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Defining geometry within a cartesian coordinate system

  1. Jul 5, 2013 #1

    Just as a warning before anyone reads my question I am not a mathematician, just an engineer with moderate math skills he wants to expand.

    So I'm writing some engineering software which involves defining/interation/modification of geometry within a cartesian system but I currently lack the geometry knowledge to pull it off. I know geometry is a big subject and my research up until now has left overwhelmed so I'm trying to narrow down what I should be learning to get to grips with the ideas I need to do what I want to do.

    To give you an idea of what type of things I'm trying to build upto I currently know how to define a straight line using the standard formula (which I store in the software as 2 points but can use the formula to define new points etc.). What I want to be able to do is define curves (curcular, clothoid, parabola etc.) in a way in which an accurate definition can be store within he program of a specific curve. I know some of the standad formulas for curves such as for a parabola on a graph but my curves will be set at specific coordinates at a certain orientation.

    I also want to learn how I can solve inerations between geometry such as where a line crosses a curve etc. I've tried to understand calculating whether and where 2 straight lines cross but I haven't found it easy on my own!

    I know my math skills are not great and my question is pretty vague but any rough idea of useful topics/skills to learn or any starting points would be appriciated. I will also understand if the answer is "You're jumping into something well over my head". I know the basic geometry you learn at school but that is about it.

  2. jcsd
  3. Jul 5, 2013 #2
    It seems that linear algebra is the answer here.

    Finding where two lines cross is very easy using linear algebra. You just need to solve a system of two equations. Finding where two arbitrary curves cross is a bit more difficult though, but you can also do it using systems of equations.

    Then, it seems that you know how to graph things like ##y=x^2## which is a parabola opening upward. You might want to know a formula for a parabola opening in some diagonal direction. Is that it?

    Well, take the parabola ##y=x^2##. Arbitrary points on there are of the form ##(t,t^2)## (thus, any ##t## will give you a point). We can rotate this parabola by multiplying with a rotation matrix. Such a matrix has the form

    [tex]\left(\begin{array}{cc}\cos(\theta) & \sin(\theta)\\
    - \sin(\theta) & \cos(\theta)\end{array}\right)[/tex]

    The new parabola becomes ##(\cos(\theta)t + \sin(\theta)t^2, -\sin(\theta)t + \cos(\theta)t^2)##. For example, if you want to rotate your parabolia 90 degrees, then ##\theta = \frac{\pi}{2}##. Thus we get ##(t^2,-t)##.

    You should probably go through a book on basic applied linear algebra.
  4. Jul 5, 2013 #3
    Thanks for the reply.

    I get the geist that for 2 lines it is just a matter of solving the 2 equations but my main problem is (and from my experience the same problem most people have) when I was taugh these things we were taught in such a set way that I didn't fully understand it and don't know how to actually use it to solve things for myself. Such as testing whether 2 lines cross at all? what is the coordinate of the cross over?

    I got the feeling someone would use a tranformation matrix. I covered matrices to a very basic level but I can't say I know how to use them for anything! Yes that is an example and to take it another step further how could I define a parabolic curve which passes through 2 specific coordinates? eg. (2,3) and (5,8). I'm guessing I will need to learn certain matrices operations.

    I hope that's clear.
  5. Jul 5, 2013 #4
    Are we talking about lines in a plane? The equations for two lines is given by ##\alpha X + \beta Y = \gamma## and ##aX + bY = c##. The lines cross if and only if

    [tex]\textrm{det}\left(\begin{array}{cc} \alpha & \beta\\ a & b\end{array}\right) \neq 0[/tex]

    The coordinate of the cross can be calculated by solving the system

    [tex]\left\{\begin{array}{l} \alpha X + \beta Y = \gamma\\ aX + bY = c\end{array}\right.[/tex]

    There are multiple parabolic curves going through those two points. Let's try to find a parabolic curve opening upwards. This has the form ##y = ax^2 + bx + c##. The point ##(2,3)## should be a solution, thus the following should be true ##4a + 2b + c = 3##. Similarly, we want ##25a + 5b + c = 8##. We solve this for ##a## and ##b## by solving the system

    [tex]\left\{\begin{array}{l} 4a + 2b + c = 3\\ 25a + 5b + c = 8\end{array}\right.[/tex]

    This system will have multiple solutions. Each of the solutions will give you a parabola opening upward.
  6. Jul 5, 2013 #5
    I'm starting to get the feeling I'm out of my depth and have alot to learn before I get anywhere. Although I can roughly follow what you are trying to do, most of it makes little sense to me. I keep looking through geometry books and websits but there is just so much information I just don't even know where to start...

    What about this example: Say I manage to draw part of a parabolic curve. It's not perpendicular with either axis (lets say it's 23 degress rotated or whatever) and it does not sit on 0,0 but at 12,19. Would a formula for such a curve be difficult to derive?
  7. Jul 5, 2013 #6
    I derived the formula above:

    Just let ##\theta## be 23 degees. Of course, this is a parabola sitting on (0,0). If you want to move the "base pont", then you need to add ##(12,19)##. So the formula is:

    [tex](12 + \cos(\theta)t + \sin(\theta)t^2,19 -\sin(\theta)t + \cos(\theta)t^2)[/tex]

    where ##\theta## is 23 degrees.
  8. Jul 5, 2013 #7
    You make it seem so easy it's genius! I'm not entirely sure how you managed to derive the formula but that is the type of thing I want to be able to work out.

    When it comes to crossing geometry eg. a line crossing the curve. Is it generally the same idea as for the crossing straight lines?
  9. Jul 5, 2013 #8
    It's the same idea, except that the system is harder to solve.
  10. Jul 5, 2013 #9
    I keep reading through what you have put and I think it's starting to click.
    How did you come up with the cos, sin etc. matrix?

    Is there a name for the this type of mathematics I could use to search for material to learn from?
  11. Jul 5, 2013 #10

    Search for linear algebra. Or analytic geometry.
  12. Jul 5, 2013 #11
    Thats great thanks! Now I know what I'm searching for things should be much easier.
    You also mentioned plane curve which led me to this:

    I understand as a function, you put x in and it gives you y but I don't fully understand what the other 2 mean ie. Implicit equation and Parametric equation.

    Hopefully I have enough now to get me going.
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