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Ambiguous Objects using Mathematical Functions

 Ambiguous: open to more than one interpretation; having a double meaning (according to the Oxford Dictionary). You typically use this adjective in an English or art class rather than a math class because literature and artwork can sometimes have multiple meanings or viewpoints. But not math. Two plus two is always four. The square root of nine is always three. And a function always results in that specific graph. I think that’s why I’ve always had a certain liking to math because it was always direct and to the point with not a lot of room for interpretation. However, by this week’s project, my mind has been blown by the ability to use math functions to create an ambiguous object.

Before we get into the math for the object, I’d also like to point out a real-world example of ambiguous objects, optical illusions. They use the same method that we are going to use by find a shared visual point or line that forces the audience to view one thing or the other. Have you ever been left starting at an optical illusion wondering why you can’t see both options? That’s why.

The goal of this ambiguous object is to view the object from one side to see one function, then view it from the other side to view the other function. But how does that work? We are going to use two functions, their parameterization, and perspective to build this ambiguous object.

First, we start with our two parameterized curves.

 







 






Observer 1 thinks they will see and observer 2 thinks they will see . We assume that each view will be looking at the object at a 45-degree angle and far enough away to treat the entire curve as a specific point. We assume the parameterization of these curves, r(t), will give the floating curve which will allow the object to be seen differently from the two different perspectives. Let’s find an expression for r(t) for the floating curve that will combine the two points of vision for these curves.

We will need to find the equation ra(s) through and parallel to <0,1,1> and the parameterization. We will also need to find the same thing for the second curve; the equation rb(t) through and parallel to <0,-1,1> . The results are and .

We now set these equations equal to each other and solve for the point of intersection.

Then plug s and t back into the parameterizations and get our final parameterization of the floating curve. This will represent our top curve and will form each function depending on the viewpoint.

We continue to build the function into and object.

If you look, you can see what function each observer’s perspective will see. Observer one will see the curve for f(x).


Observer 2 will see the curve for g(x). 


(Note: the rectangle at the bottom is used as a platform for the object and tells the observers to always view it from the longer side.)

I chose this object and these functions because I enjoy working with trig functions as they always surprise me with the different curves they can make. I also wanted to see how two completely different curves would show in a 3D object and how distinct they will be for each viewer. For example, one curve has distinct directional changes and is duplicated in all four quadrants, while the other curve is a single line function. 




 


 

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