Computer Graphics IIgarryowen.csisdmz.ul.ie/~cs4085/resources/cs4085-lect08.pdf · 2020. 12....

3-D Rotations (contd.)An Alternative Take on Rotations

Computer Graphics II

P. Healy

CS1-08Computer Science Bldg.

tel: [email protected]

Autumn 2020–2021

P. Healy (University of Limerick) CS4085 Autumn 2020–2021 1 / 15


Outline

1 3-D Rotations (contd.)Rodrigues Rotation Formula and QuaternionsDetermining the Axis and Angle from the Matrix

2 An Alternative Take on Rotations



Rodrigues Rotation Formula and QuaternionsDetermining the Axis and Angle from the Matrix

Outline






Amalgamating Steps 2, 3 & 4

Quaternions are a more compact way of solving the taskand lend themselves very easily to rotating through anangle θ one or more vectors (and, therefore, points) aboutan axis uLike previous two methods a translation of the axis to theorigin is required firstThe method of using quaternions is based on

formula for rotation of a vector around anaxis(See also, here )


http://en.wikipedia.org/wiki/Rodrigues%27_rotation_formulahttp://garryowen.csisdmz.ul.ie/~cs4085/resources/oth8.pdf



Rodrigues Rotation Formula

A one-step method that rotates a vector around anarbitrary axis coincident with the originReminder: rotating a unit-length vector

#»s = 1#»

i + 0#»

j

through a positive angle θ gives the vector

#»

s′ = cos θ#»

i + sin θ#»

j

#»

j

#»

i1 =⇒

(cos θ, sin θ)

θ

1




Rodrigues Rotation Formula (contd.)

Task: Rotate the vector #»p aroundthe axis #»r through aθ

(assuming || #»r || = 1)project #»p onto centre axis

#»u = ( #»p · #»r ) #»r

it is #»v = #»p − #»u that we need torotate in plane normal to #»r#»v = #»p − ( #»p · #»r ) #»r

#»r #»p

p

qθ







#»u = ( #»p · #»r ) #»r


#»r

θ

#»p







#»u = ( #»p · #»r ) #»r


#»p

#»u







#»u = ( #»p · #»r ) #»r


#»p

#»u

#»v





#»v lies in defined by #»p and #»rThe vector #»s = #»r × #»p is ⊥ to thisplane; it is also ⊥ to #»v#»w has same orientation as #»s , andso is a scaled multiple of #»s ; #»w = k #»sBecause | #»r | = 1

#»w = #»r × #»p = #»s| #»w | = | #»v |

So #»v rotated through θ becomes#»v cos θ + #»w sin θ

#»p






#»w = #»r × #»p = #»s| #»w | = | #»v |


#»s

#»p






#»w = #»r × #»p = #»s| #»w | = | #»v |


#»v

#»w






#»w = #»r × #»p = #»s| #»w | = | #»v |


θ





Rotation of #»p through θ around #»r is

R(θ, #»r , #»p ) = #»u + cos θ #»v + sin θ #»w= (

#»p · #»r ) #»r + cos θ( #»p − ( #»p · #»r ) #»r ) + sin θ #»r × #»p= cos θ

#»p + ( #»p · #»r )(1− cos θ) #»r + sin θ #»r × #»p

By noting that #»v = − #»r × ( #»r × #»p ) and #»p = #»u + #»v we canwrite R(θ, #»r , #»p ) as

R(θ, #»r , #»p ) = #»u + cos θ #»v + sin θ #»w= (

#»p − #»v ) + cos θ #»v + sin θ #»r × #»p=

#»p + (cos θ − 1) #»v + sin θ #»r × #»p=

#»p + (cos θ − 1)(− #»r × ( #»r × #»p )) + sin θ #»r × #»p=

#»p + (1− cos θ) #»r × ( #»r × #»p ) + sin θ #»r × #»p(1)





Graphical realisation of expressions:

First expressionSecond, more useful, form – eqn.(1) previously

#»p

#»q





Graphical realisation of expressions:

First expressionSecond, more useful, form – eqn.(1) previously




Euler-Rodrigues Rotation Formula

Using trig. identities:sin θ = 2 cos θ2 sin

θ2

cos θ = 1− 2 sin2 θ2 ≡ 1− cos θ = 2 sin2 θ

2

With #»r unit-length (as always), and lettinga = cos θ2ωt = sin θ2

#»r = (b, c,d)t

(not too difficult to show that a2 + b2 + c2 + d2 = 1)

Can now tidy up as follows

R(θ, #»r , #»p ) = #»p + (1− cos θ)( #»r × ( #»r × #»p )) + sin θ( #»r × #»p )R(a, #»ω, #»p ) = #»p + 2( #»ω × ( #»ω × #»p )) + 2a( #»ω × #»p )




Euler-Rodrigues Rotation Formula (contd.)

We have seen that#»n = #»u × #»v

can be viewed as the #»u acting on #»v , “sending” it to thevector #»n . From this point of view #»u operates on #»v and inour case we can write

#»s = #»ω × #»p = M #»p

in terms of matrix multiplication where

M =

(0 −d cd 0 −b−c b 0

), (b, c,d)t = sin

θ

2#»r

Then #»ω × ( #»ω × #»p ) = M(M #»p ) = M2 #»p

R(a, #»ω, #»p ) = I #»p + 2M2 #»p + 2aM #»p = R #»p




Euler-Rodrigues Rotation Formula (contd.)

R(a, #»ω, #»p ) = R #»p

where

R =

a2 + b2 − c2 − d2 2(bc − ad) 2(bd + ac)2(bc + ad) a2 + c2 − b2 − d2 2(cd − ab)2(bd − ac) 2(cd + ab) a2 + d2 − b2 − c2

Note

tr(R) = 3a2 − (b2 + c2 + d2)= 4a2 − 1 = 2(2a2 − 1) + 1

= 2(2 cos2θ

2− 1) + 1

= 2 cos θ + 1 3

To verify: detR = 1 ...P. Healy (University of Limerick) CS4085 Autumn 2020–2021 12 / 15



Outline






Going the other Way

We’ve seen how to construct the rotation matrixGiven a matrix, M how do we determine its axis and angle?We can firstly confirm that it is a 3-D rotation matrix bychecking that determinant is 1To find #»u , the axis, we observe that #»u is the only vectorwhen rotated that is itself

M #»u = #»u(M− I) #»u = 0

Solving for this gives #»uRecall in 3-D, s, the sum of the main diagonal elements(the trace) of the matrix sums to s = 1 + 2 cos θ

s = M11 + M22 + M33 = 1 + 2 cos θ

θ = cos−1(s − 1)/2



xkcd’s

xkcd’s Matrix Transform


https://xkcd.com/184/

3-D Rotations (contd.)Rodrigues Rotation Formula and Quaternions Determining the Axis and Angle from the Matrix

An Alternative Take on Rotations

Computer Graphics IIgarryowen.csisdmz.ul.ie/~cs4085/resources/cs4085-lect08.pdf · 2020. 12....

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