I already got the set of points that define that convex hull. How can I add a kind of padding to them, so that all the points is inside? I am trying to draw the red line
What I'd try to create an outline offsetted from a polygon by distance d:
Consider a straight line segment between each pair of consecutive points (the original convex hull).
Offset each segment by d along its perpendicular to the outside. The offsetted segments have gaps between ends.
For each pair of consecutive offsetted line segments, add a segment of a circle such that it both segments are its tangents; its radius should be d. It's easy to do: draw perpendiculars at the ends of both segments; their intersection is the center of the circle.
Now you have a sequence of interspersed straight lines and arcs. It should be the outline you're seeking.
To work comfortably, you'd need these functions:
Find the equation of the straight line by two points (store it as an object with 3 float fields);
Find an equation of a perpendicular line by a line equation and a point;
Find the point of intersection of two lines, given their equations.
Related
I have an ArrayList of some Point-s. It's guaranteed that the Points are part of a convex polygon.
How can I calculate the perimeter of this convex polygon?
Update: The Points in the ArrayList are out of any order
Update 2: All the points are part of the convex polygon's edge
Are the points in order? If so, you just need to sum up the distance from each vertex to the next
Sum the distance between each two consecutive points.
If the points are not ordered, its impossible without determining the correct order. Thats because if there are more than 3 points there is more than one polygon they could form.
I'm not entirely sure the constraint that the points are forming a convex polygon is sufficient to determine a canonic shape from the point cloud.
My guess is that by taking a random point from the list, and then looking for the nearest remaining point you can build a canonic order. From there its just summing up the length of the lines formed by consecutive points.
Edit: On second thought, scratch that idea. It won't work for all cases. That leaves you with permuting the points and checking if the formed polygon is indeed convex.
The question on how to check if a polygon is convex has been asked and answered here: How do determine if a polygon is complex/convex/nonconvex?
I need to plot a group of points based on distances. I have three unknown points X, Y, and Z. I then get another unknown point (A) and its distances from the originals (AX, AY, AZ). I will continue getting points and distances (B, BX, BY, BZ; C, CX, CY, CZ) etc.
My question is whether its possible to plot all of the points. If so, how many points would I need for an exact plot map? What about an approximate map?
This is similar to this question but I get a different set of distances and am not limited to the original number of points.
Also, if it would help I could add more points to the X, Y, Z group which would give me more distances for A, B, etc.What I don't know until it's been somehow calculated are any of the Distances XY, XZ, YZ, AB, AC, etc.
I am not sure exactly what you mean by exact plot map or approximate plot map. I think I might know but I am not sure. But plotting all points in this case, to me is not possible if the user can continue to "add more points to the XYZ group", which is dynamic. It would sound like you need to know what the user is going to plot before he does. Now if all this is static, it is possible
I assume you use 2D space
If it is 1D then 2 points are enough (not identical !!!).
If 2D then 3 distances is enough but the points used must not lay on the same line !!!
position/orientation of the plot
for relative plot are above conditions enough if you want also the exact orientation and position then you need to know exact position of first 3 points otherwise your plot will look the same but can be offseted,rotated and mirrored to original geometry.
knowing 1 point eliminates offset
knowing 2 point eliminates rotation
knowing 3 point eliminates mirroring
[notes]
you need n+1 points for n-D coordinate system
[edit1] equations
original question text did not contain any equations need but comments requires it so here are some:
You will need intersection point between two hyperspheres (in 2D circles, in 3D spheres,...) so look here:
circle-circle intersection
Cast circle from each point as center with radius equal to the distance from that point. Find out intersection point that is the same between all combinations of circles (0,1),(0,2),(1,2)
Yellow intersection is the same in all 3 combinations so that is the next point or for 2D just solve this system:
(x-x0)^2+(y-y0)^2=l0^2
(x-x1)^2+(y-y1)^2=l1^2
(x-x2)^2+(y-y2)^2=l2^2
where x,y is the intersection point, xi,yi are center of circle and li is distance from that point.
The first option should be simpler and more accurate if done right but need some knowledge on vector and trigonometry math. You will need to add rotation or compute on vectors and use perpendicular vector feature in 2D
V(x,y) -> V0(+y,-x),V1(-y,+x)
where V0,V1 are perpendicular to V
Essentially I need to return a boolean if a line and a line segment intersect. For the line the information I have is the slope, xy coords for a random point, and xy for a y intercept. For the line segment I have the line segment and the two end point xy coordinates. Any ideas?
The line through (Xr, Yr) with slope S has equation D(X, Y):= (Y - Yr) - S (X - Xr) = 0.
Just check that D(Xa, Ya) and D(Xb, Yb) have opposite signs.
Implementing this is not terribly difficult, it's the conceptual part that seems the most difficult. I might code something up for this for fun later but this should be enough to get you started. Also, note that this is might be a really bad way to solve it (time/space wise) but it will definitely work.
If you want to find a good solution, use convert the line to vectors and use the implementation in the answer in the link below all my text.
Calculate the slope of the second line. If it's equal to the first, they're parallel and never intersect.
Calculate where the two lines would intersect if they do intersect. They can only intersect at one point. This is how you would check if they will intersect:
Find 2 distinct coordinates for each line at some arbitrary point.
Determine the distance between the two lines at each point.
Whichever point results in a smaller distance between the lines represents the direction you need to travel in to move closer to the intersection.
Keep checking the distance between the two lines until they start increasing (which means you need to reverse the direction again) or until the distance is 0. The xy coord at distance = 0 is the intersection.
If the x-value of the point where the two lines intersect is in between the two x-values for your line segment, the line and line fragment intersect.
This should be easier for you because you already have two xy coords for the first line and two end point coordinates for the line segment.
Check this answer out for a really nice solution with some examples in the comments: https://stackoverflow.com/a/565282/2142219
So, I'm working on a 2D physics engine, and I have an issue. I'm having a hard time conceptualizing how you would calculate this:
Take two squares:They move, collide, and at some vector based off of the velocity of the two + the shape of them.
I have two vector lists(2D double lists) that represent these two shapes, how does one get the normal vector?
The hit vector is just (s1 is the first shape, s2 the second) s2 - s1 in terms of the position of the center of mass.
Now, I know a normal vector is one perpendicular to an edge, and I know that you can get the perpendicular vector of a line by 90 degrees, but what edge?
I read in several places, it is the edge a corner collided on. How do you determine this?
It just makes no sense to me, how you would mathematically or programmatically determine what edge.
Can anyone point out what I'm doing wrong in my understanding? Sorry for providing no code to explain this, as I'm having an issue writing the code for it in the first place.
Figure1: In 2D the normal vector is perpendicular to the tangent line:
Figure2: In 3D the normal vector is perpindicular to the tangent plane
Figure3: For a square the normal vector is easy if you are not at a corner; It is just perpendicular to the side of the square (in the image above, n = 1 i + 0 j, for any point along the right side of the square).
However, at a corner it becomes a little more difficult because the tangent is not well-defined (in terms of derivatives, the tangent is discontinuous at the corner, so perpendicular is ambiguous).
Even though the normal vector is not defined at a corner, it is defined directly to the left and right of it. Therefore, you can use the average of those two normals (n1 and n2) as the normal at a corner.
To be less technical, the normal vector will be in the direction from the center of the square to the corner of the collision.
EDIT: To answer the OP's further questions in the chat below: "How do you calculate the normal vector for a generic collision between two polygons s1 and s2 by only knowing the intersecting vertices."
In general, you can calculate the norm like this (N is total verts, m is verts inside collision):
vcenter = (∑N vi) / N
vcollision = (∑m vi) / m
n = vcollision - vcenter
Fig. 1 - vcollision is only a single vertex.
Fig. 2 - vcollision is avg of two verts.
Fig. 3 - vcollision for generic polygon intersection.
I need an algorithm to intersect a (potentially non-convex) polygon with a rectangle. The rectangle will be parallel to the xy-plane, but the polygon could be any orientation.
Furthermore, I don't just need a true/false result, but also the exact points where the polygon intersects the rectangle, so that I can draw lines where the polygon overlaps the rectangle. For non-convex polygons, this could result in two or more lines of intersection.
This would be for a section-cut module that can slice a set of polygons and create a 2D "cut" where the shapes intersect the "plane," specified by a z-value.
I'm developing in Java, so if Java3(2)D has any built-in methods to help, that would be ideal.
Any help/pointers in the right direction would be greatly appreciated!
Here's a picture... I want the red line as a result of the intersection:
This should find all the intersecting segments, for any arbitrary polygon.
Consider the polygon as an ordered collection of edges AB,BC,CD,etc, where the 'direction' from each edge's first point to its second point is 'clockwise'. That is, if we're looking at point A, point B is the next point when moving clockwise.
The method is to find an edge of the polygon that crosses the plane and then find the next segment, moving clockwise, that crosses back to the original side of the plane. The two points where these segments intersect the plane form the endpoints for an intersecting segment. This is repeated until all the polygon's edges have been checked.
Be advised not all segments are necessarily within the polygon if it is concave.
let P be any point on the polygon.
TOP:
while (P has not been checked)
mark P as having been checked.
let f be the point following P, clockwise.
if (P and f are on opposite sides of the plane) then
Continuing from f clockwise, find the next point Y that is on
the same side of the plane as P.
Let z be the point counter-clockwise from Y.
(note - Sometimes z and f are the same point.)
let S1 be the point where P,f intersects the plane
let S2 be the point where Y,z intersects the plane
if (segment (S1,S2) is inside the polygon)
add (S1,S2) to a 'valid' list.
let P = Y
else
let P = f
endif
else
let P = f
endif
endwhile
This algorithm is worth what you paid for it. :-)