I noticed it back when I was writing on my nifty lil MSVC++ 6.0 compiler and the DX8 SDK: I could rotate an object such that two axes canceled each other out. I read up on quaternions (of all places, the Allegro toolkit) and discovered a program that explains that quaternions prevent this "cancelling out", or Gimbal lock, but I'm quite lost as to discover why and how. Please help?
203688
Egads.  It's been quite awhile.  I remember some papers at SIGGRAPH a decade or so ago on this topic.\n\nAlright, here's a stab at the explanation.\n\nWith Euler mechanics or Euler angles, there are three reference angles: yaw, pitch, and roll. It is possible to rotate an object using these three reference angles such that two line up causing the inability to do an arbitrary rotation.  Say you roll to the right 90 degrees and this lines up the yaw axis with your pitch axis -- gimble lock results.\n\nWith quaternions, you don't have three reference angles.  You have only one plane and an arbitrary axis of rotation.  Each move creates a new plane and axis of rotation.  As such, there are no additional axes of rotation with which to collide because there is only one axis of rotation and it's position is arbitrary where as with Euler angles, the reference angles are fixed.\n\nI don't know if I've done it justice, but maybe this explanation and the references below will help you stitch together an understanding.
500349
I suspect it's something simple.  I think because of the way your arm joints work, when you release the board with your right hand, for example, the board will have a very slight horizontal inclination to the left, which will eventually cause it to turn in the air and veer to the left.\n\nIt will be interesting to hear what someone with a knowledge of biology or physics will say though.  :)
566920
Spinning flights were really designed to prevent flight deflection on tightly grouped darts. Instead of bouncing off the other dart the flight will rotate to allow another dart next to it.
481336
yes. the cartesian system itself is a mirror image.\n\nit is a mirror image about the x-axis and the y-axis. \n\nin the cartesian coordinate system:\n\nevery line has a mirror image.\n\nevery parabola has a mirror image.\n\nnow, if these have a mirror image, they must also have an algebraic equation to go with it. this is where one learns about inverse functions...\n\ngeneral descriptions here as this takes a lot of math to show...
672037
if you can picture a gyro spinning, picture it spinning out of control. or take a quarter and drop it on a table and watch it before it stops moving. it's all about speed. the rate of inertial is so grate the human eye can not see and regulate it at the same time.
437148
You haven't told us what point it's pivoting about.\nThis is a Physical Pendulum problem:\n\nT = 2π√(I/mgL)......for small angular oscillations....\n\nI = mR²  where R is the mean radius of the hoop\nL = the distance from the pivot to the center of mass of the hoop\n\nGOTTA have L............
213860
They are a set of Matricies that map one set of {x,y,z} to another set of {x',y',z'} through a rotation angle.  In oder to use them you need to know your starting set of coordiantes and the three angles through which you want to rotate.\n\nOne such set of angles that forms a perfectly good rotation basis are the Euler angles.\n\nIf you know the rotations, you can actually write the whole transformation as one matrix (usefull if you have to apply it over and over again.
369736
Given a rotating hub or wheel all the centrifugal forces are along the radii from every point on the hub to the center. Because of circular symmetry, all the these forces will cancel; one pointing up is cancelled by one pointing down. Therefore the net external force is zero; there is no way the linear motion is influenced.
266220
Although merry-go-rounds (normally) do not experience any linear motion, they do experience ANGULAR, or rotational motion. Angular equations are the same set-up as linear equations, except for their variables. Here's a table to help:\nd ==> theta (angle in radians)\nv ==> omega (velocity in radians/second)\na ==> alpha (radians/second^2)\n= = = = = = = = = = = = = = = = = = = = = = = =\nThe relationship between linear and angular motion is:\nd = theta*r\nv = omega*r\na = alpha*r\nWhere r is the distance from the center of rotation. (This relationship is not the same for other variables, such as force or momentum)
188727
because one side cancels the other since it's symmetrical
668861
If you have one point on the wheel negatively charged, and move a positively charged magnet around it, it would move. \nOr, if you want something less manual, you can motorize one wheel w/ alternationg positive and negative charges around the edges, and then have another identical wheel on an axle right next to it, so as the first wheel turned, the second one would have to turn so that the opposite charges would line up.\nI'm not sure if the second one would work, but i tried. Good luck! =D
90848
You finally have xy=0 so x=0 or y=0 (unless it is a problem of modular algebra...)\n\n----\n\nNotice that xy doesn't "cancel": you have \n45xy=148xy\n\nBring all terms on the left:\n(45-148)xy = 0 ----> -103xy = 0\n\ndivide by -103:\nxy=0\n\nby the law of cancellation, the last equation is true for\n(x=0 and y whatever) OR (x whatever and y=0)\n\nSo there are infinite solutions for the couple (x,y)\n\n\nGraphically, the solutions are given by all the points of the x and y axes
499862
Here are two (see refs.). Note that the second one is a rotating 3-D model.
543246
e, because of the Newton's 1st law: "An object at rest will remain at rest unless acted upon by an external and unbalanced force . An object in motion will remain in motion unless acted upon by an external and unbalanced force". Assuming inertial frame of reference.\nThe object can still have angular acceleration if there are two forces cancelling each other but are applied to an object at different points. \n\nIn real life there is always something that will slow down the object even in space.
384126
inverse tangent of ( y / x ) = atan(40/30) = 53.13 degrees from the horizontal.\n\nEdit: the answer is not the y-direction because if it is, then the original momentum in the x-direction would have disappeared from the system.
381693
Think of the wheels as spinning masses or revolving weights, in fact gyroscopes. \nEach wants to turn left or right while spinning but is constrained by it's axis,\nHowever both drift overtime.\nNow add another pair further back and wider apart, and this pair are not on the same line of travel.\nWith four wheels all wanting to drift the resultant direction becomes a straighter line.
180777
Some additional material to add to the excellent answer #1. In case you're not familiar with transformation matrices, this is an answer I gave to a question on rotations in graphics.\nYou 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 2x2 matrix. And rotation is about only one basis axis. To expand to a 3x3 matrix that accommodates rotation about any arbitrary axis is not difficult. The ref. is a good starting point.