Dynamics of the Weight StackMachine

Dynamics of the Weight Stack Machine

Introduction

Weight stack machines use weights to create the load (resistance). Since the weight is moving during the exercise the actual load experienced by the subject will vary depending on the acceleration of the weight mass. This can e.g. be demonstrated by measuring the resistance for static conditions, as well as for slow and rapid dynamic conditions. In this paper we present an overview of the physics involved and discuss some measurements on a leg extension/curl weight stack machine in order to illustrate this effect of inertia and acceleration on the resistance. Such dynamical responses make it difficult to achieve a desired resistance for the weight stack machines.

Physical background

The principal mechanical setup of the leg extension/curl weight stack machine is shown in the adjoining figure (extension mode). The weight W causes a moment M (via the system of

transmission) which the subject will try to overcome by exerting a force F or its equivalent moment denoted Mf ; ms is the mass of the leg support system, and rs is the distance of the center of mass of the leg supports system from the axis of rotation (g is the acceleration of gravity). (The weight stack machines are also equipped with a balancing weight so that the gravity term can be eliminated. ) According to mechanics, the equation of motion of the leg support becomes

M

F W

θ y

T

S

(1)

I¹

h

¹t

= M

− M − m

gr

sin(h)

where I is the moment of inertia of the leg support system visavi the axis of rotation (I includes also the moment of inertia of the balancing weight if such one is used). We can write this equation as

(2)

M

= M + m

gr

sin(h) + I ¹

h

¹t

The moment M exerted by the weight mass depends on the tension T of the belt which further depends on the acceleration and mass m of the weight W.

(3)

T = mg + m ¹

y

¹t

The effect of friction may be described using a friction parameter γ > 1 by replacing M in (1) and (2) with γM. (The friction parameter can be related to the inverse of the efficiency η of the machine for (slowly) lifting weights defined by g=mgDy/M_fDh = (change in potential energy of the weight)/(work expended).) The moment M will be T times a geometrical factor r(θ) which depends on the angle θ. Also the height y of the weight mass may be expressed as a function y(θ) of the angle θ. Thus, if we neglect the effects of friction (and the kinetic energy of the pulleys etc) we can finally write (2) on the form

(4)

M

= m 

 g + ¹

y

¹h



 ¹h

¹t 



2

+ ¹y

¹h ¹

h

¹t



 r(h) + m

gr

sin(h) + I ¹

h

¹t

= mgr(h) + m

gr

sin(h ) + m ¹

y

¹h



 ¹h

¹t 



2

r(h) +  I + m ¹y

¹h r(h) 

 ¹

h

¹t

According to this, the resistance Mf experienced by the subject is a function of the angle θ and its first and second time derivatives (angular velocity and acceleration) and can be divided into a static part Mstat and a dynamic part Mdyn.

(5)

M

= mgr(h) + m

gr

sin(h) M

= mr(h) ¹

y

¹h



 ¹h

¹t 



2

+  I + m ¹y

¹h r(h) 

 ¹

h

¹t

Mstat is only a function of the angle θ and the weight mass m and can be determined through static measurements for a set of angles thus giving the static resistance curve Mstat(θ). The function y(θ) can also be determined in a straightforward manner by measuring y for a set of angles θ. The moment of inertia can be calculated from the masses and dimensions of the mechanical parts of the leg support system. It can also be estimated using a “drop test”; that is, with the machine in the

flexion mode we lift the weight by turning the leg support into the θ = 0 position whereafter it is released and we measure θ as a function of time. In the “drop test” one should, however, use a heavy load enough in order to reduce the relative contribution of friction. If we neglect friction and the kinetic energy of the pulleys, and if we use the balancing weight to offset the gravity term of the leg support (rs = 0 in the equations), we get from the conservation of energy,

(6)

mgy(h(0)) = mgy(h(t)) + 1 2 m 

 ¹y

¹h ¹h(t)

¹t 



2

+ 1 2 I 

 ¹h(t)

¹t 



2

= mgy(h(t)) + 1 2



  m 

 ¹y

¹h 



2

+ I 

   ¹h(t)

¹t 



2

Thus, by measuring θ(t) and calculating the angular velocity, and knowing y as a function of θ, we can calculate I using equ (6). In this case I is the sum of two parts,

(7)

I = I

+ I

(Strictly speaking, if we measure I through (6) it will include the contributions from the pulleys and the lever S, since their kinetic energy is proportional to the square of the angular velocity also.) That is, I is the sum of the moment of inertia Is of the leg support, and the moment of inertia Ib of the balancing weight. Since Ib can be easily calculated from the geometry and the mass distribution of the balancing weight (or measured using the “pendulum method”), we can from (7) also determine the individual contribution of Is.

A toy model

In order to illustrate the theoretical development above we may consider a “toy model” of a weight stack machine. We will assume for simplicity that the leg support is balanced so that the gravity term of the leg support system is eliminated from the equations.

First we observe, that if we may neglect friction, and if we consider a slow turn of the leg support through a small angle ∆θ (“virtual displacement”), then from the conservation of energy we obtain,

(8)

M

( h)Dh = mg ¹y

¹h Dh u mg ¹y

¹h = M

( h)

which is equivalent to

(9)

¹h ¹y = r(h)

Thus, in our model, by defining the function Mstat(θ) (for a given weight mass m) the function y(θ) will also be determined. A very simple model is one where we have a linear relation

(10)

y(h) = kh

then equ (4) reduces to

(11)

M

= mgk + (I + mk

) ¹

h

¹t

which for a constant driving momentum Mf has a solution of the form (initial velocity is zero)

(12)

h(t) = 1 2 at

If we suppose that the leg swings through 90 degrees in about 0.4 second we may set the angular acceleration a to π/0.16 (rad/s² ) in (12). From a weight stack machine we obtain that it lifts the weight about 0.3 m during a 90 degree turn of the leg support giving k = 0.6/π (m/rad) in (10).

Using the “drop test” with a weight stack machine we estimated I to be around 1.2 kg m².

Substituting these values into (11) we find that Mf exceeds the static value mgk (in the range of 37 - 75 Nm) by about 58 - 48 % for m in the range 20 to 40 kg. From this it should already be evident that the resistance of the weight stack machine during dynamic exercise is significantly affected by the acceleration. In real cases the acceleration may be still higher because it takes place only during the initial phase of the movement. This also means that maximum resistance is shifted nearer to the beginning of the motion the higher the value of the initial acceleration, as can been seen from measurement data to be presented below.

For real machines one can measure y(θ) for a set of angles and then use polynomial curve fitting in order to get a smooth approximation of y(θ) (this function depends on the form of the oval shaped lever S). The adjoining figure shows the graphing (height y in cm) of such measurements (points) for a David weight stack machine together with a polynomial fitting (3d degree). We have also drawn the resistance curve based on equ (8) (for m = 20 kg) using the polynomial

approximation of y(θ).

40.701339

0.033791 HEIGHT

k INTHk

MYk

120

3 ANGLEk

0 50 100

10 0 10 20 30 40 50

y height polyn. fitting mg dy/dv

angle v

If y(θ) is 3rd degree curve then Mstat(θ) will according to (8) be a parabolic curve whose function can be writtten as

(13) Mstat(h)=Mmax$

1−a(h−hmax)²



where θmax is the angle for which the resistance reaches maximum and α is a parameter which determines the curvature of the parabola. One could also model Mf above as a function of θ and angular velocity e.g. using a Hill-model for the muscle, and then solve the closed system of equation of motion (through numerical integration), but we defer the discussion of these points to another place.

We may also observe that the relation (8) can be used in order to estimate the coefficient of friction γ for a real machine. Indeed, we first determine y(θ) and use (8) to calculate Mstat that will be compared with the real value Mstat /γ obtained from measurements of the resistance at steady, slow movements. (This can be tricky. In practice one has to repeat the measurement several times and then average the data.)

Measurement setup

In the measurements we used a David leg extension/curl weight stack machine equipped with a potentiometer (inclinometer) for measuring the rotation angle of the leg support, and a force

transducer (strain gauge, range ca 0 - 1000 N) for measuring the force exerted by the leg on the leg support system. Data were sampled using a computer and a 16-bit resolution DAQ-card (65536 values over the measurement range of 5 V). The sampling rate was set to 2000 S/s. The raw data was compressed by a factor of 10 by taking 10-point averages, reducing the effective sampling rate to 200 S/s for the final data. The inclinometer was calibrated using a waterlevel, and the force transducer by loading it with a known mass. In order to study the difference between static and dynamic resistance we performed measurements for (extension) slow and rapid (MVC) kicks with a given load (weight mass).

Measurement: “slow” vs “fast”

The adjoining graph shows the result of a series of “fast” kicks (m = 40 kg) and the graph below shows the result of a series of “slows” kicks (m = 20 kg). Force is in Newton and the angle is the knee joint angle (= 180 when the leg is straightened). As we can see in the “rapid” test the resistance falls off much steeper after initial maximum than in the “slow” test.

300

10

forcei

85.610722 angle 180

i

80 100 120 140 160 180

100 200 300

300

10

forcei

86.576959 angle 180

i

80 100 120 140 160 180

100 200 300