## Abstract

This paper develops a mathematical model describing the motion through the air of an American football. The model is based on established equations used to describe spinning projectiles. While the equations are applicable to general motions, the emphasis of the paper is on the spiral pass and punt. Separate sections introduce formulas for the forces and moments understood to act on spun projectiles. The discussion of each force and moment includes an assessment of how well available experimental data characterizes the force or moment for an American football. For each force or moment, there is a description of how it affects the motion and trajectory. While the equations are valid for arbitrary motions, the available aerodynamic data is not. In parallel with the derivation of the nonlinear mathematical model, a linearized dynamics model is developed. The linearized model is used to help explain the behavior of the nonlinear model and to provide insight into the underlying physics. The linearized model is also used to derive a relationship between linear and angular velocity that ensures that the gyroscopic motion of a football is stable. The paper provides physical insights into what causes the apparent “wobble” of a spiral pass and what the character of the wobble says about the quality of the pass. Among the physical insights provided are the reason some passes have a rapid wobble and some slow, why a pass exhibits a lateral swerve, and why the Magnus effect may be neglected. The results are applicable to rugby footballs.

## 1 Introduction

Rae [1] appears to have published the first numerical study of the motion of an American football during a spiral pass. That paper showed results derived from video analysis of passes during games and under controlled conditions. The paper derived a mathematical model that considered the spin momentum of the football along with four aerodynamic effects: (1) the drag force, (2) the lift force, (3) the overturning moment, and (4) the Magnus lift force. The paper presented results from a numerical simulation using those equations based on the wind-tunnel test data described in Ref. [2] for several initial conditions. The paper also included a linearized analysis and a consideration of stability. This paper extends Rae’s mathematical model for a football to include all of the terms understood to be necessary to describe spinning projectiles in flight [3,4]. While those references focus on projectiles like bullets and artillery shells, the same forces and moments need to be characterized to describe the flight of a football. This paper extends Rae’s results [1] by considering additional aerodynamic effects and a more detailed investigation of the equilibrium conditions. In addition, this paper considers each aerodynamic effect in turn and its effect on ball’s motion. There does not appear to be direct experimental data related to the aerodynamic terms that result in damping due to angular motion. This paper provides estimates for those terms. In this paper, the word spin refers to angular rotation about the symmetry axis of the ball, and rotation refers to angular motion perpendicular to that axis. Because of its frequent occurrence, the angular momentum will be referred to simply as momentum.

In addition to the work on footballs by Rae [1] and Rae and Streit [2], similar work has been done by Seo et al. [5] on rugby balls. The rugby screw kick or torpedo punt is similar to the spiral pass, the primary differences being that the linear velocity of the rugby ball is somewhat lower while the angular spin of the ball is significantly lower. While rugby balls are similar in shape to American footballs, it is notable that there are significant quantitative differences between the reported aerodynamic data. Other authors measured the drag on American footballs [6,7]. These authors obtain results that differ substantially from each other as well as Ref. [2]. The data in Ref. [6] indicate that the spin of the football affects the measured results leading to the conclusion that it should be included when making measurements. The results of Ref. [7] showed differences in drag between the balls used in the National Collegiate Athletic Association (NCAA) and the National Football League (NFL). The numerical results presented in this paper primarily make use of the data from Ref. [2] and would be noticeably different if other researchers’ data were used to parameterize the mathematical models. This should be kept in mind when considering the quantitative results.

The next section of the paper describes the reference frames that are needed to model the dynamics of a spiral pass. The subsequent section provides a basic set of nonlinear differential equations for describing the trajectory. There are four equations for each of the vector quantities that need to be modeled. One pair of equations describes the translational motion in terms of the position and velocity of the football. The second pair of equations describes the orientation of the football using a unit vector aligned with the symmetry axis of the football to represent its orientation in space, and a quantity related to the angular momentum. The following section develops a set of equations that approximate the nonlinear dynamics as linear differential equations. The forces and moments are considered in the next section. For each mathematical model of a force or moment, an expression is given for representing it in the nonlinear equations and the linear equations. The effect on the flight of the spiral pass of each term is illustrated by simulation results. The simulation used to produce the results in this paper has been validated against McCoy’s equations [4] generally and the equations presented in Ref. [1] for linear coefficients. The equations formulated in this paper do not contain the divide-by-zero when the symmetry axis of the ball and the velocity vector are aligned that is present in Rae’s equations [1]. Because American football is played on a field measured in English units, English units are used throughout the paper with SI units given parenthetically.

## 2 Defining the Location and Orientation of the Football

When simulating the motions of airplanes, missiles, ships, or footballs, it is necessary to define a reference frame or set of reference frames that are used to characterize the motion. A judicious choice of frames can make a problem easier to model and then interpret results. In this paper, the “base” reference frame is located at a point on the field where the quarterback is standing when the ball is released. The 1-axis of the frame is pointed toward the opposing goal line and parallel to the sideline. The 2-axis is oriented upward and perpendicular to the surface of the field. The 3-axis completes a right-handed reference frame and points toward the offense’s right and is parallel to the yard lines. The image in Fig. 1 illustrates these definitions. The three components of the location of the football $x→=[x1,x2,x3]T$ are shown relative to the base frame.

Fig. 1
Fig. 1
Close modal

The “ball” reference frame is located at the geometric center of the ball and translates with the ball. Figure 2 shows the ball reference frame and its orientation relative to the three axes of the base frame. The axes of the ball frame are denoted by $s^i$, with $s^1$ oriented along the ball’s symmetry axis. The angles associated with the orientation of the ball are defined relative to the base frame by rotating the ball about the 2-axis through an angle −θ and then about the $s^3$ axis by ϕ. The angles −θ and ϕ are the first two angles, respectively, of a 2–3–1 Euler angle sequence. The usual convention is that Euler angles are defined positive relative to the axis about which they are taken. In this case, the negative sign is introduced so that positive θ’s correspond to directions to the right of the quarterback (i.e., positive x3). Since the ball spins and is (almost) rotationally symmetric, the single vector $s^1$ is sufficient to define its orientation. From this point on, the subscript will be dropped. Note that the $s^3$ axis of the ball frame always lies in a horizontal plane parallel to the 1–3 plane of the field and will be normal to the plane containing 2 and $s^$. The vector $s^2$ is oriented so that it is normal to the plane defined by $s^$ and $s^3$; therefore, it lies in the same plane as that defined by the small triangle in Fig. 2. Equivalently, $s^2$ lies in the plane containing the 2-axis and $s^$.

Fig. 2
Fig. 2
Close modal
In the equations developed in the next section, the orientation of the ball is defined by the unit vector $s^$ along the ball’s axis and its components [s1, s2, s3]T. The Euler angles have been introduced to provide a way of interpreting the orientation physically. The first angle θ is the angle between the axis of the ball, $s^$, and the direction straight down field. Noting that this angle is the angle between that plane and the projection of $s^$ onto the horizontal plane, the angle can be computed from $sinθ=s2/‖s^‖=s2$. The angle ϕ is the angle between the horizontal plane and the axis of the ball given by tan ϕ = s3/s1. Given the angles, the components of $s^$ can be computed by first computing s2, then the magnitude of the projection using the fact that the $s^$ has unit magnitude, and then multiplying that by cos θ and sin θ to obtain s1 and s3, respectively. For a 2–3 Euler angle sequence, (−a, b), the matrix that transforms vectors expressed in the base frame to the rotated frame is
$C23(a,b)=[cosacosbsinbsinacosb−cosasinbcosb−sinasinb−sina0cosa]$
(1)
This matrix already has the negative sign for the first rotation included, so a is assumed to have the sense shown in Fig. 2. This matrix will transform any vector expressed in the base frame into the basis of a reference frame rotated by 2–3 Euler angles (a, b). The matrix that does the inverse transform is the transpose $C23T$. Because of the assumption that the ball has rotational symmetry about the 1-axis, the rotation about this axis is ignored in our model.

## 3 Describing the Football’s Translation and Rotation

This section presents a derivation of the differential equations that describe the translational and rotational motion of the football. The translational motion is defined by the location $x→$ and velocity $v→$ of the center of the football either at an instant time or over some period of time. The differential equation for the position is simple since in our non-rotating base reference frame, the time derivative of position is equal to the velocity.
$x→˙=v→$
(2)
The effect of Earth’s rotation on the trajectory of the ball will be described at the end of the section where the forces and moments are modeled. The time derivative of the velocity is the acceleration, so the differential equation describing the time evolution of the velocity is obtained from Newton’s law as
$v→˙=∑i∈g,L,D,MFim+aωE+aC=1m[Fg+FL+FD+FM]+aωE+aC$
(3)
for a football of mass m. The Fi are the set of forces acting on the football. The subscripts denote the forces associated with gravity, Fg, lift, FL, drag, FD, and the Magnus force, FM. The accelerations $aωE$ and aC denote apparent accelerations due to the centripetal acceleration of the base frame and the Coriolis acceleration. This paper will adopt the same value m = 0.0282 slug (0.411 kg) used in Ref. [1]. The sum is taken over all of the forces assumed to be acting on the body through the center of mass. The initial force assumed to be acting on the body is that due to the gravitational attraction of the Earth
$Fg=[0,−mg,0]T$
(4)
A value of g = 32.157 ft/s2 (9.8015 m/s2) will be used in all numerical results.
So far, these equations are easy to solve if we know the initial conditions. The symbol $x→0=[x10,x20,x30]T$ denotes the initial position of the ball. While most of the examples in the paper assume that the initial location of the ball is the origin of the base frame, $x→0=[0,0,0]T$, it may be useful in certain cases to offset the initial condition to account for the height and displacement of the point of release of the ball relative to the origin of the base frame. The symbol $v→0$ denotes the initial velocity. This vector will be defined by two Euler angles Θ0 and Φ0. The angle between the velocity vector and the horizontal plane will be denoted Φ and the direction of the velocity relative to the vertical plane aligned with the length of the field is Θ. The Euler angles (−Θ, Φ) define a new reference frame co-located with the ball frame that will be referred to as the “velocity” frame. The 1-axis of the velocity frame is always aligned with the velocity vector, the 3-axis is perpendicular to the velocity vector in the horizontal plane, and the 2-axis completes a right-handed set. If the magnitude of the velocity at the point of release is V0, then the initial velocity vector is
$v→0=[v10,v20,v30]T=V0C23T(Θ0,Φ0)[1,0,0]T=V0[cosΘ0cosΦ0,sinΦ0,sinΘ0cosΦ0]T$
(5)
Acting solely under the influence of gravity, the trajectory of the football is described by the solution
$v→(t)=[v10,v20−gt,v30]Tx→(t)=[v10t,v20t−g2t2,v30t]T$
(6)

As a first example, the results in Fig. 3 show the effect of height of release on the distance traveled by the ball. The simulation shows a ball released at a velocity of V0 = 61.3 mph (27.4 m/s), an angle of Φ0 = 30 deg, and a spin rate of 600 rpm. These initial conditions will be used repeatedly in the examples and are the same as those used by Rae [1]. The difference between the two trajectories is that one ball is released at a height of 7 ft and the other zero. The effect of the change in height of release is that the ball hits the ground almost 4 yards further downfield. This indicates that if we want to model the flight of a football for some practical purpose, knowing the actual height of the release point is important. The curves also have arrows that show the orientation of the ball relative to the trajectory. The figure shows that initially the ball is tangent to the trajectory and that the ball maintains this orientation throughout its flight.

Fig. 3
Fig. 3
Close modal
Following the standard practice in the external ballistics literature [3,4], the rotation of the football is modeled using the vector $s^$ for orientation and the momentum divided by the transverse moment of inertia $h→=L/It$ to characterize angular rate. The transverse moment of inertia is measured about an axis perpendicular to the axis of symmetry and through the center of the ball. The value It = 0.00237 slug − ft2 (0.00321 kg m2) is the value used in Ref. [1]. The moment of inertia along the spin axis taken from that reference is It = 0.00143 slug − ft2 (0.00194 kg m2). The references in Rae [1] suggest that these are measured values. The differential equation that describes the time evolution of the orientation vector is
$s^˙=h→×s^$
(7)
The differential equation for the time derivative of the modified momentum based on Newton’s law for rotating bodies is
$h→˙=∑i∈α,M,p,qτiIt=τα+τM+τp+τqIt$
(8)
where the τi denotes the set of torques acting on the body. The subscripts denote the torques associated with the overturning moment, $τα$, the Magnus torque, τM, the damping due to spin, τp, and the rotational damping associated with angular velocity perpendicular to the spin axis, τq.
It will be useful to define the initial orientation of the football $s^0$ relative to the velocity vector. In order to make the definition of the initial orientation more intuitively meaningful, the concepts of angle-of-attack, α, and sideslip, β, are introduced [8]. The matrix $C23T(β,α)$ transforms the vector $s^$ from its representation in the ball frame to the velocity frame, and the matrix $C23T(Θ0,Φ0)$ transforms that vector into the base frame; therefore, the initial value of the vector $s^$ in the base frame is
$s^0=C23T(Θ0,Φ0)C23T(β0,α0)[1,0,0]T$
(9)
Readers of Ref. [4] should be aware that the definitions of α, β, θ, ϕ introduced here are different than those used in that reference where the angles are a 3–2–1 sequence.
Finally, we need to define the initial value of the modified momentum. When the ball leaves the quaterback’s hand, it has spin predominantly along its symmetry axis, but it may also have angular velocity about axes perpendicular to the spin axis. The angular velocity will be denoted as $ω→0=[ω10,ω20,ω30]T$ with $ω10$ denoting the axial spin. Since this vector is defined in the ball frame, the matrices that transformed $s^$ to the base frame will do the same for the angular velocity so that $Ω→0=C23T(Θ0,Φ0)C23T(β0,α0)ω→0$ represents the initial angular velocity in the base frame. Using this definition, the initial modified angular momentum in the base frame is
$h→0=IsItω10s^0+s^0×s^˙0=IsIt(Ω→0⋅s^0)s^0+s^0×Ω→0×s^0$
(10)
The first term is a dot product to get the projection of the angular velocity on $s^0$ and the second is the component perpendicular to the spin axis.

## 4 Linearized Dynamics

This section derives a set of linear differential equations that approximate the nonlinear equations of the previous section. A disadvantage of using linearized equations is that they are not applicable to the entire flight of the football; however, studying the linear equations is a way to gain insight into the nature of the motions at a point along a trajectory. The linear assumption is that the orientation vector, the momentum vector, and the velocity vector all remain within about 15 deg of each other and their starting values over the entire period of time considered. Under this assumption, the results from a linear analysis will reasonably accurately model the nonlinear analysis. The angle is a rule of thumb, and a larger or smaller value may be appropriate based on the context. The approach taken here is to write the equations in the velocity frame and retain first-order (linear) terms. This allows the linearized analysis to be conducted for any initial velocity vector or at an arbitrary point along a trajectory. For example, the nonlinear equations can be simulated up to a given point in the pass, and then the linear analysis is applied to those values of the state vector at that point. Typically, an equilibrium state is used as the origin for the linearized dynamics; however, the paper will show that while there is zero aerodynamic torque on the ball when the velocity and symmetry axis are aligned, this is not an equilibrium condition. By defining the linearized dynamics relative to the velocity vector, rather than the equilibrium orientation relative to that vector, we believe that the physical interpretation of the states is more intuitively meaningful. A slight improvement on the accuracy of the linearized model could be obtained by recomputing the aerodynamic coefficients at the equilibrium state, but we do not do that here.

The linear dynamics equations are obtained by transforming initial conditions from the base frame into the velocity frame. This is done by simply applying the transformation matrix C23 (Θ, Φ) to the components of the state in the base frame about which we desire to linearize the model. If $(v→,s^,h→)$ is an arbitrary initial state, then the equivalent state for the linearized equations is
$v~0=C23(Θ,Φ)v→=V[1,0,0]Ts~0=C23(Θ,Φ)s^=C23(Θ,Φ)C23T(Θ,Φ)×C23T(β,α)[1,0,0T]≈[1,α0,β0]Th~0=C23(Θ,Φ)h→$
(11)
The tilde ($~$) is introduced to be clear whenever we are discussing the dynamics of the linear model. The position is not included in the linearized dynamics because it will not be of particular interest in this context. Also, the definitions of the angle of attack and the sideslip angle have been used to show that they are approximately equal to the second and third components of $s~$. These angles will be used in the linearized dynamics because of their intuitive interpretation as angles.

Since the velocity vector is always tangent to the tragectory, Fig. 3 shows that the effect of gravity is to continuously turn the velocity vector downward. Similarly, any force that has a component perpendicular to the direction of travel will cause the velocity vector to rotate. The angular velocity of the velocity frame is equal to $ωv=(v~/V)×(v~˙/V)$. Because the one axis of the velocity vector is aligned with the velocity, the second and third components of the velocity in the linear frame are always zero. It follows that the first component of ωv is always zero and we can write $ωv=[0,ωv2,ωv3]T$. So far, we have introduced the acceleration due to gravity $v~˙=−g[sinΦ,cosΦ,0]T$ which results in an angular velocity for the velocity vector with components $ωv2=0$ and $ωv3=−gcosΦ/V$. As other forces are introduced, their contributions will be incorporated into the angular velocity vector when their contributions to the linear dynamics are discussed.

Taking the time derivative of the velocity vector for an arbitrary ωv yields in the linearized frame
$v~˙=∑iF~im−ω~v×v~=∑iF~im+[ωv3v2−ωv2v3−ωv3v1ωv2v1]≈∑iDF~im+[0−ωv3v1ωv2v1]$
(12)
The cross-product with the ωv term appears in the equation because the velocity frame in which the equations are written is assumed to be rotating at a constant angular velocity [9]. The symbol D denotes the derivative (Jacobian) of a function evaluated at the values for the linearized state. For the orientation vector, the time derivative is
$s~˙=h~×s~−ω~v×s~=[h~1β−h~1αh~3−h~1β−h~2+h~1α]+[ωv3α−ωv2β−ωv3ωv2]≈[0h~3−h~1β−h~2+h~1α]+[0−ωv3ωv2]$
(13)
Finally, the time derivative of the modified momentum vector is
$h~˙=∑iτ~iIt−ω~v×h~=∑iτ~iIt+[ωv3h2−ωv2h3−ωv3h1ωv2h1]≈∑iDτ~iIt+[0−ωv3h1ωv2h1]$
(14)
For the case of the rotational dynamics described by Eqs. (13) and (14), it is not necessary to carry forward the differential equations associated with the first components $s~1$ and $h~1$ of those vectors because their time derivatives are assumed to be zero. We have not yet addressed the challenge of expressing the forces and moments in the linearized frame. In the next section, expressions for the aerodynamic forces and moments will be derived. A vector notation will be used such that the expressions are independent of the reference frame in which they are expressed. The only requirement is that all vector quantities be expressed in the same frame.

## 5 Modeling the Aerodynamic Forces and Moments

This section describes the aerodynamic forces and moments that have been found necessary to model the dynamics and flight trajectories of spinning projectiles [3,4]. Each subsection considers an individual force or moment and how it is modeled mathematically. The models are added one-at-a-time to the equations developed in the previous two sections, and simulation results illustrate the component’s effect. Forces and moments are reported relative to a “wind” that notionally defines a reference frame for an imaginary wind tunnel. The 1-axis is aligned with the direction of airflow (assume right-to-left). The 2-axis is oriented upward and the 3-axis is pointed toward a viewer of the football in the imaginary tunnel. In this reference frame, the force of drag will be positive, an upward lift force will be positive, and a positive rotation about the 3-axis will be counter-clockwise as will a positive torque (right-hand rule relative to the 3-axis).

Each force or moment arises from a specific interaction between the ball and the air flowing around it, and each can be described by a three-dimensional vector representing its action in three-dimensional space. One approach to modeling a force or moment is to separate it into a scalar quantity representing its magnitude multiplied by a vector quantity derived from the geometry of the flow to determine the orientation of the vector-valued force or moment. The scalar quantities are the forces and moments measured in the wind frame made non-dimensional by dividing the forces by the product of the dynamic pressure 1/2ρV2 and some reference area S. Moments are divided by an additional linear dimension. Here the linear dimension is the maximum diameter of the football and S is the corresponding area [3,4]. These are called aerodynamic coefficients and will be represented by an upper case C with a subscript to identify the force or moment. A sub-subscript is a notational convention to indicate that the quantity is a stability derivative $CLδ=∂CL/∂δ$ [10]. For axisymmetric bodies, the only relevant geometric quantity is the angle between $s^$ and $v^$ which will be denoted by δ. This angle is related to the velocity and orientation by $cosδ=v^Ts^$ and to angle of attack and sideslip by the formula sin2δ = sin2β + cos2β sin2α. Experimental data for axisymmetric bodies is usually reported by specifying C(δ) only. The mathematical formulas presented in the following sections are correct for arbitrarily large values of δ, provided the methods used to represent the aerodynamic coefficient’s dependence on δ are correctly implemented and accurate for those large angles.

### 5.1 Overturning Moment.

The overturning moment occurs when the symmetry axis of the football is not aligned with the velocity vector. The vector axis of the moment acts perpendicular to the plane containing the velocity and symmetry axis. The moment has the characteristic that it tends to get larger as δ gets larger and acts in the direction to increase δ. It is therefore destabilizing in the sense that the torque associated with the moment wants to increase the angle between the two vectors which should lead to tumbling. The ball does not tumble because the interaction between this moment and the angular momentum of the spin causes the football to roughly align itself with the velocity vector during a pass. Namely, nose-up at the release and nose-down at the catch. The interaction between this torque and the momentum is also the cause of the ball’s wobble in flight. This section describes the modeling of the overturning moment and how it causes these particular motions of the spinning football.

The model for the overturning moment used here is [3,4]
$τα=12ρSD‖v→‖CMαv→×s^=12ρSD‖v→‖2CMαv^×s^≈ItMα[0,−β,α]TMα=ρSDV2CMα2It$
(15)
Dimensional coefficients such as $Mα$ are introduced to simplify the notation in the linearized dynamics equations. The symbol V will represent the velocity’s magnitude. For Eq. (15) to be valid for arbitrary values of δ, it is necessary to assign a different meaning to the symbol $CMα$ than the standard one. Noting that $‖v^×s^‖=sinδ$, letting $CMα=CM/sinδ$ makes Eq. (15) valid for arbitrary δ.

The results in this paper use the same models that Rae [1] used based on the experiments [2]. The experiment collected data up to an angle of attack of about 55 deg and the formula $CM=0.165sin(2δ)$ is a reasonable fit over that range, and $CMα$ = 0.33. Some caution should be used in applying this model at larger angles of attack. The area and aspect of the football projected onto a plane normal to the flow are substantially different at 0 deg and 90 deg, and it is expected that the magnitude of the slope of CM will be smaller at a right-angle to the flow than when parallel. For comparison, Seo et al. [5] recommend a model of CM = 0.771δ − 0.494δ2. This model has been corrected for angular units and the use of volume in the normalization is therefore comparable to Rae’s model. The factor of two discrepancy in the slope at small angles for balls that are relatively similar in size seems large. Neither paper attempts to make corrections for the effect on the measured forces and moments of the mounting arrangement for the ball [11].

For the linear analysis of the dynamics with the overturning moment, we are only interested in the equations associated with the rotation of the body (Eqs. (13) and (14)). We will also assume that $h~1=Isω10/It≜P$. Recalling that we are neglecting the first rows of these two equations, and combining with the linear expression from Eq. (15) yields the following system of differential equations in first-order matrix form
$(α˙β˙h~˙2h~˙3)=[0−P01P0−100−Mα00Mα000](αβh~2h~3)+[0−1100−PP0](ωv2ωv3)$
(16)
The first two equations for $α˙$ and $β˙$ above can be differentiated, and substituting the second pair of equations along with $ωv2=0$ yields a pair of coupled second-order equations for α and β.
$α¨=−Pβ˙+Mααβ¨=Pα˙+Mαβ−Pωv3$
(17)
Multiplying the second equation by $i=−1$, substituting γ = α + , and − = − + β yields the single equation
$γ¨−iPγ˙−Mαγ=−iωv3P$
(18)
The complementary solution of this differential equation has characteristic exponents $0.5(iP±−P2+4Mα)$. The exponents are purely complex if $Mα; otherwise, the solution is unstable. This is called the gyroscopic stability criterion because of its similarity to the stability condition for a gyroscope. It is apparent by inspection that a particular solution to this equation is given by $α=α˙=β˙=0$ and
$βr=ωv3PMα=PgcosΦVMα=2Isω10gcosΦρSDCMαV3$
(19)
This angle is referred to as the “yaw of repose” in the external ballistics literature [3,4]. Substituting (19) back into (18) and assuming $γ¨=γ˙=0$ show that γ = r is an equilibrium solution for the linear equations. Multiplying this equation through by It shows that the torque due to βr is exactly equal to the rate of change of the spin angular momentum rotating at the same angular velocity as the velocity vector. The spiral pass rotates at the same rate as the angular velocity vector because it is an instantaneous equilibrium condition. For a left-handed passer, P is negative, the sideslip angle is negative, and the nose of the ball will be rotated to the left of the velocity vector for a left-handed spiral pass. This explains why the nose of the ball remains roughly aligned with the velocity vector during a pass.

The results in Fig. 4 were produced by simulating the nonlinear equations of motion for the same initial conditions as the first example (Fig. 3). In that figure, only gravity acted, but now with the effect of the overturning moment included. The curve labeled “perfect” corresponds to the ball being released with the spin axis and the velocity vector aligned. This would seem to be the ideal condition since the aerodynamic torque is absent. The figure plots the angle of attack versus the sideslip angle during the pass which lasts about 2.8 s. An alternate interpretation of this figure is that (α, β) are the projections of the unit vector $s^$ onto a plane through the center of the ball and perpendicular to the velocity vector. This implies (α, β) ≈ (s2, s3) if the angles are expressed in radians. The angle of repose is about 3.5 deg at the release, and the first half-orbit of the curve is centered roughly about this point. The second full-orbit corresponds to the smaller circle. At the apex of the pass, the angle of repose is about 6.3 deg and during the middle-half of the pass, the motion is centered on this point. During the final quarter of the pass, the axis of the ball orbits roughly about the same point as the first quarter except in the lower half-plane. The figure shows that the ball travels downfield with the nose pointed slightly to the right because of the right-handed spin and the yaw of repose (Eq. (19)).

Fig. 4
Fig. 4
Close modal

While the release with the velocity and spin axis aligned corresponds to zero aerodynamic torque, it is evident that this does not correspond to an equilibrium condition. The second curve on the plot in Fig. 4 corresponds to the ball being released at the equilibrium state for the release conditions; namely, the ball is released with a sideslip angle corresponding to the yaw of repose. It is evident that amplitudes of the motion are much smaller than for the “perfect” release.

From watching spiral passes, we know that some passes exhibit very slow wobbling in flight, while others exhibit a pronounced high-frequency motion. The pitch-angle data in Fig. 1 [1] exhibit such an oscillation at about 5 Hz. Rae’s interpretation of this motion as precession is probably correct. The result in Fig. 4 gives only a hint of the presence of motion at that frequency. In order to understand the discrepancy, consider a homogeneous solution to Eq. (18):
$γ(t)=α+iβ=C1exp0.5(ih1+−h12+4Mα)t+C2exp0.5(ih1−−h12+4Mα)t=C1e32.01it+C2e5.958it≈C1e(2π)5it+C2e(2π)it$
(20)
It is encouraging that our model predicts that 5 Hz should be a characteristic frequency of the ball’s motion. The complex constants C1 and C2 are chosen to satisfy the initial conditions. Assume for a moment that C1 = 0. The solution γ(t) traces out a circle in the complex plane at the frequency ωs = 6 of about 1 Hz. If we assume that the horizontal axis is complex and recall that s2α and s3β, then we can interpret this motion as pure precession of $s^$ at the slow frequency. Precession is understood to be the first Euler angle associated with rotation about an axis. The axis in this case is a vector aligned with the equilibrium state. If C1C2, then the motion remains predominantly precession at the slow frequency, with a small periodically varying complex number added to the complex number associated with the vector rotating at the slow frequency. This additional motion causes nutation at the fast frequency if nutation is understood to be a second Euler angle. The nutation angle is $α2+(β−βr)2$ is the angle between the equilibrium axis and the symmetry axis. While nutation happens at the fast frequency, the precession rate is modulated, but on average equals the slow frequency. This is essentially the motion shown in Fig. 4.

So what has to happen to have precession at the fast frequency? Basically, the requirement is that C1 > C2. Following the discussion of the previous paragraph, the precession will proceed on average at the fast frequency and nutation at the slow frequency. The magnitudes of the complex constants Ci contribute equally to the magnitude of the initial angle of the ball relative to the velocity axis. This is not true for the initial angular velocity of the ball relative to the velocity vector. It is evident from the derivative of (20) that the magnitude of C1 has five times the effect on the initial angular velocity of the ball. Basically, if the ball is precessing at the fast angular frequency, the quarterback released the ball with significant angular velocity perpendicular to the spin axis of the ball. The results in Figs. 5 and 6 illustrate these ideas. In Fig. 5, the ball is initialized with its axis $s^0$ 10 deg above the velocity vector and with zero rotational angular velocity. There is pronounced nutation, but the precession frequency is still the slow frequency. In Fig. 6, the ball is initialized with the same initial α0 = 10 deg and with an initial angular velocity $β˙0=α0ωf$. In this case, the precession frequency is now ωf and the nutation frequency is ωs. When we have completed the modeling of all of the aerodynamic forces, the paper will briefly consider the effect of the release on the distance of the pass. Until then, the focus of the paper is on how each aerodynamic coefficient affects the motion of the ball. This is best understood when the motions are slow.

Fig. 5
Fig. 5
Close modal
Fig. 6
Fig. 6
Close modal

### 5.2 Lift.

The flow of air over the ball when the symmetry axis and velocity are not aligned causes lift. The lift varies with the angle between the axis of the ball and the velocity vector. In the previous section, we saw that the overturning moment interacting with the spin causes the nose of the ball to precess about the velocity vector. The force associated with lift acts in a direction perpendicular to the velocity vector and is always oriented towards $s^$. The equation for modeling the lift is [3,4]
$FL=12ρSCLα[‖v‖2s^−(v→⋅s^)v→]=12ρS‖v→‖2CLα(s^−(v^⋅s^)v^)=12ρS‖v→‖2CLα(v^×s^×v^)≈mLαV[0,α,β]TLα=12mρSVCLα$
(21)
The interpretation of the vector (Eq. (21)) is that the line of action of the force is the difference between $s^$ and its projection onto $v^$.
As with the moment, the only source for a numerical value for $CLα$ appears to be the data from Ref. [2]. Those authors report their data in terms of normal and axial force coefficients parallel to and perpendicular to the sting mount. In order to convert these to lift and drag coefficients, it is necessary to transform their data into a reference frame whose axes are parallel and normal to the wind. This was done by scanning the images for axial and normal force in the reference, digitally sampling their data points, and then using an Euler angle transformation to convert their data into the standard wind-tunnel lift and drag frame. A least-squares fit to the lift data yields for the lift coefficient
$CL(δ)=0.6528sinδ−0.1229sin3δ−0.2528sin5δ$
(22)
Seo et al. report a value of 0.06 for rugby balls, and once again the discrepancy with Rae and Streit’s data seems exceptional. The result in Fig. 7 shows the trajectory of the football with only the moment of the previous section and the lift force included. The range of the pass is not significantly affected because the lift force rotates around the velocity vector and the net effect averages out in the vertical. The main difference is the lateral drift. This is caused by the positive sideslip angle which is evident in Fig. 8.
Fig. 7
Fig. 7
Close modal
Fig. 8
Fig. 8
Close modal
The lift force does not appear directly in the linearized rotational dynamics equations, but it does affect the rotation through its effect on the angular velocity of the ball. From the linear approximation in Eq. (21), the resulting angular velocity resulting from $v~˙/V=F1/m/V$ is $[0,−Lαβ,Lαα]T$. When combined with the effect of gravity, the angular velocity of the velocity vector is
$ω~v=[0,−Lαβ,Lαα−gcosΦV]$
(23)
and the resulting linear model is
$(α˙β˙h~˙2h~˙3)=[−Lα−P01P−Lα−10−PLα−Mα00Mα−PLα00](αβh~2h~3)+[10P0]gcosΦV$
(24)
The way that the lift force appears on the diagonals of the two left quadrants of the matrix in Eq. (24) reflects the fact that the force causes the velocity vector to rotate. The second-order complex form for this equation is
$γ¨+(Lα−iP)γ˙−(Mα+iPLα)γ=−iPgcosΦV$
(25)
and the corresponding equilibrium condition is
$αr=PLαMαβrandβr=P(Lα+(gcosΦ/V))Mα$
(26)
The equilibrium condition now includes an angle of attack, and the sideslip angle associated with the angle of repose is larger. We noted that β is positive for a right-handed throw and negative for a left-handed throw in the previous section. Because P has the same sign as βr, αr will always be positive causing the nose of the ball to tend to remain above the velocity vector. This positive angle of attack results in a torque due to the moment coefficient which acts to prevent the ball from rotating with the velocity vector. The increase in βr is exactly the amount required to overcome this torque, and the ball will rotate in an average sense at the same angular velocity as the velocity vector.

The insight from the linear analysis can be applied to interpret the result in Fig. 8. The effect of adding the lift force increases the maximum sideslip angle, and the nose of the ball tends to stay farther to the right throughout the trajectory. On average, the nose of the ball remains above the velocity vector for most of the duration of the pass. Based on the discussion about the equilibrium condition, this should have been expected. Equation (26) indicates the lift tends to increase the equilibrium sideslip at all conditions. For this simulation, the equilibrium value for βr varies from 10.5 deg at the beginning and end to a maximum of 14.5 deg at the apex of the pass. The corresponding values for the equilibrium αr are 1.3 deg and 2.1 deg. The result in Fig. 7 shows that the pass travels further downfield and drifts to the right. The longer range is a result of the lift due to the positive angle of attack αr. This lift force counters the effect of gravity and keeps the ball in the air for a longer time. The drift to the right is the result of the lift associated with the nose of the ball being to the right of the velocity vector due to the positive βr during the pass. Because a left-handed pass is associated with a negative βr, the corresponding drift will be to the left.

### 5.3 Magnus Force and Moment.

The Magnus force and moment are due to the spin on the ball. The force is caused by the fact that the spin causes more air to flow around one side of the ball, and the moment arises because the fore-aft asymmetry of the flow results in the force not acting through the center of mass. The equation describing the Magnus force is [3,4]
$FM=12ρS‖v→‖2CLpα(ω1D‖v→‖)v^×s^=12ρSD‖v→‖CLpαv^×s^(h→⋅s^)ItIs≈mLpαVP[0,−β,α]TLpα=ρSDCLpα2m$
(27)
Based on the conventions for the wind frame, the Magnus force coefficient will be negative. From the formula given in Ref. [2], $CLpα=−0.006684$. The Magnus force does not appear directly in the linearized rotational dynamics equations, but it does affect the rotation through its effect on the angular velocity in the same way that the lift force does. The equation corresponding to Eq. (23) for the Magnus force is
$ω~v=[0,−LpαPα,−LpαPβ]T$
(28)
The equation describing the Magnus moment is [3,4]
$τM=12ρSD‖v→‖2CMpα(ω1D‖v→‖)s^×v^×s^=12ρSD2‖v→‖CMpαs^×v^×s^(h→⋅s^)ItIs≈ItMpαP[0,−α,−β]TMpα=ρSD2VCMpα2It$
(29)
If the center-of-pressure for the Magnus force is forward of the geometric center of the ball, the Magnus moment coefficient will be negative. The linearized dynamics equations for the football including the Magnus force and moment are
$(α˙β˙h~˙2h~˙3)=[−Lα−P+PLpα01P−PLpα−Lα−10−P(Lα+Mpα)−Mα+P2Lpα00Mα−P2Lpα−P(Lα+Mpα)00](αβh~2h~3)+[10P0]gcosΦV$
(30)

Rae and Streit [2] report that their data indicated that there was no measurable Magnus moment. For spherically symmetric objects like baseballs, golfballs, and tennis balls, the absence of a Magnus moment would be expected. Since the football is symmetrically tapered fore-and-aft and the Magnus force depends on the diameter of the ball, it is expected that its effect would be concentrated near the center of the ball. Also, the football has an aspect ratio of less than 2:1 which is not very dissimilar from a sphere. These considerations suggest that the absence of a Magnus moment for the football is reasonable.

For the Magnus force and the conditions of the spiral pass we have been considering, L = −0.0025. Even when multiplied by P2, the corresponding value of −3.6 is small when compared with $Mα=191$. Repeating the simulation of the previous section including the Magnus force changes the distance traveled from Fig. 7 by less than 5 cm, a change of less than 0.1%. Taken together, this suggests that the Magnus effect is negligible in the case of footballs and passes. There may be situations, say after a bad punt or when the quarterback is hit at the release, where the ball has substantial spin and the spin axis and velocity vector are nearly perpendicular that this assumption does not hold. For the remainder of the paper, the effect of the Magnus force will not be included in the model. While the lateral swerve of the football is substantial during a long pass, the swerve is caused by the lift on the ball and not the Magnus effect. Noting that Eq. (27) implies that the Magnus force is perpendicular to the plane containing $v^$ and $s^$, and that the equilibrium condition (Eq. (26)) places $s^$ above and to the right of $v^$ for a right-handed pass, the small effect of the Magnus force acts to reduce the amount of lateral swerve due to lift. The Magnus force acts in the wrong direction to be the cause of the swerve.

### 5.4 Drag.

The drag force acting on the football is the primary force causing the ball to decelerate and loose kinetic energy. The drag force generally increases as the projected area normal to the airflow increases. Drag by definition acts in the direction opposite to the velocity of the ball. The equation for modeling the drag is [3,4]
$FD=−12ρSCD‖v→‖2v^=−mDV2v^D=12mρSCD$
(31)
The Reynolds number is a scaling parameter that provides an indication of the structure of the flow near the surface of a body. Low Reynolds numbers correspond to flow where there is a minimal wake behind the body and laminar flow near the surface of the object, and large Reynolds numbers correspond to flow where there is a significant wake and a turbulent boundary layer. A drop in drag coefficient of an order of magnitude is common when the transition from laminar to turbulent occurs. This is generally in the range 105 − 106 for streamlined bodies. Assuming the conditions for the long spiral pass in most of our examples yields a Reynolds number of 500,298, right in the middle where the transition should occur. Because of the spin in the ball, roughness of the surface, and the potential for the laces to act as a boundary layer trip, it seems reasonable to expect that the flow characteristics around the ball are those above the laminar-to-turbulent transition and that the actual Reynold’s number is not indicative of the character of the flow.
There appear to be three sources of drag data for a football. The highest number for the drag coefficient comes from a set of wind-tunnel tests [7]. These tests indicate a minimum drag coefficient of 0.2 at zero angle of attack, 0.5 at $45deg$, and a maximum of 0.8 at 90 deg. It is important to note that the ball was not spinning during these tests. A second series of tests [2] made measurements on a spinning ball and reported a zero angle of attack drag coefficient of 0.14. The axial force coefficient data reported there decreases to zero at about 55 deg angle of attack. This result is suspected as it suggests that the force acting along the sting mount becomes negative at larger angles. The authors do not attempt to correct their data for the area blocked by the sting or for the possibility that the pressure inside the ball is varying with angle and thus affecting this measurement. It is certain that not correcting the drag for the sting results in a low measurement [11]. At this point, it seems reasonable to conclude that the zero angle of attack drag coefficient is in the range 0.14–0.2. The measurements by Watts and Moore [6] suggest a value of about 0.055. While this low value may be questioned due to the fact that a real football was not used, it is notable that they found a reduction in drag due to spin. Figure 22 on pp. 6–16 of Ref. [12] has a plot that shows drag coefficients for ellipsoidal shapes, including one for a length to diameter ratio of 1.8 which is close to 1.6 of the football. Hoerner’s information suggests that the drag coefficient referenced to surface area is likely in the range 0.006–0.009. Using a value of 4.8 for the ratio of surface area to cross section for a football suggests a value in the range 0.029–0.043. This is probably a reasonable estimate of the lowest possible value for the drag coefficient. The value for zero angle drag coefficient in Ref. [5] is 0.0024 for a rugby ball. This number is an order of magnitude less than our estimate of a lowest reasonable value. This is likely because the mount for the rugby ball experiments has a large area, and because a correction is not applied to account for this [11]. We will essentially adopt the same value as Rae and Streit measured except transformed into a drag coefficient. Following the same procedure described prior to Eq. (22) yields
$CD(δ)=0.1418−0.6169sin2δ−0.2659sin4δ$
(32)
One final note on the available data is that the data presented in Ref. [7] showed significant quantitative differences between NFL and NCAA footballs. The drag data for the NFL ball were relatively flat near zero angle of attack, while the NCAA was curved, concave-up. For the angles of attack associated with the examples in this paper, the drag on the NFL ball will be relatively constant while the drag on the NCAA ball will increase at larger angles of attack. This suggests that the reduction in distance and velocity associated with a bad spiral in the college game will be more pronounced than in the professional game.

Figures 9 and 10 show simulation results now including the drag force. The drag force is acting in addition to the overturning moment and lift force from the previous two sections. The first figure shows a reduction in the distance traveled by the ball of over ten yards. While the total lateral drift is less at the end of the path, the trajectory followed indicates more drift for a given distance downfield. One of the reasons for this is that the drag reduces the velocity, so the rate-of-change of the linear velocity (i.e., the angular velocity of the velocity vector) is larger for a given side force. The side force is also larger because as shown in Fig. 10, the yaw of repose is larger. The reason for this can be seen by referring to Eq. (19) and noting that the yaw of repose is proportional to the inverse cube of the velocity; namely, small reductions in velocity have a strong effect on the angle. The linearized model we have proposed will not exhibit a dependence on the drag.

Fig. 9
Fig. 9
Close modal
Fig. 10
Fig. 10
Close modal

### 5.5 Damping Moments.

The overturning moment acting on the football causes the football to both precess and nutate. Much like drag, there are rotational motions that can result in a loss of energy due to aerodynamic friction. One of these acts to slow the spin rate of the ball and therefore acts along the symmetry axis of the ball. The ball also has an angular velocity that is perpendicular to the spin axis of the ball, and there is also energy dissipation associated with this motion. The second torque acts perpendicular to the longitudinal axis of the ball. The first of these torques is referred to as spin damping and is modeled using the equation [3,4]
$τp=12ρSD‖v→‖2CMp(ω1D‖v→‖)s^=12ρSD2‖v→‖CMpω1s^=12ρSD2‖v→‖CMps^(h→⋅s^)ItIs=IsMp[ω1,0,0]TMp=12IsρSD2VCMp$
(33)
The introduction of the term $D/‖v→‖$ is done to non-dimensionalize the angular velocity and is a standard practice. The product of the ratio of moments of inertia times the dot product in Eq. (33) is the projection of the angular velocity onto the symmetry axis of the football and is equal to the spin ω1. The remaining damping term is due to the angular velocity perpendicular to the spin $s^×s^˙$. Using this definition and the convention that it be non-dimensionalized as above leads to the model [3,4]
$τq=12ρSD‖v→‖2CMq(D‖v→‖)s^×s^˙=12ρSD2‖v→‖CMqs^×h→×s^=12ρSD2‖v→‖CMq(h→−(h→⋅s^)s^)≈ItMq[0,h~2−αh~0,h~3−βh~0]TMq=ρSD2VCMq2It$
(34)
References [3,4] make a distinction between angular velocity in space and the angular velocity relative to the velocity vector, and usually write the above equation with two coefficients that are summed. In practice, it is very difficult to make this distinction experimentally. Chapter 9 of McCoy [4] presents spin damping data for a bullet and an artillery shell. The coefficient for a bullet is (−0.0150) and for an artillery shell is (− 0.0178). While shells and bullets are generally smooth surfaced, they have greater relative surface area and rifling cuts grooves that result in drag. Something of a guess suggests that these numbers are 2–3 times higher than the appropriate number for a football and value of −0.005 would be a reasonable guess for a football. Subsequent to these estimates, we realized that Fig. 1 in Ref. [1] has an image indicating the times that the ball completed a rotation. After digitizing the image and determining the times of the vertical marks, the angular velocity can be calculated by dividing 2π by the difference in the times of neighboring marks, and the time associated with each difference taken as the mid-point between the marks. Taking the natural logarithm of the angular velocities and fitting a line to that data yield an estimate −0.0347 of the exponent associated with the decay of angular velocity. Converting this to a spin damping coefficient yields a value of −0.00383, very close to the estimate based on ballistics data. This number will be used in the simulation results. We were unable to find any data or methodology to make an estimate of the pitch damping coefficient. A value of $CMq=−0.5$ seems to be too large in that it suppresses oscillations that experience suggests are sustained in real passes. Using this limited observation as a guide, a value of $CMq=−0.1$ will be used.

The change in the spin velocity due to spin damping during the flight of the pass we have been considering in the previous examples is illustrated in Fig. 11. The response is the simple exponential decay associated with a stable, first-order, linear, differential equation. The change in spin rate during the pass is about 5% reduction. Due to the linearity of the model, the relative change will be the same for any initial spin rate. The effect of the pitch/yaw damping coefficient on the long pass used in most of the examples is shown in Fig. 12. The dashed line on the plot is the curve from Fig. 10. Since the energy dissipation associated with this damping term is proportional to the square of the angular velocity, this term will have a more pronounced effect on the pass with a high initial rotational angular velocity shown in Fig. 6. The effect of this term will also be more significant for a place kick where the ball has very large rotational angular velocity (i.e., angular velocity perpendicular to the symmetry axis).

Fig. 11
Fig. 11
Close modal
Fig. 12
Fig. 12
Close modal
Adding the pitch/yaw damping coefficient to our linear model of the spiral pass yields the set of equations
$(α˙β˙h~˙2h~˙3)=[−Lα−P01P−Lα−10−P(Lα+Mq)−MαMq0Mα−P(Lα+Mq)0Mq]×(αβh~2h~3)+[10P0]G$
(35)
The main thing to note is the appearance of Mq on the diagonal of the matrix in the lower right-hand corner of the matrix. Recall from their definition that the second and third components of the modified angular momentum correspond to angular velocities. Since Mq is negative, the resulting torque will always have a sign opposite to the angular velocity that causes it. Since the product of the torque times the angular velocity has units of power, it follows that this power will always be negative, and like drag, will always result in a loss of kinetic energy from the ball.

### 5.6 Coriolis and Centripetal Accelerations.

The Coriolis effect is an acceleration that is apparent to an observer in a rotating reference due to the velocity of the football. If the football is in constant velocity motion in a non-rotating reference frame, the football will appear to an observer in the rotating frame to be accelerating at a rate of $−2ω→E×v→$. In this case, our rotating reference frame is the base frame and it is rotating at the angular velocity of the Earth $ω→E$. In order to incorporate this term into the equations of motion, we need to obtain an expression for it in the base frame. Starting with a base frame located at the equator, with its 1-axis pointing north, imagine moving this frame north along a line of constant longitude to the latitude of its location on the field. Then imagine rotating it about the up or 2-axis clockwise until the 1-axis is parallel to the sideline. This corresponds to a 3–2–1 Euler angle sequence, except that according to the normal convention a positive latitude (L) corresponds to a negative rotation about the 3-axis, and the change in azimuth (A) to align with the field also corresponds to a negative rotation relative to 2. The transformation matrix associated with this rotation is
$C32(L,A)=[cosAcosL−cosAsinLsinAsinLcosL0−sinAcosLsinAsinLcosA]$
(36)
Denoting the Earth’s angular velocity as [ωE, 0, 0]T, the Earth’s angular velocity in the base frame of our football field is $ω→E=ωE[cosAcosL,sinL,sinAcosL]T$ and the adjustment that we need to add to the right-had side of Eq. (3) is
$a→C=−2ωE→×v→=2ωE(−v3sinL−v2sinAcosLv1sinAcosL+v3cosAcosL−v2cosAcosL+v1sinL)$
(37)
Because the base frame is attached to the field and the field is on the surface of the Earth, the frame is under constant centripetal acceleration toward the axis of rotation. Referring to the WGS84 [13] datum, the radius of the prime vertical at latitude L and altitude h is $a2/a2cos2L+b2sin2L+h$ where a and b are the semi-major and semi-minor axes of the reference ellipsoid. The distance between the rotation axis and this location is just the product of the radius times the cosine of the latitude. Since the direction of this acceleration is parallel to the plane of the equator in the base reference frame on the field, the apparent centripetal acceleration is
$a→ωE2=(−cosAsinLcosLsinAsinL)(ωE2(6378137+h)cosLcos2L+0.9933056sin2L)$
(38)
In this formula, the values of a and b are taken from Ref. [13] as ωE = 7.292115e − 05 rad/s. Running the simulation for the initial conditions of our long pass at the equator and with an azimuth angle of 90 deg changes the location of the end of the pass by less than one-tenth of a percent. This is the same order-of-magnitude as the Magnus force, so we conclude that these two effects can be neglected.

## 6 Stability of the Spiral Pass

The stability of a pass can be determined by evaluating the eigenvalues of the matrix (Eq. (35)). The four eigenvalues can be computed using the formula
$λ1…4=12[Mq−Lα±iP±4Mα−P2−i2P(Lα+Mq)+Lα2+2LαMq]$
(39)
The linear model is stable if the real parts of all of the eigenvalues are negative. One of the pair of eigenvalues with the largest real part is computed by evaluating the above equations with the two ±’s replaced by +. Setting this equation equal to zero, and solving for the angular velocity gives the relationship between pass velocity in meters-per-second and angular velocity in radians-per-second that defines the stability boundary
$ωcrit=V4ρS(CLαIy−CqD2m)2(8CMαIym+ρSCqD(4CLαIy−CqD2m))CLαCqDIx2Iym2=kcrit(δ)V;kcrit(0)=7.39$
(40)
For a given value of V in mph, the pass is stable if the angular velocity is faster than ωcrit. In units of rpm and meters-per-second, the constant in Eq. (40) is 1.73. The notation kcrit(δ) is included as a reminder that since some of the aerodynamic coefficients depend on δ, then so does kcrit. The correct value of delta to use should be that corresponding to the equilibrium angle; namely, a sideslip equal to the yaw of repose and the corresponding angle of attack for the actual flight conditions. For a well behaving pass with sufficient spin, using δ = 0 is reasonably accurate. Since the linear velocity decreases after the release of the pass and the angular velocity remains relatively constant, a pass that is released with an angular velocity equal to the critical value will become stable.

To illustrate what happens as the initial spin rate is reduced to near and then below the critical value, the results of six simulation are presented in Figs. 13 and 14. The initial release velocity is reduced slightly to 60 MPH. The initial release angle is 30 deg and the starting sideslip and angle of attack are zero. For the results in Fig. 13, the angular velocity at the release of the ball is above the critical value. The initial spin rates and ratio of angular velocity to linear velocity are shown in the legends. As the rpm is reduced in the first figure, the first three curves show that the angle of repose is decreasing due to the reduced spin rate and the nutation is becoming more pronounced. For the case nearly at the critical value, there is an interesting change in behavior. Using Eq. (39) to calculate the expected frequencies of the motion yields eigenvalues with negative real parts and imaginary parts corresponding to frequencies of 0.75 Hz and 3.25 Hz. The simulation shows that the ball precesses about 1.25 periods at the slower frequency and then precesses at the fast frequency. It seems reasonable to expect that under certain conditions, the reverse transition might happen. The results in Fig. 14 show results for an initial spin rate below the critical value. These results show precession at a frequency corresponding to the fast frequency that remains bounded. Part of the reason for the bounded behavior is that the reduction in linear velocity and the increase in δ both tend to reduce the magnitude of the overturning moment ($CMα(δ$) decreases with δ). The behavior here is similar to a precessing gyroscope as the spin rate drops. As the magnitude of the gyroscope’s angular moment decreases, the angle between the gyroscope axis and the vertical increases until the gyroscope falls over. Eventually, the football will do the same thing. Because the available aerodynamic data are not valid at angles much larger than those shown in Fig. 14, simulation results showing a transition to tumbling or some other motion would be unreliable and speculative.

Fig. 13
Fig. 13
Close modal
Fig. 14
Fig. 14
Close modal

Figure 15 plots the simulated trajectories of each of the passes in the previous two figures. If drag were the only contributing factor, the results in Figs. 13 and 14 would suggest that because of the larger sideslip angle that the largest spin rate case would have higher drag and shorter distance. At least when compared to the stable throws, the reason this pass is longer is that the larger average sideslip angle results in a larger angle of attack according to Eq. (26), and therefore, comparatively greater average lift and slightly longer flight time. The group of passes associated with 375–500 rpm travels relatively similar distances downfield, with the change in drift being only about a yard. Note that in all these cases, the initial velocity is the same, just the rpm is changing; reducing the initial rpm by a little less than half reduces the distance of the pass by nearly $6yards$.

Fig. 15
Fig. 15
Close modal

## 7 Effect of Drag Model and Release on Trajectory

In the final set of simulation results, we consider the effect of changing the drag model and the importance of the release on the motion of the ball and the distance traveled. In the first two cases shown in Figs. 16 and 17, the drag model used is the one based on Rae’s data [11]. The difference between these two cases is that in the first curve, the ball is released with the symmetry-spin axis and velocity vector aligned, and in the second the ball is released at the equilibrium state. The wind angle plots begin at different points, but from the mid-point of the pass onward are nearly identical. There is also almost no difference in the trajectories. The next two curves repeat the initial conditions but with a drag model derived from the data for the NCAA football [7]. In contrast to the data from Ref. [11] and the NFL football [7] where the drag curves are relatively flat near zero angle of incidence, the NCAA data show a dependence on incidence at low angles. The most notable effect is shown in Fig. 17 where the change in the distance of the throw is about 5 yards. This is an effect just due to changing the source of the data for drag. The implication is that much better information about the aerodynamics is needed to be able to use simulation models to make accurate predictions about football trajectories. The corresponding pair of curves in the wind angle plot (Fig. 16) shows nearly identical behavior shortly after the release. The higher drag results in lower linear velocities, so the sideslip angle is larger when compared to the results using Rea’s drag data. The final two curves in the figure correspond to the ball being released at a 10 deg angle above the velocity vector. In the first curve, the angular velocity of the ball is zero and in the second it is $50degs−1$. These are the same release conditions in Figs. 5 and 6. The wind angle plots associated with these two initial conditions are markedly different from the other curves in Fig. 16. The effect on the length of the pass and the drift is not very significant, only 1–2 yards. This suggests that if the angle and velocity of pass are the same, the angle and angular velocity of the ball relative to the velocity vector are not that important. This does not imply that the mechanics of the pass are not important, it is possible that for the poor release conditions considered here, a quarterback may not be able to impart the same spin and linear velocity.

Fig. 16
Fig. 16
Close modal
Fig. 17
Fig. 17
Close modal

## 8 Conclusion

This paper developed a mathematical model for the dynamics of a football. The equations are valid for arbitrary motions of the ball. The equations include parameters that characterize the relationship between the forces and moments on the ball and motion of the ball relative to the air. The available experimental data do not allow the characterization of these parameters for arbitrary motions, so we have limited our examples to cases where the ball has substantial spin, like a spiral pass, and does not tumble end-over-end. We noted in our discussion that there are significant questions about the accuracy of the available experimental data, and in one case our estimate is an educated guess. An obvious recommendation is that there needs to be work done to better characterize the football’s aerodynamic coefficients. One possibility is additional wind-tunnel experiments. If these experiments are performed, it is essential that the experiment design allows for accurate measurement of the effect of the mounting arrangement of the ball on the measured data. An alternative is to use free-flight data. The data reduction procedure is more complicated and measurements need to be made to identify the characteristic frequencies of the motion of the ball as well as the changes in these frequencies. It is also expected that there will be ball-to-ball variations in the aerodynamic coefficients. It is essential to know the magnitude of these variations prior to attempting to make simulation-based predictions about the flights of real footballs.

While there are significant questions about the quality of the existing aerodynamic data for footballs, there is a dearth of information that could be used to validate the simulation results. A specific set of aerodynamic data is only good if it allows one to predict the trajectory and dynamic motion of a football. A quality data set with information about initial conditions (or conditions at some point on the trajectory), trajectories, impact points, velocities, and frequencies of motion does not seem to exist. At this time, this seems to be the greatest deficiency of information. Without this information, there can be little certainty in any ability to make predictions.

The paper discussed how each aerodynamic coefficient affects the motion of a pass. The discussion showed that the wobbling of the ball was due to the interaction between the momentum of the spin and an aerodynamic moment that acts to turn the nose of the ball away from the velocity vector. The ball precesses and nutates at two characteristic frequencies. The precession may occur at either frequency, but if it is the higher frequency, it follows that the ball was released with a relatively large rotational angular velocity and was probably not a good pass. Good passes always precess at the slow frequency. The analysis showed that there is an interaction between the angular velocity of the velocity vector and the momentum that causes the ball to rotate with the angular velocity: nose-up at release and nose-down at the end of the pass. The nose of the ball will on average lie to the right of the velocity vector for a right-handed quarterback and to the left for a left-handed quarterback. This angle of repose interacts with the lift on the ball to cause a lateral drift; right for a right-handed pass and left otherwise. The force associated with the lift causes the velocity vector to rotate and this rotation cases a pass to travel with a slight nose-up, regardless of the hand throwing the pass. While it is tempting to compare this drift with the motion of a slider or curve ball, the Magnus effect does not affect footballs in a significant way and can be neglected. We have left open the possibility that there may be cases where this might not be true, but these cases involve the ball traveling with the spin axis at right angles to the velocity. A tumbling ball like a place kick does not satisfy this condition. The paper discussed the effect of drag on the linear velocity and the angular velocities. Finally, the paper derived a stability condition for a spinning football and presented some examples showing simulated motions of unstable and nearly unstable cases.

As a closing observation, we would like to note that the paper made many assumptions about the motion of a football when it leaves a quarterback’s hand. It is expected that the mechanics of passing impose their own constraints on the relationships between angles and linear and angular velocity at the point of release. It may not be possible to throw some of the passes simulated here, but more importantly, maybe it should not be tried by a person.

## Conflict of Interest

This article does not include research in which human participants were involved. Informed consent not applicable. This article does not include any research in which animal participants were involved.

## Data Availability Statement

The authors attest that all data for this study are included in the paper.

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