How an ambiguous object works
Ambiguous objects are exactly what they sound like - ambiguous! Imagine you and your friend are out on a hike and find an ambiguous object in the wild. You decide to stand on opposite sides of it (and 45 degrees up in the air). You and your friend look something like this from where you are floating:
Your friend says they see a circle from where they're standing, but you disagree. You see a square! How is that possible?
You are standing along the y-axis, and the lines from your eyes to the curves of the object are perpendicular to the vectors \(\langle 0,1,1 \rangle\) and \(\langle 0,-1,1 \rangle\). The equation through \((x,f(x),0)\) and parallel to \(\langle 0,1,1 \rangle\) is \(r_{A} (s) = (x,f(x)+s,s)\). The equation through \((x,g(x),0)\) and parallel to \(\langle 0,-1,1 \rangle\) is \(r_{B} (t) = (x,g(x)-t,t)\). This is when you are observer A and your friend is observer B. We can find the point of intersection by setting \(r_{A} (s) = r_{B} (t)\), which gives us \[x=x\] \[f(x)+s=g(x)-t\] \[s=t.\] Now, we can use the third equation to make a substitution. \[f(x)+s=g(x)-s\] \[s=\frac{1}{2} (g(x)-f(x)).\] Use another substitution to obtain \[(x, \frac{1}{2}(g(x)+f(x),\frac{1}{2}(g(x)-f(x).\] This is the parameterization of something like an "average" curve of \(f(x)\) and \(g(x)\) from two dimensions into three. In order for this to work, \(f(x)\) and \(g(x)\) must be of the form \((x,f(x),0\) and \(x,g(x),0)\). The x-coordinates must match. The interesting thing about this "average" curve in three dimensions is that now observer A sees what appears like one of the curves but observer B sees the other. This optical illusion is an ambiguous object.
What curves I chose
For my object, I chose to use the curves \(f(x)=(cos(x)^3, sin(x)+cos(x), 0)\) and \(g(x)=(cos(x)^3, cos(x), 0)\). These curves look like this:
The "average" of these curves looks like the blue:
Consider the perspectives from observer A and B. Observer A (the green ray) sees this:
This appears to be \(g(x)\). However, observer B (the red ray) sees this:
This appears to be \(f(x)\). How weird!
Why I chose this design
I chose this design for a couple of reasons. Firstly, I was limited to what I could figure out on OpenSCAD. The four-piece code ended up hurting my brain, so I went with something simpler. Next, I kept the \(cos(x)^3\) from the provided code as my \(x\) parameterization simply because I was not observant enough to notice it was there. I finally got nice curves to work, and I like the ones I ended up with because there is very little confusion about what is going on. When looking at other examples, it was easy to get lost, so-to-speak. I had to keep the observer rays in place in order to orient myself. If someone is learning this topic for the first time (me, for example), this is extra hard to understand. I like how my curves simplify such a tricky illusion.
My print is going to be around 2.5 by 4.5 inches, which are the dimensions of the rectangular base I added. To see the illusion properly, make sure you are viewing it from the skinnier end of the base.
Word count: 538
Ambiguous objects are exactly what they sound like - ambiguous! Imagine you and your friend are out on a hike and find an ambiguous object in the wild. You decide to stand on opposite sides of it (and 45 degrees up in the air). You and your friend look something like this from where you are floating:
You are standing along the y-axis, and the lines from your eyes to the curves of the object are perpendicular to the vectors \(\langle 0,1,1 \rangle\) and \(\langle 0,-1,1 \rangle\). The equation through \((x,f(x),0)\) and parallel to \(\langle 0,1,1 \rangle\) is \(r_{A} (s) = (x,f(x)+s,s)\). The equation through \((x,g(x),0)\) and parallel to \(\langle 0,-1,1 \rangle\) is \(r_{B} (t) = (x,g(x)-t,t)\). This is when you are observer A and your friend is observer B. We can find the point of intersection by setting \(r_{A} (s) = r_{B} (t)\), which gives us \[x=x\] \[f(x)+s=g(x)-t\] \[s=t.\] Now, we can use the third equation to make a substitution. \[f(x)+s=g(x)-s\] \[s=\frac{1}{2} (g(x)-f(x)).\] Use another substitution to obtain \[(x, \frac{1}{2}(g(x)+f(x),\frac{1}{2}(g(x)-f(x).\] This is the parameterization of something like an "average" curve of \(f(x)\) and \(g(x)\) from two dimensions into three. In order for this to work, \(f(x)\) and \(g(x)\) must be of the form \((x,f(x),0\) and \(x,g(x),0)\). The x-coordinates must match. The interesting thing about this "average" curve in three dimensions is that now observer A sees what appears like one of the curves but observer B sees the other. This optical illusion is an ambiguous object.
What curves I chose
For my object, I chose to use the curves \(f(x)=(cos(x)^3, sin(x)+cos(x), 0)\) and \(g(x)=(cos(x)^3, cos(x), 0)\). These curves look like this:
Why I chose this design
I chose this design for a couple of reasons. Firstly, I was limited to what I could figure out on OpenSCAD. The four-piece code ended up hurting my brain, so I went with something simpler. Next, I kept the \(cos(x)^3\) from the provided code as my \(x\) parameterization simply because I was not observant enough to notice it was there. I finally got nice curves to work, and I like the ones I ended up with because there is very little confusion about what is going on. When looking at other examples, it was easy to get lost, so-to-speak. I had to keep the observer rays in place in order to orient myself. If someone is learning this topic for the first time (me, for example), this is extra hard to understand. I like how my curves simplify such a tricky illusion.
My print is going to be around 2.5 by 4.5 inches, which are the dimensions of the rectangular base I added. To see the illusion properly, make sure you are viewing it from the skinnier end of the base.
Word count: 538
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