Intro. I Motivation (Pose est )Newton's method I Iterative method Exercise 0: Pose estimation image correspondence a)What are the input and output of a pose mage at Image at estimation algorithm? Time t=to Time t=t1 b) How many images are 5 enough for pose estimation =2 c)Estimate the correspondences and fill in u=0 1 2 3 4 5 u=0 1 23 4 5 the blanks Time q1 q3 q4 uV|u|vu「「u|V t=t t=t123 Pose estimation vo.a
Intro. | Motivation | Pose est.| Newton’s method | Iterative method Exercise 0:Pose estimation & image correspondence a) What are the input and output of a pose estimation algorithm? b) How many images are enough for pose estimation? c) Estimate the correspondences and Fill in the blanks Pose estimation V0.a 16 R,T qi=1,t=0 qi=2,t=0 qi=2,t=t1 qi=1,t=t1 Image at Time t=t0 Image at Time t=t1 u=0 1 2 3 4 5 V 5 4 3 2 1 0 Time t q1 q2 q3 q4 u v u v u v u v t=t0 2 4 3 1 3 3 t=t1 2 3 u=0 1 2 3 4 5 q4 q3
Intro. I Motivation I Pose estqNewton's method) Iterative method Exercise 1: Newton s method (An iterative method First guess, Xo=2 f=df(×x)/d An iterative method for finding f(×)=-f(×0)+f(x0)*(X-×)≈0 =d(x2-5)=2X the solution of a non -inear f(xo)+f(xo)*(x-xo) So if Xo=2 system 0-f(x)(x)≈(xx f(x0)=22 Exercise 1 Find sqrt(5), same as find the non -linear function 01x25)(2*×)=AxNx f(x)=x25=0 [0-(x25/(2*x)=△ Taylor series(by definition) Take x=2,[0-(225)/(2*2)=△x f(×)=f(×0)+(X0)*(xx0)=0 z=△x f()=2Xo,So Since△X≈(x-×) f(x)=fx0)+2*x*(Xx)=0 X=new guess, xo=old_guess z≈x-2,x≈2.25 That means the next guess is xe 2.25 Exercise: Complete the steps to ohttp://ww.ugrad.mathubcca/coursedoc/math100/n find the solution otes/approx/newton. html For your reference Pose estimation Vo.a sart(5=2.2360679(by calculator)
Intro. | Motivation | Pose est.| Newton’s method | Iterative method Exercise 1: Newton’s method (An itervative method ) • An iterative method for finding the solution of a non-linear system • Exercise 1.Find sqrt(5), same as find the non-linear function. – f(x)=x2 -5=0 – Taylor series (by definition) – f(x)=f(x0 )+f’(x0 )*(x-x0 )=0 – f’(x0 )=2*x0 , so – f(x)=f(x0 )+2*x0*(x-x0 )=0 • First guess, x0=2. • f(x)=f(x0 )+ f’(x0 ) *(x-x0 )0 • 0 f(x0 ) + f’(x0 ) *(x-x0 ) • [0-f(x0 )]/f’(x0 ) (x-x0 ) • [0-(x0 2 -5)]/(2*x0 )= x (x-x0 ) • [0-(x0 2 -5)]/(2*x0 ) = x • Take x0=2, [0-(22 -5)]/(2*2) = x • ¼ = x • Since x (x-x0 ), • x=new guess, x0=old_guess • ¼ x-2, x 2.25 • That means the next guess is x x2.25. • Exercise: Complete the steps to find the solution. • For your reference: Pose estimation V0.a sqrt(5)=2.2360679 (by calculator) 17 ⚫http://www.ugrad.math.ubc.ca/coursedoc/math100/n otes/approx/newton.html f’=df(x)/dy =d(x2 -5)=2x So if x0=2, f’(x0 )=2*2
Intro. I Motivation I Pose est. I Newton's method I Iterative method The main idea of newton 's method We saw this formula before:fx×)=f8X)+f(X0)*(X-×)≈0-(i) From f(x)=-f(×)+f(×0)*(x)≈0 0≈f(X)+(x0)*(xX0) 0-f(X0)=f(X0)*(X-X0 [0-f(X)/(x)=△X=(XXo We can compute Ax=[O-f(xo)l/f(xol, then Since Ax= x-Xo), SO X=Xo+ AX That means: Xnew guess= Xolold guess)+A Pose estimation vo.a
Intro. | Motivation | Pose est.| Newton’s method | Iterative method The main idea of Newton's method • We saw this formula before: f(x)=f(x0 )+f’(x0 )*(x-x0 )0 -----(i) • From f(x)=f(x0 )+f’(x0 )*(x-x0 )0 • 0 f(x0 )+f’(x0 )*(x-x0 ) • 0 - f(x0 )= f’(x0 )*(x-x0 ) • [0 - f(x0 )]/ f’(x0 )=x=(x-x0 ) • We can compute x=[0 - f(x0 )]/ f’(x0 ), then • Since x=(x-x0 ), so x=x0+ x • That means: Xnew_guess = x0(old_guess) + x Pose estimation V0.a 18
Intro. I Motivation I Pose estqNewton's method) Iterative method Pose estimation in 3D o find pose(r,t), the model and one image is enough 3D object at to 1,t0 Camera center gi=2A qi=1.IN q=2t1 3D object at t=t1 Pose estimation vo.a
Intro. | Motivation | Pose est.| Newton’s method | Iterative method Pose estimation in 3D To find pose (R,T), the model and one image is enough Pose estimation V0.a 19 qi=1,t0 qi=2,t0 Camera center 3D object at t0 3D object at t=t1 qi=1,t1 qi=2,t1 •