For my current project, a user taps 2 locations on the X,Y plane. Once the two points are tapped, the user should then click and drag to extend 2 new points starting at the original 2 locations into a perfect rectangle (90 degree corners).
The math seems super simple, I just can't seem to get the right configuration to slide these two points along the perpendicular slope (by a certain distance).
My current attempt is to find the perpendicular slope and slide it by X distance (the distance the user has dragged), but I'm stuck on translating the perp. slope by distance.
You have points A and B. Difference vector
D = (Dx, Dy) = (Bx - Ax, By - Ay)
Normalized (unit) vector
Len = Sqrt(Dx*Dx + Dy*Dy)
(dx, dy) = (Dx / Len, Dy / Len)
Perpendicular unit vector
(px, py) = (-dy, dx)
Shift by distance L
pL = (px * L, py * L)
So shifted A will have coordinates
(a'x, a'y) = (Ax +/- px * L, Bx +/- py * L)
+ or - for two possible shift directions
Hello again first part is working like a charm, thank you everyone.
But I've another question...
As I've no interface, is there a way to do the same thing with out not knowing the radius of the circle?
Should have refresh the page CodeMonkey solution is exactly what I was looking for...
Thank you again.
============================
First I'm not a developer, I'm a simple woodworker that left school far too early...
I'm trying to make one of my tool to work with an autonomous robot.
I made them communicate by reading a lot of tutorials.
But I have one problem I cant figure out.
Robot expect position of the tool as (X,Y) but tool's output is (A,B,C)
A is the distance from tool to north
B distance to east
C distance at 120 degree clockwise from east axe
the border is a circle, radius may change, and may or may not be something I know.
I've been on that for 1 month, and I can't find a way to transform those value into the position.
I made a test with 3 nails on a circle I draw on wood, and if I have the distance there is only one position possible, so I guess its possible.
But how?
Also, if someone as an answer I'd love pseudo code not code so I can practice.
If there is a tool to make a drawing I can use to make it clearer can you point it out to me?
Thank you.
hope it helps :
X, Y are coordinate from center, Da,Db, Dc are known.
Trying to make it more clear (sorry its so clear in my head).
X,Y are the coordinate of the point where is the tool (P).
Center is at 0,0
A is the point where vertical line cut the circle from P, with Da distance P to A;
B is the point where horizontal line cuts the circle fom P, with Db distance P to B.
C is the point where the line at 120 clockwise from horizontal cuts the circle from P, with Dc distance P to C.
Output from tool is an array of int (unit mm): A=123, B=114, C=89
Those are the only informations I have
thanks for all the ideas I'll try them at home later,
Hope it works :)
Basic geometry. I decided to give up having the circle at the origin. We don't know the center of the circle yet. What you do have, is three points on that circle. Let's try having the tool's position, given as P, as the new (0,0). This thus resolves to finding a circle given three points: (0, Da); (Db,0), and back off at 120° at Dc distance.
Pseudocode:
Calculate a line from A to B: we'll call it AB. Find AB's halfway point. Calculate a line perpendicular to AB, through that midpoint (e.g. the cross product of AB and a unit Z axis finds the perpendicular vector).
Calculate a line from B to C (or C to A works just as well): we'll call it BC. Find BC's halfway point. Calculate a line perpendicular to BC, through that midpoint.
Calculate where these two lines cross. This will be the origin of your circle.
Since P is at (0,0), the negative of your circle's origin will be your tool's coordinates relative to the circle's origin. You should be able to calculate anything you need relative to that, now.
Midpoint between two points: X=(X1+X2)/2. Y=(Y1+Y2)/2.
The circle's radius can be calculated using, e.g. point A and the circle's origin: R=sqrt(sqr((Ax-CirX)+sqr(Ay-CirY))
Distance from the edge: circle's radius - tool's distance from the circle's center via Pythagorean Theorem again.
Assume you know X and Y. R is the radius of the circle.
|(X, Y + Da)| = R
|(X + Db, Y)| = R
|(X - cos(pi/3) * Dc, Y - cos(pi/6) * Dc)| = R
Assuming we don't know the radius R. We can still say
|(X, Y + Da)|^2 = |(X + Db, Y)|^2
=> X^2 + (Y+Da)^2 = (X+Db)^2 + Y^2
=> 2YDa + Da^2 = 2XDb + Db^2 (I)
and denoting cos(pi/3)*Dc as c1 and cos(pi/6)*Dc as c2:
|(X, Y + Da)|^2 = |(X - c1, Y - c2)|^2
=> X^2 + Y^2 + 2YDa + Da^2 = X^2 - 2Xc1 + c1^2 + Y^2 - 2Yc2 + c2^2
=> 2YDa + Da^2 = - 2Xc1 + c1^2 - 2Yc2 + c2^2
=> Y = (-2Xc1 + c1^2 + c2^2 - Da^2) / 2(c2+Da) (II)
Putting (II) back in the equation (I) we get:
=> (-2Xc1 + c1^2 + c2^2 - Da^2) Da / (c2+Da) + Da^2 = 2XDb + Db^2
=> (-2Xc1 + c1^2 + c2^2 - Da^2) Da + Da^2 * (c2+Da) = 2XDb(c2+Da) + Db^2 * (c2+Da)
=> (-2Xc1 + c1^2 + c2^2) Da + Da^2 * c2 = 2XDb(c2+Da) + Db^2 * (c2+Da)
=> X = ((c1^2 + c2^2) Da + Da^2 * c2 - Db^2 * (c2+Da)) / (2Dbc2 + 2Db*Da + 2Dac1) (III)
Knowing X you can get Y by calculating (II).
You can also make some simplifications, e.g. c1^2 + c2^2 = Dc^2
Putting this into Python (almost Pseudocode):
import math
def GetXYR(Da, Db, Dc):
c1 = math.cos(math.pi/3) * Dc
c2 = math.cos(math.pi/6) * Dc
X = ((c1**2 + c2**2) * Da + Da**2 * c2 - Db * Db * (c2 + Da)) / (2 * Db * c2 + 2 * Db * Da + 2 * Da * c1)
Y = (-2*X*c1 + c1**2 + c2**2 - Da**2) / (2*(c2+Da))
R = math.sqrt(X**2 + (Y+Da)**2)
R2 = math.sqrt(Y**2 + (X+Db)**2)
R3 = math.sqrt((X - math.cos(math.pi/3) * Dc)**2 + (Y - math.cos(math.pi/6) * Dc)**2)
return (X, Y, R, R2, R3)
(X, Y, R, R2, R3) = GetXYR(123.0, 114.0, 89.0)
print((X, Y, R, R2, R3))
I get the result (X, Y, R, R2, R3) = (-8.129166703588021, -16.205081335032794, 107.1038654949096, 107.10386549490958, 107.1038654949096)
Which seems reasonable if both Da and Db are longer than Dc, then both coordinates are probably negative.
I calculated the Radius from three equations to cross check whether my calculation makes sense. It seems to fulfill all three equations we set up in the beginning.
Your problem is know a "circumscribed circle". You have a triangle define by 3 distances at given angles from your robot position, then you can construct the circumscribed circle from these three points (see Circumscribed circle from Wikipedia - section "Other properties"). So you know the diameter (if needed).
It is also known that the meeting point of perpendicular bisector of triangle sides is the center of the circumscribed circle.
Let's a=Da, b=Db. The we can write a system for points A and B at the circumference:
(x+b)^2 + y^2 = r^2
(y+a)^2 + x^2 = r^2
After transformations we have quadratic equation
y^2 * (4*b^2+4*a^2) + y * (4*a^3+4*a*b^2) + b^4-4*b^2*r^2+a^4+2*a^2*b^2 = 0
or
AA * y^2 + BB * y + CC = 0
where coefficients are
AA = (4*b^2+4*a^2)
BB = (4*a^3+4*a*b^2)
CC = b^4-4*b^2*r^2+a^4+2*a^2*b^2
So calculate AA, BB, CC coefficients, find solutions y1,y2 of quadratic eqiation, then get corresponding x1, x2 values using
x = (a^2 - b^2 + 2 * a * y) / (2 * b)
and choose real solution pair (where coordinate is inside the circle)
Quick checking:
a=1,b=1,r=1 gives coordinates 0,0, as expected (and false 1,-1 outside the circle)
a=3,b=4,r=5 gives coordinates (rough) 0.65, 1.96 at the picture, distances are about 3 and 4.
Delphi code (does not check all possible errors) outputs x: 0.5981 y: 1.9641
var
a, b, r, a2, b2: Double;
aa, bb, cc, dis, y1, y2, x1, x2: Double;
begin
a := 3;
b := 4;
r := 5;
a2 := a * a;
b2:= b * b;
aa := 4 * (b2 + a2);
bb := 4 * a * (a2 + b2);
cc := b2 * b2 - 4 * b2 * r * r + a2 * a2 + 2 * a2 * b2;
dis := bb * bb - 4 * aa * cc;
if Dis < 0 then begin
ShowMessage('no solutions');
Exit;
end;
y1 := (- bb - Sqrt(Dis)) / (2 * aa);
y2 := (- bb + Sqrt(Dis)) / (2 * aa);
x1 := (a2 - b2 + 2 * a * y1) / (2 * b);
x2 := (a2 - b2 + 2 * a * y2) / (2 * b);
if x1 * x1 + y1 * y1 <= r * r then
Memo1.Lines.Add(Format('x: %6.4f y: %6.4f', [x1, y1]))
else
if x2 * x2 + y2 * y2 <= r * r then
Memo1.Lines.Add(Format('x: %6.4f y: %6.4f', [x2, y2]));
From your diagram you have point P that you need it's X & Y coordinate. So we need to find Px and Py or (Px,Py). We know that Ax = Px and By = Py. We can use these for substitution if needed. We know that C & P create a line and all lines have slope in the form of y = mx + b. Where the slope is m and the y intercept is b. We don't know m or b at this point but they can be found. We know that the angle of between two vectors where the vectors are CP and PB gives an angle of 120°, but this does not put the angle in standard position since this is a CW rotation. When working with circles and trig functions along with linear equations of slope within them it is best to work in standard form. So if this line of y = mx + b where the points C & P belong to it the angle above the horizontal line that is parallel to the horizontal axis that is made by the points P & B will be 180° - 120° = 60° We also know that the cos angle between two vectors is also equal to the dot product of those vectors divided by the product of their magnitudes.
We don't have exact numbers yet but we can construct a formula: Since theta = 60° above the horizontal in the standard position we know that the slope m is also the tangent of that angle; so the slope of this line is tan(60°). So let's go back to our linear equation y = tan(60°)x + b. Since b is the y intercept we need to find what x is when y is equal to 0. Since we still have three undefined variables y, x, and b we can use the points on this line to help us here. We know that the points C & P are on this line. So this vector of y = tan(60°)x + b is constructed from (Px, Py) - (Cx, Cy). The vector is then (Px-Cx, Py-Cy) that has an angle of 60° above the horizontal that is parallel to the horizontal axis. We need to use another form of the linear equation that involves the points and the slope this time which happens to be y - y1 = m(x - x1) so this then becomes y - Py = tan(60°)(x - Px) well I did say earlier that we could substitute so let's go ahead and do that: y - By = tan(60°)(x - Ax) then y - By = tan(60°)x - tan(60°)Ax. And this becomes known if you know the actual coordinate points of A & B. The only thing here is that you have to convert your angle of 120° to standard form. It all depends on what your known and unknowns are. So if you need P and you have both A & B are known from your diagram the work is easy because the points you need for P will be P(Ax,By). And since you already said that you know Da, Db & Dc with their lengths then its just a matter of apply the correct trig functions with the proper angle and or using the Pythagorean Theorem to find the length of another leg of the triangle. It shouldn't be all that hard to find what P(x,y) is from the other points. You can use the trig functions, linear equations, the Pythagorean theorem, vector calculations etc. If you can find the equation of the line that points C & P knowing that P has A's x value and has B's y value and having the slope of that line that is defined by the tangent above the horizontal which is 180° - phi where phi is the angle you are giving that is CW rotation and theta would be the angle in standard position or above the horizontal you have a general form of y - By = tan(180° - phi)(x - Ax) and from this equation you can find any point on that line.
There are other methods such as using the existing points and the vectors that they create between each other and then generate an equilateral triangle using those points and then from that equilateral if you can generate one, you can use the perpendicular bisectors of that triangle to find the centroid of that triangle. That is another method that can be done. The only thing you may have to consider is the linear translation of the line from the origin. Thus you will have a shift in the line of (Ax - origin, By - origin) and to find one set the other to 0 and vise versa. There are many different methods to find it.
I just showed you several mathematical techniques that can help you to find a general equation based on your known(s) and unknown(s). It just a matter of recognizing which equations work in which scenario. Once you recognize the correct equations for the givens; the rest is fairly easy. I hope this helps you.
EDIT
I did forget to mention one thing; and that is the line of CP has a point on the edge of the circle defined by (cos(60°), sin(60°)) in the 1st quadrant. In the third quadrant you will have a point on this line and the circle defined by (-cos(60°), -sin(60°)) provided that this line goes through the origin (0,0) where there is no y nor x intercepts and if this is the case then the point on the circle at either end and the origin will be the radius of that circle.
I would like to write a function that takes a 3d triangle (as 3 points (vector3ds)) and returns a 2d triangle (as 3 points (vector2ds)):
When given a 3d triangle, it should return the 2 dimensional coordinates of its points as they lie on its plane. (By 'its plane' I mean the plane that all three points lie on).
I can think a long winded way todo this:
rotate the triangle until its normal is equal to +z (0,0,1), then construct a triangle from the (x, y) coords of each point.
I cant help but think there must be an easier way to achieve the same thing.
If posting code examples please try not to use Greek alphabet. Some pseudo code in a C/java style language would be ideal.
From your comments I infer that you can choose the coordinate system of the plane in an arbitrary way, as long as the Euclidean metric of this coordinate system is the same as the metric induced by the Euclidean metric of your three-dimensional coordinate system. (That is, Euclidean distances will stay the same.)
One possible solution:
x0' = 0
y0' = 0
x1' = sqrt((x1 - x0)^2 + (y1 - y0)^2 + (z1 - z0)^2)
y1' = 0
x2' = ((x1 - x0) * (x2 - x0) +
(y1 - y0) * (y2 - y0) +
(z1 - z0) * (z2 - z0)) / x1'
y2' = sqrt((x2 - x0)^2 + (y2 - y0)^2 + (z2 - z0)^2 - x2'^2)
There is no singular answer to this problem.
The plane in which the triangle sits has no defined origin, and no defined orientation.
The best you can do is define one of the vertices as the origin, and one of the edges laying along the X axis:
v1 = (0, 0)
You will need to calculate vectors A (i.e. v2 - v1) and B (i.e. v3 - v1).
Vertex 2 will then be at:
v2 = (|A|, 0)
The position of vertex 3 can be worked out by using the vector cross product rule, e.g.:
A x B = |A| * |B| sin(theta)
So, work out A x B and from that you can work out the sine of the angle theta between A and B:
sin(theta) = | A x B | / (|A| * |B|)
Vertex 3 is then at coordinates:
v3 = |B| (cos(theta), sin(theta))
You can take advantage of cos(theta) = sqrt(1 - sin(theta) ^ 2) to avoid any inverse trig operations.
You should also see that |B| sin(theta) is just | A x B | / | A |
I tried using a raycasting-style function to do it but can't get any maintainable results. I'm trying to calculate the intersection between two tangents on one circle. This picture should help explain:
I've googled + searched stackoverflow about this problem but can't find anything similar to this problem. Any help?
Well, if your variables are:
C = (cx, cy) - Circle center
A = (x1, y1) - Tangent point 1
B = (x2, y2) - Tangent point 2
The lines from the circle center to the two points A and B are CA = A - C and CB = B - C respectively.
You know that a tangent is perpendicular to the line from the center. In 2D, to get a line perpendicular to a vector (x, y) you just take (y, -x) (or (-y, x))
So your two (parametric) tangent lines are:
L1(u) = A + u * (CA.y, -CA.x)
= (A.x + u * CA.y, A.y - u * CA.x)
L2(v) = B + v * (CB.y, -CB.x)
= (B.x + v * CB.y, B.x - v * CB.x)
Then to calculate the intersection of two lines you just need to use standard intersection tests.
The answer by Peter Alexander assumes that you know the center of the circle, which is not obvious from your figure http://oi54.tinypic.com/e6y62f.jpg.
Here is a solution without knowing the center:
The point C (in your figure) is the intersection of the tangent at A(x, y) with the line L perpendicular to AB, cutting AB into halves. A parametric equation for the line L can be derived as follows:
The middle point of AB is M = ((x+x2)/2, (y+y2)/2), where B(x2, y2). The vector perpendicular to AB is N = (y2-y, x-x2). The vector equation of the line L is hence
L(t) = M + t N, where t is a real number.
I have 2 points in X,Y + Rotation and I need to calculate a bezier spline (a collection of quadratic beziers) that connects these 2 points smoothly. (see pic) The point represents a unit in a game which can only rotate slowly. So to get from point A to B, it has to take a long path. The attached picture shows quite an exaggeratedly curvy path, but you get the idea.
What formulas can I use to calculate such a bezier spline?
Just saw that I misunderstood your question. Couldn't you use a single cubic hermite splines instead since you have a start and end point and two directions (tangents)? Are there any additional constraints?
To calculate the start and end tangents just use the start and end direction and scale them with the distance between the start and end points (and additionally some other constant factor like 0.5 depending on how curvy you want the path to be).:
p0 = startpoint;
p1 = endpoint;
float scale = distance(p0, p1);
m0 = Vec2(cos(startangle), sin(startangle)) * scale;
m1 = Vec2(cos(endangle), sin(endangle)) * scale;
I use this system to interpolate camera paths in a game I'm working on and it works great.
As you probably know, there are infinite solutions, even if we assume 2 control points.
I found Smoothing Algorithm Using Bézier Curves, which answers your question (see equations of Bx(t) and By(t) in the beginning):
Bx(t) = (1-t)3 P1x + 3 (1-t)2 t P2x + 3 (1-t) t2 P3x + t3 P4x
By(t) = (1-t)3 P1y + 3 (1-t)2 t P2y + 3 (1-t) t2 P3y + t3 P4y
P1 and P4 are your endpoints, and P2 and P3 are the control points, which you're free to choose along the desired angles. If your control point is at a distance r along an angle θ from point (x, y), the coordinates of the point are:
x' = x - r sin(θ)
y' = y - r cos(θ)
(according to the co-ordinate system you've used—I think I got the signs right). The only free parameter is r, which you can choose as you want. You probably want to use
r = α dist(P1, P4)
with α < 1.
I don't remember where I got it, but I have been using this:
Vector2 CalculateBezierPoint(float t, Vector2 p0, Vector2 p1, Vector2 p2, Vector2 p3)
{
float u = 1 - t;
float tt = t*t;
float uu = u*u;
float uuu = uu * u;
float ttt = tt * t;
Vector2 p = uuu * p0; //first term
p += 3 * uu * t * p1; //second term
p += 3 * u * tt * p2; //third term
p += ttt * p3; //fourth term
return p;
}
where t is ratio along path between start and end points.