28 Horizontal angle correction using straight line detection in an equirectangular image

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1 28 Horizontal angle correction using straight line detection in an equirectangular image

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3 Abstract Horizontal angle correction using straight line detection in an equirectangular image Yuto ISHII Recently, the omnidirectional camera has been widely used as a more general one because of the development of camera technology. The 3D structure recognition using omnidirectional camera is actively performed. Conventionally, rotation correction in the omnidirectional image is performed by using the acceleration sensor output. But there are some images that can not be sufficiently corrected. In addition, there are some images in which the acceleration sensor output is missing. Therefore a method for correcting horizontal angle without using the acceleration sensor output is required. In this research, we propose a rotation correction method using horizontal lines in an omnidirectional image. The method is conducted as follows 1)straight lines in the equirectangular image are detected using Hough transform, 2)straight lines nearly vertical are selected, 4)make a line group which can be considered as the same line, 5)rotation parameter of the horizontal line is calculated by voting method for the straight line group, 6)a rotation corrected image is created by applying rotation correction on the spherical surface using the correction angle. In order to demonstrate the usefulness of the proposed method, verification experiments were conducted using the proposed method. As a result, vertical lines and horizontal lines can be estimated, and corrected images can be created using the horizontal line rotation esimation. key words equirectangular image, omnidirectional camera, horizontal correction, straight line detection ii

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9 1 1.1 [1] [2] Seon [3] [4] 1

11 Google ( 2.1(a)) RICOH THETA( 2.1(b)) 360 [7] (a) Google [5] (b) RICOH THETA S[6]

12 (a) 2.3 (b) λ ϕ R (2.1) x ( ) y ( ) 2 [ ] [ ] x R (λ + π) = y R ( ) (2.1) ϕ + π 2 2.4(a) 4

13 (b) y 2.4(c) 2.4(b) (c) (a) (b) (c) [8] 5

14 2.4 ρ θ (2.2) ρ = x cos θ + y cos θ (2.2) (x y) (2.2) θ (ρ θ) (ρ θ) θ (ρ θ) (ρ θ) OpenCV 2.5 (a) 2.5 (b) (x y z) (r θ ϕ) 2.6 (2.3) (2.4) x sin θ cos ϕ y = r sin θ sin ϕ (2.3) z cos θ 6

15 r x2 + y 2 + z 2 θ = cos 1 z ϕ cos 1 x x2 +y 2 +z 2 x2 +y 2 (2.4) x y z (R x R y R z ) θ (2.5) (2.6) (2.7) R x (θ) = 0 cos θ sin θ (2.5) 0 sin θ cos θ cos θ 0 sin θ R y (θ) = (2.6) sin θ 0 cos θ cos θ sin θ 0 R z (θ) = sin θ cos θ 0 (2.7)

16 2.5 (2.8) θ ϕ ψ (R x R y R z ) θ ϕ ψ xyz R z (ψ)r y (ϕ)r x (θ) (2.8) 2.5 [9] (x y) I(x y) 4 4 (f 11 f 12 f 44 ) (2.9) f 11 f 12 f 13 f 14 h(y 1 ) I(x y) = (h(x 1 )h(x 2 )h(x 3 )h(x 4 )) f 21 f 22 f 23 f 24 h(y 2 ) f 31 f 32 f 33 f 34 h(y 3 ) f 41 f 42 f 43 f 44 h(y 4 ) (2.9) x 1 x 2 x 3 x 4 y 1 y 2 y 3 y 4 (2.10) x 8

17 2.5 x x 1 = 1 + x x x 2 = x x x 3 = x + 1 x x 4 = x + 2 x y 1 = 1 + y y (2.10) y 2 = y y y 3 = y + 1 y y 4 = y + 2 y h(t) sinc (sinc(t) = sin(πt) πt ) (2.11) t 3 2 t ( t 1) h(t) = t t 2 8 t + 4 (1 < t 2) 0 (2 < t ) (2.11) 9

18 3 Step 1 Step 2 Step 3 Step 4 Step 3 Step 5 Step [0, 0] 10

19 Step 2 y Step 2 y α 11

20 ψ W H (3.1) px 500px (3.1) ψ ( ( y = (sin H ( tan 2πx π sin 1 ψ sin tan 1 W + α) ))) + H (3.1) cos ψ

21 3.4 ψ sin ψ ψ, cos ψ 1 (3.2) y = H π ψ sin ( 2πx W ) + α + H 2 (3.2) (x y) (x y) m (x y) (3.2) (3.2) x (3.3) m = dy ( ) 2πx dx = ψ cos W + α (3.3) (3.2) (3.3) ψ ψ α (3.4) α ( ( )) π y h α = tan 1 2 2πx Hm W (3.4) α ψ (x y) α ψ α ψ α ψ

22 3.5 α ψ α ψ

24 4 4.1 RICOH THETA S

30 4.2 (a) (b) 5 (c) 10 (d) 15 (e) 20 (g) 30 (i) 40 (k) 50 (l) 60 (m) 70 (n) 80 (o)

31 ( ) ±10 y = 100 y = 1700 ( )

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40 [1] J. Gaspar. et.al. Vision-based navigation and environmental representations with an omnidirectional camera Roboties and Automation IEEE Transactions Vol.16 6 pp [2] ITE Technical Report Vol. 39 No. 8 pp [3] Seon Ho Oh. et.al. A Great Circle Arc Detector in Equirectangular Images Proceedings of the International Conference on Computer Vision Theory and Applications Vol.1:VISAPP pp [4] Seon Ho Oh. et.al. RANSAC-based Orthogonal Vanishing Point Estimation in the Equirectangular Image Joumal of Korea Multimedia Society Vol.15 No.12 pp [5] Google Google Street View trekker/ :2017/2/11 [6] RICOH RICOH THETA :2017/2/11 [7] Regis Cosnier Small Planet androidsmallplanet :2017/2/28 [8] Vol.9 No.3 pp [9] [ ] pp

41 A xyz A.1 z z ( ) x y ( ) 2 A (x y z) (A.1) ϕ θ 33

42 ϕ = [ π π] θ = 0 x cos θ cos ϕ y = cos θ sin ϕ z sin θ (A.1) (x y z ) ψ (2.5) (x y z ) (A.2) x x x y = R x y = y cos ψ z sin ψ (A.2) z y sin ψ + z cos ψ z Φ Θ z = 0(θ = 0) (A.3) x cos ϕ y = sin ϕ cos ψ z sin ϕ sin ψ Φ (A.4) Θ (A.5) ( ) y Φ = tan 1 x = tan 1 ( sin ϕ cos ψ cos ϕ = tan 1 (cos ψ tan ϕ) Θ = sin 1 (z ) ) = sin 1 (sin ϕ sin ψ) (A.3) (A.4) (A.5) Φ x (A.4) ϕ W 1 Φ = 2πx W (A.6) x x ( tan 2πx ϕ = tan 1 W cos ψ ) (A.6) (A.6) y (A.7) H H 2 0 H 2 34

43 α ( ( y = (sin H ( tan 2πx π sin 1 ψ sin tan 1 W + α) ))) + H (A.7) cos ψ 2 35

44 B 3.4 (B.1) ( ( y = (sin H ( tan 2πx π sin 1 ψ sin tan 1 W + α) ))) + H cos ψ 2 (B.1) (B.1) (B.2) dy dx = ( cos2 ψ cos 2 ( 2πx W ( sin ψ cos ψ cos (tan 1 + α) + sin 2 ( 2πx W + α)) tan( 2πx W +α) cos ψ )) 1 sin 2 ψ sin 2 (tan 1 ( tan( 2πx W +α) cos ψ )) (B.2) (x y) (B.1) f 1 (B.2) f 2 (B.3) f 1 (ψ α) = 0 f 2 (ψ α) = 0 (B.3) f(t) = 0 t = (ψ α) T f(t) = (f 1 (t) f 2 (t)) k t k 1 36

45 (B.4) f(t k+1 ) = f(t k ) + J(t k )(t k+1 t k ) (B.4) J(t) (B.5) J(t) = [ df1 dψ df 2 dψ df 1 dα df 2 dα ] (B.5) t δ = t k+1 t k J(t k )δ = f(t k ) δ = (δ ψ δ α ) (B.6) δ δ = J(t k ) T f(t k ) (B.7) df 1 dψ δ ψ + df 1 dα δ α = f 1 (ψ k α k ) df 2 dψ δ ψ + df 2 dα δ α = f 2 (ψ k α k ) (B.6) δ ψ = δ α = df 2 dα f 1 df 1 dα f 2 J(t k ) df 2 dψ f 1 + df 1 dψ f 2 J(t k ) (B.7) (B.8) ψ α ψ k+1 = ψ k + δ ψ α k+1 = α k + δ α (B.8) 37

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22 A person recognition using color information 1110372 2011 2 13 ,,.,.,,.,.,.,.,,.,..,,,, i Abstract A person recognition using color information Tatsumo HOJI Recently, for the purpose of collection of