The Physics of Putting

Connect the Stroke, Impact, and Green

A putting model has three connected stages: the golfer moves the putter, impact establishes the ball’s velocity and spin, and contact with the green changes both. A reproducible stroke does not guarantee a reproducible putt if the impact or surface changes. Conversely, the same rollout can result from different strokes and launch conditions. We derive simple models for each stage, state their limits, and keep calculated examples separate from measurements of golfers.

Rolling Versus Sliding Dynamics

Initial spin is an outcome of impact, not universally zero. Loft, impact location and contact conditions can also produce a short airborne or bouncing phase. Penner explicitly distinguishes those possibilities before adopting pure rolling as an approximation for his path calculations (Penner 2002). The following contact model begins after any flight: a rigid spherical ball remains on a stationary, level plane, with constant kinetic friction and no rolling-resistance moment during the sliding interval.

Let $R$ be ball radius, $I=\kappa mR^2$ its central inertia, and choose the positive angular rate so forward rolling satisfies $v=R\omega$. A uniform solid sphere has $\kappa=2/5$; a layered golf ball need not have exactly that value. The signed slip speed is $s=v-R\omega$. For $s\ne0$, \[ \dot v=-\mu_k g\,\operatorname{sgn}(s),\qquad \dot\omega=\frac{\mu_k g}{\kappa R}\operatorname{sgn}(s). \] Friction opposes slip, not necessarily translation: an overspinning ball can accelerate forward while losing total kinetic energy. The slip sign stays fixed until it reaches zero. Integrating gives \[ t_r=\frac{\kappa|v_0-R\omega_0|}{(1+\kappa)\mu_k g},\qquad v_r=\frac{v_0+\kappa R\omega_0}{1+\kappa}, \] \[ \begin{aligned} x_r&=v_0t_r-\frac12\mu_k g\,\operatorname{sgn}(s_0)t_r^2,\\ E_0-E_r&=\frac{m\kappa}{2(1+\kappa)}s_0^2. \end{aligned} \] Here $x_r$ is signed displacement, not path length if translation reverses. Pure rolling initially gives $t_r=0$. These formulas require $\mu_k>0$ and do not cover a bouncing, deforming or intermittently contacting ball. The combined rotational and translational energy is $E=\tfrac12mv^2+\tfrac12I\omega^2$; during slip, $\dot E=-\mu_kmg|s|$.

For the illustrative values $v_0=2.0$ m/s, $\omega_0=0$, $\mu_k=0.25$ and $g=9.81$ m/s$^2$, the uniform-sphere model gives \[ \begin{aligned} t_r&=0.2330 \mathrm{s},\\ x_r&=0.3994 \mathrm{m},\\ v_r&=1.4286 \mathrm{m/s}. \end{aligned} \] It dissipates $2/7$ of the initial kinetic energy. This example already contradicts treating every skid as a few centimeters or a fixed $0.1$ s. Its fraction of the whole putt also depends on the subsequent rolling resistance and target distance; no universal 20% follows from the equations.

Putt Paths on Slopes

After rolling begins, model the green locally as a fixed plane with slope angle $\alpha$, downslope unit vector $\hat{\mathbf s}$ and sufficient static traction to maintain rolling. Represent resistance by a moment of magnitude $c_rNR$ opposing the rolling angular velocity, where $N=mg\cos\alpha$ and $c_r\ge0$ is dimensionless. This is a declared effective contact law, not a kinetic-sliding coefficient. Neglect aerodynamic forces and spin about the plane normal.

Eliminating the tangential contact force between Newton’s and Euler’s equations, using the no-slip relation, gives \[ \begin{aligned} \ddot{\mathbf r} &=\frac{g\sin\alpha}{1+\kappa}\hat{\mathbf s}\\ &\quad-a_0\cos\alpha\,\frac{\dot{\mathbf r}}{\|\dot{\mathbf r}\|},\\ a_0&=\frac{c_rg}{1+\kappa},\quad \|\dot{\mathbf r}\|>0. \end{aligned} \tag{1}\] Here $a_0$ is the level-green deceleration in this model. The factor $1/(1+\kappa)$ accounts for rotational inertia. With $\kappa=2/5$ and no resistance, downslope acceleration is $(5/7)g\sin\alpha$, not $g\sin\alpha$. Static friction still supplies the torque needed to spin up the ball; zero rolling resistance does not mean zero tangential contact force. The Newton–Euler derivation of no-slip rolling is reviewed in Moebs et al. (2016b).

At rest the velocity direction is undefined. Equation 1 must end at a stopping event or be completed by a static contact/resistance law; assigning an arbitrary direction at zero speed does not determine whether the ball stays at rest on a slope. The rolling assumption must also be rejected when the required tangential force exceeds available traction. Real turf introduces anisotropy, speed dependence, local slope changes and stochastic deviations.

For initial motion across the fall line, set $\mathbf r(0)=0$ and $\dot{\mathbf r}(0)=(V,0)$, with $y$ downslope. Locally, \[ x(t)=Vt-\frac12a_0\cos\alpha\,t^2+O(t^3), \] \[ y(t)=\frac{g\sin\alpha}{2(1+\kappa)}t^2+O(t^3). \] Thus the leading small-time relation is approximately parabolic, $y\simeq g\sin\alpha\,x^2/[2(1+\kappa)V^2]$. It is not an exact finite-putt parabola when resistance changes the velocity direction and speed. Do not extrapolate this expansion to the stopping time to predict the aim point of a long putt. Hole capture additionally depends on arrival speed and offset; crossing the hole’s location is not a capture model.

The Stimpmeter and Effective Deceleration

A Stimpmeter reading is a rollout distance, conventionally reported in feet, rather than a velocity. The USGA procedure uses repeated rolls in opposite directions on a sufficiently level area and averages the distances; its instructions also distinguish the standard and alternate notches (United States Golf Association n.d.). It is a surface-comparison procedure, not a direct measurement of a microscopic friction coefficient.

If the established-rolling entry speed $v_s$ is known and the subsequent level deceleration is constant, elementary kinematics gives \[ \begin{aligned} a_0&=\frac{v_s^2}{2L_s},\\ \mu_{\mathrm{eff}}&=\frac{a_0}{g},\\ c_r&=(1+\kappa)\mu_{\mathrm{eff}}. \end{aligned} \] The last equality belongs to the resistance-moment model above. $\mu_{\mathrm{eff}}$ is a deceleration ratio; it is neither $\mu_k$ nor a universal static-friction coefficient. Energy is dissipated during rollout, so this is not an assertion of mechanical-energy conservation.

Penner uses a historical entry-speed estimate of $1.83$ m/s (Penner 2002). Using that value as an illustrative calibration, an 8-foot rollout gives $a_0=0.687$ m/s$^2$, whereas 13 feet gives $0.423$ m/s$^2$. Their ratio is $13/8=1.625$, not two. These selected distances do not define universal course categories. Device/notch choice, the launch-to-roll transition and surface conditions must be included in a measured calibration rather than assigning $1.83$ m/s to every modern setup.

A Compound Pendulum Is a Conditional Stroke Model

For a rigid assembly rotating about a fixed horizontal pivot, let $I_p$ be inertia about that pivot and $d$ the pivot-to-COM distance. With a point torque $\tau(t)$ and no other losses, \[ I_p\ddot\theta+mgd\sin\theta=\tau(t),\qquad L_{\mathrm{eff}}=\frac{I_p}{md}. \] The equivalent pendulum length $L_{\mathrm{eff}}$ is not generally the actual pivot-to-putterhead radius $\ell_h$. Linearizing near the downward equilibrium and setting $\tau=0$ gives \[ \omega_n=\sqrt{g/L_{\mathrm{eff}}},\qquad T=2\pi\sqrt{L_{\mathrm{eff}}/g}. \] For $L_{\mathrm{eff}}=0.8$ m, $T=1.794$ s, $T/2=0.897$ s and $T/4=0.449$ s. Comparing the full period with a measured half-cycle is a factor-of-two error. Larger amplitude changes the passive period; $T/T_0=1+\theta_{\max}^2/16+O(\theta_{\max}^4)$ for angles in radians. Compound-pendulum assumptions and equivalent length are described in Moebs et al. (2016a).

A passive bottom-to-extremum-to-bottom excursion has equal outgoing and return durations. It does not derive a universal 2:1 backswing-to-downswing ratio. Starting from rest at the bottom requires an input to initiate motion at all. A real golfer can move the pivot, change joint configuration and stiffness, and apply time-dependent torque; event definitions also affect reported ratios. Grober proposes a driven oscillator and fixed timing/velocity-shape factors as a distance-control model (Grober 2011). Such a conditional account should not be replaced by a claim that an unforced pendulum proves optimal human tempo or minimal neuromuscular noise.

Connect Putter Speed to Ball Speed and Accuracy

For passive release from angle $\theta_0$, the ideal assembly’s head speed at its lowest point is \[ U_h=\ell_h\sqrt{\frac{2g}{L_{\mathrm{eff}}}(1-\cos\theta_0)} \simeq\ell_h\omega_n|\theta_0|. \] This is putterhead speed, not ball launch speed. In an ideal one-dimensional central collision of a moving free putter mass $M$ with a stationary ball of mass $m$, coefficient of restitution $e$ gives $v_b^+=(1+e)M U_h/(M+m)$. Real impact geometry, inertia, shaft/hand coupling, spin and the subsequent contact transition require a richer model or measured calibration. Do not silently identify $M$ with total putter mass or treat the central-collision formula as a complete golf impact law.

For a declared impact-to-established-roll speed gain $k_{\mathrm{imp}}$, write $v_r=k_{\mathrm{imp}}U_h$. Combining this calibration with level constant deceleration yields \[ D=\frac{k_{\mathrm{imp}}^2U_h^2}{2a_0}, \] where $D$ starts at established rolling, not necessarily at the impact point. Holding the impact transfer, surface and timing fixed explains why stroke amplitude can control distance in a small-angle model. It does not prove that golfers must keep timing fixed.

Direction and speed errors must also be separated from aiming and green reading. Karlsen, Smith and Nilsson studied 1,301 approximately 4-m putts from 71 skilled players and distinguished stroke direction from the aim line (Karlsen et al. 2008). Their stroke-specific variability estimates should not be promoted to universal hole-out probabilities. Standard deviation has units of degrees; angular variance has units of degrees squared. Even a deterministic $0.5^\circ$ direction error gives approximately $6.096\tan(0.5^\circ)=0.0532$ m lateral displacement at 20 feet before slope or speed effects. That calculation alone cannot establish capture, perceptual thresholds, or a coaching prescription.

Problems and Thought Experiments

Spin and Slip: Reproduce the zero-spin example above. Then set $R\omega_0=3.0$ m/s while keeping $v_0=2.0$ m/s. Show that translation speeds up during contact slip while total kinetic energy decreases. Identify which assumptions fail if the ball bounces.
A Local Slope Prediction: Use $\kappa=2/5$, $V=2.0$ m/s, $\alpha=2^\circ$, $a_0=0.15$ m/s$^2$ and $t=0.10$ s. Compute the second-order $x,y$ approximation and compare it with a numerical integration of Equation 1. Explain why this is not an aim recommendation for a 20-foot putt with zero arrival speed.
Separate the Three Stages: For illustrative Stimp distances 9 and 12 feet, use $v_s=1.83$ m/s to calculate $a_0$. For $D=15$ feet of established rolling, obtain $v_r$, then $U_h$ using a supplied $k_{\mathrm{imp}}=1.5$. Use $L_{\mathrm{eff}}=0.8$ m and $\ell_h=0.9$ m to compute small-angle release amplitudes. Compare their ratio and the exact passive-energy correction. Which measurements would be needed before transferring this calculation to a golfer?

On This Site

Impact Mechanics and Ball Flight — Connect putterhead motion to ball launch while keeping the impact and subsequent rolling stages distinct.
Strokes Gained: Assumptions, Limits and Proper Interpretation — Separate stroke mechanics from outcome statistics and the context needed to interpret putting performance.

Cite this page

References

Grober, Robert D. 2011. Resonance as a Means of Distance Control in Putting. arXiv:1103.2827. https://doi.org/10.48550/arXiv.1103.2827.

Karlsen, Jon, Gerald Smith, and Johnny Nilsson. 2008. “The Stroke Has Only a Minor Influence on Direction Consistency in Golf Putting Among Elite Players.” Journal of Sports Sciences 26 (3): 243–50. https://doi.org/10.1080/02640410701530902.

Moebs, William, Samuel J. Ling, and Jeff Sanny. 2016a. University Physics Volume 1: Pendulums. OpenStax, Section 15.4. https://openstax.org/books/university-physics-volume-1/pages/15-4-pendulums.

Moebs, William, Samuel J. Ling, and Jeff Sanny. 2016b. University Physics Volume 1: Rolling Motion. OpenStax, Section 11.1. https://openstax.org/books/university-physics-volume-1/pages/11-1-rolling-motion.

Penner, A. Raymond. 2002. “The Physics of Putting.” Canadian Journal of Physics 80 (2): 83–96. https://doi.org/10.1139/p01-137.

United States Golf Association. n.d. Stimpmeter Instruction Booklet. https://www.usga.org/content/dam/usga/pdf/imported/StimpmeterBookletFINAL.pdf.

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--- description: "Putting mechanics connecting stroke motion, impact, slip, rolling resistance, slope and distance calibration." css: ../../../css/putting-chapter.css --- # The Physics of Putting {#sec-putting} ::: {.callout-note} ## Connect the Stroke, Impact, and Green {#lay:putting} A putting model has three connected stages: the golfer moves the putter, impact establishes the ball's velocity and spin, and contact with the green changes both. A reproducible stroke does not guarantee a reproducible putt if the impact or surface changes. Conversely, the same rollout can result from different strokes and launch conditions. We derive simple models for each stage, state their limits, and keep calculated examples separate from measurements of golfers. ::: ## Rolling Versus Sliding Dynamics {#sec-putting_roll_slide} Initial spin is an outcome of impact, not universally zero. Loft, impact location and contact conditions can also produce a short airborne or bouncing phase. Penner explicitly distinguishes those possibilities before adopting pure rolling as an approximation for his path calculations [@penner2002putting]. The following contact model begins after any flight: a rigid spherical ball remains on a stationary, level plane, with constant kinetic friction and no rolling-resistance moment during the sliding interval. Let $R$ be ball radius, $I=\kappa mR^2$ its central inertia, and choose the positive angular rate so forward rolling satisfies $v=R\omega$. A uniform solid sphere has $\kappa=2/5$; a layered golf ball need not have exactly that value. The signed slip speed is $s=v-R\omega$. For $s\ne0$, $$ \dot v=-\mu_k g\,\operatorname{sgn}(s),\qquad \dot\omega=\frac{\mu_k g}{\kappa R}\operatorname{sgn}(s). $$ Friction opposes *slip*, not necessarily translation: an overspinning ball can accelerate forward while losing total kinetic energy. The slip sign stays fixed until it reaches zero. Integrating gives $$ t_r=\frac{\kappa|v_0-R\omega_0|}{(1+\kappa)\mu_k g},\qquad v_r=\frac{v_0+\kappa R\omega_0}{1+\kappa}, $$ $$ \begin{aligned} x_r&=v_0t_r-\frac12\mu_k g\,\operatorname{sgn}(s_0)t_r^2,\\ E_0-E_r&=\frac{m\kappa}{2(1+\kappa)}s_0^2. \end{aligned} $$ Here $x_r$ is signed displacement, not path length if translation reverses. Pure rolling initially gives $t_r=0$. These formulas require $\mu_k>0$ and do not cover a bouncing, deforming or intermittently contacting ball. The combined rotational and translational energy is $E=\tfrac12mv^2+\tfrac12I\omega^2$; during slip, $\dot E=-\mu_kmg|s|$. For the illustrative values $v_0=2.0$ m/s, $\omega_0=0$, $\mu_k=0.25$ and $g=9.81$ m/s$^2$, the uniform-sphere model gives $$ \begin{aligned} t_r&=0.2330 \mathrm{s},\\ x_r&=0.3994 \mathrm{m},\\ v_r&=1.4286 \mathrm{m/s}. \end{aligned} $$ It dissipates $2/7$ of the initial kinetic energy. This example already contradicts treating every skid as a few centimeters or a fixed $0.1$ s. Its fraction of the *whole putt* also depends on the subsequent rolling resistance and target distance; no universal 20% follows from the equations. ## Putt Paths on Slopes {#sec-putting_slopes} After rolling begins, model the green locally as a fixed plane with slope angle $\alpha$, downslope unit vector $\hat{\mathbf s}$ and sufficient static traction to maintain rolling. Represent resistance by a moment of magnitude $c_rNR$ opposing the rolling angular velocity, where $N=mg\cos\alpha$ and $c_r\ge0$ is dimensionless. This is a declared effective contact law, not a kinetic-sliding coefficient. Neglect aerodynamic forces and spin about the plane normal. Eliminating the tangential contact force between Newton's and Euler's equations, using the no-slip relation, gives $$ \begin{aligned} \ddot{\mathbf r} &=\frac{g\sin\alpha}{1+\kappa}\hat{\mathbf s}\\ &\quad-a_0\cos\alpha\,\frac{\dot{\mathbf r}}{\|\dot{\mathbf r}\|},\\ a_0&=\frac{c_rg}{1+\kappa},\quad \|\dot{\mathbf r}\|>0. \end{aligned} $$ {#eq-putting_rolling} Here $a_0$ is the level-green deceleration in this model. The factor $1/(1+\kappa)$ accounts for rotational inertia. With $\kappa=2/5$ and no resistance, downslope acceleration is $(5/7)g\sin\alpha$, not $g\sin\alpha$. Static friction still supplies the torque needed to spin up the ball; *zero rolling resistance does not mean zero tangential contact force*. The Newton–Euler derivation of no-slip rolling is reviewed in @OpenStaxRollingReview. At rest the velocity direction is undefined. @eq-putting_rolling must end at a stopping event or be completed by a static contact/resistance law; assigning an arbitrary direction at zero speed does not determine whether the ball stays at rest on a slope. The rolling assumption must also be rejected when the required tangential force exceeds available traction. Real turf introduces anisotropy, speed dependence, local slope changes and stochastic deviations. For initial motion across the fall line, set $\mathbf r(0)=0$ and $\dot{\mathbf r}(0)=(V,0)$, with $y$ downslope. Locally, $$ x(t)=Vt-\frac12a_0\cos\alpha\,t^2+O(t^3), $$ $$ y(t)=\frac{g\sin\alpha}{2(1+\kappa)}t^2+O(t^3). $$ Thus the leading small-time relation is approximately parabolic, $y\simeq g\sin\alpha\,x^2/[2(1+\kappa)V^2]$. It is not an exact finite-putt parabola when resistance changes the velocity direction and speed. Do not extrapolate this expansion to the stopping time to predict the aim point of a long putt. Hole capture additionally depends on arrival speed and offset; crossing the hole's location is not a capture model. ## The Stimpmeter and Effective Deceleration {#sec-putting_stimp} A Stimpmeter reading is a rollout *distance*, conventionally reported in feet, rather than a velocity. The USGA procedure uses repeated rolls in opposite directions on a sufficiently level area and averages the distances; its instructions also distinguish the standard and alternate notches [@USGAStimpmeterReview]. It is a surface-comparison procedure, not a direct measurement of a microscopic friction coefficient. If the established-rolling entry speed $v_s$ is known and the subsequent level deceleration is constant, elementary kinematics gives $$ \begin{aligned} a_0&=\frac{v_s^2}{2L_s},\\ \mu_{\mathrm{eff}}&=\frac{a_0}{g},\\ c_r&=(1+\kappa)\mu_{\mathrm{eff}}. \end{aligned} $$ The last equality belongs to the resistance-moment model above. $\mu_{\mathrm{eff}}$ is a deceleration ratio; it is neither $\mu_k$ nor a universal static-friction coefficient. Energy is dissipated during rollout, so this is not an assertion of mechanical-energy conservation. Penner uses a historical entry-speed estimate of $1.83$ m/s [@penner2002putting]. Using that value as an *illustrative calibration*, an 8-foot rollout gives $a_0=0.687$ m/s$^2$, whereas 13 feet gives $0.423$ m/s$^2$. Their ratio is $13/8=1.625$, not two. These selected distances do not define universal course categories. Device/notch choice, the launch-to-roll transition and surface conditions must be included in a measured calibration rather than assigning $1.83$ m/s to every modern setup. ## A Compound Pendulum Is a Conditional Stroke Model {#sec-putting_pendulum} For a rigid assembly rotating about a fixed horizontal pivot, let $I_p$ be inertia about that pivot and $d$ the pivot-to-COM distance. With a point torque $\tau(t)$ and no other losses, $$ I_p\ddot\theta+mgd\sin\theta=\tau(t),\qquad L_{\mathrm{eff}}=\frac{I_p}{md}. $$ The equivalent pendulum length $L_{\mathrm{eff}}$ is not generally the actual pivot-to-putterhead radius $\ell_h$. Linearizing near the downward equilibrium and setting $\tau=0$ gives $$ \omega_n=\sqrt{g/L_{\mathrm{eff}}},\qquad T=2\pi\sqrt{L_{\mathrm{eff}}/g}. $$ For $L_{\mathrm{eff}}=0.8$ m, $T=1.794$ s, $T/2=0.897$ s and $T/4=0.449$ s. Comparing the full period with a measured half-cycle is a factor-of-two error. Larger amplitude changes the passive period; $T/T_0=1+\theta_{\max}^2/16+O(\theta_{\max}^4)$ for angles in radians. Compound-pendulum assumptions and equivalent length are described in @OpenStaxPendulumReview. A passive bottom-to-extremum-to-bottom excursion has equal outgoing and return durations. It does not derive a universal 2:1 backswing-to-downswing ratio. Starting from rest at the bottom requires an input to initiate motion at all. A real golfer can move the pivot, change joint configuration and stiffness, and apply time-dependent torque; event definitions also affect reported ratios. Grober proposes a *driven* oscillator and fixed timing/velocity-shape factors as a distance-control model [@Grober2011PuttingReview]. Such a conditional account should not be replaced by a claim that an unforced pendulum proves optimal human tempo or minimal neuromuscular noise. ## Connect Putter Speed to Ball Speed and Accuracy {#sec-putting_tempo} For passive release from angle $\theta_0$, the ideal assembly's head speed at its lowest point is $$ U_h=\ell_h\sqrt{\frac{2g}{L_{\mathrm{eff}}}(1-\cos\theta_0)} \simeq\ell_h\omega_n|\theta_0|. $$ This is *putterhead speed*, not ball launch speed. In an ideal one-dimensional central collision of a moving free putter mass $M$ with a stationary ball of mass $m$, coefficient of restitution $e$ gives $v_b^+=(1+e)M U_h/(M+m)$. Real impact geometry, inertia, shaft/hand coupling, spin and the subsequent contact transition require a richer model or measured calibration. Do not silently identify $M$ with total putter mass or treat the central-collision formula as a complete golf impact law. For a declared impact-to-established-roll speed gain $k_{\mathrm{imp}}$, write $v_r=k_{\mathrm{imp}}U_h$. Combining this calibration with level constant deceleration yields $$ D=\frac{k_{\mathrm{imp}}^2U_h^2}{2a_0}, $$ where $D$ starts at established rolling, not necessarily at the impact point. Holding the impact transfer, surface and timing fixed explains why stroke amplitude can control distance in a small-angle model. It does not prove that golfers must keep timing fixed. Direction and speed errors must also be separated from aiming and green reading. Karlsen, Smith and Nilsson studied 1,301 approximately 4-m putts from 71 skilled players and distinguished stroke direction from the aim line [@karlsen2008putting]. Their stroke-specific variability estimates should not be promoted to universal hole-out probabilities. Standard deviation has units of degrees; angular variance has units of degrees squared. Even a deterministic $0.5^\circ$ direction error gives approximately $6.096\tan(0.5^\circ)=0.0532$ m lateral displacement at 20 feet before slope or speed effects. That calculation alone cannot establish capture, perceptual thresholds, or a coaching prescription. ::: {.callout-note} ## Problems and Thought Experiments {#ex:putting} 1. **Spin and Slip:** Reproduce the zero-spin example above. Then set $R\omega_0=3.0$ m/s while keeping $v_0=2.0$ m/s. Show that translation speeds up during contact slip while total kinetic energy decreases. Identify which assumptions fail if the ball bounces. 2. **A Local Slope Prediction:** Use $\kappa=2/5$, $V=2.0$ m/s, $\alpha=2^\circ$, $a_0=0.15$ m/s$^2$ and $t=0.10$ s. Compute the second-order $x,y$ approximation and compare it with a numerical integration of @eq-putting_rolling. Explain why this is not an aim recommendation for a 20-foot putt with zero arrival speed. 3. **Separate the Three Stages:** For illustrative Stimp distances 9 and 12 feet, use $v_s=1.83$ m/s to calculate $a_0$. For $D=15$ feet of established rolling, obtain $v_r$, then $U_h$ using a supplied $k_{\mathrm{imp}}=1.5$. Use $L_{\mathrm{eff}}=0.8$ m and $\ell_h=0.9$ m to compute small-angle release amplitudes. Compare their ratio and the exact passive-energy correction. Which measurements would be needed before transferring this calculation to a golfer? ::: ::: {.callout-note} ## On This Site - **[Impact Mechanics and Ball Flight](../../impact-mechanics-and-ball-flight.html)** — Connect putterhead motion to ball launch while keeping the impact and subsequent rolling stages distinct. - **[Strokes Gained: Assumptions, Limits and Proper Interpretation](../../strokes-gained-limitations.html)** — Separate stroke mechanics from outcome statistics and the context needed to interpret putting performance. :::