Can anyone tell me how to draw an Ellipse using the CodeWorld package of Haskell? I want it to be like the rectangle function where I give two arguments for length and width. I have tried using solidClosedCurve-am I heading in the right direction?
Using a closed curve, you can set the 4 vertices like so:
ellipse'(a, b) = closedCurve([(a,0),(0,b),(-a,0),(0,-b)])
Another way to do it is to say that an ellipse is a circle rescaled in one direction.
ellipse(a, b) = scaled(circle(1), a, b)
https://code.world/#Ps8tKc4KH4v8Z4iq91NZsew
Related
I am trying various visualizations for an Igraph in R (version.3.3.1).
Currently my visualizing is as shown as below, 2 nodes (blue and green) in circular layout.
Circular Layout
visNetwork(data$nodes,data$edges) %>% visIgraphLayout(layout="layout_in_circle")
Now I want to have a semicircle structure instead of a full circle as in the pic. All blue nodes form a semicircle, green nodes another semicircle. Each semicircle separated by a small distance as well. How can i achieve this. I found grid package has an option for semicircle, but i couldnt make it work with igraph. Please provide some pointers.
The layout argument accepts an arbitrary matrix with two columns and N rows if your graph has N vertices; all you need to do is to create a list of coordinates that correspond to a semicircle. You can make use of the fact that a vertex at angle alpha around a circle with radius r centered at (0, 0) is to be found at (r * cos(alpha), r * sin(alpha)). Since you are using R, alpha should be specified in radians, spaced evenly between 0 and pi (which corresponds to 180 degrees).
I need to find 4 points in Latitude/Longitude format surrounding a given center point and a resulting algorithm (if possible).
Known information:
Equal distances for each "bin" from center of point (Radar) outward.
Example = .54 nautical miles.
1 Degree beam width.
Center point of the "bin"
This image is in Polar coordinates (I think this is similar to Radial coordinates???):
I need to convert from Polar/Radial to Cartesian and I should be able to do that with this formula.
x = r × cos( θ )
y = r × sin( θ )
So now all I need to do is find the "bin" outline coordinates (4 corners) so I can draw a polygon in a Cartesian coordinate space.
I'm using Delphi/Pascal for coding, but I might be able to convert other languages if you have a sample algorithm.
Thanks for any suggestions or sample algorithms.
Regards,
Bryan
You need to convert everything to the same coordinate system and then impose the distance criteria as follows:
Convert your center point from geographic coordinates to polar coordinates to yield (rC, θC)
Convert your center point from polar to Cartesian coordinates using your equations yielding (xC, yC)
The corner points on the right side of the center points (xR, yR) satisfy the equation
(xR - xC)2 + (yR - yC)2 = D2
[rRcos(θC+0.5o) - xC]2 + [rRsin(θC+0.5o) - yC]2 = D2
where D=distance between the center point and corner points
Everything is known in the above equation except rR. This should yield a quadratic equation with two solutions which you can easily solve. Those are your two corner points on the right side.
Repeat step 3 with angle θC-0.5o to get the corner points on the left side.
Given:
A point P, circle 1 and circle 2's positions and radii
What is:
The equation for T, the 'mix level' between color 1 and 2 (a value between 0 to 1)
Many radial gradient equations only apply to concentric circles or circles that share a position. I'm looking for something that matches the image below, created using Quartz (Core Graphics). I am writing a GLSL shader, but I need to understand the math first.
If this is in 2D, then you can write the parameters of the circle that your point lies on as:
x3=T*x1+(1-T)*x2
y3=T*y1+(1-T)*y2
r3=T*r1+(1-T)*r2
EDIT: Of course, that circle can be represented as:
(x3-xP)^2+(y3-yP)^2=r3^2
You can substitute the first 3 equations into the last (and remember that (xP, yP) is your point) to get 1 equation with only T as a variable that is quadratic in T, so it is easy to solve for T. Doing so gives us:
T=(-r2*(r1-r2)+(x1-x2)*(x2-xP)+(y1-y2)(y2-yP)
{+-}sqrt(r2^2*((x1-xP)^2+(y1-yP)^2)-2*r1*r2*((x1-xP)*(x2-xP)
+(y1-yP)*(y2-yP))+r1^2*((x2-xP)^2+(y2-yP)^2)
-(x2*y1-xP*y1-x1*y2+xP*y2+x1*yP-x2*yP)^2))
/((r1-r2)^2-(x1-x2)^2-(y1-y2)^2)
I know that that is a bit hard to read, but it is not actually that bad mathematically. It is just addition, multiplication, and squaring (which is really just multiplication).
I'm trying to find the best way to get the most distant point of a circle from a specified point in 2D space. What I have found so far, is how to get the distance between the point and the circle position, but I'm not entirely sure how to expand this to find the most distant point of the circle.
The known variables are:
Point a
Point b (circle position)
Radius r (circle radius)
To find the distance between the point and the circle position, I have found this:
xd = x2 - x1
yd = y2 - y1
Distance = SquareRoot(xd * xd + yd * yd)
It seems to me, this is part of the solution. How would this be expanded to get the position of Point x in the below image?
As an additional but optional part of the question: I have read in some places that it would be possible to get the distance portion without using the Square Root, which is very performance intensive and should be avoided if fast code is necessary. In my case, I would be doing this calculation quite often; Any comments on this within the context of the main question would be welcome too.
What about this?
Calculate A-B.
We now have a vector pointing from the center of the circle towards A (if B is the origin, skip this and just consider point A a vector).
Normalize.
Now we have a well defined length (the length is 1)
If the circle is not of unit radius, multiply by radius. If it is unit radius, skip this.
Now we have the correct length.
Invert sign (can be done in one step with 3., just multiply with the negative radius)
Now our vector points in the correct direction.
Add B (if B is the origin, skip this).
Now our vector is offset correctly so its endpoint is the point we want.
(Alternatively, you could calculate B-A to save the negation, but then you have to do one more operation to offset the origin correctly.)
By the way, it works the same in 3D, except the circle would be a sphere, and the vectors would have 3 components (or 4, if you use homogenous coords, in this case remember -- for correctness -- setting w to 0 when "turning points into vectors" and to 1 at the end when making a point from the vector).
EDIT:
(in reply of pseudocode)
Assuming you have a vec2 class which is a struct of two float numbers with operators for vector subtraction and scalar multiplicaion (pretty trivial, around a dozen lines of code) and a function normalize which needs to be no more than a shorthand for multiplying with inv_sqrt(x*x+y*y), the pseudocode (my pseudocode here is something like a C++/GLSL mix) could look something like this:
vec2 most_distant_on_circle(vec2 const& B, float r, vec2 const& A)
{
vec2 P(A - B);
normalize(P);
return -r * P + B;
}
Most math libraries that you'd use should have all of these functions and types built-in. HLSL and GLSL have them as first type primitives and intrinsic functions. Some GPUs even have a dedicated normalize instruction.
I need to calculate a chord`s starting and ending point, now I have the starting point which lies on the circumfrence of the circle and I also have the angle between starting point and ending point, but I cannot seem to find a way to determine the end of point of chord as it should lie on the circumfrence , I also have the centre and radius of circle, the methods I have looked over the internet all give chord length. So i in short I need to find the other end of a chord when one point and the angle between both points is given, any idea of links would be appreciated
thanks
Imran
The arbitrary point on the circle with center (X0,Y0) and radius R has coordinates
x = Xo+R*cos(a)
y = Y0+R*sin(a), 0<=a<2*Pi
knowing X,Y,X0,Y0 and R you can easily find angle a:
a = acos((x-X0)/R) (may be + Pi depending on sign of the y-Y0).
Then you can calculate the angle for the second chord endpoint (you'll have two solutions actually) - a+angle and a-angle. Then put angle you got into circle equation and you'll get your required points coordinates.