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0. THE SHORT VERSION:



How do you solve this kind of parametric system of equations ? Can we say it is linear ?

  • Solving for $s,t$. We are given $r \in R$:

$$\cos t = \cos s \cdot r$$

$$\sin t = \sin s \cdot r$$

Solutions may exist in $Z$, maybe Euler Formula introducing $e^{i\theta}$ and $e^{i\phi}$ with $t=\theta$ and $s=\phi$ can be of some help here ?

Thanks a lot, longer version of the problem below if you wanna dive into it.





Although having searched extensively on various boards about my issue, I couldn't seem to solve my problem which is the following:



1. Defining the Goal


Given two arbitrary ellipsis, I would like to identify the cases and find their 1, 2, 3, 4 or infinitely many intersection points. I would like to solve this using parametric equations of the ellipsis.

I can define my ellipsis with parametric equation in the 2D plane as such:

  • Ellipsis A:
  • A(t):
  • $x(t) = \cos t \cdot a + h$
  • $y(t) = \sin t \cdot b + k$
  • $t$ is the parameter I want to find, $t \in [ 0, 2\pi ]$ or whatever
  • $a$ is let's say the radius on the minor axis of the ellipsis A
  • $b$ is let's say the radius on the major axis of the ellipsis A
  • $( h, k )$ is the vector from origin to the center of my ellipsis A.

  • Ellipsis B:
  • B(s):
  • $x(s) = \cos s \cdot c + i$
  • $y(s) = \sin s \cdot d + j$
  • $s$ is the parameter I want to find, $s \in [ 0; 2\pi ]$ or whatever
  • $c$ is let's say the radius on the minor axis of the ellipsis B
  • $d$ is let's say the radius on the major axis of the ellipsis B
  • $( i, j )$ is the vector from origin to the center of my ellipsis B.


2. Apologies


I'm not quite the mathematician myself, I'm doing this as a recreational exercise, and I wouldn't be too disappointed if it wasn't possible to solve but I think it is an interesting thing to think about.

One thing I didn't mention is the ellipsis could also be rotated by arbitrary angles theta & phi on top of that but solving without these rotations would be enough to satisfy my curiosity :)



3. Trying to get everything done anyway


Ok so now for the things I tried:

I know how to solve linear parametric systems for line intersections, but I understood that would not help me here because this system is not linear anymore.

One thing I tried is to simplify the cases up to trivial problems and try to solve adding known constants by iteration:

  • Solving Intersection of two unit circles A & B at origin ( trivial )
  • Solving Intersection of two unit circles A & B, A at origin and B with center at $( 1, 0 )$
  • Solving Intersection of two unit circles A & B, A at origin and B with center at $( h, 0 )$, h known for the general case

These seemed linear

  • Solving Intersection of two unit circles A & B, A at origin and B with center at $( h, k )$, $h$ & $k$ known for the general case

This one seemed linear too but I'm not sure of my solution



4. The simplified case where part of the problem lies


  • Solving Intersection of two circles A & B, A unit circle at origin and B circle of Radius r centered at origin
  • Solving Intersection of two circles A & B, A unit circle at origin and B circle of Radius r and B with center at ( h, k )

I could not solve these cases. I see these cases as special cases of my wider ellipsis problem, trying to solve one case at a time but I am already stuck at the point of the the two circles A & B at origin with A unit circle and B circle of Radius R. Solving this one will help me go further in this goal.

Defining parametric equations for this case gives us:

  • Circle A:
  • A(t):
  • $x(t) = \cos t$
  • $y(t) = \sin t$
  • $t$ is the parameter I want to find, $t \in [ 0; 2\pi ]$ or whatever


  • Circle B:

  • B(s):
  • $x(s) = \cos s \cdot r$
  • $y(s) = \sin s \cdot d$
  • $s$ is the parameter I want to find, $s \in [ 0; 2\pi ]$ or whatever
  • $r$ is the radius of the circle B, it is given.


5. Formalisation


  • plugging $A = B$

  • => $A( t ) = B( s )$

  • $\cos t = \cos s \cdot r$

  • $\sin t = \sin s \cdot r$

While it is geometrically trivial, it is analytically really hard to grasp, I don't have the tools to solve this, yet I am only doing the primary cases of my broader problem. The thing is I don't think i know the right tools in order to solve this. I guess my question, in the end, is the following:

  • Is this solvable ?
  • How to identify wheter an equation system is linear or not ?
  • How to identify the polynomial degree of an equation system ? I guess ellipsis intersection dives into degree 4, yet circle equation are degree 2, how to know for sure ?
  • What approach do you recommend in order to solve this
  • Do you recommend working with parametric equations ? (I do have a weakness for them when working with lines, interpolations and geometry in general. They proved helpful in the case of Bézier curve manipulation & more.

Thanks for your time, I guess my question is pretty long but i wanted to be clear despite being straightforward. I don't have the kind of mind it takes to do advanced maths but i am really eager to understand how to do that kind of magic. Thanks again, looking forward to your responses.

KingLogic
  • 1,441
Layl
  • 78
  • I doubt whether many will red the whole post: way tooooooo long. Try to shorten it a lot and focus on one single question....whatever it ibe. – DonAntonio Jan 19 '18 at 23:19
  • Thanks for the tip, i didn't even realize. I Wrote a short question on top which summarizes the whole thing i guess :) – Layl Jan 19 '18 at 23:26
  • I recommend switching from parametric equations to matrices. There’s a tedious, but straightforward, algorithm for finding the intersection of two conics using this representation. It involves solving a cubic equation, but even that part has a straightforward set of formulas to let you crank out the answer. See this question for a description of the algorithm. – amd Jan 19 '18 at 23:47
  • Ok thanks a lot for this i will dive into this solution right now ! Although one thing i would like to understand is why switching tools like using matrices will change radically the possibility of solving things since the amount of information i start with is the same ? – Layl Jan 19 '18 at 23:59
  • @Layl Please use Matjax. I formatted a part of your question as a sample. Have a look. – Gaurang Tandon Jan 20 '18 at 01:18
  • @Layl In the short version of the question, do you have to solve for $r$, $s$, $t$, or all three? – Gaurang Tandon Jan 20 '18 at 01:19
  • Ok i'll have a look and edit ( noob here ), for the short version we want to solve for s and t, while r is known – Layl Jan 20 '18 at 11:38
  • The choice of representation can make a huge difference in the methods that can be applied and their ease. If you’re comfortable with solving systems of equations that involve trig functions, then go for it. The method that I cited is an algorithm that gets you the solutions by a direct calculation. Even the “solve a cubic” part of it proceeds by plugging values from the specific instance into a set of formulas. If you’re trying to automate the process, say, the latter approach is quite nice for that. – amd Jan 24 '18 at 20:03
  • A simple example of the difference in approach is line-plane intersection. A common method is to parametrize the line, solve the resulting equation for the value of the parameter that satisfies the equation of the plane, and then plug it back in. If you instead use the Plücker matrix of the line, which is easily constructed from two points on it or the point and direction vector from the parametric equation, a direct computation gets you the intersection point without having to solve any equations at all. Same information, different types and amount of work involved in the solution. – amd Jan 24 '18 at 20:07

1 Answers1

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First of all let rearrange your equations to $$x(t) = a\cdot cos t + h$$ and $$y(t) = b\cdot sin t + k$$ and for the second ellipse $$x(s) = c\cdot cos s + i$$ and $$y(s) = d\cdot sin s + j $$

In order to find the potential points of intersection you need to solve a system of four equations. $$a\cdot cos t + h =d\cdot cos s +j$$ , $$ b\cdot sin t + k= d\cdot cos s +j$$, $$ cos^2(t)+sin^2(t)=1$$ and $$ cos^2(s)+sin^2(s)=1$$ for s and t.

I strongly recommend numerical method for solving such a system.

  • Thanks a lot for your response. Could expand on why you introduce $\cos^2(t)+\sin^2(t)=1$, how can it help since it is true for any $t$ and any $s$ ? – Layl Jan 20 '18 at 12:01
  • @Layl Thanks for your attention. We have 4 unknown and we need four equations. $cos^2(t)+\sin^2(t)=1$, and $ cos^2(s)+\sin^2(s)=1$, fill in the gap. – Mohammad Riazi-Kermani Jan 20 '18 at 12:57
  • Ok thanks again i'm gonna see if i can handle it this way, cheers. – Layl Jan 20 '18 at 13:00
  • Ok thanks a lot that unlocked me for this part, i'll keep going. It seems silly but i didn't have the intuition for that ... :) I'll mark your answer as the solution and thanks again, it is nice to see people are ready to help on that kind of matter, hope i can give back someday ! – Layl Jan 20 '18 at 13:09
  • Thanks, you just returned the favor by your good words and deed. – Mohammad Riazi-Kermani Jan 20 '18 at 13:19