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I am trying to produce a distribution for points within a circle where there are more towards the center, but for a donut shape
I can produce a uniformly distributed donut shape (1), or a circle where there are more points towards the center (2) but not a donut shape that has so many points towards its inside boundary.
(1)
r = sqrt(runif(250, min = 0.25, max =1))
p = runif(250)
theta = p *2*pi
a = r * cos(theta) #coordinates for circle
b = r * sin(theta)
plot(a,b)
(2)
r = runif(250)
p = runif(250)
theta = p *2*pi
a = r * cos(theta)
b = r * sin(theta)
plot(a,b)
My closest attempt is modifying program (2) where r is bound between 0.5 and 1 but this removes most of the points closest to the center and does not have as many around its inside.
As #RobertDodier said, try to use some advanced distribution for radius. What I would like to propose is Beta distribution.
First, it is naturally in the [0...1] range, no need to truncate, accept/reject etc.
Second, it has two parameters (a,b) which could be used to get single peak, zero at 0, and slope to 1. Look at graphs in the wiki page. And last, it is implemented in R. a being smaller than b means peak is on the lef of 0.5, a being larger than b means peak is on the right of 0.5.
Along the lines
N = 10000
r = rbeta(N, 7.0, 5.0)
theta = 2.0*pi*runif(N)
a = r * cos(theta)
b = r * sin(theta)
plot(a,b)
will produce plot like that
Does it look like at donut?
UPDATE
This one is with clear hole at the center and shape proposed by #RobertDodier
N = 10000
hole = 0.25
r = hole + (1.0-hole)*rbeta(N, 1.0, 3.0)
theta = 2.0*pi*runif(N)
a = r * cos(theta)
b = r * sin(theta)
plot(a,b)
Another one with clear hole in the center and symmetric shape, like a true donut
r = hole + (1.0-hole)*rbeta(N, 2.0, 2.0)
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.
What is best way or an algorithm for generating a random 3d point [x,y,z] inside the volume of the circular cylinder if radius r and height h of the cylinder are given?
How about -- in Python pseudocode, letting R be the radius and H be the height:
s = random.uniform(0, 1)
theta = random.uniform(0, 2*pi)
z = random.uniform(0, H)
r = sqrt(s)*R
x = r * cos(theta)
y = r * sin(theta)
z = z # .. for symmetry :-)
The problem with simply taking x = r * cos(angle) and y = r * sin(angle) is that then when r is small, i.e. at the centre of the circle, a tiny change in r doesn't change the x and y positions very much. IOW, it leads to a nonuniform distribution in Cartesian coordinates, and the points get concentrated toward the centre of the circle. Taking the square root corrects this, at least if I've done my arithmetic correctly.
[Ah, it looks like the sqrt was right.]
(Note that I assumed without thinking about it that the cylinder is aligned with the z-axis and the cylinder centre is located at (0,0,H/2). It'd be less arbitrary to set (0,0,0) at the cylinder centre, in which case z should be chosen to be between -H/2 and H/2, not 0,H.)
Generate a random point inside the rectangular solid circumscribing the cylinder; if it's inside the cylinder (probability pi/4), keep it, otherwise discard it and try again.
Generate a random angle (optionally less than 2π), a random r less than the radius, and a random z less than the height.
x = r * cos(angle)
y = r * sin(angle)
The z axis is easy: -0.5 * h <= z <= 0.5 * h
The x and y are equal to a circle will be:
x^2 + y^2 <= r^2
Buth math is long ago for me :-)
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 am working with a hexagonal grid. I have chosen to use this coordinate system because it is quite elegant.
This question talks about generating the coordinates themselves, and is quite useful. My issue now is in converting these coordinates to and from actual pixel coordinates. I am looking for a simple way to find the center of a hexagon with coordinates x,y,z. Assume (0,0) in pixel coordinates is at (0,0,0) in hex coords, and that each hexagon has an edge of length s. It seems to me like x,y, and z should each move my coordinate a certain distance along an axis, but they are interrelated in an odd way I can't quite wrap my head around it.
Bonus points if you can go the other direction and convert any (x,y) point in pixel coordinates to the hex that point belongs in.
For clarity, let the "hexagonal" coordinates be (r,g,b) where r, g, and b are the red, green, and blue coordinates, respectively. The coordinates (r,g,b) and (x,y) are related by the following:
y = 3/2 * s * b
b = 2/3 * y / s
x = sqrt(3) * s * ( b/2 + r)
x = - sqrt(3) * s * ( b/2 + g )
r = (sqrt(3)/3 * x - y/3 ) / s
g = -(sqrt(3)/3 * x + y/3 ) / s
r + b + g = 0
Derivation:
I first noticed that any horizontal row of hexagons (which should have a constant y-coordinate) had a constant b coordinate, so y depended only on b. Each hexagon can be broken into six equilateral triangles with sides of length s; the centers of the hexagons in one row are one and a half side-lengths above/below the centers in the next row (or, perhaps easier to see, the centers in one row are 3 side lengths above/below the centers two rows away), so for each change of 1 in b, y changes 3/2 * s, giving the first formula. Solving for b in terms of y gives the second formula.
The hexagons with a given r coordinate all have centers on a line perpendicular to the r axis at the point on the r axis that is 3/2 * s from the origin (similar to the above derivation of y in terms of b). The r axis has slope -sqrt(3)/3, so a line perpendicular to it has slope sqrt(3); the point on the r axis and on the line has coordinates (3sqrt(3)/4 * s * r, -3/4 * s * r); so an equation in x and y for the line containing the centers of the hexagons with r-coordinate r is y + 3/4 * s * r = sqrt(3) * (x - 3sqrt(3)/4 * s * r). Substituting for y using the first formula and solving for x gives the second formula. (This is not how I actually derived this one, but my derivation was graphical with lots of trial and error and this algebraic method is more concise.)
The set of hexagons with a given r coordinate is the horizontal reflection of the set of hexagons with that g coordinate, so whatever the formula is for the x coordinate in terms of r and b, the x coordinate for that formula with g in place of r will be the opposite. This gives the third formula.
The fourth and fifth formulas come from substituting the second formula for b and solving for r or g in terms of x and y.
The final formula came from observation, verified by algebra with the earlier formulas.