I really was trying to find an answer on this very basic (at first sight) question.
For simplicity depth test is disabled during further discussion (it doesn’t have a big deal).
For example, we have triangle (after transformation) with next float4 coordinates.
top CenterPoint: (0.0f, +0.6f, 0.6f, 1f)
basic point1: (+0.4f, -0.4f, 0.4f, 1f),
basic point2: (-0.4f, -0.4f, 0.4f, 1f),
I’m sending float4 for input and use straight VertexShader (without transforms), so I’m sure about input. And we have result is reasonable:
But what we will get if we'll start to move CenterPoint to point of camera position. In our case we don’t have camera so will move this point to minus infinity.
I'm getting quite reasonable results as long as w (with z) is positive.
For example, (0.0f, +0.006f, 0.006f, .01f) – look the same.
But what if I'll use next coordinates (0.0f, -0.6f, -1f, -1f).
(Note: we have to switch points or change rasterizer for culling preventing).
According to huge amount of resource I'll have test like: -w < z < w, so GPU should cut of that point. And yes, in principle, I don’t see point. But triangle still visible! OK, according to huge amount of other resource (and my personal understanding) we'll have division like (x/w, y/w, z/w) so result should be (0, 0.6, 1). But I'm getting
And even if that result have some sense (one point is somewhere far away behind as), how really DirectX (I think it is rather GPU) works in such cases (in case of infinite points and negative W)?
It seems that I don't know something very basic, but it seems that nobody know that.
[Added]: I want to note that point w < 0 - is not a real input.
In real life such points are result of transformation by matrices and according to the math (math that are used in standard Direct sdk and other places) corresponds to the point that appears behind the camera position.
And yes, that point is clipped, but questions is rather about strange triangle that contains such point.
[Brief answer]: Clipping is essentially not just z/w checking and division (see details below).
Theoretically, NDC depth is divided into two distinct areas. The following diagram shows these areas for znear = 1, zfar = 3. The horizontal axis shows view-space z and the vertical axis shows the resulting NDC depth for a standard projective transform:
We can see that the part between view-space z of 1 and 3 (znear, zmax) gets mapped to NDC depth 0 to 1. This is the part that we are actually interested in.
However, the part where view-space z is negative also produces positive NDC depth. However, those are parts that result from fold-overs. I.e., if you take a corner of your triangle and slowly decrease z (along with w), starting in the area between znear and zfar, you would observe the following:
we start between znear and zfar, everything is good
as soon as we pass znear, the point gets clipped because NDC depth < 0.
when we are at view-space z = 0, the point also has w = 0 and no valid projection.
as we decrease view-space z further, the point gets a valid projection again (starting at infinity) and comes back in with positive NDC depth.
However, this last part is the area behind the camera. So, homogeneous clipping is made, such that this part is also clipped away by znear clipping.
Check the old D3D9 documentation for the formulas and some more illustrative explanations here.
I seem to have searched the whole internet trying to find an implementation of checking if a 3d point is within an elliptical cone defined by (origin, length, horizontal angle, vertical angle). Unfortunately without success as I only really found one math solution which I did not understand.
Now I am aware on how to use implement it using a normal cone:
inRange = magnitude(point - origin) <= length;
heading = normalized(point - origin);
return dot(forward, heading) >= cos(angle) && inRange;
However there the height detection is far too tall. I would really like to implement a more realistic vision cone for the AI for a game but this requires having the cone shaped more like a human field of view being more wide than tall.
Thanks a lot for any help:)
Given a 3D elliptic cone, with base at B=(x_B,y_B,z_B), height h along the cone axis k=(k_x,k_y,j_z), major base radius a, minor base radius b and direction along the major axis i=(i_x,i_y,i_z) you need to find if a point P=(x,y,z) lies inside the cone. It is your choice on how to parametrize the major axis direction and I think your are trying to use spherical coordinates with two angles.
Here are the steps to take:
Establish a coordinate system with origin on the base B and with the local x axis along your major axis i. The local z axis should be towards the tip along k. Finally the local y axis should be
j=cross(k,i)=(i_z*k_y-i_y*k_z, i_x*k_z-i_z*k_x, i_y*k_x-i_x*k_y)
j=normalize(j)
Your 3×3 rotation matrix is defined by the columns E=[i,j,k]
Transform your point P=(x,y,z) into the local coordinates with
P2 = transpose(E)*(P-B) = (x2,y2,z2)
Now establish how far along the axis of the cone is with s=(h-z2)/h where s=0 at the tip and s=1 at the base.
If s>1 or s<0 then the point is outside
Otherwise if s>0 you need to check that (x2/(s*a))^2+(y2/(s*b))^2<=1 for the point to be inside.
If s=0 then check that x2=0 and y2=0 for the point being exactly at the tip.
If you cannot do basic vector algebra, like cross products, 3D transformations and normalization that I suggest you have some reading to do before you can understand what is going on here.
Note:
// | i_x i_y i_z |
// transpose(E) = | j_x j_y j_z |
// | k_x k_y k_z |
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.
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.
I'm an artist involved with building various sorts of computer controlled machines. I've started prototyping a gimble-based XY painting machine and have realized that the maths needed are out of my reach. I'm a decent enough programmer but not strong in math- esp. 3D math.
To get a sense of what I'm needing to do, it might be helpful to look at the rig:
Early prototype:
http://roypardi.com/gimble/gimbleSmall.MOV (small video)
http://roypardi.com/gimble/gimbleLarge.mov (larger video)
The two inner rings represent the X/Y axes and are controlled by stepper motors. I want to be able to use both raster images and vector data (gcode). So I need to be able to address a point in 2D space on the paper/from my data and have the gimble figure out what orientation it needs to be at in order to get there (i.e. how much to step each motor).
I've been searching out 2D > 3D projection, Euler angles, etc. but I'm out of my depth. Any pointers, pushes in the right direction, or code snippets would be most welcome. I can make sense of most programming languages.
Very nice machine you have made, I hope this works for you I believe it is correct.
The way I see it, is to get one angle is simple, but the other is slightly harder to visualise as we have tilted the axis which it turns upon.
I'm going to avoid using tan, as when programming this could result in a division by 0, which could be frustrating. Also Z is going to be the height of the origin above the paper.
YAxis = arcsin( X / sqrt(X² + Z²))
XAxis = arcsin( Y / sqrt(Y² + X² + Z²))
or we could use
XAxis = arcsin(Y / sqrt(Y² + Z²))
YAxis = arcsin( X / sqrt(X² + Y² + Z²))
Also, I'd very much like to see a video of this plotting, if it works.
Edit:
After thinking about it i believe only one solution will work it depends on which axis is affected by the other. Is the YAxis in the Middle or the Xaxis?
I think it's a problem of simple http://en.wikipedia.org/wiki/Trigonometry
Let's say that the distance from the centre of your rings to the nearest point on the paper (which I'll call point 'O' for 'Origin') is distance X.
Take another point P directly north of O, whose distance from O is Y.
To paint this point, you need the angle alpha such that tan(alpha)=Y/X, i.e. you can calculate alpha using the formula "arctan(Y/X)" [arctan is sometimes also known as atan]. Arctan is a trignometric function, which I think you'll probably find defined in the API of a general purpose math library.
The above is the simplest case.
The only other case that I can think of is when the point P isn't due north. Instead of being due north, let's say that its distance is Y1 to the north, and Y2 to the east. The solution is two angles (one angle for each of two rings), one of which is "arctan(Y1/X)" and the other of which is "arctan(Y2/X)".
Perhaps I misunderstand, but I don't believe a gimbal will do what you want. A gimbal can point in any 3D direction, but it cannot move to arbitrary points in 3D space. If the plane of the paper intersects the volume swept by the pen held in the gimbal, the pen might be able to draw a circle, but nothing more. Even drawing a circle is not a sure thing, since in this case the paper would also intersect the volume swept by the gimbal rings; trying to orient the pen would make a ring hit the paper.
I think what you want is a plotter, not a gimbal.