# [GiF] Simulation of a Camshaft

Posted 5 years ago
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 I helped to answer a camshaft question and thought I'd share the code here too. A camshaft is a shaft to which a cam is fastened or of which a cam forms an integral part. The goal is to make the valve tap the cam throughout the course and produce an image similar to the one below.Here is the cam provided by the OP. r1 = 15; r2 = 8; c = 50; ? = ArcSin[(r1 - r2)/c] // N; l1 = (2 ? + ?) r1; l2 = (? - 2 ?) r2; l3 = 2 c Cos[?]; L = 2 c Cos[?] + (2 ? + ?) r1 + (? - 2 ?) r2; cam = {Circle[{0, 0}, r1, {Pi/2 - ?, ((3*Pi))/2 + ?}], Circle[{c, 0}, r2, {Pi/2 - ?, -(Pi/2) + ?}], Line[{{Cos[Pi/2 - ?]*r1, Sin[Pi/2 - ?]*r1}, {Cos[Pi/2 - ?]*r2 + c, Sin[Pi/2 - ?]*r2}}], Line[{{Cos[((3*Pi))/2 + ?]*r1, Sin[((3*Pi))/2 + ?]*r1}, {Cos[-(Pi/2) + ?]*r2 + c, Sin[-(Pi/2) + ?]*r2}}], PointSize[0.03], Point[{0, 0}]}; g = Graphics[cam]; rValve = 4; posição = 15 + rValve; valve = Graphics[ { Red, Thickness[0.008], Circle[{0, posição}, rValve, {0, -Pi}], Line[{{-rValve, posição}, {-rValve, posição + 20}}], Line[{{rValve, posição}, {rValve, posição + 20}}], Line[{{-rValve, posição + 20}, {rValve, posição + 20}}] } ]; Show[g, valve] To solve this, RegionDistance can be useful here. cambd = RegionUnion[ Circle[{0, 0}, r1, {Pi/2 - ?, ((3*Pi))/2 + ?}], Circle[{c, 0}, r2, {-(Pi/2) + ?, Pi/2 - ?}], Line[{{Cos[Pi/2 - ?]*r1, Sin[Pi/2 - ?]*r1}, {Cos[Pi/2 - ?]*r2 + c, Sin[Pi/2 - ?]*r2}}], Line[{{Cos[((3*Pi))/2 + ?]*r1, Sin[((3*Pi))/2 + ?]*r1}, {Cos[-(Pi/2) + ?]*r2 + c, Sin[-(Pi/2) + ?]*r2}}] ]; posiçãoVal[?_?NumericQ] := With[{cambd? = TransformedRegion[cambd, RotationTransform[?]]}, y /. Quiet[FindRoot[RegionDistance[cambd?, {0, y}] == rValve, {y, 60}]] ] Table[posiçãoVal[?], {?, 0, 2? - ?/6, ?/6}]  {19., 20.4853, 30.8282, 62., 30.8282, 20.4853, 19., 19., 19., 19., 19., 19.}  frames = Table[ Graphics[{ GeometricTransformation[cam, RotationMatrix[?]], posição = posiçãoVal[?]; {Red, Thickness[0.008], Circle[{0, posição}, rValve = 4, {-Pi, 0}], Line[{{-rValve, posição}, {-rValve, posição + 20}}], Line[{{rValve, posição}, {rValve, posição + 20}}], Line[{{-rValve, posição + 20}, {rValve, posição + 20}}]} }, PlotRange -> {{-59, 59}, {-59, 83}} ], {?, 0, 2? - ?/12, ?/12} ]; Export["Desktop/cam.gif", frames]; Here's a plot of the valve height relative to the cam center point: Plot[posiçãoVal[?], {?, 0, 2?}, PlotRange -> {15, 66}, PerformanceGoal -> "Speed"] In polar coordinates, we can see we're indeed tracing the isocurve of distance 4 from the cam: Show[ PolarPlot[posiçãoVal[?+?/2], {?, 0, 2?}, PolarAxes -> {True, False}, PolarTicks -> {"Degrees", Automatic}, PerformanceGoal -> "Speed"], Graphics[{Red, Thick, cam}] ] 
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Posted 5 years ago
 Thanks for sharing! Might be useful at some point! Very courageous of using "çã" in a function name!
Posted 5 years ago
 Ha, thanks! I just copied the variable name OP used. It looks like it translates to position in Portuguese.
 Sorry about that! I've added the code from Stack Exchange to this post.As for the speed and memory consumption, generating the plot takes me about 94 seconds and uses about 12MB of RAM: AbsoluteTiming[MaxMemoryUsed[Plot[posiçãoVal[?], {?, 0, 2?}, PlotRange -> {15, 66}]]]  {94.1281, 11679216}  Adding the option PerformanceGoal -> "Speed" improves things in a noticeable way and does not seem to make the plots appear any different. I've gone ahead and added that option in the post. On my machine, it's now about 18 seconds and 3MB of RAM: AbsoluteTiming[MaxMemoryUsed[Plot[posiçãoVal[?], {?, 0, 2?}, PlotRange -> {15, 66}, PerformanceGoal -> "Speed"]]]  {18.4705, 2907720}  Let me know if you're still having issues.