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Many university texts on Mechanics deal with the problem of the effect of the air drag force, using as example the slowing down of a parachute. Hardly no one discuss what happens when the drag force is proportional to both u and u2u^2. In this paper we deal with a real problem to illustrate the effect of both terms in the speed of a runner: a theoretical model of the performance of the 100 m world record sprint of Usain Bolt during the 2009 World Championships at Berlin is developed, assuming a drag force proportional to u and to u2u^2. The resulting equation of motion is solved and fitted to the experimental data obtained from the International Amateur of Athletics Federations that recorded Bolt's position with a LAVEG (laser velocity guard) device. It is worth to note that our model works only for short sprints.
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IOP PUBLISHING EUROPEAN JOURNAL OF PHYSICS
Eur. J. Phys. 34 (2013) 1227–1233 doi:10.1088/0143-0807/34/5/1227
On the performance of Usain Bolt in
the 100 m sprint
JJHern
´
andez G ´
omez, V Marquina and R W G´
omez
Facultad de Ciencias, Universidad Nacional Aut´
onoma de M´
exico, Circuito Exterior CU,
M´
exico DF, 04510, Mexico
E-mail: jorge_hdz@ciencias.unam.mx,marquina@unam.mx and rgomez@unam.mx
Received 16 May 2013, in final form 21 June 2013
Published 25 July 2013
Online at stacks.iop.org/EJP/34/1227
Abstract
Many university texts on mechanics consider the effect of air drag force, using
the slowing down of a parachute as an example. Very few discuss what happens
when the drag force is proportional to both uand u2. In this paper we deal
with a real problem to illustrate the effect of both terms on the speed of a
runner: a theoretical model of the world-record 100 m sprint of Usain Bolt
during the 2009 World Championships in Berlin is developed, assuming a drag
force proportional to uand to u2. The resulting equation of motion is solved
and fitted to the experimental data obtained from the International Association
of Athletics Federations, which recorded Bolt’s position with a laser velocity
guard device. It is worth noting that our model works only for short sprints.
(Some figures may appear in colour only in the online journal)
1. Introduction
In Z¨
urich, Switzerland, 21 June 1960, the German Armin Harry astounded the sports world
by equalling what was considered an insuperable physiological and psychological barrier for
the 100 m sprint: the 10 s race. It was not until 20 June 1968, at Sacramento, USA, that Jim
Hines ran 100 m in 9.9 s, breaking through this barrier. Many sprinters have run this distance
in under 10 s subsequently, but 31 years were needed to lower Harry’s record by 0.14 s (Carl
Lewis, 25 August 1991, at Tokyo, Japan). The current world record of 9.58 s was established
by Usain Bolt (who also held the 200 m world record of 19.19 s up to 2012) in the 12th
International Association of Athletics Federations (IAAF) World Championships in Athletics
(WCA) at Berlin, Germany in 2009.
The performance of Usain Bolt in the 100 m sprint is of physical interest because he can
achieve speeds and accelerations that no other runner has been able to. Several mathematical
models to fit the position and velocity (or both) of a sprinter have been proposed [16].
Recently, Helene et al [6] fitted Bolt’s performance during both the summer Olympics in 2008
at Beijing and the world championships in 2009 at Berlin, using a simple exponential model
for the time dependence of the speed of the runner.
0143-0807/13/051227+07$33.00 c
2013 IOP Publishing Ltd Printed in the UK & the USA 1227
1228 JJHG
´
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et al
2. Theoretical model
The important forces acting during the race are the horizontal force that Bolt exerts and a drag
force that depends upon the horizontal velocity (speed). Other factors affecting the mechanics
of his motion, such as humidity, altitude above sea level (36 m), oxygen intake and his turning
his head to watch other runners, are not taken into account. Based on the fact that Bolt’s 200 m
time is almost twice that for 100 m, our main assumption is that in the 100 m sprint he is able
to develop a constant horizontal force F0during the whole race. The drag force, D(u),isa
function of Bolt’s horizontal speed with respect to the ground u(t), with or without wind. This
force causes a reduction of his acceleration, so his speed tends to a constant value (terminal
speed). Thus, the equation of motion is
m˙u=F0D(u). (1)
This equation can readily be cast as a quadrature,
tt0=mu
u0
du
F0D(u).(2)
The integral above does not have an analytical solution for a general drag function; however,
the drag force can be expanded in Taylor series,
D(u)D(0)+dD(u)
du
0
u+1
2
d2D(u)
du2
0
u2+O(u3). (3)
The constant term of the expansion is zero, because the runner experiences no drag when at
rest. The second and third terms must be retained. While the term proportional to the speed
represents the basic effects of resistance, the term proportional to the square of the speed takes
into account hydrodynamic drag, obviously present due to the highly non-uniform geometry
of the runner. In general, for relatively small speeds it suffices to take only the first three terms
of the expansion.
Renaming the uand u2coefficients as γand σ, respectively, the equation of motion (1)
takes the form
m˙u=F0γuσu2,(4)
whose solution follows straightforwardly from equation (2),
u(t)=AB 1ekt
A+Bekt ,(5)
where the coefficients are related by σ=km/(A+B),F0=kmAB/(2A+2B)and
γ=km(AB)/(A+B).
The position can be obtained by integrating equation (5),
x(t)=A
kln A+Bekt
A+B+B
kln Aekt +B
A+B,(6)
while the acceleration can also be calculated by deriving equation (5),
a(t)=ABk(A+B)ekt
A+Bekt 2.(7)
On the performance of Usain Bolt in the 100 m sprint 1229
0510
0
25
50
75
100
x (m)
t (s)
Measured data
Theoretical position fit
0510
0
4
8
12
v (m/s )
t (s)
Measured data
Theoretical speed fit
(a) (b)
Figure 1. Position (a) and speed (b) of Bolt in the 100 m sprint at the 12th IAAF WCA. The
dotted (blue) line corresponds to the experimental data while the solid (red) one corresponds to the
theoretical fitting.
Tab le 1. Fitted values of the parameters A,Band k.
Parameter Position fitting Velocity fitting
A(m s1) 110.0 110.0
B(m s1) 12.2 12.1
k(1 s1) 0.9 0.8
Tab le 2. Values of the physical parameters F0,γand σ.
Constant Value
F0(N) 815.8
γ(kg s1) 59.7
σ(kg m1) 0.6
3. Experimental data fitting
The experimental data we used were from the 12th IAAF WCA, which were obtained from
[7], and consist of Bolt’s position and speed every 1/10 s. To corroborate the accuracy of the
data obtained from [7], we reproduced the velocity versus position plot given in [8] with them,
obtained by the IAAF using a laser velocity guard device. The parameters A,Band kwere fitted
using a least-squares analysis via Origin 8.1 (both position and speed data sets) considering a
reaction time of 0.142 s [6]. In figures 1(a) and (b) we show such fittings, together with the
experimental data.
The parameter values for both fittings are shown in table 1. We do not report errors,
because the standard error of the fitting on each parameter lies between the second and the
third significant digit, which is finer than the measurement error in the data.
The accuracy of the position and velocity fittings is R2
p=0.999 and R2
v=0.993,
respectively, so we use the results of the parameters A,Band kfrom the position fitting
henceforth. The computed values of the magnitude of the constant force, F0, and the drag
coefficients, γand σ, are shown in table 2, taking Bolt’s mass as 86 kg [9].
1230 JJHG
´
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et al
0510
0
5
10
a (m/s2)
t (s)
Figure 2. Theoretical acceleration of Bolt in the 100 m sprint at the 12th IAAF WCA.
We also show in figure 2the plot of the magnitude of the acceleration we obtained; no
fitting was made because there are no experimental data available.
4. Results
As for any mechanical system subject to drag, the runner experiments a terminal velocity uT
that is formally obtained when ˙u=0 in the equation of motion (1); that is, by solving the
equation
F0=D(uT)(8)
for uT. Nevertheless, the solution of the equation for the terminal velocity can also be found
when t→∞in equation (5), and it turns out to be uT=B. Therefore, in this model the runner
acquires a terminal speed of uT=12.2ms
1, which is physically feasible (see figure 1(b)).
According to the data obtained from [7] the average speed in the second half of the sprint
(surprisingly equal to 99% of the maximum speed recorded [7]) is 12.15 m s1. Moreover, the
initial acceleration of Bolt is a(0)=9.5ms
1, which is of the order of the acceleration of
gravity, g; this value for the initial acceleration is reasonable, considering that the acceleration
a man must exert in order to be able to jump half of his own height should be slightly greater
than g. Furthermore, the value of the constant force in table 2,F0=815.8 N, is entirely
consistent with the fact that one expects that the maximum constant (horizontal) force he
could exert should be of the order of his weight, i.e. w=842.8N.
Now, σ=0.5ρCdArepresents the hydrodynamic drag, where ρis the density of air, Cd
the drag coefficient of the runner and Ahis cross section area. The density of air at the time
of the spring can be approximated as follows. Berlin has a mean altitude of 34 m above sea
level, and an average mean temperature for the month of August1[10]of18.8C. Bearing
in mind that the race took place at night, we consider an average temperature between the
average mean temperature and the mean daily minimum temperature for August in Berlin,
which is 14.3 C. Thus, the density of air is ρ=1.215 kg m3and the drag coefficient of Bolt
1The sprint took place 16 August 2009.
On the performance of Usain Bolt in the 100 m sprint 1231
0510
0
500
1000
1500
2000
2500
P (W)
t (s)
Figure 3. Theoretical power of Bolt in the 100 m sprint at the 12th IAAF WCA.
is Cd=2σ/ρA=1.2, where the cross section area of Bolt2was estimated as A=0.8m
2.
This value of Cdlies in the typical range for human beings reported in the literature (between
1.0 and 1.3) [1113].
The instantaneous power that Bolt develops, considering the drag effect, is simply
P(t)=Fu =m˙uu =mABk(A+B)(1ekt )ekt
A+Bekt 3.(9)
In figure 3we plot the power of the sprint for Bolt and the drag. It is remarkable that the
maximum power of Pmax =2619.5 W (3.5 HP) was at time tPmax =0.89 s, when the speed
u(tPmax )=6.24 m s1was only about half of the maximum speed. The fact that the maximum
instantaneous power arises in such a short time indicates the prompt influence of the drag
terms in the dynamics of the runner.
The effective work (considering the effect of the drag force) is then
WEff =τ
0
P(t)dt=τ
0
1
2mdu2=1
2mu2(τ ), (10)
where τis the running time (the official time of the sprint minus the reaction time of the
runner). The effective work is the area under the curve of figure 3, and it is WEff =6.36 kJ.
On the other hand, as Bolt is assumed to develop an essentially constant force, his mechanical
work is simply WB=F0d=81.58 kJ, where dis the length of the sprint (100 m). This
means that from the total energy that Bolt develops only 7.79% is used to achieve the motion,
while 92.21% is absorbed by the drag; that is, 75.22 kJ are dissipated by the drag, which is an
incredible amount of lost energy.
5. Discussion
As mentioned in section 2, a central assumption in our model is that a 100 m sprinter (not
only Bolt) is able to develop a constant force during a race (except in the initial few tenths
of a second, where he pushes himself against the starting block). To delimit how good this
2To calculate such a cross section area, we used a similar procedure to that used in [9], where instead of a circle
we estimated the area of the head with an ellipse. We averaged several scaled measures from Bolt pictures taken
from [14].
1232 JJHG
´
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et al
0246810
1000
800
t (s)
600
F0 (N)
Figure 4. Force exerted by the runner during the race. The red line is calculated with the
experimental data, the dash-dot-dot (green) line is the average force of 818.3 N, while the short-
dash-dot (black) line is the value of the force F0obtained from the adjustment (815.7 N).
assumption is we use the experimental values of u, the calculated acceleration and the fitted
values of the constants γand σto compute F0. The result is shown in figure 4. It is interesting
to note that the average value of the force obtained from this figure is 818.3 N, which is very
close to the value obtained from the fitting of the data, 815.7 N. The high value of the force in
the first part of the race is due do the acceleration he obtains when he pushes himself from the
starting block.
At first glance, observing the values of the drag coefficients in table 2, one is impelled
to argue that because σγthe hydrodynamic drag could have been neglected. However,
one can calculate the drag terms in the equation of motion at the terminal speed uT, attaining
γuT=725.59 N and σuT
2=90.18 N. Thus, from the total drag γuT+σu2
T, 11.05%
corresponds to turbulent drag, which turns to be an important contribution.
If we would like to make predictions considering different wind corrections, this can be
done as follows. Once a runner acquires the wind speed (almost instantly), the second term in
the right side (γu) of equation (1) behaves as if the sprinter were running in still air, because γ
is proportional to the air viscosity, which is independent of air pressure. However, that is not
the case for the third term in (1), (σu2), which arises from the collisions per unit time of the air
molecules against the sprinter and it is proportional to the speed of the runner with respect to
the ground. In a simple model, the hydrodynamical drag force is DH=σ(ρ)(v+vw)2, where
vis the speed achieved by the runner without wind and vwis the speed of the wind. The value
of σdepends on the number of molecules that impact on the runner per unit time and should
be different in still air conditions. Then, the equation of motion (1) can be rewritten as
m˙u=m˙v=F0γvσ(v +vw)2,(11)
and without wind as
m˙v=F0γvσv2.(12)
Subtracting (11) and (12), we obtain
σv2+2vvw+v2
w=σv2,(13)
so then
σ=σ1+2vw
v+v2
w
v2σ1+2vw
v,(14)
where the third term in the second expression has been neglected (vwv). To estimate the
value of σ, we consider vas the terminal speed of Bolt, uT. With these conditions, σ=0.69
with still air (vw=0ms
1) and σ=0.49 with a tailwind of vw=2ms
1. It should be
clear that this calculation is only a crude way to estimate the differences of running time with
On the performance of Usain Bolt in the 100 m sprint 1233
Tab le 3. Predictions of the running time for Bolt without tailwind, and with a tailwind of 2 m s1.
vw(m s1) Estimated running time (s)
0 9.68
0.9 9.58
2 9.46
and without wind. The results, which are close to the values reported in the literature [15], are
summarized in table 3.
Although this is a simple way to calculate a correction due to wind, it turns out to be a
good proposal for it. A more realistic assumption would be to modify equation (14)tobe
σ=σ1+αvw
uT,(15)
with the parameter αlying between 1 and 2.
The results obtained, together with the facts indicated in this discussion, show the
appropriateness and quality of the model developed in this paper. We look forward to the
next IAAF WCA, which will be held in Moscow, Russia, 10–18 August 2013, to test our
model with the experimental data obtained from such sprints, as well as waiting to see if the
fastest man on earth is able to beat his own world record again.
Acknowledgment
This work was partially supported by PAPIIT-DGAPA-UNAM Project IN115612.
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