**M**y neighbour is an half-retired entrepreneur who still runs his electric engine company. A few weekends ago, he came to me with the following physics question related with one of those engines: given a primary disk rotating at the angular speed of *ω _{0}* and a secondary disk located on the first one with a centre

*O*, a distance

_{1}*r*between both centres, a radius of

_{0}*r*and a relative angular speed of

_{1}*ω*, what is the absolute speed of a point

_{1}*M*on the periphery of the secondary disk? Since I could not make sense of the solutions given in Wikipedia, I wrote a small (and crude) R code to show the location of M and to derive an approximation to the speed… Any suggestion for improvement is welcome!

#location of the second centre A a=function(t,R=c(2,1),om=c(1,13)){ sum(R)*c(cos(om[2]*t),sin(om[2]*t))} #location of the peripheral point M b=function(t,R=c(2,1),om=c(1,13)){ a(t,R,om)+R[1]*c(cos(om[1]*t),sin(om[1]*t))} #plot of the location of M as above draw_position=function(R=c(1,2), om=c(3,1), phi=c(0,0) ){ position=function(t,index,R=c(1,2),om=c(3,1),phi=c(0,0)){ relative_position=function(){ return(list(R[index]*cos(om[index]*t+phi[index]),R[index]*sin(om[index]*t+phi[index]))) } if(index == 1){ return(relative_position() ) } else { p1=position(t,index-1,R=R,om=om,phi=phi) p2=relative_position() return(list(p1[[1]]+p2[[1]],p1[[2]]+p2[[2]])) }} tes=seq(0,2*pi,length=10**3) xy_range=c(-1,1)*sum(abs(R)) plot(0,0,pch="x",xlim=xy_range,ylim=xy_range,axes=F,xlab="",ylab="") pA=position(tes,1,R,om,phi) pB=position(tes,2,R,om,phi) lines(pB[[1]],pB[[2]],pch=19,cex=.4,col="tomato") lines(pA[[1]],pA[[2]],pch=19,cex=.2,col="blue") } draw_position(om=c(1,30)) #angle at time t the=function(t,R=c(2,1),om=c(1,13)){ bb=b(t,R=R,om=om) bb=bb/sqrt(sum(bb^2)) theta=acos(bb[1]) if (bb[2]<0) theta=2*pi-theta theta} #angular speed at time t #by very crude differenciating dthe=function(t,R=c(2,1),om=c(1,13)){ dtes=mean(diff(tes)) (the(t+dtes)-the(t-dtes))/(2*dtes)} #new plot tes=seq(0,2*pi,le=5555) plot(apply(as.matrix(tes),1,dthe,om=c(-55,2)),type="l",ylim=c(-2*pi,2*pi))

**A**ntoine Dreyer actually contributed to improve the above code from an earlier version and he also derived the (Cartesian) components of the speed for me:

and

if *O _{1}(0)* has polar coordinates

*(r*) and

_{0},φ_{0}*M(0)*has polar coordinates

*(r*) with respect to

_{1},φ_{1}*O*.

_{1}(0)