From c864b4bb1a5fdf95ed93daed0991de2d242f545d Mon Sep 17 00:00:00 2001 From: emiliie Date: Thu, 12 Mar 2015 16:42:50 +0100 Subject: [PATCH 1/3] circle --- report/subsections/circles.tex | 7 +++++-- 1 file changed, 5 insertions(+), 2 deletions(-) diff --git a/report/subsections/circles.tex b/report/subsections/circles.tex index 776a432..e29cd44 100644 --- a/report/subsections/circles.tex +++ b/report/subsections/circles.tex @@ -5,6 +5,10 @@ \paragraph{Characteristic circles computation} We take a point sampled on the spline. It gives us information about the sphere located on this point : C(t) the sphere center coordinates and r(t) its radius. We also have their derivatives : C'(t) and r'(t). The idea is to find the intersection between this sphere and the characteristic plane, which would give us the characteristic circle. + If a point P is on the circle, then we can write this formula : + - r'(t)r(t) = 0 +and then we calculate the center of the characteristic circle with this formula : + Cp(t) =C(t) – r'(t)r(t)/norm(C'(t))² * C'(t) \begin{figure}[H] \begin{center} @@ -17,5 +21,4 @@ After obtaining characteristic circles we sample them. Then we need to find matching points. We decided to make points sampling start in one direction of our mark. When this is done we project the mark on the new circle and sample again. Then we link each point to the point of the same index on the next circle. - -$TODO$ +%TODO From 7475c05da348fb429d20e47336917ace93b62618 Mon Sep 17 00:00:00 2001 From: emiliie Date: Thu, 12 Mar 2015 16:45:28 +0100 Subject: [PATCH 2/3] circles --- report/subsections/circles.tex | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/report/subsections/circles.tex b/report/subsections/circles.tex index e29cd44..b9365f8 100644 --- a/report/subsections/circles.tex +++ b/report/subsections/circles.tex @@ -6,9 +6,9 @@ We take a point sampled on the spline. It gives us information about the sphere located on this point : C(t) the sphere center coordinates and r(t) its radius. We also have their derivatives : C'(t) and r'(t). The idea is to find the intersection between this sphere and the characteristic plane, which would give us the characteristic circle. If a point P is on the circle, then we can write this formula : - - r'(t)r(t) = 0 + $ - r'(t)r(t) = 0$ and then we calculate the center of the characteristic circle with this formula : - Cp(t) =C(t) – r'(t)r(t)/norm(C'(t))² * C'(t) + $Cp(t) = C(t) – r'(t)r(t)/norm(C'(t))² * C'(t)$ \begin{figure}[H] \begin{center} From 3a9a24f2ccbff9146627b0e779c5568fe36c4755 Mon Sep 17 00:00:00 2001 From: Thomas FORGIONE Date: Thu, 12 Mar 2015 17:13:24 +0100 Subject: [PATCH 3/3] Tests --- Tests/Test.tex | 11 ++++++++++- 1 file changed, 10 insertions(+), 1 deletion(-) diff --git a/Tests/Test.tex b/Tests/Test.tex index 4bf9a2a..bf9ffc9 100644 --- a/Tests/Test.tex +++ b/Tests/Test.tex @@ -18,14 +18,23 @@ \part{First part : segmentation, camera calibration, skeletonization, detection and matching of keypoints} \chapter{Segmentation} \chapter{Calibration} -\section{Internal calibration} \section{External calibration} +To test the external calibration, the idea is to have two pictures of a scene +containing two chessboards : one blue and one red. We will compute the +homography that transforms the blue chessboard on the first picture to the blue +chessboard of the second picture. Then, we will apply this transformation to +the red chessboard on the first picture, and if the calibration has a good +quality, the points found should be on the red chessboard. \begin{figure}[H] \centering \includegraphics[scale=0.45]{images/externalCalibration/result.png} \caption{Test of the external calibration} \end{figure} +As you can see, the points computed are slightly displaced from the corners of +the chessboard. We can explain this by the difficulty we had to have a very +plane chessboard : the ink from the printer humidify the paper, and we were not +able to have a really plane chessboard. \chapter{Detection and matching of keypoints} \section{Detection of keypoints}