Anamorphic Art with a Tilted Cylinder

Transcription

1 Anamorphic Art with a Tilted Cylinder Erika Gerhold and Angela Rose Math and Computer Science Salisbury University 1101 Camden Avenue Salisbury Maryland USA Faculty Advisor: Dr. Don Spickler Abstract Proceedings of The National Conference On Undergraduate Research (NCUR) 2012 Weber State University, Ogden Utah March 29 31, 2012 Anamorphic art is created by taking a distorted image and reflecting it off a mirrored surface so that from the viewer's vantage point the distorted image appears correct. The term anamorphic comes from two words that mean "change again" and dates back to as early as the Renaissance. Leonardo da Vinci created the earliest known examples of anamorphic art. Current day artists, like Kurt Wenner and Julian Beever, use anamorphic art to create three dimensional illusions with sidewalk drawings 1. This project focused on catoptrics or mirror anamorphous. In mirror anamorphous, the distorted image is placed around a mirrored shape to alter a distorted flat image into a correct three dimensional picture on the object. The techniques used are mainly vector calculus and inverse ray tracing to develop a series of formulas to achieve the illusion. Specifically, start with a mirrored cylinder whose central axis was not perpendicular to the horizontal plane and a viewing position above the plane. For each pixel in the original image, imagine a ray was traced from the viewer to the pixel. Then calculate the induced reflections to both the plane and a containing box, thus resulting in the distorted image. In this paper, the methods and mathematical details of the catoptric anamorphic process for a tilted cylindrical mirror will be discussed. Keywords: Anamorphic, Reflection, Vector-calculus 1. Introduction Anamorphic art can be divided into two main types perspective anamorphous and mirror anamorphous. Perspective anamorphous is where an image is drawn onto a surface in such a way that when the viewer observes the image from a particular position the image appears normal, that is, in an undistorted manner. Most perspective anamorphous is achieved by simple elongation of the image in the direction of the viewer s line of sight. While this is very impressive when done by an artist, it does not produce a very interesting mathematical problem. Mirror anamorphous, on the other hand, is created by placing the distorted image around a mirrored object such as a cylinder or sphere so that the image reflects correctly onto the surface of the mirror. In 2010, Nicole Massarelli developed an algorithm that, given an undistorted image, produced the necessary distorted image for a cylinder whose central axis was perpendicular to the horizontal plane 4. She also developed procedures, but no specific algorithms, for the reflection onto any convex mirrored surface. The main result in this paper is a natural continuation of Nicole s work, specifically looking at a rotation of a mirrored cylindrical surface. That is, creating specific algorithms for mirrored anamorphous on a cylindrical surface whose central axis is not perpendicular to the horizontal plane.

2 2. The General Process Take a vertical cylinder with the central axis perpendicular to the xy-plane and rotate it by an angle,, around the x- axis (Figure 1). Position the image strictly inside the cylinder along its central axis. The viewer s position will be V = (x 0, y 0, z 0 ). Take a pixel on the image inside the cylinder, denoted by P = (x p, y p, z p ). The goal is to find where the point A should be placed so that, from the viewer s perspective, its reflection off of the cylinder is directly over P. To do this, take the line of sight from the viewer to the pixel and find the intersection, I, of this line with the cylinder. This is the position on the cylinder that the viewer sees the pixel (Figure 2). Then calculate the reflection line at this point and see where the vector intersects the enclosing box. This intersection point, A, is where the viewer would see the point P when reflected in the cylinder, so this is where a point should be plotted that is the same color as the pixel P. Figure 1: The set-up for a mirrored cylinder. More specifically, let v = (a, b, c) be the vector from the pixel, P, to the viewer, V, which is the line of sight. Using the parametric equations of the tilted cylinder, find the point of intersection, I, of this vector and the cylinder. Even though the cylinder is a curved surface, at a single point the reflection calculation is identical to a reflection on a flat surface. So to find the reflection vector, v, take the normal (or perpendicular) vector, n, to the surface at the point I and calculate v as the vector whose angle with n is identical to the angle between n and v. The final step is to find where r intersects the bounding environment. In this case, place the cylinder inside a rectangular box so we simply calculate the point of intersection, A, with the closest face of the box. To produce the entire mapping of the image in the bounding box, continue this process with every pixel on the initial image. 450

3 Figure Rotation of the Cylinder and Image The first step is to derive the level surface equation for the tilted cylinder. To do so take the level surface of a vertical cylinder, + =, and transform it by rotation about the x-axis. The standard rotation matrix for rotation about the x-axis from the y-axis to the z-axis is 3, R x,θ = 0 cos() sin() 0 sin() cos() Let (x, y, z) be any point in the space, then the rotated point is given by, = R x,θ = cos() sin() sin()+cos() Substituting the transformed system into the surface equation gives the equation of the rotated cylinder, +(cos() sin()) = The calculation of the position of each pixel, P =,, on the original image is done in a similar manner. Before the image is rotated it is on the xz-place and has the coordinates (x, 0, z). Thus the transformed pixel position is 3, R x,θ 0 = cos() = cos() 4. Intersection of the Viewing Vector with the Surface To find the intersection point, I, of the line of sight with the cylinder, start with the parametric equations of the line of sight 6 = + 451

4 = + = + where (a, b, c) is the vector from the pixel to the viewer. Substituting these into the equation of the rotated cylinder gives, ( +) +(( +)cos() ( +)sin()) = Solving this equation for t finds where the viewer s line of sight intersects the cylinder. Figure 3 The solution is the scalar multiple of the viewing vector, v, that when added to the pixel position, P, gives the intersection point, I. Since the equation is quadratic in t, there will be two solutions. This is also evident from the geometry since a line can intersect a cylinder in two points (Figure 3). Since our viewing vector (a, b, c) is from the pixel position to the viewer, there will be one positive and one negative solution for t. The positive solution is the one closest to the viewer and hence is the desired solution. Using the quadratic equation, our solutions are, where, = ± 4λυ 2λ λ = + cos () 2cos()sin()+ sin () µ = 2 +2 cos () 2 cos()sin() 2 cos()sin()+2 sin () ν = + + cos () 2 cos()sin()+ sin () To find the coordinates of I, substitute the positive solution for t, call it t I, into the parametric equations dof the viewing vector, I = (,, ) = +, +, + 452

5 5. Calculation of the Reflection Vector To calculate the reflection vector, first find the normal vector to the surface at I, which is the gradient vector to the surface at I. n =,, =(2,2 cos () 2 sin()cos(), 2 cos()sin()+2 sin ()) Denote the normal vector as, =(,, ) One approach would be to calculate the angle between n and v and derive r to make the same angle with n. Instead, use a technique from computer graphics 2. We project v on to n giving the vector n, calculating a by taking the difference between n and v and finally finding the reflection vector by adding 2a to v (Figure 4). Using the projection formula to project v onto n we have 6 Figure 4: The vector triangle. where and =o = = (,, )(,,) = + + = 2 +2 cos () 2 sin()cos() 2 cos()sin()+2 sin () gives = + + = 4 +(2 cos () 2 sin()cos()) +( 2 cos()sin()+2 sin ()) =o = = (,, )=,, Then and = =,, (,,)=,, r = (,,)+2,, = 2,2,2 = (,,) 453

6 This gives parametric equations for the reflection line, x = + y = + z = + 6. Intersection of the Reflected Vector with the Bounding Box To find where the point would reflect onto the xy-plane, set=0 and solve for t. 0 = + = Substituting = into the parametric equations for the reflection line gives the point (x A,y A z A ) where, = + = + = + =0 So (x p, y p, z p ) would appear to the viewer at (x A, y A, 0) on the xy-plane. Since the cylinder is inside a bounding box and the reflection could possibly intersect the other faces of the box, the intersection of the reflection line and the other five faces need to be found. In general the bounding box could be represented as,,, and following the above process for the xy-plane the solutions for each of the six faces of the box would be In most cases the system will be sitting on the xy-plane which simplifies =0. After completing these calculations for each pixel on the original image, the reflection image on each of the six bounding planes placed in the box will be constructed. This will complete the entire map and the viewer will be able to observe the correct image reflected in the tilted cylinder. 454

7 7. Algorithm Implementation When implementing the algorithm on the computer we chose to write the program in Mathematica. Our reasoning for using Mathematica was because Mathematica has matrix and vector manipulation capabilities as well as an image processing package. These facilities made using Mathematica far easier than writing the program from scratch in a general programming language, such as Java or C++. We will not include the complete code here but you can download the Mathematica notebook from Dr. Don Spickler s personal web site 5. To use the command, first load an image into a variable img and then invoke the following command. where cylinderconversionbox[image, viewposy, viewposz, cylradius, imagewidth, imagewidth, imagecenterheight, filename, finalwidth, rot, lx, ux, ly, uy, lz, uz] image - the image to be converted. viewposy - the horizontal distance the viewer is from the origin, where the center of the base of the mirrored cylinder is located. viewposz - the vertical distance the viewer is from the origin. cylradius - the radius of the mirrored cylinder. imagewidth - the physical width of the image to be viewed inside the cylinder. It is assumed that the image width is small enough to be completely contained inside the cylinder. imagecenterheight - the height of the center of the image from the origin. filename - the output base filename. The filenames for your six images will all have this name at the start and _LX, _UX, _LY, _UY, _LZ, and _UZ appended to the end to let you know the face for the image. finalwidth - the width, in pixels, of the transformed images, the height will automatically be calculated. rot - The rotation of the cylinder in degrees, note that negative rotation rotates the cylinder away from the viewer which is the usual direction. lx - x value of the vertical plane to the left of the cylinder. ux - x value of the vertical plane to the right of the cylinder. ly - y value of the vertical plane behind the cylinder. uy - y value of the vertical plane in front of the cylinder. lz - z value of the horizontal plane at the base of the cylinder. uz - z value of the horizontal plane above the cylinder. 8. Generalizations In the future, we would like to continue the research to include curved bounding surfaces such as cylinders. The process will be primarily the same, but the calculations will be altered due to the differences in the surface. Our next investigation will be reflections off a sphere into a cylindrical bounding surface. 9. Acknowledgements The authors wish to thank faculty advisor Dr. Don Spickler for his aid and direction throughout this project. They would also like to acknowledge their appreciation to Salisbury University for support through the Student Research Award. 455

8 10. References 1. 3 Amazing 3D Graffiti Artists: Street Painting and Sidewalk Chalk Art. Web Urbanist, Buss, Samuel, 3D Computer Graphics: A Mathematical Introduction With Open GL: (Cambridge: Cambridge University Press, 2003) Lay, David, Linear Algebra: (Boston: Pearson Education, 2003) Massarelli, Nicole. Mathematics Behind Anamorphic Art. Proceedings of the National Conference On Undergraduate Research (2010): Spickler, Don, Erika Gerhold, and Angela Rose, Tilted Cylinder Reflection Generator in Mathematica, Salisbury University, 6. Steward, James, Calculus: Early Transcendentals: (Belmont: Thomas Learning, 2003)