This is a TypeScript function which is based on the excellent accepted answer by Sergey Illinsky above (which ends somewhat halfway through, IMHO). It calculates the center of an ellipse with given radii, given the condition that both provided points a and b must lie on the circumference of the ellipse. Since there are (almost) always two solutions to this problem, the code choses the solution that places the ellipse "above" the two points:
(Note that the ellipse must have major and minor axis parallel to the horizontal/vertical)
/**
* We're in 2D, so that's what our vertors look like
*/
export type Point = [number, number];
/**
* Calculates the vector that connects the two points
*/
function deltaXY (from: Point, to: Point): Point {
return [to[0]-from[0], to[1]-from[1]];
}
/**
* Calculates the sum of an arbitrary amount of vertors
*/
function vecAdd (...vectors: Point[]): Point {
return vectors.reduce((acc, curr) => [acc[0]+curr[0], acc[1]+curr[1]], [0, 0]);
}
/**
* Given two points a and b, as well as ellipsis radii rX and rY, this
* function calculates the center-point of the ellipse, so that it
* is "above" the two points and has them on the circumference
*/
function topLeftOfPointsCenter (a: Point, b: Point, rX: number, rY: number): Point {
const delta = deltaXY(a, b);
// Sergey's work leads up to a simple system of liner equations.
// Here, we calculate its general solution for the first of the two angles (t1)
const A = Math.asin(Math.sqrt((delta[0]/(2*rX))**2+(delta[1]/(2*rY))**2));
const B = Math.atan(-delta[0]/delta[1] * rY/rX);
const alpha = A + B;
// This may be the new center, but we don't know to which of the two
// solutions it belongs, yet
let newCenter = vecAdd(a, [
rX * Math.cos(alpha),
rY * Math.sin(alpha)
]);
// Figure out if it is the correct solution, and adjusting if not
const mean = vecAdd(a, [delta[0] * 0.5, delta[1] * 0.5]);
const factor = mean[1] > newCenter[1] ? 1 : -1;
const offMean = deltaXY(mean, newCenter);
newCenter = vecAdd(mean, [offMean[0] * factor, offMean[1] * factor]);
return newCenter;
}
This function does not check if a solution is possible, meaning whether the radii provided are large enough to even connect the two points!