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.
Related
I'm trying to generate a mesh from a sphere of radius r. My goal is to create a UV sphere such that every point on the polyhedron has distance from the sphere smaller than tol.
The following code creates a grid of points on the sphere. How can I compute parallels_count and meridians_count so that all the point of the mesh are within tolerance?
for j in parallels_count:
parallel = PI * (j+1) / parallels_count
for i in meridians_count:
meridian = 2.0 * PI * i / meridians_count
return spherical_to_cartesian(meridian, parallel)
The code comes from here, and this is a picture of the UV sphere:
The distance between each face of the mesh and the sphere will be maximum around the center of the face.
So, for the distance between a face and the sphere to be smaller than tol it is not sufficient that the distances between the edges of the face and the corresponding circumferences are smaller than tol.
This picture is out of context but helps me explaining what I mean.
the biggest distance between points is on equator so use circle circumference to obtain angular step if I am not mistaken it should be...
dangle = tol/r; //[rad]
where r is sphere radius in the same units as tol you can use smaller step to be sure like dangle*=0.75; use this for both parallel and meridian angles.
If you want your counts instead then try:
meridians_count = (2.0*PI*r/tol)+1; // ceil or +1 just to be sure
parallels_count = 0.5*meridians_count;
It is still early here so I hope I did not make any silly math mistake (the easiest stuff is the worst for silly bugs).
Also take a look at few related QA's of mine:
Applying map of the earth texture a Sphere
Make a sphere with equidistant vertices
Sphere triangulation
[Edit1] well your new definition of tol changes everything
I see it like this:
sin(da/2) = (r-tol)/r
da = 2.0*asin((r-tol)/r)
If you convert to sphericalsurface than max difference is in center of uv grid cell which represents sqrt(2)*dadiagonal so try to use:
da = sqrt(2.0)*asin((r-tol)/r)
so your angle step should be a bit smaller than that ...
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.
I'm trying to find the best way to calculate this. On a 2D plane I have fixed points all with an instantaneous measurement value. The coordinates of these points is known. I want to predict the value of a movable point between these fixed points. The movable point coodinates will be known. So the distance betwwen the points is known as well.
This could be comparable to temperature readings or elevation on topography. I this case I'm wanting to predict ionospheric TEC of the mobile point from the fixed point measurements. The fixed point measurements are smoothed over time however I do not want to have to store previous values of the mobile point estimate in RAM.
Would some sort of gradient function be the way to go here?
This is the same algorithm for interpolating the height of a point from a triangle.
In your case you don't have z values for heights, but some other float value for each triangle vertex, but it's the same concept, still 3D points.
Where you have 3D triangle points p, q, r and test point pt, then pseudo code from the above mathgem is something like this:
Vector3 v1 = q - p;
Vector3 v2 = r - p;
Vector3 n = v1.CrossProduct(v2);
if n.z is not zero
return ((n.x * (pt.x - p.x) + n.y * (pt.y - p.y)) / -n.z) + p.z
As you indicate in your comment to #Phpdevpad, you do have 3 fixed points so this will work.
You can try contour plots especially contour lines. Simply use a delaunay triangulation of the points and a linear transformation along the edges. You can try my PHP implementations https://contourplot.codeplex.com for geographic maps. Another algorithm is conrec algorithm from Paul Bourke.
Given a description of an arc which has a startpoint and endpoint (both in Cartesian x,y coordinates), radius and direction (clockwise or counter-clockwise), I need to convert the arc to one with a start-angle, end-angle, center, and radius.
Is there known algorithm or pseudo code that allows me to do this? Also, is there any specific term to describe these kinds of transformations?
You can find a center solving this equation system:
(sx-cx)^2 + (sy-cy)^2=R^2
(ex-cx)^2 + (ey-cy)^2=R^2
where (sx,sy) are coordinates of starting point, (ex,ey) for ending point, unknowns cx, cy for center.
This system has two solutions. Then it is possible to find angles as
StartAngle = ArcTan2(sy-cy, sx-cx)
EndAngle = ArcTan2(ey-cy, ex-cx)
Note that known direction doesn't allow to select one from two possible solutions without additional limitations. For example, start=(0,1), end=(1,0), R=1 and Dir = clockwise give us both Pi/2 arc with center (0,0) and 3*Pi/2 arc with center (1,1)
I'd propose a different approach than MBo to obtain the centers of the two circles, which have the given radius and pass to both start and end point.
If P and Q are start and end point of the arc, the center of each of the two circles lies on the line L which is orthogonal to PQ, the line from P to Q, and which bisects PQ. The distance d from the centers to L is easily obtained by Pythagoras theorem. If e is the length of PQ, then d^2 + (e/2)^2 = r^2. This way you avoid to solve that system of equations you get from MBo's approach.
Note that, in case you have a semicircle, any approach will become numerically unstable because there is only one circle of the given radius with P and Q on it. (I guess I recall the correct term is 'the problem is ill posed' in that case. It happens when P and Q are precisely 2r apart, and to figure out whether this actually true you need to check for equality of two doubles, which is always a bit problematic. If, for some reason, you know you have a semicircle you are better of to just calculate the center of PQ).
I have some lines that their intersection describes a polygon, like this:
I know the order of the lines, and their equations.
To find the internal angles, I found each lines orientations. But I've got confused as subtracting two lines orientation would give two different angles, even if I do it in the order of polygon's sides.
For example, in the following image, if I just subtract the orientation of the lines, I would get any of the following angles:
What made me more confused, is when the polygon is not convex, I will have angles greater than 180, and using my approach I don't get the correct angle at all:
And I found out that this way of approaching the problem is wrong.
So, What is the best way of finding the internal angles using just the lines? I know for a convex polygon, I may find vectors and then find the angle between them, but even for P6 in my example the vector approach fails.
Anyway, I prefer a method that won't include a conditional case for solving that concavity problem.
Thanks.
With ordered lines it is possible to find points of intersection (polygon vertexes) in clockwise order. Then you can calculate internal angles:
Angle[i] = Pi + ArcTan2(V[i] x V[i+1], V[i] * V[i+1])
(crossproduct and dotproduct of incoming and outgoing vectors for every vertex)
or
Angle[i] = Pi + ArcTan2( dx_in*dy_out-dx_out*dy_in, dx_in*dx_out+dy_in*dy_out2 )
Note: change plus sign after Pi to minus for anti-clockwise direction.
Edit:
Note that crossproduct and dotproduct are scalars, not vectors.
Example for your data:
dx1 = 5; dy1 = -15; dx2 = -15; dy2 = 5
Angle = Pi + ArcTan2(5*5-15*15, -5*15-5*15) = Pi - 2.11 radians ~ 59 degrees
Example for vectors:
(0,-1) (1,0) (L-curve)
Angle = Pi + ArcTan2(1, 0) = 270 degrees