What is a quadric surface?
A quadric surface is the graph of a 2nd degree equation in 3 variables. The general form of quadric surfaces is \[Ax^2+By^2+Cz^2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0\] For different values of the variables from the general form, we get 6 quadric surfaces. Ellipsoid, elliptic paraboloid, hyperbolic paraboloid, cone, hyperboloid of one sheet, and hyperboloid of two sheets.
The quadric surfaces are created by different equations from the general form and can be sliced by a thin plane with respect to different axis's to display different lines of functions called traces. These traces show what that part of the quadric surface intersected with the plane would look like in 2-D. Below you can see the different surface's equations and what traces you get by cutting into different axis's with a plane.
Stil confused? Me too. Look at this picture of a cone (a quadric surface) being cut into by a thin square (plane) and see what the resulting line (trace) looks like for a better idea of what it going on.
Look at how a circle is created by cutting into the cone horizontally, or an elipse by cutting into the cone horizontally at an angle, or a parabola by cutting vertically into one of its bases, or a hyperbola by cutting vertically into both of its bases. Starting to make more sense? We have a quadric surface, and we analayze what the behavior of the surface is at the intersection of some plane going through it.
Now, something cool is that you can essentially recreate a quadric surface by using traces in different planes and putting them together. The more traces you have the more clear the picture of what you are trying to create will be. You can also eliminate a plane by setting its corresponding variable to 0, and then solving to find what the trace looks like in that plane for the shape you're looking at. If you substitute a value for the variable you are taking out, you will see a trace in a plane paralell to that plane at the level of the constant you subtituted the variable for. To find a trace simply find a plane you want to use, set one of the variables to 0 and solve for a remaining variable. Then plug that new value in to the original eqaution and you will see what the resulting function is for the quadric surface at that intersection of the plane you made. For example, to find the xy trace, put z=0 into the equation. Put z=k to get a trace in a plane paralell to the xy plane at level k. To find the yz trace put x=0 in the equation, and put x=k to get traces at level k in a plane paralell to yz. To find xz trace's put y=0 in the equation and put y=k to get a trace in a plane parallel to the xz plane at level k.
Too much information at once? I get it. Let's look do this with a cone. If we look at a cone, we see that the trace in the xy plane is a point (the origin) and the traces in planes parallel to the xy plane are elipses. The traces in the yz ans xz planes are pairs of lines intersecting at the origin. The traces in planes parallel to these are hyperbolas.
This makes sense because when you take out the z plane you are isolating the xy plane and you get the equation of a circle, \(x^2+y^2=0\) where the radius is 0, so it is simply a point at the origin. The intersecting lines make sense too because when you take out the x plane you isolate the yz plne and are left with \(y^2-z^2\) which is the exact same function as the xz trace \(x^2-z^2\) just in a different plane. These are shown on desmos below.
To get a slice and view its trace in that plane, there are a few steps you have to follow.
1.) Write down the original equation for the quadric surface. For a cone, this will be \[0=x^2+y^2-z^2\]
2.) Find the plane you want to use and solve for a variable. For this example, we want to see what kind of trace lives in the xz plane with the plane \(0=6x+4z\) shown below, that is, we want to see what behavior occurs at the intersection of the quadric surface and this plane.
\[0=6x+4z\]
\[-4z=6x\]
\[-2z=3x\]
\[z= (\frac{-3x}{2})x\]
3.) Substitute the value you found into the original equation and analyze what type of function it is. You can plug it into a 2-D graphing calculator to see the behavior for yourself. \[0=x^2+y^2-(\frac{-3x}{2})x^2\] \[0=(\frac{2y^2-2x^2}{2})\] which is a a pair of intersecting lines. When we set k equal to something other than 0, we see that this is a hyperbola. Shown below on desmos with k=1. This is because we are setting the point in whatever axis/plane we aren't using as a point for the plane to intersect at. When we choose k=0 we are choosing to see what the function of the trace looks like in 2D at the origin. This is simply an intersection of lines as mentioned before. However, when we work our way up or around the origin at different points, we are no longer perfectly intersecting the center of the cones any more and we enter the shapes that make a cone a cone rather than its center point which is just a circle with raidus equal to 0. The farther we go out we see that the hyperbolas are farther and farther away from the origin just like the actual shape and behavior of the cone. So at k=0 the intersection is a pair of intersecting lines, when we choose something other than 0 for k we get hyperbolas at this intersection between the cone and the plane.
In openscad you can create your own planes that intersect a quadric at different points and use the difference feature to see the trace that is left behind when you remove that plane from the quadric surface.
There's our cut with an xz plane similar to before but not exactly through the origin so that we get hyperbolas instead of intersecting lines \(0.1=6x+4z\).
And there are the hyperbolas, neat!
I chose the cone because it has the most versatility in what shapes you can create from its form. It's itneresting to me that such a simple quadric surface can have so many different concepts inside of it including parabolas, hyperbolas, elipse, and a circle. I chose the plane I did because I wanted to show that the top and bottom parts of the cone are hyperbolas if you cut them straight or at an angle. I also chose the plane and intersection I did because it is easy to see that it is a hyperbola when you look it it at an angle from top to bottom.
A quadric surface is the graph of a 2nd degree equation in 3 variables. The general form of quadric surfaces is \[Ax^2+By^2+Cz^2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0\] For different values of the variables from the general form, we get 6 quadric surfaces. Ellipsoid, elliptic paraboloid, hyperbolic paraboloid, cone, hyperboloid of one sheet, and hyperboloid of two sheets.
Now, something cool is that you can essentially recreate a quadric surface by using traces in different planes and putting them together. The more traces you have the more clear the picture of what you are trying to create will be. You can also eliminate a plane by setting its corresponding variable to 0, and then solving to find what the trace looks like in that plane for the shape you're looking at. If you substitute a value for the variable you are taking out, you will see a trace in a plane paralell to that plane at the level of the constant you subtituted the variable for. To find a trace simply find a plane you want to use, set one of the variables to 0 and solve for a remaining variable. Then plug that new value in to the original eqaution and you will see what the resulting function is for the quadric surface at that intersection of the plane you made. For example, to find the xy trace, put z=0 into the equation. Put z=k to get a trace in a plane paralell to the xy plane at level k. To find the yz trace put x=0 in the equation, and put x=k to get traces at level k in a plane paralell to yz. To find xz trace's put y=0 in the equation and put y=k to get a trace in a plane parallel to the xz plane at level k.
Too much information at once? I get it. Let's look do this with a cone. If we look at a cone, we see that the trace in the xy plane is a point (the origin) and the traces in planes parallel to the xy plane are elipses. The traces in the yz ans xz planes are pairs of lines intersecting at the origin. The traces in planes parallel to these are hyperbolas.
1.) Write down the original equation for the quadric surface. For a cone, this will be \[0=x^2+y^2-z^2\]
2.) Find the plane you want to use and solve for a variable. For this example, we want to see what kind of trace lives in the xz plane with the plane \(0=6x+4z\) shown below, that is, we want to see what behavior occurs at the intersection of the quadric surface and this plane.
3.) Substitute the value you found into the original equation and analyze what type of function it is. You can plug it into a 2-D graphing calculator to see the behavior for yourself. \[0=x^2+y^2-(\frac{-3x}{2})x^2\] \[0=(\frac{2y^2-2x^2}{2})\] which is a a pair of intersecting lines. When we set k equal to something other than 0, we see that this is a hyperbola. Shown below on desmos with k=1. This is because we are setting the point in whatever axis/plane we aren't using as a point for the plane to intersect at. When we choose k=0 we are choosing to see what the function of the trace looks like in 2D at the origin. This is simply an intersection of lines as mentioned before. However, when we work our way up or around the origin at different points, we are no longer perfectly intersecting the center of the cones any more and we enter the shapes that make a cone a cone rather than its center point which is just a circle with raidus equal to 0. The farther we go out we see that the hyperbolas are farther and farther away from the origin just like the actual shape and behavior of the cone. So at k=0 the intersection is a pair of intersecting lines, when we choose something other than 0 for k we get hyperbolas at this intersection between the cone and the plane.
I chose the cone because it has the most versatility in what shapes you can create from its form. It's itneresting to me that such a simple quadric surface can have so many different concepts inside of it including parabolas, hyperbolas, elipse, and a circle. I chose the plane I did because I wanted to show that the top and bottom parts of the cone are hyperbolas if you cut them straight or at an angle. I also chose the plane and intersection I did because it is easy to see that it is a hyperbola when you look it it at an angle from top to bottom.
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