By James Smith, Global Sport Concepts and Athletic Consulting
In 2012, NFL star running back Chris Johnson challenged multi-World and Olympic Champion Sprinter Usain Bolt to race in a 40yd dash; that ultimately never happened. This article explores a theoretical matchup.
Note that data will be pulled from each athlete’s fastest recorded times (Bolt’s 100m WR in 2009 and Johnson’s NFL Combine in 2008).
A Kinematic Discussion of Sprinting as Rectilinear Motion
It must be noted that while electronic timers are staged at the combine, the times you see posted during coverage of the combine are hand times. In 2010, NFL Scout Mark Gorscak (who manages the start of the 40yd dash at Indianapolis) told me that the electronic timers are only set up as a means of checks and balances. Additionally, and consistent with what Mark told me, in 2009 I was made privy to the actual NFL data sheets for the combine and what was made evident is the arbitrary means of arriving at “official times” for players in the 40yd dash. Each player runs the 40 two times. On the data sheet, there were two columns for the hand times and two columns for the electronic times. Surprisingly, however, the time listed for the “official time” was neither a product of calculation nor was whatever method used to arrive at it consistent for each player. For example, one player may have had:
- Hand Time 1: 4.50sec
- Hand Time 2: 4.52sec
- Electronic Time 1: 4.63sec
- Electronic Time 2: 4.65sec
Yet the ‘official’ time was 4.58sec
Whereas, another player with identical times would have an ‘official time’ of 4.52sec.
Regardless, however, the entire business of ‘official times’ is predominantly something for the fans to talk about because all team personnel at the combine from coaches, to managers, to scouts have their stopwatches in their hands. Whatever the number says on their stopwatch is what they record on their own data sheets. Thus, at the end of the day, the stopwatch continues to provide a reference for speed in the NFL.
The following article is an excerpt from my upcoming book on the Governing Dynamics of Coaching. The information discussed here was taken from the section on Biodynamics and pertains to the kinematic domain of displacement- specifically rectilinear motion by way of the translatory realm.
Displacement is a vector quantity and is characterized by magnitude and direction (different from the scalar quantity of distance which only concerns magnitude). Displacement regards the change in position regarding initial and final states.
The SI (Système International d’unités), or international system of units, is the metric system. As a consequence, the international unit of displacement is the meter.
The motion of all components of a body (particles) that move through an equal distance in equal time is defined as translatory motion. There are two kinds of translatory motion:
Consider the profile view of a sprinter and their path of travel during a flying sprint between cones. This is rectilinear motion in which the lines connecting the sprinter’s position crossing the first cone (initial state) and the sprinter’s body as it crosses the second cone (final state) are parallel.
When calculating the average velocity of rectilinear displacement one may use the following equation:
v = ∆x ÷ ∆t
where ∆ (delta) is change, x is displacement, and t is time
For example: when Usain Bolt set his WR 9.58 in the 100m in 2009 in Berlin, his fastest 10m split (.81) was between the 60m and 70m mark. His 60m split was 6.29, and his 70m split was 7.10, therefore:
v = (x2 – x1) ÷ (t2 – t1)
v = (70m – 60m) ÷ (7.10sec – 6.29sec)
v = 10m ÷ 0.81sec
v = 12.35m/s
And when calculating for acceleration:
a = Δv ÷ ∆t
Staying with the Bolt 9.58 example, we know that his average velocity at 60m was 12.20m/s and, as shown above, 12.35m/s at 70m, therefore:
a = (v2 – v1) ÷ (t2 – t1)
a = (12.35m/s – 12.20m/s) ÷ (7.10sec – 6.29sec)
a = (0.15m/s ÷ 0.81sec)
a = 0.19m/s2
This extremely small value for acceleration may seem odd to readers who are not familiar with the stages of acceleration just before attaining maximum velocity. In this case, acceleration is nearing 0 due to the sprinter drawing imminently close to attaining maximal velocity (in which acceleration ceases to occur).
Alternatively, if we take time 0 at the start and calculate that along with Bolts time, minus reaction time of 0.146, (1.74sec) and velocity at the 10m mark (5.29m/s), we have:
a = (5.29m/s – 0m/s) ÷ (1.74sec – 0sec)
a = 5.29m/s ÷ 1.74sec
a = 3.04m/s2
As acceleration defines the rate of change of velocity, we know that it is one of the most fundamentally important locomotive qualities for nearly all team sport athletes. Curiously, however, the language of acceleration in SI units is relatively foreign to coaches, athletes, media, and the viewing audience. Further, one may observe the magnitude of subject matter incompetence that exists amongst coaches and media who suggest that athletes attain their top speed in very short distances. In American football, for example, one may routinely note the instances in which coaches or media describe how a running back hits the hole at top speed.
NFL Running Back Chris Johnson recorded some of the fastest sprint times in the history of the NFL combine with a 1.40sec split for the 10yd and a 4.24sec split for the 40yd. Keep in mind, however, that NFL combine timing is done by hand (despite the presence of electronic timing equipment) and the accepted margin of error, regarding short sprints, is 0.24sec when hand timing. Thus, a plausible fully automatic time conversion (not including a reaction time) for Johnson’s 4.24 is 4.48. Never the less, the metric conversion for 10yd and 40yd is 9.144m and 36.576m respectively.
From this, we may calculate Johnson’s velocity and acceleration at the 10yd and 40yd marks and provide education to any coach or media person who has mistakenly operated under the false knowledge that a running back, or any other athlete, is hitting their maximal velocity in a distance as short as 7yds, or 6.4m, (the distance the running back lines up at relative to the line of scrimmage).
Johnson’s 10yd kinematic quantities of average velocity and acceleration:
v = ∆x ÷ ∆t
v = (9.144m -0m) ÷ (1.40sec – 0sec)
v = 6.53m/s
a = Δv ÷ ∆t
a = (6.53m/s – 0m/s) ÷ (1.40sec – 0sec)
a = 4.66m/s2 (remander of the hand timing and 10yd mark)
Johnsons 40yd kinematic quantity of average velocity between 40yd and 10yd. It must be pointed out, however, that there is no available data of Johnson’s 30yd split. For this reason, the calculable data set of 10yd and 40yd is quite broad and thus renders the conservative value listed below:
v = (36.576m – 9.144m) ÷ (4.24sec – 1.40 sec)
v = 9.66m/s
From this, and despite the error associated with hand timing, one may definitively state that 9.66m/s is a greater value for average velocity than 6.53m/s (at the 10yd mark). Thus, one of the fastest running backs (and players in general) to have ever played in the NFL is incapable of achieving his maximum velocity in a distance of 10yd or less. (again, note that the value for average velocity would be greater and more accurate if we had a 30yd split for Johnson; thereby showing a more accurate picture of average velocity over a smaller displacement).
Those interested in Bolt’s hand timed acceleration over 10yd may and convert the 1.74 time for 10m to hand time (-.24) we get a corrected time of 1.50.
a = (5.29m/s – 0m/s) ÷ (1.50sec – 0sec)
a = 5.29m/s ÷ 1.50sec
a = 3.53m/s2
From here we can use the kinematic equation to solve for Bolt’s time to 9.144m in which we will subtract 9.144 from 10 to arrive at a displacement of -0.856m:
d = vit + 1/2at2
(where d= displacement, vi = initial velocity, a= acceleration, and t = time)
-0.856 = 5.29t + 1/2 • 3.53 • t2
Using the invaluable online resource Wolfram Alpha, we are given two solutions for t:
Taking the solution closest to 0 we subtract 0.17 from 1.50 to arrive at Bolt’s 10yd time, converted for hand time, of 1.33sec.
While we’re at it, we may estimate Bolt’s average velocity at the 40yd mark in his 9.58 world record. We know that his average velocity at 30m was 11.11m/s and at 40m it was 11.63m/s, and his time at 30m was 3.78sec and 4.64sec at 40m.
From this we may calculate his acceleration between 30 and 40m:
a = (11.63m/s – 11.11m/s) ÷ (4.64sec – 3.78sec)
a = 0.60m/s2
If we take an acceleration of 0.60m/s2 (while not 100% uniform, for this example approximation we’ll allow for it to be in order to use kinematic equations), and we know that his 30m split was 3.78 at an average velocity of 11.11m/s, and the 40yd mark (36.576m) is displaced by 6.576m from 30m, we can select the most appropriate equation from the kinematic equations that most well suits the available data. Of the data we have, what we must find is time (t) for Bolt to sprint from the 30m mark to the 36.576m mark then we may add that time to his 30m split.
The appropriate equation is the same one listed earlier.
d = vit + 1/2at2
When we plug in our known, and relevant, data we arrive at:
6.576 = 11.11t + 1/2 • 0.6t2
From here we may use Wolfram Alpha again. To arrive at two solutions for t:
By taking the positive solution (0.58273) and adding it to Bolt’s time at 30m (3.78), we arrive at a sum of 4.33273 rounded to 4.33.
Further, we must subtract Bolt’s reaction time (to the starting gun) in that particular race (0.146) which brings the electronic time down to 4.18.
Lastly, we subtract our known correction for hand timing (.24) and Bolt’s fully electronic conversion to hand timing is 3.94 compared to Chris Johnson’s hand-timed 4.24.
Running Surface and Starting Blocks
To be fair, the remaining variables to be discussed are the surfaces on which each athlete is sprinting in addition to the fact that the T&F sprinter is using starting blocks and subject to environmental conditions.
While the running surface at Lukas Oil Stadium (the site of the NFL combine) is relatively fast artificial field turf, it in no way provides the level of stiffness such as the modern T&F Regupol Compact surface that is featured at the Olympiastadion in Berlin.
The field in Lukas Oil Stadium is not subject to the elements, however. Thus, wind and temperature have no bearing on the sprint performances at Indianapolis. Alternatively, in that moment in 2009 in Berlin, Bolt had a very modest tailwind of 0.9m/s. Engineering efforts have determined that a tailwind between 0ms and 1m/s results in a -0.05sec advantage at sea level for up to 100m. Therefore, we may add 0.05 to Bolt’s electronic 4.18 to arrive at 4.23 and from here we subtract the hand time correction of 0.24 to arrive at 3.99. Likewise, if we add 0.05 to the calculated 10yd hand time of 1.33, we arrive at 1.38.
For these reasons, we understand that Johnson’s hand-time 40yd dash would be faster on the Regupol Compact surface (even without starting blocks and provided he wasn’t facing a considerable headwind) and Bolt’s converted fully electronic time would be slower on the Lukas Oil field surface.
One final example, without estimating what each athlete’s time differentials would be on either surface, when comparing 3.99 to 4.24 we may calculate the distance separating Bolt and Johnson at the finish line. In using the following kinematic equation.
vf = vi + at
We may compute Bolt’s velocity at the finish line (again, allowing for acceleration to be uniform in order to use these kinematic equations even though acceleration is not uniform when sprinting).
vf = 11.11m/s + 0.6m/s/s • 0.58273sec
Our solution is 11.46m/s
We know that:
v = ∆x ÷ ∆t
∆x = ∆tv
∆x = 3.99 – 4.24 • 11.46
∆x = -2.865m
According to these calculations, and acknowledging that the time differential would not be as substantial if both athletes were racing on the same surface, Bolt would beat Johnson in a 40yd dash by 2.87m or 9.42 feet. Given a margin this substantial, Bolt would be afforded a significant leeway in reacting slower than Johnson at the start.
In closing, if there was any question in anyone’s mind regarding the outcome of a 40yd race between these two athletes (in their prime), despite the variability in running surfaces and environmental conditions, one may state with confidence that if Usain Bolt and Chris Johnson were to race 40yds 100 times, that Johnson, despite being much more competitive over 10yd, would lose every race. Further, the only athlete who would be hitting their max V in a distance less than 10m would have the power to weight ratio of three-year-old child.
Please share so others may benefit.
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- The Kinematic Equations
- NFL Combine Records
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