A Tale of Two Coefficients of Restitution

Most engineers are familiar from their dynamics course (assuming, of course, it was required) with the coefficient of restitution during the impact of two bodies. Then, if they’re introduced to pile dynamics, they’re told that certain parts of the pile-hammer-soil system have a coefficient of restitution of their own. Most of the time they’re solemnly told that the two different “coefficients of restitution” are different. I’ve said this over the years.

But are they that different? The answer is “not really,” but to understand this we need to delve into the physics of impact. The way that wave equation routines such as the WEAP family model use the coefficient of restitution is based on classical impact physics but the way it’s implemented is different from the way the dynamic formulae used it.

In this monograph we’re going to compare the classical impact model with a simple spring-mass model to see if they’re really the same.

Basics of the Coefficient of Restitution

Meriam (1975) defines the coefficient of restitution e as “…the ratio of the impulse during the period of restoration to the impulse during the period of deformation.” The same reference also has the following warning:

The coefficient of restitution is often considered a constant for given geometries and a given combination of contacting materials. Actually it depends upon the impact velocity and approaches unity as the impact velocity approaches zero. A handbook value for e is generally unreliable. Impact is a complex phenomenon involving a loss of initial elastic waves within the bodies, and the generation of sound energy.

That being the case, in most presentations of the wave equation the coefficient of restitution is shown as the way by which the cushion and other inextensible interfaces respond to compression. A schematic of this is shown below.

Bilinear Response of Cushion (from Holloway (1976))

To demonstrate the relationship between this and simple impulse-momentum impact mechanics, we will consider this problem in two ways: a) a spring-mass model with the ram, pile and hammer cushion and b) the classic impulse-momentum impact method.

Spring-Mass Model

Let’s start with a spring-mass model. A diagram is shown below.

Schematic of Ram-Cushion-Pile System (modified from from Holloway (1976))

The equations of motion are as follows:

M{\frac {d^{2}}{d{t}^{2}}}{\it x_r}(t)=-{\it k}\,\left ({\it x_r}(t)-{\it x_p}(t)\right ) (1)

m{\frac {d^{2}}{d{t}^{2}}}{\it x_p}(t)={\it k}\,\left ({\it x_r}(t)-{\it x_p}(t)\right ) (2)


  • t = time
  • x_r = displacement of ram
  • x_p = displacement of pile
  • k = spring constant/stiffness of cushion
  • M = mass of ram
  • m = mass of pile

We should first define the following:

m' = \frac{M}{m} (3)

where m' = ratio of ram to pile mass. This is different from the way this ratio is usually defined, but it makes the math simpler.

This problem is similar in many ways to that of Warrington (2021) and can be solved with Laplace transforms in a similar fashion. We need to define the initial conditions at first impact as follows:

v_r(0) = V_0 (4)

v_p(0) = 0 (5)

x_r(0) = x_p(0) = 0 (6)


  • V_0 = impact velocity of ram = \sqrt{2 g_c e_{ff} s}
  • g_c = gravitational constant
  • e_{ff} = mechanical efficiency of the hammer
  • s = stroke (of a single acting hammer) or equivalent stroke of a hammer

Because the spring is bilinear, i.e., it has two stiffnesses (one in compression and the other in rebound,) we need to determine the point at which the spring compression is at a maximum and thus the difference between the ram and pile deflection is at a minimum, at the beginning of rebound. Solving for the two deflections yields

x_r (t) = \frac{m' t \omega_1 + \sin (\omega_1 t) V_o}{\omega_1 (m' +1)}  (7)

x_p (t) = \frac{ (\omega_1t - \sin (\omega_1 t))m' V_o}{\omega_1 (m' +1)} (8)


\omega_1 = \sqrt{\frac{k(m'+1)}{m' m}} (9)

In this case \omega_1 = natural frequency in compression, rad/sec. The spring deflection is the difference between the two, or

\Delta x (t) = \frac {\sin(\omega_1 t) V_0}{\omega_1} (10)

where \Delta x(t) = compression of the cushion. The minimum value of this takes place when

t = \frac{\pi}{2 \omega_1} (11)

and the spring deflection at that point is

\Delta x( \frac{\pi}{2 \omega_1}) = \frac{V_0}{\omega_1} (12)

The velocities of the ram and pile are the same:

v_p  (\frac{\pi}{2 \omega_1}) = v_r ( \frac{\pi}{2 \omega_1}) = V = \frac{m' V_0}{m'+1} (13)

where V is the velocity of both ram and pile at maximum compression of the spring. It’s possible to continue to use the same coordinate system, but to simplify the math we will do the following:

  • Reset the time to zero, t'=0 .
  • Reset the ram displacement x'_p (t' = 0) = 0
  • Reset the pile displacement to be the maximum displacement of the rebounded spring, thus x'_r (t'=0) = \frac{e^2 V_0}{\omega_1}
  • Leave the initial velocities the same as the final velocities from the compression phase.

The equations of motion in the rebound phase are

{\it m'}\,m{\frac {d^{2}}{d{t}^{2}}}{\it x'_r}(t)=-{\frac {{\it k}\,\left ({\it x'_r}(t)-{\it x'_p}(t)\right )}{{e}^{2}}} (14)

m{\frac {d^{2}}{d{t}^{2}}}{\it x'_p}(t)={\frac {{\it k}\,\left ({\it x'_r}(t)-{\it x'_p}(t)\right )}{{e}^{2}}} (15)

where e is the coefficient of restitution. The solution to these equations with the noted initial conditions is

x'_r (t') = \frac{m' \omega_1 t' + m' e^2 + e^2 \cos(\frac{\omega_1 t'}{e}) V_0}{\omega_1 (m'+1)} (16)

x'_p (t') = \frac{m'(\omega_1 t' +e^2 - e^2 \cos(\frac{\omega_1 t'}{e})) V_0}{\omega_1 (m'+1)} (17)

The deflection of the rebound spring is

\Delta x'(t') = \frac {\cos(\frac{\omega_1 t'}{e}) e^2 V_0}{\omega_1} (18)

and the rebound state ends when the rebound spring returns to zero deflection at

t' = \frac{e \pi}{2 \omega_1} (19)

At this point the ram is travelling at a velocity of v_{1f} and the pile v_{2f} , which can be obtained by taking the derivative of Equations (14) and (15) and substituting the result of Equation (19). To simplify the results we can define the following ratios:

R' = \frac{v_{1f}}{V_0} (20)

R = \frac{v_{2f}}{V_0} (21)


  • R' = stroke speed restoration coefficient of the ram
  • R = stroke speed restoration coefficient of the pile
  • v_{1f} = ram velocity at separation from the cushion
  • v_{2f} = pile velocity at separation from the cushion

Making the appropriate substitutions yields for each

R' = \frac{m'-e}{m'+1} (22)

R = \frac{m'(1+e)}{m'+1} (23)

The result of Equation (22) is the same as discussed in Warrington and Erofeev (1995). It is interesting to note that the results are independent of the spring system between the ram and the pile.

Standard Rigid Body Model

So is this different from the rigid body model used in the dynamic formulae? The rigid body model is given in dynamics textbooks such as Meriam (1975) and Seely and Ensign (1921). For pile driving the problem is a little simpler because a) the motion and impact of the ram and pile is co-linear (or at least should be) and b) the initial velocity of the pile is taken to be zero. With both of those conditions, the coefficient of restitution of impact is

e = \frac{v_{2f}-v_{1f}}{V_0} (24)

As was the case before, the impact is in two stages, a compression stage and a rebound stage. At the transition point between the two stages, the ram and pile velocity is the same. By conservation of momentum,

M V_0 = (m + M) V (25)

At the point where the ram and pile separate, again by conservation of momentum,

(m + M) V = M v_{1f} + m v_{2f} (26)

If we combine Equations (3), (24), (25) and (26) and solve for v_{1f} and v_{2f} , we obtain the following:

  • The intermediate velocity is the same as given in Equation (13).
  • The final velocities of the ram and pile are the same as given in Equations (22) and (23), using the notation of Equations (20) and (21).

The basic result is the same. So what’s the point of going through what we did with the spring mass method? This exercise has accomplished two things:

  • We have two different methods which use the coefficient of restitution in different ways which give the same results, one of the goals of the investigation.
  • Including the spring can give us an estimate of the ram force, which is a major defect in using simple impulse-momentum assumptions. This was a major defect in many earlier estimates of this, as we discussed in our post So What is the Force of the Ram?

A worked example will clarify things.


Let us consider a Vulcan #1 hammer driving a HP 14×102 pile 50′ long. The hammer is using a micarta and aluminium cushion material in a standard pot (for an explanation of this, see our post Vulcan Onshore Tip #69: Micarta and Aluminum Cushion Material.) We will use all values in this example from the WEAP86 manual. The modulus of elasticity of this material is 350 ksi and the coefficient of restitution is 0.8; spring constant formulae can be found in the WEAP86 manual as well.

Let us begin by considering strictly the rigid body impact model. From this, making the appropriate substitutions the results are as follows:

  • The mass/weight ratio between the ram and the pile is m' = \frac{5000}{50 \times 102} = 0.98 (Equation (3))
  • Using the standard efficiency, the striking velocity V_0 = 11.38 ft/sec.
  • The “intermediate” velocity at maximum compression V = 5.63 ft/sec. (Equation (13))
  • The final velocities of the ram and pile are v_{1f} = 1.04 and v_{2f} = 10.13 ft/sec respectively. (Equations (20) and (21))
  • The stroke speed restoration coefficients are thus R' = 0.09 and R = 0.89 respectively. (Equations (20) and (21))

If we consider the effects of the intermediate spring, the following results can be added:

  • The spring constant for the hammer cushion is k = \frac{350000 \times 12 \times 99.4}{7.375} = 56.6 \times 10^6 lb/ft.
  • The natural frequency of the compressed system is \omega_1 = 849.7 rad/sec (Equation (9).) In U.S. units, the mass needs to be expressed in slugs.
  • The maximum compression of the spring is \Delta x_{max} = 0.013' = 0.16" (Equation (12).)
  • The maximum force in the spring is F_{max} = k \Delta x_{max} = 758,159 lbs. = 758 kips.

It is interesting to note that the maximum g-force on the ram is about 152 g’s, which is a more reasonable value than could be obtained using the “old Vulcan method” of ram force estimation.


  • Seely, F.B., and Ensign, N.E. (1921) Analytical Mechanics for Engineers. New York: John Wiley and Sons.
  • Meriam, J.L. (1975) Dynamics (SI Version.) Second Edition. New York: John Wiley and Sons.

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