Converting rotary movement to linear movement?
954062
put an arm perpindicular on the shaft, put another arm pinned to the first, and slide the second through a sleeve.\n\njust like an old fasioned steam engine turned linear movement (piston) into rotary(drive wheel
51955
rolling wheels with a stick, kick the can, etc
748663
The answer hippoterr... gave you was OK till the last step. It's always is wise to check the units and see if they make sense. His result is in sec/rev.\n\n2 x Pi x11.5 / 30 = 2.4 He left out the units in the math part, but if you include them you see that the result is in sec/rev.\n\nUnit conversion can be confusing. A better way - you know you want it to be something per sec so start with:\n\n30 meter/sec. Then to convert to rev/sec multiply by 1, in a special form.\n\nIn the form 1 rev/(2 x Pi x11.5 meters). This = 1, right? The top and bottom are equivalent. So back to \n\n30 meter/sec * 1 rev/(2 x Pi x11.5 meters) = .415 rev/sec\n\nYou understand why weight/mass wasn't needed? It's the acceleration that results from the rotation,  and  a = F/m.
803641
http://images.google.com\n\nSearch for whatever types of motion you are interested in.
775545
If you mean rotate vertically like the square is doing backflips, I recommend effect->3d->rotate, and then select "front" from position, then pick a few degrees to rotate around the x-axis by, then rotate, save, rotate, save, etc. \n\nIf you mean rotate like the square is "rolling" like a wheel, I do the same thing, but manipulate the z-axis instead.\n\nGood luck.
1087615
degrees of freedom is not like 14.6* but rather how many directions it can move.  Pin joints can hinge in one degree of freedom (X, Y, or Z) Slider can do two (combo of X, Y or Z) and ball can move 6 ways, X, Y, Z, and the rotations on those axes
34052
I do not know a preprogrammed utility to do this, but can point you in the direction that you might need to go.  See the reference below for the math and an on-line converter:
750239
In half the rotation, the direction of velocity will become perpendicular to original direction and will be equal in magnitude to the original as no force acts in  direction of motion(if the air-resistance is negligible).\nsuppose initial direction is along positive z-axis\n=>/&#92; V=V2-V1\n=>/&#92; V=100(-k^) -100(k^)=200(-k^)\n/&#92;V=delta V(change in velocity)\nk^=unit vector along z-axis(axis perpendicular to the x-y plane)
824183
just go 2 google search & write the non linear edition tutorial  \nn u will get nice tutorials
521634
There are two types of motion; translational (linear) and rotational motion.\n\nFor linear motions, the momentum (product of mass and velocity) will not change unless an external force acts on it. (Newton’s law)\n\nSimilarly for rotational motion the angular momentum (moment of linear momentum; product of linear momentum and the radius of the circular path along which the body rotates) will not change unless an external torque (Moment of force)\nacts on it. \n\nTherefore, if a body rotates about an axis with certain angular speed it will be rotating for ever unless a torque acts on it.\n\nAbout centrifugal force, there is no such force acting on the body which is in rotation. Only centripetal force acts on it.\n\nCentrifugal force is the force which is acting on the body which makes the body to rotate.\n\nIf a stone is tied to one end of a rope and if we make the stone revolve round a circle, \nthe force which pulls the stone toward the center of the circle is the centripetal force and it acts on the stone.\n\nThe centrifugal force is the force which we experience in our hand; the force which is pulling our hand toward the stone and this force is not acting on the stone.
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.
145461
since pulley size vs. rpm is linear, you can just use a simple relation. 2.5/3450 = ?/4000
998322
Lets make an enormous assumption to fill in the gaps.\nYou have a thing on a string of length L and spin it round at a certain rate of rotation. Keeping the rate of roataion what would the tangential velocity of the thing be if the string was shortened to L/2?\nIn this case the tangential velocity (= the radius x the angular velocity) will also halve.
482229
Have you considered a DC electric linear actuator? Lots of manufactures on the web such as http://www.dcactuators.com although there are others.
525309
1) Convert rpm to revolutions per second, by dividing rpm by seconds (so the minutes cancel).\n\n2)Convert revs per sec to radians per second, by multiplying rev/sec *2pi rad/rev (the revs cancel).\n\n3 Acceleration is in the form of distance/(time^2), so dividing radians per second by the time required to stop will give you the magnitude of acceleration over that time period. \n\nFor your example:\n\nrotational acceleration = \n            1000 rev/min *1 min/60 sec* 2Pi rad/rev *1/2 sec\n           =  52.4 rad/sec^2\n                 \n\nPart 2:"\n\nIf by distance you mean radial distance traveled by the tip of a spinning blade, then you know that \n\n1 revolution = 2*pi*r \n\nso to find the number of revolutions, multiply the distance by 1 rev/(2*pi*r)- the units of linear distance will cancel (see a pattern?).\n\n(distance) * 1rev/(2*pi*r) = number of revolutions
175247
Not sure what you mean by linear, but linear acceleration can be described:  acceleration = force / mass\n\nArea of circle = pi r^2\n\nVolume of cube = length x width x height\n\nHope it helps
438745
try wingfields art of mechanical motion, doubleday i think.