I have problems in camera calibration. I need tutorial on"vanishing points geometric mathematical formulation"
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The Geometry of Perspective Drawing on the Computer\nhttp://www.math.utah.edu/~treiberg/Perspect/Perspect.htm
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There is several way to solve the problem, but I think it is easies, to find the parametric representation, and then use, that the scalar-product of the orientations-vectors is zero, if the line are perpendicular, and the orientations-vectors is linear depended if parallel.\n\nL1:  y = -½x-3 => (x,y)= (1,-½)*t+(0,-3)\n4x-2y+5 = 0 <=> y=2x+5/2=>\nL2: (x,y) = (1,2)*t + (0+5/2)\n\nbecause (1,-½)*(1,2)=1-1=0 the lines are perpendicular
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http://mathworld.wolfram.com/PoincareConjecture.html
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The Pythagorean Theorem definitely: a^2+b^2=c^2 \n\nPoint slope form: y - y1 = m(x-x1) \nStandard y equation: y=mx+b\n...Sorry, cant think of any more. \nSee, the rest of the equations that i could give you are not the ones EVERYONE should know :)
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Let the two foci be at (a, b) and (c, d) which we assume are distinct -- if they are identical then you have a circle and a much easier problem. Afterwards we can check the result to see if relaxing that restriction would introduce a division by zero.\n\nLet the sum of the distances from the two foci (in the case of a circle, the diameter) be s.\n\nLet theta be the angle from (a, b) following the usual convention that an angle of zero goes from (a, b) to (a + infinity, b) and theta increases counterclockwise.\n\nThe ray going out from (a, b) in the angle theta direction can be parameterized as the curve\n\n(x(t), y(t)) = (a + t cos theta, b + t sin theta) for t >= 0\n\nWe want to find t such that (x(t), y(t)) satisfies that the sum of its distances from the two foci is s:\n\nThe distance from (a, b) is\n\nsqrt((x(t) - a)^2 + (y(t) - b)^2)\n = sqrt((t cos theta)^2 + (t sin theta)^2)\n = sqrt( t^2 (cos^2 theta + sin^2 theta) )\n = sqrt( t^2 )\n = t\n\nImagine that. :-)  (We don't need to say the absolute value of t because we know t >= 0).\n\nSo we want to find t such that\n\nt + sqrt((a + t cos theta - c)^2 + (b + t sin theta - d)^2) = s\n\nsqrt((a + t cos theta - c)^2 + (b + t sin theta - d)^2) = s - t\n\n(a + t cos theta - c)^2 + (b + t sin theta - d)^2 = (s - t)^2\n\nt^2 cos^2 theta + 2 (a - c) t cos theta + (a - c)^2 + t^2 sin^2 theta + 2 (b - d) t sin theta + (b - d)^2 = s^2 - 2 s t + t^2\n\nWell look at that, on the left hand side the quadratic term in t is t^2 (cos^2 theta + sin^2 theta) = t^2 whic is also the quadtratic term in t on the right.  So we can drop those and we get\n\nt (2 (a - c) cos theta + 2 (b - d) sin theta + 2 s) = s^2 - (a - c)^2 - (b - d)^2\n\nIf s is at least the distance between the two foci then the right hand side is >= 0 because it is s^2 minus the square of the distance between the two foci.\n\nSo let T = (s^2 - (a - c)^2 - (b - d)^2) / (2 (a - c) cos theta + 2 (b - d) sin theta + 2 s)\n\n(assuming the denonimator is nonzero) and the point you want is\n\n(x(T), y(T)) = (a + T cos theta, b + T sin theta)
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Didn't got the question do you want the solution for all the problems of goodman's Introduction to Fourier optics book? Well I do not think I have them any more if you want a specific problem I can help you. I can even try to get a chapter solution for you. Just tell me each one, I can try to get it for you, I do not garantee but I think I may have a way to got them just send me an email (I have a yahoo email). If you have the solutions and you don't understand tell me each problem I can do it in more detail probably, not garantee at 100% but at a good 90% .
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D'un point de vue mathématique, calculer la vitesse de la balle revient à résoudre un système de n équations différentielles avec conditions initiales. C'est un peu laborieux, donc dans la pratique on utilise effectivement un radar à effet Doppler, le plus souvent sous forme de canon.\n\n Ce radar est pourvu d'un émetteur à radio-fréquences (dans la gamme des micro-ondes) dont les ondes sont réfléchies par la balle en déplacement. Le délai de retour de l'onde est proportionnel à la vitesse de la balle. Ce délai est amplifié, filtré et analysé par une puce DSP qui pratique une correction d'erreurs et envoie les données analogiques à l'afficheur. \n\nA noter que les joueurs amateurs peuvent eux aussi chronométrer leur coup droit, grâce à un module de type Speed Trac qui s'accroche sur le filet. Ludique et pas trop cher.
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You can write the plain eqn in two forms.\n1) point + a(direction vector 1) + b(direction vector2)\n--You can obtain 'direction vector 1' by subtracting two of the points and 'direction vector2' by subtracting a different two points.\n'a' and 'b' are variable parameters which define all the points along the vector..\n\n2) take the cross product of 'direction vector 1' and 'direction vector 2' to get the normal to the plane..as already mentioned above.
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sphere points: A, B, C as vectors from sphere center O.\n\ncompute:\n\nX = A x B\nY = A x C\n\n\nU = X x (A + B)\nV = Y x (A + C)\n\ncenter points:\n\nP = ± U x V\n\n(if necessary normalize P to sphere radius)\n\n--------------\n\ncross product: A x B\n\nsketch:\nhttp://static.flickr.com/48/129027255_d097e01a2d_o.jpg\n\nfurther reading: spherical Voronoi Diagrams, projective geometry\n\nthanks Neodiogenes. my solution was incomplete. i corrected it.
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You can describe rotations and the effect they have on each point in the figure mathematically by means of simple trigonometric equations which can be condensed into a matrix. Imagine (or better, draw) a point on the plane in coordinate system X0,Y0, located at X0=1, Y0=0. Draw a line to the point from the origin. Now draw another coordinate system X1,Y1 with the same origin as X0,Y0 but rotated CW by an angle Theta. (Make Theta small, say 30 degrees.) What are the coordinates of our point in the new system? The answer is a function of Theta. We find that the point is above the X1 axis and nearer the Y1 axis. Specifically, X1 = X0*cosine(Theta) and Y1 = X0*sine(Theta). \nNow if you repeat this with a new point at X0=0, Y0=1, you find that X1 = -Y0*sine(Theta) and Y1 = Y0*cosine(Theta).\nNow the good part. You can put these four equations together to find X1,Y1 for any value of X0,Y0 (not just points located on one axis or the other). The result is\nX1=X0*cosine(Theta) - Y0*sine(Theta)\nY1=X0*sine(Theta) + Y0*cosine(Theta)\nNext you form a rotation matrix R from the Theta-dependent parts of these equations:\nR=[cosine(Theta)___ -sine(Theta)]\n__[sine(Theta) ___ cosine(Theta)] \nNow when you multiply any vector P0, representing any point in X0,Y0 coordinates, by this matrix R using the rules of matrix multiplication, you are actually using the above two equations, and the result is a vector P1 which is the same point represented in X1,Y1 coordinates. Expressed compactly, P1 = R*P0. P0 is said to be transformed by R to P1.\nOf course, this is a 2-dimensional transformation, and R is a 2-by-2 matrix. Sophisticated 3-D computer graphics use a 3-by-3 matrix implementing very similar equations to do 3-D transformations. This is how you can see convincing 3-D models of objects in rotation.
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Rather, by drawing a ray diagram you will see tt since the object is placed BETWEEN the focal point and the 'second focal point' (which is 2 x the focal length from the verticle), the image formed will be real, magnified and inverted.\n\nYou may want to check out the other variations for objects placed BEFORE the focal point, AT the focal point, AT 2 focal points away, or BEYOND 2 focal points as well.\n\nI dunno of any equation but i doubt there are any. the answer to ur qn is more of how does the object looks like. and to find out the size of the image it very much depends on the lens used. and perhaps you can apply some trigo.
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Try these;\nhttp://www.mathleague.com/help/geometry/geometry.htm\nhttp://www.coolmath.com/home.htm\nhttp://www.ics.uci.edu/~eppstein/geom.html
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from the edge of the object.. put your compass see to it, it is zero. then mark the 50th degrees. remove your compass then draw a line from the center and put a line to it. then do the same. but your compass is on the line you draw.
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Hey, here's the setup for these types of problems.\nYou want the equation in the form of y = mx + b\nsince we know that x = 2, y = -7, and m = -1/5, we can plug these values into that equation to find the equation of the line that passes through the desired point. SO:\n-7 = (-1/5)(2) + b\n-7 = (-2/5) + b\nb = (-7)+(2/5)\nb = -33/5\nso the equation for the line is:\ny = (-1/5)x - (33/5)\n\nhope that helps!!
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This site may have some useful information for you on how to use the program.\n\nhttp://www.digitalreview.ca/cams/NikonCapture4.shtml
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Because the focal point is half the distance of the radius of curvature, the focal point is 4 cm.\n\n1/f = [1/d(image)]  + [(1/d(object)]\n\n1/4 = 1/17 + 1/d(image)\n\n1/d(image) = 1/4 - 1/17 = 17/68 - 4/68 = 13/68\n\nd(image) = 68/13 = 5.23 cm
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I think I get it.  You're given the two circles (which I assume do not intersect and one is not inside the other) and then you have to come up with the equations for two tangential lines to both of these circles.  \n\nOnce you have the two lines this becomes a linear inequality system.  What you're asking is if the point is in the area bounded by one line on top and the other line on the bottom, so write these two lines as two inequalities.   Example: \n\ny1 < mx1 + b\n\ny2 > mx2 + b\n\nI think this is the format, it's been a while since I did stuff like this.  You can graph it by shading the area between the lines.\n\nAnyway, once you have this it's pretty easy: if a point satisfies both inequalities then it's inside the area.  If not, then it's outside.\n\nOkay, how to draw the tangent lines?  Well, this is a little trick to explain, it would be easier if I could graph it.  But the thing to remember is that by definition a tangent line is perpendicular to the circle at that point.  Thus if you draw a line connecting the centers of the two circles, and then draw one tangent line to the circles, then draw a line connecting the center of each circle to the tangent point on each circle, you will see that you have two similar right triangles that share the same angle.\n\nAt this point I don't think you need to use trignometry to find the equations of the tangent line ... but my mind is slipping a bit at this point and I need a moment to think it out.\n\nupdate:  I found the solution to finding the tangent to 2 circles knowing only the center points and the radii, but I'd have to draw you a picture to show you how.  It's complicated, as it involves a bunch of similar right triangles.