I have a complicated problem and it involves an understanding of Maths I'm not confident with.
Some slight context may help. I'm building a 3D train simulator for children and it will run in the browser using WebGL. I'm trying to create a network of points to place the track assets (see image) and provide reference for the train to move along.
To help explain my problem I have created a visual representation as I am a designer who can script and not really a programmer or a mathematician:
Basically, I have 3 shapes (Figs. A, B & C) and although they have width, can be represented as a straight line for A and curves (B & C). Curves B & C are derived (bend modified) from A so are all the same length (l) which is 112. The curves (B & C) each have a radius (r) of 285.5 and the (a) angle they were bent at was 22.5°.
Each shape (A, B & C) has a registration point (start point) illustrated by the centre of the green boxes attached to each of them.
What I am trying to do is create a network of "track" starting at 0, 0 (using standard Cartesian coordinates).
My problem is where to place the next element after a curve. If it were straight track then there is no problem as I can use the length as a constant offset along the y axis but that would be boring so I need to add curves.
Fig. D. demonstrates an example of a possible track layout but please understand that I am not looking for a static answer (based on where everything is positioned in the image), I need a formula that can be applied no matter how I configure the track.
Using Fig. D. I tried to work out where to place the second curved element after the first one. I used the formula for plotting a point of the circumference of a circle given its centre coordinates and radius (Fig. E.).
I had point 1 as that was simply a case of setting the length (y position) of the straight line. I could easily work out the centre of the circle because that's just the offset y position, the offset of the radius (r) (x position) and the angle (a) which is always 22.5° (which, incidentally, was converted to Radians as per formula requirements).
After passing the values through the formula I didn't get the correct result because the formula assumed I was working anti-clockwise starting at 3 o'clock so I had to deduct 180 from (a) and convert that to Radians to get the expected result.
That did work and if I wanted to create a 180° track curve I could use the same centre point and simply deducted 22.5° from the angle each time. Great. But I want a more dynamic track layout like in Figs. D & E.
So, how would I go about working point 5 in Fig. E. because that represents the centre point for that curve segment? I simply have no idea.
Also, as a bonus question, is this the correct way to be doing this or am I over-complicating things?
This problem is the only issue stopping me from building my game and, as you can appreciate, it is a bit of a biggie so I thank anyone for their contribution in advance.
As you build up the track, the position of the next piece of track to be placed needs to be relative to location and direction of the current end of the track.
I would store an (x,y) position and an angle a to indicate the current point (with x,y starting at 0, and a starting at pi/2 radians, which corresponds to straight up in the "anticlockwise from 3-o'clock" system).
Then construct
fx = cos(a);
fy = sin(a);
lx = -sin(a);
ly = cos(a);
which correspond to the x and y components of 'forward' and 'left' vectors relative to the direction we are currently facing. If we wanted to move our position one unit forward, we would increment (x,y) by (fx, fy).
In your case, the rule for placing a straight section of track is then:
x=x+112*fx
y=y+112*fy
The rule for placing a curve is slightly more complex. For a curve turning right, we need to move forward 112*sin(22.5°), then side-step right 112*(1-cos(22.5°), then turn clockwise by 22.5°. In code,
x=x+285.206*sin(22.5*pi/180)*fx // Move forward
y=y+285.206*sin(22.5*pi/180)*fy
x=x+285.206*(1-cos(22.5*pi/180))*(-lx) // Side-step right
y=y+285.206*(1-cos(22.5*pi/180))*(-ly)
a=a-22.5*pi/180 // Turn to face new direction
Turning left is just like turning right, but with a negative angle.
To place the subsequent pieces, just run this procedure again, calculating fx,fy, lx and ly with the now-updated value of a, and then incrementing x and y depending on what type of track piece is next.
There is one other point that you might consider; in my experience, building tracks which form closed loops with these sort of pieces usually works if you stick to making 90° turns or rather symmetric layouts. However, it's quite easy to make tracks which don't quite join up, and it's not obvious to see how they should be modified to allow them to join. Something to bear in mind perhaps if your program allows children to design their own layouts.
Point 5 is equidistant from 3 as 2, but in the opposite direction.
Imagine a photo, with the face of a building marked out.
Its given that the face of the building is a rectangle, with 90 degree corners. However, because its a photo, perspective will be involved and the parallel edges of the face will converge on the horizon.
With such a rectangle, how do you calculate the angle in 2D of the vectors of the edges of a face that is at right angles to it?
In the image below, the blue is the face marked on the photo, and I'm wondering how to calculate the 2D vector of the red lines of the other face:
example http://img689.imageshack.us/img689/2060/leslievillestarbuckscor.jpg
So if you ignore the picture for a moment, and concentrate on the lines, is there enough information in one of the face outlines - the interior angles and such - to know the path of the face on the other side of the corner? What would the formula be?
We know that both are rectangles - that is that each corner is a right angle - and that they are at right angles to each other. So how do you determine the vector of the second face using only knowledge of the position of the first?
It's quite easy, you should use basic 2 point perspective rules.
First of all you need 2 vanishing points, one to the left and one to the right of your object. They'll both stay on the same horizon line.
alt text http://img62.imageshack.us/img62/9669/perspectiveh.png
After having placed the horizon (that chooses the sight heigh) and the vanishing points (the positions of the points will change field of view) you can easily calculate where your lines go (of course you need to be able to calculate the line that crosses two points: i think you can do it)
Honestly, what I'd do is a Hough Transform on the image and determine a way to identify the red lines from the image. To find the red lines, I'd find any lines in the transform that touch your blue ones. The good thing about the transform is that you get angle information for free.
Since you know that you're looking at lines, you could also do a Radon Transform and look for peaks at particular angles; it's essentially the same thing.
Matlab has some nice functionality for this kind of work.
I have an interesting problem here I've been trying to solve for the last little while:
I have 3 circles on a 2D xy plane, each with the same known radius. I know the coordinates of each of the three centers (they are arbitrary and can be anywhere).
What is the largest triangle that can be drawn such that each vertex of the triangle sits on a separate circle, what are the coordinates of those verticies?
I've been looking at this problem for hours and asked a bunch of people but so far only one person has been able to suggest a plausible solution (though I have no way of proving it).
The solution that we have come up with involves first creating a triangle about the three circle centers. Next we look at each circle individually and calculate the equation of a line that passes through the circle's center and is perpendicular to the opposite edge. We then calculate two intersection points of the circle. This is then done for the next two circles with a result of 6 points. We iterate over the 8 possible 3 point triangles that these 6 points create (the restriction is that each point of the big triangle must be on a separate circle) and find the maximum size.
The results look reasonable (at least when drawn out on paper) and it passes the special case of when the centers of the circles all fall on a straight line (gives a known largest triangle). Unfortunate i have no way of proving this is correct or not.
I'm wondering if anyone has encountered a problem similar to this and if so, how did you solve it?
Note: I understand that this is mostly a math question and not programming, however it is going to be implemented in code and it must be optimized to run very fast and efficient. In fact, I already have the above solution in code and tested to be working, if you would like to take a look, please let me know, i chose not to post it because its all in vector form and pretty much impossible to figure out exactly what is going on (because it's been condensed to be more efficient).
Lastly, yes this is for school work, though it is NOT a homework question/assignment/project. It's part of my graduate thesis (abet a very very small part, but still technically is part of it).
Thanks for your help.
Edit: Heres a new algorithm that i came up with a little while ago.
Starting at a circle's centre, draw a line to the other two centres. Calculate the line that bisects the angle created and calculate the intersections between the circle and the line that passes through the centre of your circle. You will get 2 results. Repeat this for the other two circles to get a total of 6 points. Iterate over these 6 points and get 8 possible solutions. Find the maximum of the 8 solutions.
This algorithm will deal with the collinear case if you draw your lines in one "direction" about the three points.
From the few random trials i have attempted using CAD software to figure out the geometries for me, this method seems to outperform all other methods previously stated However, it has already been proven to not be an optimal solution by one of Victor's counter examples.
I'll code this up tomorrow, for some reason I've lost remote access to my university computer and most things are on it.
I've taken the liberty of submitting a second answer, because my original answer referred to an online app that people could play with to get insight. The answer here is more a geometric argument.
The following diagram illuminates, I hope, what is going on. Much of this was inspired by #Federico Ramponi's observation that the largest triangle is characterized by the tangent at each vertex being parallel to the opposite side.
(source: brainjam.ca)
The picture was produced using a trial version of the excellent desktop program Geometry Expressions. The diagram shows the three circles centered at points A,E, and C. They have equal radii, but the picture doesn't really depend on the radii being equal, so the solution generalizes to circles of different radii. The lines MN, NO, and OM are tangent to the circles, and touch the circles at the points I,H, and G respectively. The latter points form the inner triangle IHG which is the triangle whose size we want to maximize.
There is also an exterior triangle MNO which is homethetic to the interior triangle, meaning that its sides are parallel to that of IHG.
#Federico observed that IHG has maximal area because moving any of its vertices along the corresponding circle will result an a triangle that has the same base but less height, therefore less area. To put it in slightly more technical terms, if the triangle is parameterized by angles t1,t2,t3 on the three circles (as pointed out by #Charles Stewart, and as used in my steepest descent canvas app), then the gradient of the area w.r.t to (t1,t2,t3) is (0,0,0), and the area is extremal (maximal in the diagram).
So how is this diagram computed? I'll admit in advance that I don't quite have the full story, but here's a start. Given the three circles, select a point M. Draw tangents to the circles centered at E and C, and designate the tangent points as G and I. Draw a tangent OHN to the circle centered at A that is parallel to GI. These are fairly straightforward operations both algebraically and geometrically.
But we aren't finished. So far we only have the condition that OHN is parallel to GI. We have no guarantee that MGO is parallel to IH or that MIN is parallel to GH. So we have to go back and refine M. In an interactive geometry program it's no big deal to set this up and then move M until the latter parallel conditions are met (by eyeballs, anyways). Geometry Expressions created the diagram, but I used a bit of a cheat to get it to do so, because its constraint solver was apparently not powerful enough to do the job. The algebraic expressions for G, I, and H are reasonably straightforward, so it should be possible to solve for M based on the fact that MIHG is a parallelogram, either explicitly or numerically.
I should point out that in general if you follow the construction starting from M, you have two choices of tangent for each circle, and therefore eight possible solutions. As in the other attempted answers to the question, unless you have a good heuristic to help you choose in advance which of the tangents to compute, you should probably compute all eight possible triangles and find the one with maximum area. The other seven will be extremal in the sense of being minimal area or saddle points.
That's it. This answer is not quite complete in that it leaves the final computation of M somewhat open ended. But it's reduced to either a 2D search space or the solution of an ornery but not humongous equation.
Finally, I have to disagree with #Federico's conclusion that this confirms that the solution proposed by the OP is optimal. It's true that if you draw perpendiculars from the circle centers to the opposite edge of the inner triangle, those perpendiculars intersect the circle to give you the triangle vertex. E.g. H lies on the line through A perpendicular to GI), but this is not the same as in the original proposed solution (which was to take the line through A and perpendicular to EC - in general EC is not parallel to GI).
I've created an HTML5 canvas app that may be useful for people to play with. It's pretty basic (and the code is not beautiful), but it lets you move three circles of equal radius, and then calculates a maximal triangle using gradient/steepest descent. You can also save bitmaps of the diagram. The diagram also shows the triangle whose vertices are the circle centers, and one of the altitudes. Edit1: the "altitude" is really just a line segment through one of the circle centers and perpendicular to the opposite edge of the triangle joining the centers. It's there because some of the suggested constructions use it. Edit2: the steepest descent method sometimes gets stuck in a local maximum. You can get out of that maximum by moving a circle until the black triangle flips and then bringing the circle back to its original position. Working on how to find the global maximum.
This won't work in IE because it doesn't support canvas, but most other "modern" browsers should work.
I did this partially because I found some of the arguments on this page questionable, and partially because I've never programmed a steepest descent and wanted to see how that worked. Anyways, I hope this helps, and I hope to weigh in with some more comments later.
Edit: I've looked at the geometry a little more and have written up my findings in a separate answer.
Let A, B, C be the vertexes of your triangle, and suppose they are placed as in your solution.
Notice that the key property of your construction is that each of the vertexes lies on a tangent to its circle which is parallel to the opposite side of the triangle. Obviously, the circle itself lies entirely on one side of the tangent, and in the optimal solution each tangent leaves its circle on the same side as the other vertexes.
Consider AB as the "base" of the triangle, and let C float in its circle. If you move C to another position C' within the circle, you will obtain another triangle ABC' with the same base but a smaller height, hence also with a smaller area:
figure 1 http://control.ee.ethz.ch/~ramponif/stuff/circles1.png
For the same reason, you can easily see that any position of the vertexes that doesn't follow your construction cannot be optimal. Suppose, for instance, that each one of the vertexes A', B', C' does not lie on a tangent parallel to the side connecting the other two.
Then, constructing the tangent to the circle that contains (say) C', which is parallel to A'B' and leaves the circle on the same side as A'B', and moving C' to the point of tangency C, it is always possible to construct a triangle A'B'C which has the same base, but a greater height, hence also a greater area:
figure 2 http://control.ee.ethz.ch/~ramponif/stuff/circles2.png
Since any triangle that does not follow your construction cannot be optimal, I do believe that your construction is optimal. In the case when the centers of the circles are aligned I'm a bit confused, but I guess that it is possible to prove optimality along the same lines.
I believe this is a convex optimization problem (no it's not, see below), and hence can be solved efficiently using well known methods.
You essentially want to solve the problem:
maximize: area(v1,v2,v3) ~ |cross((v2-v1), (v3-v1))|
such that: v1 in C1, v2 in C2, v3 in C3 (i.e., v_i-c_i)^2 - r_i^2 <= 0)
Each of the constraints are convex, and the area function is convex as well. Now, I don't know if there is a more efficient formulation, but you can at least use an interior point method with derivatives since the derivative of the area with respect to each vertex position can be worked out analytically (I have it written down somewhere...).
Edit: grad(area(v1,v2,v3))(v_i) = rot90(vec(vj,vk)), where vec(a,b) is making a 2D vector starting at a and ending at b, and rot90 means a positive orientation rotation by 90 degrees, assuming (vi,vj,vk) was positively oriented.
Edit 2: The problem is not convex, as should be obvious considering the collinear case; two degenerate solutions is a sure sign of non-convexity. However, the configuration starting at the circle centers should be in the globally optimal local maximum.
Not optimal, works well when all three are not colinear:
I don't have a proof (and therefore don't know if it's guaranteed to be biggest). Maybe I'll work on one. But:
We have three circles with radius R with positions (from center) P0, P1, and P2. We wish to find the vertices of a triangle such that the area of the triangle is maximum, and the vertices lie on any point of the circles edges.
Find the center of all the circles and call that C. Then C = (P0 + P1 + P2) / 3. Then we find the point on each circle farthest from C.
Find vectors V0, V1, and V2, where Vi = Pi - C. Then find points Q0, Q1, and Q2, where Qi = norm(Vi) * R + Pi. Where norm indicates normalization of a vector, norm(V) = V / |V|.
Q0, Q1, and Q2 are the vertices of the triangle. I assume this is optimal because this is the farthest the vertices could be from each other. (I think.)
My first thought is that you should be able to find an analytic solution.
Then the equations of the circles are:
(x1-h1)^2 + (y1-k1)^2 = r^2
(x2-h2)^2 + (y2-k2)^2 = r^2
(x3-h3)^2 + (y3-k3)^2 = r^2
The vertices of your triangle are (x1, y1), (x2, y2), and (x3, y3). The side lengths of your triangle are
A = sqrt((x1-x2)^2 + (y1-y2)^2)
B = sqrt((x1-x3)^2 + (y1-y3)^2)
C = sqrt((x2-x3)^2 + (y2-y3)^2)
So the area of the triangle is (using Heron's formula)
S = (A+B+C)/2
area = sqrt(S(S-A)(S-B)(S-C))
So area is a function of 6 variables.
At this point I realize this is not a fruitful line of reasoning. This is more like something I'd drop into a simulated annealing system.
So my second thought is to choose the point on circle with centre A as follows: Construct line BC joining the centres of the other two circles, then construct the line AD that is perpendicular to BC and passes through A. One vertex of the triangle is the intersection of AD and circle with centre A. Likewise for the other vertices. I can't prove this but I think it gives different results than the simple "furthest from the centre of all the circles" method, and for some reason it feels better to me. I know, not very mathematical, but then I'm a programmer.
Let's assume the center of the circles to be C0,C1 and C2; and the radius R.
Since the area of a triangle is .5*base*height, let's first find the maximum base that can be constructed with the circles.
Base = Max {(|C0-C1|+2R),(|C1-C2|+2R,(|C2-C0|+2R}
Once the base length is determined between 2 circles, then we can find the farthest perpendicular point from the base line to the third circle. (product of the their slopes is -1)
For special cases such as circles aligned in a single line, we need to perform additional checks at the time of determining the base line.
It appears that finding the largest Apollonius circle for the three circles and then inscribing an equilateral triangle in that circle would be a solution. Proof left as an exercise ;).
EDIT
This method has issues for collinear circles like other solutions here, too and doesn't work.
Some initial thoughts.
Definition Call the sought-after triangle, the maximal triangle. Note that this might not be unique: if the circles all have the same centre, then there are infinitely many maximal triangles obtained by rotation around the center, and if the centres are colinear, then there will be two maximal triangles, each a mirror image of the other.
Definition Call the triangle (possibly, degenerately, either a point or a line) whose vertices are the centres of the circles the interior triangle.
Observation The solution can be expressed as three angles, indicating where on the circumference of each circle the triangle is to be found.
Observation Given two exterior vertices, we can determine a third vertex that gives the maximal area: draw the altitude of the triangle between the two exterior vertices and the centre of the other circle. This line intersects the circumference in two places; the further away point is the maximising choice of third vertex. (Fixed incorrect algorithm, Federico's argument can be adapted to show correctness of this observation)
Consequence The problem is reduced to from a problem in three angles to one in two.
Conjecture Imagine the diagram is a pinboard, with three pins at the three centres of the circles. Imagine also a closed loop of string of length equal to the perimiter of the interior triangle, plus the radius of a circle, and we place this loop around the pins. Take an imaginary pen and imaginarily draw the looping figure where the loop is always tight. I conjecture that the points of the maximal triangle will all lie on this looping figure, and that in the case where the interior triangle is not degenerate, the vertices of the maximal triangle will be the three points where the looping figure intersects one of the circle circumferences. Many counterexamples
More to follow when I can spare time to think about it.
This is just a thought, no proof or math to go along with the construction just yet. It requires that the circle centers not be colinear if the radii are the same for each circle. This restriction can be relaxed if the radii are different.
Construction:
(1) Construct a triangle such that each side of the triangle is tangent to two circles, and therefore, each circle has a tangent point on two sides of the triangle.
(2) Draw the chord between these two tangent points on each circle
(3) Find the point on the boundary of the circle on the extended ray starting at the circle's center through the midpoint of the chord. There should be one such point on each of the three circles.
(4) Connect them three points of (3) to fom a triangle.
At that point I don't know if it's the largest such triangle, but if you're looking for something approximate, this might be it.
Later: You might be able to find an approximate answer for the degenerate case by perturbing the "middle" circle slightly in a direction perpendicular to the line connecting the three circles.
Given a list of points that form a simple 2d polygon oriented in 3d space and a normal for that polygon, what is a good way to determine which points are specific 'corner' points?
For example, which point is at the lower left, or the lower right, or the top most point? The polygon may be oriented in any 3d orientation, so I'm pretty sure I need to do something with the normal, but I'm having trouble getting the math right.
Thanks!
You would need more information in order to make that decision. A set of (co-planar) points and a normal is not enough to give you a concept of "lower left" or "top right" or any such relative identification.
Viewing the polygon from the direction of the normal (so that it appears as a simple 2D shape) is a good start, but that shape could be rotated to any arbitrary angle.
Is there some other information in the 3D world that you can use to obtain a coordinate-system reference?
What are you trying to accomplish by knowing the extreme corners of the shape?
Are you looking for a bounding box?
I'm not sure the normal has anything to do with what you are asking.
To get a Bounding box, keep 4 variables: MinX, MaxX, MinY, MaxY
Then loop through all of your points, checking the X values against MaxX and MinX, and your Y values against MaxY and MinY, updating them as needed.
When looping is complete, your box is defined as MinX,MinY as the upper left, MinX, MaxY as upper right, and so on...
Response to your comment:
If you want your box after a projection, what you need is to get the "transformed" points. Then apply bounding box loop as stated above.
Transformed usually implies 2D screen coordinates after a projection(scene render) but it could also mean the 2D points on any plane that you projected on to.
A possible algorithm would be
Find the normal, which you can do by using the cross product of vectors connecting two pairs of different corners
Create a transformation matrix to rotate the polygon so that it is planer in XY space (i.e. normal alligned along the Z axis)
Calculate the coordinates of the bounding box or whatever other definition of corners you are using (as the polygon is now aligned in 2D space this is a considerably simpler problem)
Apply the inverse of the transformation matrix used in step 2 to transform these coordinates back to 3D space.
I believe that your question requires some additional information - namely the coordinate system with respect to which any point could be considered "topmost", or "leftmost".
Don't forget that whilst the normal tells you which way the polygon is facing, it doesn't on its own tell you which way is "up". It's possible to rotate (or "roll") around the normal vector and still be facing in the same direction.
This is why most 3D rendering systems have a camera which contains not only a "view" vector, but also "up" and "right" vectors. Changes to the latter two achieve the effect of the camera "rolling" around the view vector.
Project it onto a plane and get the bounding box.
I have a silly idea, but at the risk of gaining a negative a point, I'll give it a try:
Get the minimum/maximum value from
each three-dimensional axis of each
point on your 2d polygon. A single pass with a loop/iterator over the list of values for every point will suffice, simply replacing the minimum and maximum values as you go. The end result is a list that has the "lowest" X, Y, Z coordinates and "highest" X, Y, Z coordinates.
Iterate through this list of min/max
values to create each point
("corner") of a "bounding box"
around the object. The result
should be a box that always contains
the object regardless of axis
examined or orientation (no point on
the polygon will ever exceed the
maximum or minimums you collect).
Then get the distance of each "2d
polygon" point to each corner
location on the "bounding box"; the
shorter the distance between points,
the "closer" it is to that "corner".
Far from optimal, certainly crummy, but certainly quick. You could probably post-capture this during the object's rotation, by simply looking for the min/max of each rotated x/y/z value, and retaining a list of those values ahead of time.
If you can assume that there is some constraints regarding the shapes, then you might be able to get away with knowing less information. For example, if your shape was the composition of a small square with a long thin triangle on one side (i.e. a simple symmetrical geometry), then you could compare the distance from each list point to the "center of mass." The largest distance would identify the tip of the cone, the second largest would be the two points farthest from the tip of the cone, etc... If there was some order to the list, like points are entered in counter clockwise order (about the normal), you could identify all the points. This sounds like a bit of computation, so it might be reasonable to try to include some extra info with your shapes, like the "center of mass" and a reference point that is located "up" above the COM (but not along the normal). This will give you an "up" vector that you can cross with the normal to define some body coordinates, for example. Also, the normal can be defined by an ordering of the point list. If you can't assume anything about the shapes (or even if the shapes were symmetrical, for example), then you will need more data. It depends on your constraints.
If you know that the polygon in 3D is "flat" you can use the normal to transform all 3D-points of the vertices to a 2D-representation (of the points with respect to the plan in which the polygon is located) - but this still leaves you with defining the origin of this coordinate-system (but this don't really matter for your problem) and with the orientation of at least one of the axes (if you want orthogonal axes you can still rotate them around your choosen origin) - and this is where the trouble starts.
I would recommend using the Y-axis of your 3D-coordinate system, project this on your plane and use the resulting direction as "up" - but then you are in trouble in case your plan is orthogonal to the Y-axis (now you might want to use the projected Z-Axis as "up").
The math is rather simple (you can use the inner product (a.k.a. scalar product) for projection to your plane and some matrix stuff to convert to the 2D-coordinate system - you can get all of it by googling for raytracer algorithms for polygons.