Test link: http://bit.ly/Runmah
pick one item from left side for testing.
it's rotating when he find two intersect points on the line.
i want to move red rectangle within the polygon. It shouldn't go outside of polygon.
My code is: http://pastebin.com/pRMpk81f
Edit 1: http://pastebin.com/C3j4WSC1
If you know how to find the intersection of points on the line then you should be able to find distance between wall and furniture (line and edge of rectangle), you can check this value and for example stop draging and snap,
EDIT1:
when dragged item is selected (mouse is down) than you constantly check distance (e.g. on mouse move), then you can decide that if distance is within some threshold you will stop movement (basicaly you will set calculated position- snap - instead of applying mouse position) otherwise you will follow mouse.
EDIT2:
also you can test if the point is inside the shape by calculating intersection points - even number the point is outside, odd number - the point is inside
best regards
Related
I'm writing a code to simulate a lap time on motorcycle, using GNU Octave.
The racetrack is represented by the mid line (as x-y point values) and width "B".
The points of the racetrack midline are not at the same distance, for example a straight section have just two points, a bend could have 5-10 points.
The racing line is a closed loop cubic bezier.
I would like to calculate the distance of that bezier from the midline, to check if the vehicle goes off road (if distance > B/2 I'm out of track). I'm not interested in where this happens, only if happens.
Right now to check if the code works, I've made this by calculating the distance of every point of the racing line to the midline as the height of a triangle, where the two base vertex are two points of the mid line and the top vertex is the point of the racing line.
It works but I'm looking for an alternative method because this one is very slow.
As alternative I tought to "straighten" the mid line by calculating angle of each section, and then straighten the racing line with the same "transformation" and check min and max Y value, that must be less than B/2. The problem with this approach is that racing line and mid line are not the same length, in some part of racetrack racing line could be shorter, in other could be longer, so the transformation will not be linear.
Another option could be to transform the racetrack into a matrix filled with "0" inside track, "1" outside track, and the racing line will be a matrix with the same MxN filled with "0" everywhere and "1" where there is he racing line. By sum these two matrices, if there is some "2" I'm out of track. Right now I don't find a method to made this.
I've found the answer of my problem, using the function "inpolygon".
So, I set the outside boundary of racetrack as polygon, and all points of racing line as points to check. The function tell me if some point is outside the polygon.
Then, I repeat the same, setting inside boundary as polygon. Now all racing line points must be outside the polygon.
I am using bing maps and I want to query my database to return all values inside the map bounds, so every time, the map moves, I want to query it again.
In order to make it more efficient, I want to query only the boundary I haven't query before.
So I get the previous bound and the current bound and want to get the square bound of the non shared rectangle between the previous and the current rectangles (The not shared rectangle of the current bounds).
For example, If I move the map right for 5cm and up for 2cm, I will recieve a new LocationRect of the rectangle 5cm 2cm (the not shared).
I have the map bounds:
LocationRect currentBounds = map.Bounds;
When I move the map I get a new bounds, but before I save the previous bounds:
previousBounds = currentBounds;
I want to get the new location I moved to (only the new, not the whole currentBounds).
So I want to do something like this:
LocationRect newMapBounds = currentBounds.NotSharedBounds(previousBounds);
But how can I check this? I saw there is a method of Intersects but it returns bool, and I need to get the new LocationRect...
I will be very thankfull for the helper :)
If i understand you right you have the blue rectangle(ABCD) and when you move the map
you have red rectangle(EHGF) and you know their vertexes coordinates
e
So the not common space creating 3 new rectangles for you: Green Yellow And Black.
And you need coordinates of those 3 rectangles in order to query your data, in other words you need to perform three queries to you DB in order to get the NOT common space of BLUE and RED rectangles.
You will have scenarios of RED and BLUE rectangles that you need to deal before you start the calculations:
The rectangles are coincide.
The scenario in the picture
They have no common space at all.
For example the vertexes coordinates of GREEN rectangle are(The second scenario):
T(x) = E(x), T(y) = D(y)
F is a common of RED and GREEN Rectangles
N(x) = B(x) , N(y) = G(y)
F is a common of BLUE and GREEN Rectangles
Hope it helps.
Please this fiddle I have copied my complete project in it
here if you open the fiddle in the output you can see an image, scribble on the image selecting pen,add text etc(like this perform some functions).then rotate the group using rotate button and then try to scribble you will understand what is the problem exactly.
In me view Problem is I am having a stage and a layer is added to the stage and a group is added to the layer and different elements like lines text etc are added to the group. then group is rotated the i am trying to draw the line based on the mouse position of the stage.But it is not coming correctly because the group got rotated the x and y what we are taking to draw a line is from stage.I need to take the x and y from the group not from the stage is their any way.If hav't understand please ask me or post a reply.
This should get you fairly close: http://jsfiddle.net/k4qB8/24/
// This rotates that added active line along with your group.
// This makes the draw direction correct
activeline.setRotationDeg(0-rootGroup.getRotationDeg());
// Here you'll have to figure a way to calculate how much to move the
// line over so the draw is on the correct spot
// This is as close as I got it
if(Math.abs(rootGroup.getRotationDeg()%360)==0)
activeline.move(rootGroup.getX()-375, rootGroup.getY()-175);
if(Math.abs(rootGroup.getRotationDeg()%360)==90)
activeline.move(rootGroup.getX()-175, rootGroup.getY()+375);
if(Math.abs(rootGroup.getRotationDeg()%360)==180)
activeline.move(rootGroup.getX()+375, rootGroup.getY()+175);
if(Math.abs(rootGroup.getRotationDeg()%360)==270)
activeline.move(rootGroup.getX()+175, rootGroup.getY()-375);
Also, add some more logic for counter-clockwise rotation, as this doesn't work 100%.
I think the real solution would be to just draw on separate layers for each rotation, kind of like this:
if (rotation is 90) : draw on lineLayer1;
if (rotation is 180) : draw on lineLayer2;
if (rotation is 270) : draw on lineLayer3;
if (rotation is 360 || 0) : draw on lineLayer4;
This way you could just rotate the layers which are not drawn on to simulate the feel of rotation.
I have a single point and a set of shapes. I need to know if the point is contained within the compound shape of those shapes. That is, where all of the shapes intersect.
But that is the easy part.
If the point is outside the compound shape I need to find the position within that compound shape that is closest to the point.
These shapes can be of the type:
square
circle
ring (circle with another circle cut out of the center)
inverse circle (basically just the circular hole and a never ending fill outside that hole, or to the end of the canvas is there must be a limit to its size)
part of circle (as in a pie chart)
part of ring (as above but
line
The example below has an inverted circle (the biggest circle with grey surrounding it), a ring (topleft) a square and a line.
If we don't consider the line, then the orange part is the shape to constrain to. If the line is taken into account then the saturated orange part of the line is the shape to constrain to.
The black small dots represent the points that need to be constrained. The blue dots represent the desired result. (a 1, b 2 etc.)
Point "f" has no corresponding constrained result, since it is already in the orange area.
For the purpose of this example, only point "e" is constrained to the line, all others are constrained to the orange orange area.
If none of the shapes would intersect, then the point cannot be constrained. If the constraint would consist of two lines that cross eachother, then every point would be constrained to the same position (the exact position where the lines cross).
I have found methods that come close to this, but none that I can combine to produce the above functionality.
Some similar questions that I found:
Points within a semi circle
What algorithm can I use to determine points within a semi-circle?
Point closest to MovieClip
Flash: Closest point to MovieClip
Closest point through Minkowski Sum (this will work if I can convert the compound shape to polygons)
http://www.codezealot.org/archives/153
Select edge of polygon closest to point (similar to above)
For a point in an irregular polygon, what is the most efficient way to select the edge closest to the point?
PS: I noticed that the orange area may actually come across as yellow on some screens. It's the colored area in any case.
This isn't much of an answer, but it's a bit too long to fit into a comment ...
It's tempting to think, and therefore to advise you, to find the nearest point in each of the shapes to the point of interest, and to find the nearest of those nearest points.
BUT
The area you are interested in is constructed by union, intersection and difference of other areas and there will, therefore, be no general relationship between the closest points of the original shapes and the closest point of the combined shape. If you understand what I mean. For example, while the closest point of A union B is the closest of the set {closest point of A, closest point of B}, the closest point of A intersection B is not a simple function of that same set; at least not for the general case.
I suggest, therefore, that you are going to have to compute the (complex) shape which represents the area of interest and use one of the algorithms you've already discovered to find the closest point to your point of interest.
I look forward to someone much better versed in computational geometry proving me wrong.
Let's call I the intersection of all the shapes, C the contour of I, p the point you want to constrain and r the result point. We have:
If p is in I, then r = p
If p is not in I, then r is in C. So r is the nearest point in C to p.
So I think what you should do is the following:
If p is inside of all the shapes, return p.
Compute the contour C of the intersection of all the shapes, it is defined by a list of parts (segments, arcs, ...).
Find the nearest point to p in every part of C (computed in 2.) and return the nearest point among them to p.
I've discussed this question at length with my brother, and together we came to conclude that any resulting point will always lie on either the point where two shapes intersect, or where a shape intersects with the line from that shape perpendicular to the original point.
In the case of a circular shape constraint, the perpendicular line equals the line to its center. In the case of a line shape constraint, the perpendicular line is (of course) the line perpendicular to itself. In the case of a rectangle, the perpendicular line is the line perpendicular to the closest edge.
(And the same, theoretically, for complex polygon constraints.)
So a new approach (that I'll have to test still) will be to:
calculate all intersecting (with a shape constraint or with the perpendicular line from the original point to the shape constraint) points
keep only those that are valid: that lie within (comply with) all constraints
select the one closest to the original point
If this works, then one more optimization could be to determine first, which intersecting points are nearest and check if they are valid, and then work outward away from the original point until a valid one is found.
If this does not work, I will have another look at the polygon clipping method. For that approach I've come across this useful post:
Compute union of two arbitrary shapes
where clipping complex polygons is made much easier through http://code.google.com/p/gpcas/
The method holds true for all the cases (all points and their results) above, and also for a number of other scenarios that we tested (on paper).
I will try a live version tomorrow at work.
Short question: Given a point P and a line segment L, how do I find the point (or points) on L that are exactly X distance from P, if it guaranteed that there is such a point?
The longer way to ask this question is with an image. Given two circles, one static and one dynamic, if you move the dynamic one towards the static one in a straight line, it's pretty easy to determine the point of contact (see 1, the green dot).
Now, if you move the dynamic circle towards the static circle at an angle, determining the point of contact is much more difficult, (see 2, the purple dot). That part I already have done. What I want to do is, after determining the point of contact, decrease the angle and determine the new point of contact (see 3, 4, the red dot).
In #4, you can see the angle is decreased by less than half, and the new point of contact is half-way between the straight-line point and the original point. In #7, you can see the angle is bisected, but the new point of contact moves much farther than half way back towards the straight-line point.
In my case, I always want to decrease the angle to 5/6ths its original value, but the original angle and distance between the circles are variable. The circles are all the same radius. The actual data I need after decreasing the angle is the vector between the new center of the dynamic circle and the static circle, that is, the blue line in 3, 4, 6, and 7, if that makes the calculation any easier.
So far, I know I have to move the dynamic circle along the line that the purple circle is a center of, towards the center of the static circle. Then the circle has to move directly back towards the original position of the dynamic circle. The hard part is knowing exactly how far back it has to move so that it's just touching the other circle.
To answer your short question, if you are on the Cartesian plane, then find the equation of the line L is sitting on (given the two endpoints of L, this is simple). Find the equation of the perpendicular to said line, which passes through P (this is done by taking the negative inverse of the slope, plugging in P's x and y values, and solving for the intercept). Then find the point where the two perpendicular lines intersect by using their equations as a single system of equations (with x's and y's equal). Then find the distance between the point of intersection and the point P, which is one leg of a triangle. Finally, with that distance and the distance X you are given, use Pythagorean theorem to find the distance of the other leg of the triangle. Now the point you are looking for is a point on L, and also on the line on which L sits. So using the distance you just obtained, the intersection point you had found before, and the equation of L's line, you can find the desired point's coordinates. There can only be a maximum of 2 such points, so all you have to test for is whether the coordinates of the points found are actually on L, or beyond L but still on its line. Sorry for the long answer and if you wanted a geometric explanation rather than an algebraic one.
Draw a circle with the same centre as the stationary circle and the radius of the sum of both radii. There are two intersections with the translation line of the moving circle's centre. The place of the moving circle's center at the time of contact is the closer of those two intersections.
Let the ends of your segment be A and B, and the center of your stationary circle be C. Let the radius of both circles be r. Let the center of the moving circle at the moment of collision be D. We have a triangle ACD, of which we know: the distance AC, because it is constant, the angle DAC, because that's what you are changing, and the distance CD, which is exactly 2r. Theoretically, two sides and angle should let you get all the rest of a triangle...