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Cylinder Reflections The Mathematics Behind the Images

Cylinder ReflectionsThe Mathematics Behind the Images

Dr. Don SpicklerDr. Jennifer Bergner

Salisbury University

[email protected]@salisbury.edu

Drs. Spickler & Bergner (Salisbury Univ.) Cylinder Reflections November 12, 2011 1 / 36

Anamorphic Art What is it?

Anamorphic Art

Anamorphic art is created by distorting animage so that is only revealed from a singlevantage point or from its reflection on amirrored surface. This artistic process wasfirst attempted during the Renaissance andbecame exceedingly popular during theVictorian Era.

The earliest known examples come from thenotebooks of Leonardo da Vinci. Hesuccessfully sketched an eyeball in 1485 thatcould only be discerned when looking at thedrawing from a certain angle.



Anamorphic Art

More modern artists using these techniquesinclude Julian Beever, who createsthree-dimensional illusions on sidewalksusing chalk.

Julian Beever’s Fountain



Anamorphic Art

Julian Beever’s Fishing



Anamorphic Art

Julian Beever’s Rafting

As you can see, Julian Beever likes toincorporate human subjects in all of hissidewalk art. This shows that not only canhe create the perspective shift but he can doit to scale.



Anamorphic Art

Hans Hamngren and Istvan Orosz usethe mirrored cylinder technique. Theyachieve this illusion by either drawingthe image on a distorted grid, similarto the way M. C. Escher createdmany of his illusions, or looking at themirrored image while drawing on aflat surface.

Pictured to the left is the work ofIstvan Orosz.



Anamorphic Art

More of Istvan Orosz’s work.



Anamorphic Art

This is a work by Hans Hamngren, an old firehydrant in an old fire extinguisher.


The Process Step 1: The Setup

The Process

Step 1: The Setup — Cylinder



The Process

Step 1: The Setup — Paper



The Process

Step 1: The Setup — Viewer Position



The Process

Step 1: The Setup — The Final Image



The Process

Step 1: The Setup — Place the image in the cylinder.


The Process Step 2: Draw a line from pixel to viewer.

The Process

Step 2: Draw a line from pixel to viewer.


The Process Step 3: Find the intersection with cylinder.

The Process

Step 3: Find the intersection with cylinder.


The Process Step 3: Find the intersection with cylinder.

The Process

Step 3: Consider the line from intersection to viewer.


The Process Step 4: Find the normal line from intersection.

The Process

Step 4: Find normal line from intersection.


The Process Step 5: Find the reflection line from intersection.

The Process

Step 5: Find the reflection line from intersection.


The Process Step 6: Find the intersection of reflection line and paper.

The Process

Step 6: Find the intersection of reflection line and paper.


The Process Step 6: Plot the point.

The Process

Step 6: Plot the point.


The Process Step 6: Repeat for all pixels on the image.

The Process

Step 6: Repeat for all pixels on the image.


The Mathematics Step 1: The Setup

The Mathematics

Step 1: The Setup

Assumptions: The base of the cylinder ison the xy -plane, the central axis passesthrough the origin, and the paper is on thexy -plane.

P = 〈px , py , pz〉V = 〈vx , vy , vz〉r 2 = x2 + y 2


The Mathematics Step 2: Draw a line from pixel to viewer.

The Mathematics

Step 2: Draw a line from pixel toviewer.

To do this we take the starting position tobe the pixel point P and the direction to betoward the viewer V − P. Thecorresponding formulas are,

L(t) = P + t(V − P)

= 〈px , py , pz〉+ t(〈vx , vy , vz〉 − 〈px , py , pz〉)= 〈px + t(vx − px), py + t(vy − py ), pz + t(vz − pz)〉

As t goes from 0 to 1 we trace out the linesegment from P to V.


The Mathematics Step 3: Find the intersection with cylinder.

The Mathematics

Step 3: Find the intersection withcylinder.

To do this we take the line from the pixel tothe viewer and plug the x and y expressionsinto the equation of the cylinder and solvefor t.

r 2 = x2 + y 2

= (px + t(vx − px))2 + (py + t(vy − py ))2

= p2x + p2

y + t(−2p2x − 2p2

y + 2pxvx + 2pyvy )

+ t2(p2x + p2

y − 2pxvx + v 2x − 2pyvy + v 2

y )

= at2 + bt + c



The Mathematics


Now solve the quadratic equation,

at2 + bt + c − r 2 = 0

for t and we get,

t =−b ±

√b2 − 4a(c − r 2)

2a

The one we want is

ti =−b +

√b2 − 4a(c − r 2)

2a



The Mathematics


Now we substitute this value, ti , in for t inthe line equations to get the point ofintersection.

L(ti) = P + ti(V − P)

= 〈px + ti(vx − px), py + ti(vy − py ), pz + ti(vz − pz)〉= 〈Ix , Iy , Iz〉


The Mathematics Step 4: Find the normal vector from intersection.

The Mathematics

Step 4: Find the normal vector fromintersection.

The normal vector is perpendicular to thesurface and is used in the calculation of thereflection line. For a cylinder, the normalvector will be parallel to the xy -plane andpass through the points 〈Ix , Iy , Iz〉 and〈0, 0, Iz〉. So our normal vector is thedifference between these two points,

n = 〈Ix , Iy , 0〉


The Mathematics Step 5: Find the reflection vector from intersection.

The Mathematics

Step 5: Find the reflection vector fromintersection.

This is probably the most involvedcalculation in the process. From the diagramon the right notice that the reflection vectorr = v + 2a where v = V − P is the vectorfrom the pixel to the viewer.

SinceProj = v + a

we havea = Proj− v



The Mathematics


Proj = Projnv

=n · v|n|2

n

Soa =

n · v|n|2

n− v



The Mathematics


r = v + 2a

= v + 2

(n · v|n|2

n− v

)=

2n · v|n|2

n− v

= 〈rx , ry , rz〉


The Mathematics Step 5: Find the reflection line from intersection.

The Mathematics

Step 5: Find the reflection line fromintersection.

So our reflection line is

R(t) = 〈Ix , Iy , Iz〉+ t 〈rx , ry , rz〉= 〈Ix + trx , Iy + try , Iz + trz〉


The Mathematics Step 6: Find the intersection of reflection line and paper.

The Mathematics

Step 6: Find the intersection ofreflection line and paper.

The paper is on the xy -plane so every threedimensional point on the paper has a zcoordinate of 0. We can use this fact to findhow far we must move down the reflectionvector until we hit the paper, this is thevalue of t in the reflection line formula,

R(t) = 〈Ix + trx , Iy + try , Iz + trz〉



The Mathematics


If we set the z coordinate of the reflectionline equal to 0 we can solve for t,

Iz + trz = 0

gives

t = − Izrz



The Mathematics


Substitute this value in for t in the reflectionline equation gives the intersection point⟨

Ix −Izrzrx , Iy −

Izrzry , Iz −

Izrzrz

⟩which gives⟨

Ix −Izrzrx , Iy −

Izrzry , 0

⟩


The Mathematics Step 6: Plot the Point

The Mathematics

Step 6: Plot the Point

Plot this point on the paper in the samecolor as the original pixel color and move onto the next pixel. When all of the points areplotted you have your transformed image.


The Mathematics Who did the work?

The Mathematics

Who did the work?

The calculations for this and several other anamorphic scenario weredone as undergraduate research projects by students at SalisburyUniversity.

Jennifer Larson and Kristi Martini (2004) — Cylinder and Sphere

Nicole Massarelli (2010) — Theoretical extensions to generalconvex surfaces. She also wrote the software for cylinderreflections.

Angela Rose and Erika Gerhold (2011) — Tilted Cylinder


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