what is a coordinate transformation?
108652
Check out wikepedia here:\n\nhttp://en.wikipedia.org/wiki/Coordinate_system#Transformations
692261
You need to switch to the Projection Matrix when you are specifying the projections. \ne.g.\n glMatrixMode( GL_PROJECTION );\n glLoadIdentity();\n gluPerspective( 60.0, aspectRatio, 1.0, 30.0 );\n\n\nThe model view matrix is used for the rotation, scaling and translation stuff. So, you need to do glMatrixMode(GL_MODELVIEW) before you do your transformations.
825050
Probably you mean the classical Galilei transformation.\nThe relativistic one is the Lorentz transformation.\nGrüss Ihnen.\n\nTh
864391
there are two coordinate systems, absolute and relative(INCREMENTAL), in absolute system you get ORIGIN POINT 0,0(x,y) position(it always stays the same) and everything is counted from there, in incremental system the ORIGIN POINT 0,0(x,y) is always changes(determined by x,y values that are given)\n\nex: absolute origin point x=0,y=0,(origin point x0,y0 never changes)  if you want to go 1 on x and 1 on y, x would be =1 and y=1, if you want to go again  2 on x and 2 on y, x=2,y=2.\n\nincremental: if you start with origin x=0,y=0, and you want to go 1 on x and 1 on y, then x=1, y=1(origin point changes to x1,y1), then if you want to go again 2 on x and 2 on y, then x=3, y=3...\n\nso relative coordinates means that new coordinates are related to old coordinates
170823
It's any system of axes along which something can be measured and expressed.  It can be 1-, 2-, 3- or more-dimensional.  The axes are usually all perpendicular to each other (which is hard to visualize in more than 3 dimensions) but don't have to be.  A frequent use of 3-D coordinate frames is to express a location, for example: "I live at 6th Avenue and 34th St. on the 100th floor," or "Starting at the old apple tree, take 17 paces North, 25 paces East, and dig down 50 feet."  Or a size: "Packages must be no larger than 5 feet long, 3 feet wide and 2 feet deep."
52711
shift or change\nfixed or immobile\nvector
949909
So Im guessing your have a transformation matrix.  \n\nYou have an initial state specified by matrix [A] and the linear transform specified by matrix [T].\n\nTo find the state after iteration 1 you multiply\n[A]x[T]=[y1]\n\nIf [y1] was your initial matrix you would multiply it by [T] to get the state after iteration.\n[y1]x[T]=[y2]\nby substitution\n([A] x [T]) x [T] = [y2]\nor \n[A] x [T]^2 = [y2]\n\nso what is T at the nth iteration?\n[A] x [T]^n = [yn]\n\nso what is the limit as n goes to inifinity?\n[A] x lim([T]^n,n->inf) = [y_inf]\n\nNumerical results break down, consider a rotation.  The limit as it goes to infinity is undefined.\nT(1,1)=sin(x)\nT(1,2)=cos(x)\nT(2,1)=-1*cos(x)\nT(2,2)=sin(x)\n\nWhen you square it you get\nT2(1,1)=sin(x)^2-cos(x)^2 = -cos(2*x)\nT2(1,2)=2*cos(x)*sin(x) = sin(2*x)\nT2(2,1)=-2*cos(x)*sin(x) = -sin(2*x)\nT2(2,2)=sin(x)^2-cos(x)^2 = -cos(2*x)\n\nWhen you cube it you get\nT3(1,1)=-sin(3*x)\nT3(1,2)=-cos(3*x)\nT3(2,1)=cos(3*x)\nT3(2,2)=-sin(3*x)\n\nWhen you raise it to the power of 100 (come on MatLab, dont die on me now, darnit) ... ahem.  When you raise it to the power of 10 (?)  aha\nT10(1,1)=-cos(10*x)\nT10(1,2)=sin(10*x)\nT10(2,1)=-sin(10*x)\nT10(2,2)=-cos(10*x)\n\nSee how that works?  Not bad huh?\n\nFor the pure rotation it never ends.\nWhat about a shrinking, a scaling by a matrix whose norm is less than 1? (note if you use 0.9, the norm is 1.8.)\n\nT(i,j)=0.45\nyields\nT2(i,j)=0.4050\nT3(i,j)=0.3645\nT10(i,j)=0.1743\nT1000(i,j)=0.1328*10^-4\n\nIt disappears to zero.\n\nintuitively you can also figure out that for a scaling by a matrix whose norm is larger than one the transform eventually goes to infinity as the limit is taken to infinity.\n\nyou could make a non-trigonometric rotation, like the sin(1/x)  Thats a fun expression.\n\nT(i,i)=sin(1/x) \nT(1,2)=cos(1/x)\nT(2,1)=-1*cos(1/x)\n\nAs x goes to infinity what happens?\nit becomes a reflection\nT_inf(i,i)=0\nT_inf(1,2)=1\nT_inf(2,1)=-1\n\nIm digressing.  Its a fun field with interesting applications.  Weather forecasting, deformation of solids, heat transfer, optimization, all use linear transformations to do their work.\n\nHave fun.
82197
You must first of all fix a system of coordinates
364344
emmmmmm\n\nd. its not a fuction of x in terms of y explicitly. It needs to be rearranged
664950
In the polar coordinate system r is always positive. It´s the distance from the origin to the point. \n\nThe transformation to rectangular (or rather carthesian) coordinates is as follows:\n\nx = r cos (theta)\ny = r sin (theta)\n\nSo assuming (r,theta) = (5,52 deg) I find\n\nx = 5 cos (52 deg) = 3.08\ny = 5 sin (52 deg) = 3.94 (both rounded off to two decimals)
483717
This is one that people base their script on.\nhttp://www.geodesicsolutions.com/products/index.htm
979191
When converting from rectangular coordinates to polar coordinated, remember the following.\n\n@ = Theta\nx = r*cos(@)\ny =r*sin(@)\n\nr^2 = x^2 +y^2\n\ntan(@) = y/x\n\nso your equation\nx^2 +y^2 =36 becomes\nr^2 = 36   OR\nr =6 (which is a circle with radius 6)
658659
DO YOU REFER TO CLASSICAL MECHANICS?\n\nIn classical Mechanics it may be useful to study changes from some generalized coordinates q_i and associated gen. momenta p_i to some other general coordinates Q_i and associated gen. momenta P_i. Most textbooks define a Transformation \n\n(q,p)->(Q(q,p,t),P(q,p,t)) (1) (where q is the "vector" of all q_i, etc.) as\n\nCanonical in the extended sense if and only if:\n\nFor any function H(p,q,t) fulfilling\n\ndq_i/dt=dH/dp_i and dp_i/dt= -dH/dq_i (partial derivatives on the right side)\n\nthere exists a Hamilton-function K(Q,P,t) full filling:\n\ndQ_i/dt=dK/dP_i and dP_i/dt= -dK/dQ_i (partial derivatives on the right side)\n\nIn other words: For any Hamiltonfct. of the p,q - Space there must exist a Hamiltonfct. of the P,Q- Space, (because Hamiltonfuncions of Systems  fulfill the relations above). Then and only then the transformation (1) is called canonical in ext. sense, because the form of Hamilton's equations is conserved.\n\nABOUT THE OTHER TRANSF. I DON'T KNOW ANYTHING. SEE THE LINK BELOW.
806989
The easiest way I see to do this is to rotate it first so its axis of symmetry is vertical (or horizontal), do the calculations, and rotate it back.\n\nThe transformation for rotation is best described as\nx → xcosθ - ysinθ\ny → xsinθ + ycosθ where θ is the angle of rotation\n\nSo rotate by 45° anticlockwise whence axis becomes vertical:\n\n(10, 0) \nx → xcos45° - ysin45° = 10cos45° - 0 = 5√2\ny → xsin45° + ycos45° = 10sin45° = 5√2\n\n(-2,0)  \nx → xcos45° - ysin45° = -2cos45° = -√2\ny → xsin45° + ycos45° = -2sin45° = -√2\n\n(12,4)  \nx → x cos45° - ysin45° = 12cos45° - 4sin45° = 4√2\ny → xsin45° + ycos45° = 12sin45° + 4cos45° = 8√2\n\nSo now we find the equation of the parabola with a vertical exis through (5√2, 5√2), (-2√2, -2√2) and (4√2, 8√2)\n\nWell the parabola is of the form y = ax² + bx + c\n\nWhen x = 5√2, y = 5√2\nSo 50a + 5√2b + c = 5√2 ................. (1)\n\nWhen x = -√2, y = -√2\nSo 2a - √2b + c = -√2................. ...... (2)\n\nAnd when x = 4√2, y = 8√2\nSo 32a + 4√2b + c = 8√2 ................. (3)\n\nEquation (1) + 5 x equation (2)\n\n60a + 6c = 0\nie 10a + c = 0 .................... ........... (4)\n\nEquation (3) + 4 x equation (2)\n\n40a + 5c = 4√2 ................ ........... (5)\n\nEquation (5) - 5 x equation (4)\n\n-10a = 4√2\na = -0.4√2\nThus c = 4√2\nand b = 4.2\n\nSo the rotated parabola is y = -0.4√2x² + 4.2x + 4√2  ← ***\n\nNow we need to unrotate this parabola and to do that we need to rotate it BACK by an angle of 45° ie an angle of -45°\n\nUnder this rototation\n\nx → xcos(-45°) - ysin(-45°) = x/√2 + y/√2\ny → xsin(-45°) + ycos(-45°) = -x/√2 + y/√2\n\nNow y = -0.4√2x² + 4.2x + 4√2\nThus -x/√2 + y/√2 = -0.4√2(x/√2 + y/√2)² + 4.2(x/√2 + y/√2) + 4√2\n= -0.4√2(x²/2 + 2xy/2 + y²/2) + 4.2x/√2 + 4.2y/√2 + 4√2\n= -0.2√2x² - 0.4√2xy - 0.2√2y² + 4.2x/√2 + 4.2y/√2 + 4√2\nMultiply by √2\n-x + y = -0.4x² - 0.8xy - 0.4y² + 4.2x + 4.2y + 8\n\nMultiply by 10 and collect terms:\n\n4x² + 8xy + 4y² - 52x - 32y - 80 = 0\nie x² + 2xy + y² - 13x - 8y - 20 = 0\n\nno sorry made another .... whoops!!!\nIt makes no sense ... it is correct up to line marked ← ***
768242
it means that as a (linear) transformation from the plane to the plane it reverses the orientation.\nthe way we calculate the area, is really a signed area, in the sense that it takes into consideration the orientation of the 2 vectors.\nit is a bit subtle and that is why most teachers do not really explain it.\n'
982451
in a graph, the x coordinate defines distance right or left.\n\nthe y coordinate defines distance up or down.
499636
Look here:\nhttp://www.posc.org/Specifications/Epicentre_V30/DataModel/LogicalDictionary/StandardValues/projected_2d_coordinate_system.html\n\n\nhttp://www.ldeo.columbia.edu/res/pi/MB-System/html/mbmosaic.html