Impact of Slowing Spin on the Trajectory of a Baseball

Haiduke Sarafian

Pennsylvania State University
University College

Key words:  Drag Force, Magnus Effect, Time Dependent Spinning Ball, Nonlinear physics, Mathematica

Abstract

A flying baseball in the air not only is subject to gravity’s pull it is also subject to air resistance.  A spinning ball in addition to these two forces experiences a spin-dependent force.  We consider a practical scenario where the speed dependent drag force not only retards the motion it slows the spin as well.  The analysis of the kinematic and dynamic of the proposed scenario entails solving a super nonlinear coupled ODE.  Utilizing Mathematica NDSolve we solve these equations numerically.  We compile numeric and graphic lists of quantities of interests vs. their comparative counter parts of non-retarded case.

■Physics of a projectile in the air; the impact of air resistance and the spin

We consider a ball projected in the air at an arbitrary initial angle above the horizontal.  In addition to gravity, the ball encounters air resistance.  Irrespective of its speed, the drag force (the air resistance) acts in the opposite direction of the motion retarding its movement.  In Fig 1 a snapshot of a flying ball with its instantaneous velocity and the drag force are shown with (Overscript[v, ->]) and (Subscript[(Overscript[F, ->]), D]) , respectively.  In practice a batted ball also spins; it may spin backward (backspin) or forwards (topspin).  We quantify the spin by its angular velocity (Overscript[ω, ->]).  We consider a spinning ball with angular velocity vector perpendicular to the vertical plane.  Hence, a top-spanned ball orients its (Overscript[ω, ->]) parallel to the ground and inward to the vertical plane.  Conversely, a back-spun ball orients its (Overscript[ω, ->]) parallel and outward to the ground.  Figure 1 displays a top-spinning (clockwise rotating) ball.  The cross at the center of the ball represents the tail of the (Overscript[ω, ->]) arrow piercing the xz-plane.  For the chosen scenario, the spinning ball is subject to an additional force; a spin-dependent force, namely the Magnus force [2].  This force is in proportion to (Subscript[(Overscript[F, ->]), M])~ (Overscript[ω, ->])×(Overscript[v, ->]).  Accordingly, this force also lies in the vertical plane making the three active forces namely, (Overscript[w, ->]), (Subscript[(Overscript[F, ->]), D]) and (Subscript[(Overscript[F, ->]), M]) a trio coplanar.  For the given scenario the ball stays in the vertical plane, and therefore, the analysis of the problem becomes two-dimensional.

(Subscript[(Overscript[F, ->]), D])=(1/2)(Subscript[C, D])ρ A v (-(Overscript[v, ->]))
(1/(m ))(Subscript[F, D])=f(v) (v^2)
f(v)≡(1/(2 m))(Subscript[C, D])ρ A=0.0039+(0.0058/(1+(e^((v-(Subscript[v, d]))/Δ))))
(Subscript[v, d])=35and Δ=5  both in(m/s)units.

Figure 2.  Variation of the drag coefficient for a baseball vs. speed.  The lower horizontal axis is in (m/s); the upper horizontal axis is the corresponding speed in MPH.

(1/m)(Subscript[(Overscript[F, ->]), M])=B (Overscript[ω, ->])×(Overscript[v, ->])
m=143 g, B=4.1 (10^(-4))

■Formulation of the physics of a batted baseball

As we explained in the previous section a batted top(back)-spun baseball with its initial angular velocity vector parallel to the horizontal flies and stays in a vertical plane.  A vertical plane is the one that contains the initial velocity vector.  In other words, the motion of the ball occurs in a 2D space.  One such plane with a Cartesian xz coordinate is shown in Fig 1.  We utilize the active forces depicted in Fig 1 and compose the equations conducive to the analysis of the motion of the batted ball.  Utilizing Newton’s second law we write (Subscript[(Overscript[F, ->]), net])=m (Overscript[a, ->]).  This equation explicitly reads, m (Overscript[r, (Overscript[.., ->])])=(Overscript[W, ->])+(Subscript[(Overscript[F, ->]), D])+(Subscript[(Overscript[F, ->]), M]), where the over double-dots means the second order time derivative.  We substitute the forces from the previous section yielding

m (Overscript[r, (Overscript[.., ->])])= m (Overscript[g, ->])+m f(v) v (-(Overscript[v, ->]))+S (Overscript[ω, ->])×(Overscript[v, ->]).  Since the velocity stays always in the vertical plane we write (Overscript[v, ->])={(OverDot[x]),0,(OverDot[z])}.  On the other hand, as we discussed the angular velocity sustains its character, meaning, its value and its orientation stay the same.  In a right handed coordinate system shown in Fig 1, we write (Overscript[ω, ->])={0,+ω,0}.  This yields, (Overscript[ω, ->])×(Overscript[v, ->])={0,+ω,0}×{(OverDot[x]),0,(OverDot[z])}={+ω (OverDot[z]),0,-ω (OverDot[x])}.  The x and the z components of the force equation yields,

Equation of Motion

(Overscript[x, ..])=-f(v) v (OverDot[x])+B ω (OverDot[z])
(Overscript[z, ..])=-g-f(v)v (OverDot[z])-B ω (OverDot[x]) 

(Overscript[x, ..]) =-( α + (β/(1+(e^((1/Δ)((Sqrt[((OverDot[x])^2)+((OverDot[z])^2)])-(Subscript[v, d])))))))(Sqrt[((OverDot[x])^2)+((OverDot[z])^2)]) (OverDot[x])+B ω (OverDot[z])
(Overscript[z, ..]) =-g -( α + (β/(1+(e^((1/Δ)((Sqrt[((OverDot[x])^2)+((OverDot[z])^2)])-(Subscript[v, d])))))))(Sqrt[((OverDot[x])^2)+((OverDot[z])^2)])(OverDot[z])-B ω (OverDot[x])

values={α->0.0039, β->0.0058,vd->35.,Δ->5.,ρ->1.23,R->36.4 (10^(-3)),m->0.145,g->9.8,B->4.1 (10^(-4)),
ω0->+1800. (π/30),α0->(*100. (π/30)*)0.,v0->37.}

v:=(Sqrt[((x'[t])^2)+((z'[t])^2)])
f[v_]=α+(β/(1+(E^((v-vd)/Δ))))/.values;
ω[t_]=ω0-α0 t/.values
eqx[αD_,βM_]=x''[t]+αD f[v] v x'[t]-βM B ω[t] z'[t]/.values;
eqz[αD_,βM_]=z''[t]+g+ αD f[v] v z'[t]+βM B ω[t] x'[t] /.values;

tabθ=Table[Table[{θ,solxz=NDSolve[{eqx[αD,βM]==0,eqz[αD,βM]==0,x[0]==0,z[0]==0.,
x'[0]==v0 Cos[θ Degree]/.values,z'[0]==v0 Sin[θ Degree]/.values},{x[t],z[t]},{t,0.001, 10.}],
First[FindRoot[First[z[t]/.NDSolve[{eqx[αD,βM]==0,eqz[αD,βM]==0,x[0]==0,z[0]==0.,
x'[0]==v0 Cos[θ Degree]/.values,z'[0]==v0 Sin[θ Degree]/.values},{x[t],z[t]},{t,0.001, 10.}]]==0.,
{t,4.}]]},{θ,10,90,5}],{αD,0,1},{βM,0,1}];

Table[{n,m,k,tabθ[[n,m,k,3]],tabθ[[n,m,k,1]],x[t]/.First[tabθ[[n,m,k,2]]]/.tabθ[[n,m,k,3]]},
{n,1,2},{m,1,2},{k,1,17}];

Table[{tabθ[[n,m,k,1]],x[t]/.First[tabθ[[n,m,k,2]]]/.tabθ[[n,m,k,3]]},{n,1,2},{m,1,2},{k,1,17}];

tabplottabθrange=Table[ListPlot[Table[{tabθ[[n,m,k,1]],x[t]/.First[tabθ[[n,m,k,2]]]/.tabθ[[n,m,k,3]]},
{k,1,17}],PlotStyle->{If[n==1&&m==1,{Black,PointSize[0.02]},If[n==1&&m==2,{Blue,PointSize[0.02]},
If[n==2&&m==1,{Green,PointSize[0.02]},{Red,PointSize[0.02]}]]]},AxesLabel->{"(θ^°)","range,m"},
GridLines->Automatic,PlotLabel-> StringJoin["{αD,βM} =",{ToString[n-1],ToString[m-1]}],
PlotRange->{{0,90},{0,180}}],{n,1,2},{m,1,2}];

interpolateθrange=Table[Interpolation[Table[{tabθ[[n,m,k,1]],x[t]/.First[tabθ[[n,m,k,2]]]/.
tabθ[[n,m,k,3]]},{k,1,17}]],{n,1,2},{m,1,2}];

tablerange=Table[Show[{Plot[interpolateθrange[[n,m]][θ],{θ,10,90},GridLines->Automatic,
PlotRange->{{0,90},{0,180}},PlotStyle->Black,AxesLabel->{"(θ^°)","range,m"},PlotLabel-> 
StringJoin["{αD,βM} =",{ToString[n-1],ToString[m-1]}]],tabplottabθrange[[n,m]]},ImageSize->250],
{n,1,2},{m,1,2}];

Grid[{{tablerange[[1,1]],tablerange[[1,2]]},{tablerange[[2,1]],tablerange[[2,2]]}}]

({{, }, {, }})

Figure 3.  The dots are the numeric solutions of eqns (1,2) corresponding to the switch setting {αD,βM}.  The solid curve in each plot is the corresponding interpolated functions.

({{, "{αD,βM}=00", "{αD,βM}=01", "{αD,βM}=10", "{αD,βM}=11"}, {"max range, m -->", 139.6938775510203`, 122.99499123510147`, 78.28322235551542`, 73.1088595961045`}, {"(θ^°) -->", θ->45.000000000000064`, θ->50.35583858969162`, θ->39.42807073554649`, θ->43.15684232674055`}})

({{, }, {, }})

Figure 4.  Plots of time of flights vs. the initial angles for four cases of interest.

({{, }, {, }, {, }, {, }})

Time dependent spin

(Overscript[x, ..])=-f(v) v (OverDot[x])+B ω(t) (OverDot[z])
(Overscript[z, ..])=-g-f(v)v (OverDot[z])-B ω(t) (OverDot[x]) 
ω(t)=ω0-α0 t

(*values={α->0.0039, β->0.0058,vd->35.,Δ->5.,ρ->1.23,R->36.4 (10^(-3)),m->0.145,g->9.8,B->4.1 (10^(-4)),
ω0->+1800. (π/30),α0->100. (π/30),v0->37.}*)

({{, "{αD,βM}=00", "{αD,βM}=01", "{αD,βM}=10", "{αD,βM}=11"}, {"max range, m -->", 139.6938775510203`, 125.0189694967104`, 78.28322235551542`, 73.63382769325277`}, {"\!\(\*SuperscriptBox[\(θ\), \(°\)]\) -->", θ->45.000000000000064`, θ->50.4507815126606`, θ->39.42807073554649`, θ->43.239506400028965`}})

For back spin, we get the “famous” (35^°)  angle.  


({{, "{αD,βM}=00", "{αD,βM}=01", "{αD,βM}=10", "{αD,βM}=11"}, {"max range, m -->", 139.6938775510203`, 162.57900443412842`, 78.28322235551542`, 84.09500907751338`}, {"\!\(\*SuperscriptBox[\(θ\), \(°\)]\) -->", θ->45.000000000000064`, θ->38.39731461784826`, θ->39.42807073554649`, θ->35.17922071601326`}})