Computer Vision Project-1

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1 University of Utah, School Of Computing Computer Vision Project- Singla, Sumedha ( February, 205 Theoretical Problems. Pinhole Camera (a A straight line in the world space is projected onto a straight line at the image plane. Prove by geometric consideration (qualitative explanation via reasoning. Assume perspective projection Refer the figure. Consider a fronto-parallel plane (π 0 (ie. no depth difference between them in the world space. Consider a line on the plane passing through points P and Q as shown in figure. The 3-d points on the world plane are projected on to the 2d image plane (π as P and Q. They are inverted and we need to prove that they are also forming a straight line. Consider the triangle POQ and P OQ. P, O, P are co-linear points as the pin hole camera keeps the object point and image point in a straight line. We know that P (x,y,z in world space is P (x, y in image space with relation x = mx y = my Where m = f z 0

2 Figure.: Weak perspective projection: A straight line in the world space is projected onto a straight line at the image plane Using this we can say that OP = m OP, OQ = m OQ and P Q = m P Q. Hence the triangles POQ and P OQ are similar hence, P and Q also lies on a straight line. (bshow that, in the pinhole camera model, three co-linear points in 3-D space are imaged into three co-linear points on the image plane (show via a formal solution. Consider 3 co-linear points in the world space all lying on a plane with coordinates P (x, y, z, Q (x 2, y 2, z 2 and R some where on the line joining P and Q, So R = (- tp + tq for any t [0, ]. Their corresponding image coordinates are P (x, y, Q (x 2, y 2 and R. From the equations of perspective projection we have We can write z 3 = ( tz + tz 2 R = R = P = f z P Q = f z 2 Q R = f z 3 R f ( tz + tz 2 ( tp + tq ( tz f tz 2 f P + Q ( tz + tz 2 z ( tz + tz 2 z 2 R = ( tz P tz 2 + Q ( tz + tz 2 ( tz + tz 2 2

3 Substituting s = tz 2 ( tz +tz 2 Hence P, R and Q are also co-linear. R = ( sp + sq.2 Perspective Projection (a Prove geometrically that the projections of two parallel lines lying in some plane Π appear to converge on a horizon line H formed by the intersection of the image plane with the plane parallel to Π and passing through the pinhole. Figure.2: Perspective Projection Refer the figure.2. Let us consider two parallel lines l and l 2 lying in the plane Π (in Blue color and also consider a line l 0 passing through the pinhole that is parallel to l and l 2.The lines l 0 and l dene a plane Π (in yellow colorand the lines l 0 and l 2 define a second plane l 2 (in orange color. Now consider the image plane I (in shaded red color. The plane Π intersects the image plane I as a line δ on image plane I (in Red Color. The plane Π 2 intersects the image plane I as a line δ 2 on image plane I (in Red Color. These two lines (δ, δ 2 intersect at the point p where l 0 intersects I. This point is the vanishing point associated with the family of lines parallel to l 0, and the projection of any line in the family appears to converge on it. 3

4 Now let us consider two other parallel lines l and l 2 (in blue color lying on plane Π and define as before the corresponding line l 0 and vanishing point p. The lines l 0 and l 0 line in a plane parallel to Π that intersects the image plane along a line H passing through p and p. This is the horizon line, and any two parallel lines in Π appears to intersect on it. They appear to converge there since any image point above the horizon is associated with a ray issued from the pinhole and pointing away from Π. Horizon points correspond to rays parallel to Π and points in that plane located at an infinite distance from the pinhole. (b Prove the same result algebraically using the perspective projection equation. You can assume for simplicity that the plane Π is orthogonal to the image plane (I. The equations of perspective projection for relating world coordinate system P(x,y,z to image coordinate P (x, y are as follow: x = f x z y = f y z Since its given the image plane I is orthogonal to the plane Π we can consider the equation for plane Π be y = c. Π : y = c Consider a line l in this plane. l : ax + bz = d P point on this line projects onto the image point defined by x = f x z = f ( d bz az y = f y z = f c z This is a parametric representation of the image δ of the line l with zas the parameter. Now consider the situation as z. x = f ( b a y = 0 This is a point on the x axis of the image plane and is actually the vanishing point for all parallel lines with the slope b/a in the plane Π. If we repeat this process for another line in the plane Π, and connect the two vanishing points with a line, we will have the horizon line H in the image. 4

5 .3 Coordinates of Optical Center Let O denote the homogeneous coordinate vector of the optical center of a camera in some reference frame, and let M denote the corresponding perspective projection matrix. Show that MO = 0. As per the extrinsic parameters of the camera we know that the perspective projection matrix M is M = K(Rt M = K ( C W R C O W The coordinate vector O of the cameras optical center in the world coordinate system is ( W O O = C Calculating MO MO = K ( C W R C (W O O C W = K[ C W R W O C + C O W ] Now if we rotate the camera coordinate to world coordinate C W R vanishes and we are left with Hence Proved. MO = K[ C O W + C O W ] MO = 0.4 Depth of Field (a An interesting and desirable property of the pinhole camera is the infinite depth of focus. Give an intuitive explanation. In a pin hole camera for every point in the world there is a ray of light joining the point passing through the pin hole and reaching the image screen. For each such ray as shown in image, it doesn t matter were the point lies on the ray, the final point will always be P on the image. Therefore every P independent of the distance z creates a sharp picture in the image plane. The depth of field therefore is infinite. (b Consider a camera equipped with a thin lens, with its image plane at position z and the plane of scene points in focus at position z o. What is the size of the blur circle 5

6 c obtained by imaging a point located at position z o + δz on the optical axis? Figure.3: Depth of Field Refer figure.3. Here we need to find c. Using the similar triangles we have Using the thin lens equation Where f is the focal length. solving further Using similar triangles we have C δz = d z o + δz z o z = f z o z o = f z o + f c C = z o z o Substituting the values c = C z o z o c d f = δz z o + δz z o + f (c Use this result to derive the equation for Z 0 + similarly to the way Z 0 is derived in the chapter 2 slides on Lenses, using the thin lens assumption. Discuss the relationship of lens diameter d and object distance Z 0 to the depth-of-field Z

7 Figure.4: thin Lense Refer figure.4. We will first consider the two similar triangles shown in yellow color in figure. We get d From the above question we have Z + o = b Z o Z o Z o = f Z o + f We are solving for Z + o. Putting the value of Z o we get Now Z + o = d b f Z o Z o + f Z + 0 = Z+ o Z o Substituting the value Z + 0 = d b f Z o Z o + f Z o Z + 0 ( f d = Z b o Z o + f 7