How do you get the points of lines tangent to two circles, where both starting point and end point of line are exactly touching their respective circles?
Please help. Thanks.
The angles at p4 and p2 are 90 degrees. If you conceptually put them at the base of a quadrilateral p2-p4-c2-c1, you can get the angle at C2 (from p4-c1) and at c1 (from p2-c2). With that angle, you can easily construct it.
Related
Is there an analytical solution to find the intersection point between the placed grain and grain 1 if the placed grain is lowered [in the reference frame of the image] along the dotted line? We know the radius of both circles. We have graphical figured out the intersection point and labelled it for reference in the image.
Assuming
a right angle between the dotted line and the line through the center of g1 and g2
the dotted line is a tangent of g1
you can use the following:
Consider the situation when g1 and gp touch. In that case, the length of the line segment between center g1 and gp is equal to radius(g1) + radius(gp). Which is also the hypotenuse of a right-angled triangle for which radius(g1) is a cathetus. arccos(radius(g1)/(radius(g1) + radius(gp))) gives you then the angle between the hypotenuse and cathetus around the center of g1.
This angle and the length radius(g1) are polar coordinates of the intersection point relative to the center of g1.
I have 5 {x,y} points randomly placed on a grid
Each of the points do not know the {x,y} coordinates of the other points
Each of the points do know the distance of each of the other points from their {x,y} position
Each of the points exchanges this distance information with every other point
So every point knows every distance of every other point
Using this distance information every point can calculate (by finding the angles) triangles for every other point using itself as a reference point
Example, point 1 can calculate the following triangles:
1-2-3,
1-2-4,
1-2-5,
1-3-4,
1-3-5,
1-4-5,
and using the distance data recieved from the other points it can also calculate
2-3-4,
2-3-5,
2-4-5,
3-4-5
I would like to build a map of the location of every other point relative to a single point
How should I go about doing this? I am asuming it would be some kind of triangulation algorithm but these mainly seem to compute the location of a point from three other points, not the other way around where the other points {x,y} coordinates are discovered based on only the distance information.
I have tried plotting the two possible triangles for every 3 triangle points and then rotating them on a fixed known point to try and align them, but I think this avenue will end up with too many possibilities and errors
Ultimately I would like every point to end up with {x,y} coordinates of every other point relative to itself
You know the distance from one point to every other, dij. Thus, point 2 lies in a circumference of center point 1 and radius = d12. Point 3 lies in a circumference of center point 1 and R=d13 and it also lies in another circumference of center point 2 and R=d23.
See this picture:
I've set point 2 in X-axis for simplicity.
As you see, point 3 is on the intersection of two cicrcumferences centered at P1 and P2. There is a second intersection, P3a. Let's choose the one that is upwards and continue.
For P4 we can use three circumferences, centered at P1, P2 and P3. Again we get two solutions.
The same process can be done with the rest of points. For Pn you have n-1 circumferences.
I'm sure you can find the maths for circle-circle intersection.
Some remarks must be observed:
1) The construction is simpler if you first sort the points by distance to P1.
2) Not all distances generate a solution. For example, increase d13 an there's no intersection between the two circumferences for P3. Or increase d14 and now the three circumferences don't intersect in just the two expected points 4 and 4a.
3) This fact can be overworked by considering the average of intersections and the distance from each solution to this average. You can set a tolerance in these distances and tell if the average is a solution or else some dij is wrong. Since two solutions are possible, you must consider two averages.
4) The two possible triangulations are symmetric, over X-axis in the case I've drawn.
The real solution is obtained by a rotation around P1. To calculate the angle of rotation you need the {x,y} coordinates of another point.
I have two problem first how to calculate the minimum distance between two lines.
For details I am attaching an image here.
In the image describe the line with start and end point. I already have start and end point for both lines but I am not getting any idea how to calculate the minimum distance between two lines.
Another problem is that how to calculate the minimum distance between two arc.
I am attaching another image here
For arc I have the start, end and center point also I have the start and end angle.
This link has lot more things - Shortest distance between a point and a line segment
I also got help from above link.
This is the another issue what I am facing now. In this scenario how to calculate minimum distance between two arc?
Any idea how solve this two issue?
Let the two segments be AB and CD. Their parametric equations can be written
P = A + u AB, Q = C + v CD, with u, v in [0, 1].
You want to minimize the (squared) distance
PQ² = (CA + u AB - v CD)², under the given constraints,
and you can cancel the first derivatives with
(CA + u AB - v CD).AB = 0
(CA + u AB - v CD).CD = 0
After resolution of the 2x2 system you get a pair (u, v). If both variables fall in [0,1], there is an intersection and the distance is 0.
Otherwise, clamp u and/or v to the relevant bound of range [0, 1] and compute the corresponding distance.
If one variable was clamped, the distance was between an endpoint and a segment; if two were clamped, it's between two endpoints.
A similar approach can be adopted for the arcs (using trigonometric functions), and lead to an optimization problem under linear constraints. Less easy to handle as the objective function is non-linear, though.
We can also proceed as follows:
find the points that make the shortest distance between the whole circles. There are two cases:
the circles intersect, at two places
the circles do not intersect; the shortest distance is between the intersection points of the two circles and the center line.
then check if these points do belong to the arcs by angle comparison. If yes, you are done (the distance is either 0 or the distance between the intersections).
otherwise, consider the endpoints of an arc against the other circle. The closest point is the intersection of the circle with the line through the point and the center. If the intersection belongs to the arc, keep the distance between the point and the intersection. Repeat this for all four combinations endpoint/arc and keep the closest pair.
If no valid pair was found, keep the shortest endpoint/endpoint distance.
The picture shows the distances that can be considered. In green, endpoint/circle; in red, endpoint/endpoint. In this case, the circle/circle distance is zero as they intersect. A distance can be considered if it joins two points inside the arcs.
Distance between two intersecting lines is 0
Otherwise, what you want is to calculate the endpoint of one of the line to the other one.
If you want to calculate the distance between a point and a line :
It is the length of the line segment which joins the point to the line and is perpendicular to the line.
For arcs I think arcs are part of circles, distance between them is the distance between their center points and minus their radiuses.
Probably you can find more here: Calculate the minimum distance between two given circular arcs
Example
I have the following information about a 2D triangle:
The length of all sides
The coordinates of 2 points
I need to find the coordinates of the last point.
The triangle is not necessarily a right triangle.
If you know point A and point B and lengths AC and BC, then finding C is a matter of doing a circle-circle intersection. The two circles in question are the circle with radius AC around A and the circle with radius BC around B. There will be two answers (which makes sense if you think about it: you can always choose which side of the AB to put C on, given the information that you have).
Given 3 2d points that form a triangle, is there a property of a triangle which does not change after any sort of rotation which would help in giving me the direction vector for that triangle. For eg if a triangle is isosceles i can get the direction vector by joining the midpoint of the base to the opposite vertex and I know which 2 points make the base by looking at the distances. Similar thing can be done for right angle triangles.
Can this be generalized to all triangles?
It seems that only an isosceles (or near-isosceles) triangle can be pointing in a certain direction.
Calculate the 3 angles of the triangle, the side surrounded by the two identical (or perhapse similar) angles is your base. Then go from there.