## Calculations (2012-present)

For calculations performed before Jan 2012, see the page ‘Calculations 2007-2011). The calculations on this page are:

1) **The equations of motion of the COG** (12 Jan 2012)

**2) Energy cost of repositioning the limbs **(5^{th} April 2012)

1) **The equations of motion of the COG** (12 Jan 2012)

The external forces acting on the runner’s body are gravity (mg); ground reaction force (GRF) to the push of the body against the ground, and wind resistance. GRF can be decomposed into vertical and horizontal components: vGRF and hGRF. Change in velocity from time 1 to time 2 is the integral over this time interval of acceleration. which in turn is force/mass. Displacement from time 1 to time 2 is the integral of velocity over time. Ignoring wind resistance, the equations of motion of the COG are:

The components of GRF can be measured using a force plate, though for the purpose of illustration, vGRF of a mid-foot runner can be approximated by a half-sine wave rising from zero at foot fall to a peak at mid-stance and falling to zero at take-off. Assuming that the push of the body against the ground is directed along the line from COG to point of support, hGRF can be derived from vGRF as in equation (5). (It should be noted that x and y are the horizontal and vertical displacement of the COG from the point of support.) The work done to reverse the braking effect of hGRF prior to mid-stance can be computed from equation (6) with F=hGRF, v(t) is horizontal velocity, t1 is mid-stance and t2 is time at take off. The work done to elevate the body from its lowest point at mid-stance to its peak height in mid-flight can be computed directly from h, the height gain computed via (3), using the formula W=mgh.

While the integral required to compute the vertical velocity (equation 1) can be computed readily in the case where vGRF is has a sinusoidal form, the computation of horizontal velocity is less straightforward because hGRF has a more complex form. It can be performed using the difference equation corresponding to (2).

Figure 1 depicts vGRF (dashed line)) and hGRF (solid line) assuming a sinusoidal vGRF with maximum value 2mg (blue) and for the case vGRFmax=4mg (ochre). The traces begin and end at the mid-airborne phase. The intermediate vertical lines represent foot-fall, mid-stance and take-off. When vGRFmax is greater, time on stance is shorter because the average value of vGRF over the entire gait cycle must be equal to mg

Figure 2 depicts the vertical velocity and the change in horizontal velocity from its value during the airborne phase. The slowing due to braking is much when vGRFmax is 4 (long time on stance).

At both long and short times on stance (vGRFmax=2mg and 4ng respectively), the energy required to overcome braking, computed using equation (6), is greater than the energy required to elevate the body from its low point at mid-stance to its high point in the airborne phase. The table shows the work required after mid-stance. A proportion of the required energy will be provided by elastic recoil,. Because muscle and tendon are visco-elastic such that elasticity is greater when a force is applied more rapidly, the proportion of energy recovered via recoil is expected to be greater when time on stance is shorter (i.e vGRFmax=4mg). At shorter time on stance the total energy requirement is less, so the additional energy required is expected to be substantially less.

**2) Energy cost of repositioning the limbs**

At the beginning of swing, the foot is stationary on the ground. During early swing as the hip flexes carrying the leg forwards the knee also flexes so that the foot comes up towards the buttocks thereby shortening the lever arm as the leg swings forward. At mid-swing, which occurs at mid-stance on the other foot, the centre of mass of the swinging leg overtakes the centre of mass of the body. In the second half of swing, the leg decelerates as the hip extensors become active to arrest the swing. In late swing the thing begins to swing back towards neutral as the knee extends, so that at foot fall the thing both hip and knee exhibit only a mild flexion.

In the first half of swing the leg accelerates forwards, and the kinetic energy of the swinging leg increases; in the second half of swing the hip extensors absorb the energy of the swing. Although the muscles are undergoing eccentric contraction as they absorb the kinetic of the swinging leg, if the main work that is done during swing is the work required to work to increase the kinetic energy of the leg between lift off and mid-swing. that is done during the swing, we need to estimate the increase in kinetic energy of the leg between lift-off and mid-swing.

The kinetic energy of a moving object is ½ m v^{2} where m is the mass and v is the velocity of the centre of mass of that object. Thus if the leg accelerates from an initial velocity v_{1} to a final velicty v_{2}, the increase in kinetic energy is ½ m (v_{2}^{2} – v_{1}^{2 }).

At lift off, the knee is straight and the leg is oriented along the line that joins the hip to the mid-foot. The centre of mass of the leg is located a little above the knee, about of the distance from foot to hip, As the foot is stationary while the hip is moving at the running velocity V, the centre of mass of the leg is moving forwards at approximately 2/3 V. At mid-swing, the velocity of the centre of mass of the leg, v_{2} is a little greater than V. We could make a crude estimate of this velocity if we assume that the COM(leg) accelerates uniformly forward to overtake the torso, but for present purposes, it will suffice to assume that v_{2} = (1+d) V were d is a small number. If we substitute the expressions for V_{1} and V_{2} into (1) we can demonstrate that the increase in kinetic energy (per step) is an expression of the form mbV^{2} where b is a constant somewhat greater than 5/18. For present purposes, we do not need to know b exactly.

Since the number of steps per Km is 1000C/60V where C is cadence in steps per minute, the cost of accelerating the swing leg is 1000 mbCV/60.

If we assume the mass of one leg is 0.2 times body mass, the cost of accelerating the swing leg per Kg of body mass per Km is given by

KE= 3.3bCV Newton-metre (2).

November 15, 2015 at 4:51 pm |

[…] rotation around the point of support. Illustrative numerical values are based on the model I presented on this site in […]