How Noah Lyles Won the Men's 100-Meter Gold by a Fraction

By Kalina Borkiewicz, Weiyi Cai, Bora Erden, Lazaro Gamio, Martín González Gómez, Malika Khurana, Bedel Saget, Joe Ward and Jeremy White

Aug. 4, 2024

But did Thompson actually win by a fraction?

Andy Long, Spring 2025
MAT229

I pulled the data from the hypertext source of the page at the top. Thanks to the New York Times for that.

I did have to scale the y-values, however; not sure what units those were in, but not in miles per hour. So, in order to make my graphs align fairly well with the NYTimes’s graph, I performed the following linear transformation:
					newspeed = 2/90 * (37.5 - speed)
					
The raw data has all the time series starting at 37.5, so the shift brings all the time series to zero. The value of 2/90 is a convenience, to make our plots line up. I processed this in lisp, which is my favorite programming language, and then just output the data in a form that Mathematica likes.
					
It doesn’t make any difference in the ensuing argument -- the transformation washes out. The negative sign is a little perplexing, but is probably a consequence of how the plotting is done in the New York Times’ software package  (how their coordinate system is oriented). 

Notice that every one of timeseries has a final repeated speed -- that doesn’t seem like real data....

Here’s Lyle’s raw data from the webpage:
‘(0.092 37.5 1.092 37.566 2.092 37.688 3.092 38.439 4.102 41.359 5.102 45.635 6.102 52.228 7.102 61.313 8.102 72.372 9.102 83.307 10.102 92.233 11.102 98.077 12.102 99.821 13.102 97.077 14.112 90.397 15.102 81.683 16.112 73.173 17.112 66.176 18.112 60.989 19.112 58.01 20.112 56.732 21.112 55.689 22.112 53.583 23.112 50.991 24.112 48.028 25.122 43.953 26.122 38.725 27.122 33.375 28.122 29.127 29.122 25.74 30.122 22.535 31.122 19.651 32.122 17.862 33.122 17.225 34.122 17.1 35.132 17.783 36.132 19.19 37.132 20.455 38.132 20.87 39.132 20.857 40.132 21.204 41.132 21.615 42.132 21.466 43.132 21.083 44.132 21.094 45.142 21.555 46.142 22.556 47.142 23.76 48.142 25.332 49.142 27.139 50.142 28.953 51.142 30.308 52.142 31.445 53.142 32.386 54.142 32.931 55.152 33.225 56.152 33.207 57.152 32.837 58.152 32.349 59.152 31.745 60.152 30.722 61.152 29.596 62.152 28.474 63.152 27.128 64.152 25.369 65.162 23.872 66.162 23.085 67.162 22.979 68.162 23.027 69.162 23.144 70.162 23.412 71.162 23.623 72.162 23.305 73.162 22.579 74.162 21.641 75.162 20.344 76.172 18.918 77.172 17.929 78.172 17.284 79.172 16.295 80.172 14.951 81.172 13.871 82.172 13.153 83.172 12.244 84.172 11.5 85.182 11.537 86.182 12.41 87.182 13.148 88.182 13.806 89.182 15.166 90.182 16.989 91.182 18.058 92.182 18.339 93.182 18.924 94.182 20.332 95.192 21.7 96.192 22.321 97.192 23.096 97.9 23.096)
Thompson:
‘(0.092 37.5 1.092 37.499 2.092 37.688 3.092 38.439 4.102 39.51 5.102 40.006 6.102 40.607 7.102 42.448 8.102 45.126 9.102 46.725 10.102 46.379 11.102 44.915 12.102 43.607 13.102 41.638 14.112 37.763 15.102 33.4 16.112 30.336 17.112 28.048 18.112 25.753 19.112 24.535 20.112 25.403 21.112 26.569 22.112 26.667 23.112 27.449 24.112 30.598 25.122 34.683 26.122 36.809 27.122 37.897 28.122 40.382 29.122 43.122 30.122 43.295 31.122 41.65 32.122 40.593 33.122 39.258 34.122 35.52 35.132 30.667 36.132 27.266 37.132 24.849 38.132 21.619 39.132 18.128 40.132 16.492 41.132 16.385 42.132 16.177 43.132 15.881 44.132 16.983 45.142 19.15 46.142 21.032 47.142 22.912 48.142 25.965 49.142 29.062 50.142 30.598 51.142 31.112 52.142 31.874 53.142 32.695 54.142 32.662 55.152 32.086 56.152 31.941 57.152 32.247 58.152 31.912 59.152 30.998 60.152 30.418 61.152 30.261 62.152 29.694 63.152 29.053 64.152 29.138 65.162 29.692 66.162 30.021 67.162 30.156 68.162 30.658 69.162 31.468 70.162 32.026 71.162 32.758 72.162 34.399 73.162 36.327 74.162 37.649 75.162 38.271 76.172 38.868 77.172 39.266 78.172 39.028 79.172 38.542 80.172 38.407 81.172 38.376 82.172 37.709 83.172 36.622 84.172 35.523 85.182 34.071 86.182 31.906 87.182 30.085 88.182 29.45 89.182 29.654 90.182 29.973 91.182 30.734 92.182 32.03 93.182 33.348 94.182 34.139 95.192 34.922 96.192 36.431 97.192 37.934 97.9 37.934)
Kerley:
‘(0.092 37.5 1.092 37.566 2.092 37.688 3.092 37.914 4.102 37.223 5.102 35.463 6.102 31.783 7.102 25.713 8.102 17.649 9.102 10.207 10.102 5.004 11.102 2.252 12.102 1.816 13.102 4.221 14.112 9.257 15.102 15.031 16.112 20.387 17.112 24.789 18.112 27.535 19.112 28.566 20.112 28.526 21.112 28.419 22.112 28.748 23.112 28.412 24.112 27.529 25.122 27.165 26.122 27.399 27.122 27.906 28.122 28.602 29.122 30.213 30.122 32.287 31.122 34.047 32.122 35.356 33.122 36.375 34.122 37.004 35.132 37.371 36.132 38.818 37.132 41.936 38.132 46.317 39.132 50.838 40.132 55.488 41.132 59.84 42.132 62.541 43.132 62.715 44.132 60.854 45.142 57.719 46.142 53.716 47.142 49.686 48.142 46.438 49.142 44.605 50.142 43.49 51.142 42.538 52.142 41.193 53.142 40.083 54.142 39.178 55.152 37.902 56.152 36.566 57.152 35.514 58.152 34.637 59.152 34.138 60.152 33.862 61.152 33.613 62.152 33.823 63.152 34.244 64.152 34.647 65.162 34.867 66.162 35.493 67.162 35.967 68.162 35.953 69.162 35.63 70.162 35.369 71.162 35.256 72.162 35.058 73.162 34.881 74.162 35.028 75.162 36.161 76.172 37.714 77.172 39.533 78.172 41.836 79.172 44.27 80.172 45.541 81.172 45.305 82.172 44.232 83.172 42.336 84.172 39.284 85.182 35.334 86.182 31.948 87.182 29.689 88.182 28.51 89.182 27.612 90.182 26.96 91.182 26.81 92.182 26.411 93.182 25.14 94.182 23.073 95.192 20.622 96.192 17.521 97.192 13.55 98.1 13.55)

Analysis:

We assume the dirt formula -- that is, that distance equals rate times time: then, since we’re given constant dt=0.1seconds, it makes sense to write the situation this way:

				100 = (∫_0^(Subscript[T, k]))(Subscript[r, k])(t)dt = (∫_0^(Subscript[T, k]))(a(t) + (Subscript[α, k])(t))dt

with (Subscript[T, k]) being the time it took runner k to reach 100 meters (i.e. their time), (Subscript[r, k])(t) their speed,  a(t) the average speed of the runners, and  (Subscript[α, k])(t) the (k^th) runner’s “off-speed” -- given by the graph (the difference from the average speed).

Each of these quantities represent a distance. Since it was a photo finish, the (Subscript[T, k]) of the top runners were almost the same. Thus

 			(∫_0^(Subscript[T, k]))(a(t) + (Subscript[α, k])(t))dt ≈  C + (∫_0^(Subscript[T, k])) (Subscript[α, k])(t)dt

Based on the Times’s graphs, and plots of the data from the hypertext, I discerned the data of the top three runners. Here are their races:

Thompson={{0.092,.0000},{1.092,.0000},{2.092,-.004},{3.092,-.021},{4.102,-.045},{5.102,-.056},{6.102,-.069},{7.102,-0.11},{8.102,-.169},{9.102,-.205},{10.1,-.197},{11.1,-.165},{12.1,-.136},{13.1,-.092},{14.11,-.006},{15.1,.0911},{16.11,.1592},{17.11,0.21},{18.11,0.261},{19.11,.2881},{20.11,.2688},{21.11,.2429},{22.11,.2407},{23.11,.2234},{24.11,.1534},{25.12,.0626},{26.12,.0154},{27.12,-.009},{28.12,-.064},{29.12,-.125},{30.12,-.129},{31.12,-.092},{32.12,-.069},{33.12,-.039},{34.12,0.044},{35.13,.1518},{36.13,.2274},{37.13,.2811},{38.13,.3529},{39.13,.4305},{40.13,.4668},{41.13,.4692},{42.13,.4738},{43.13,.4804},{44.13,.4559},{45.14,.4078},{46.14,0.366},{47.14,.3242},{48.14,.2563},{49.14,.1875},{50.14,.1534},{51.14,0.142},{52.14,0.125},{53.14,.1068},{54.14,.1075},{55.15,.1203},{56.15,.1235},{57.15,.1167},{58.15,.1242},{59.15,.1445},{60.15,.1574},{61.15,.1609},{62.15,.1735},{63.15,.1877},{64.15,.1858},{65.16,.1735},{66.16,.1662},{67.16,.1632},{68.16,0.152},{69.16,0.134},{70.16,.1216},{71.16,.1054},{72.16,.0689},{73.16,.0261},{74.16,-.003},{75.16,-.017},{76.17,-0.03},{77.17,-.039},{78.17,-.034},{79.17,-.023},{80.17,-0.02},{81.17,-.019},{82.17,-.005},{83.17,.0195},{84.17,.0439},{85.18,.0762},{86.18,.1243},{87.18,.1648},{88.18,.1789},{89.18,.1744},{90.18,.1673},{91.18,.1504},{92.18,.1216},{93.18,.0923},{94.18,.0747},{95.19,.0573},{96.19,.0238},{97.19,-0.01},{97.9,-0.01}};
ThompsonPlot=ListPlot[Thompson,Joined->True,PlotStyle->{Thick,Gray}];
Lyles={{0.092,.0000},{1.092,-.001},{2.092,-.004},{3.092,-.021},{4.102,-.086},{5.102,-.181},{6.102,-.327},{7.102,-.529},{8.102,-.775},{9.102,-1.02},{10.1,-1.22},{11.1,-1.35},{12.1,-1.38},{13.1,-1.32},{14.11,-1.18},{15.1,-.982},{16.11,-.793},{17.11,-.637},{18.11,-.522},{19.11,-.456},{20.11,-.427},{21.11,-.404},{22.11,-.357},{23.11,-0.3},{24.11,-.234},{25.12,-.143},{26.12,-.027},{27.12,.0917},{28.12,.1861},{29.12,.2613},{30.12,.3326},{31.12,.3966},{32.12,.4364},{33.12,.4506},{34.12,.4533},{35.13,.4382},{36.13,.4069},{37.13,.3788},{38.13,.3696},{39.13,.3698},{40.13,.3621},{41.13,0.353},{42.13,.3563},{43.13,.3648},{44.13,.3646},{45.14,.3543},{46.14,.3321},{47.14,.3053},{48.14,.2704},{49.14,.2302},{50.14,.1899},{51.14,.1598},{52.14,.1346},{53.14,.1136},{54.14,.1015},{55.15,0.095},{56.15,.0954},{57.15,.1036},{58.15,.1145},{59.15,.1279},{60.15,.1506},{61.15,.1756},{62.15,.2006},{63.15,.2305},{64.15,.2696},{65.16,.3028},{66.16,.3203},{67.16,.3227},{68.16,.3216},{69.16,0.319},{70.16,.3131},{71.16,.3084},{72.16,.3154},{73.16,.3316},{74.16,.3524},{75.16,.3812},{76.17,.4129},{77.17,.4349},{78.17,.4492},{79.17,.4712},{80.17,.5011},{81.17,.5251},{82.17,0.541},{83.17,.5612},{84.17,.5778},{85.18,0.577},{86.18,.5576},{87.18,.5412},{88.18,.5265},{89.18,.4963},{90.18,.4558},{91.18,0.432},{92.18,.4258},{93.18,.4128},{94.18,.3815},{95.19,.3511},{96.19,.3373},{97.19,.3201},{97.9,.3201}};
LylesPlot=ListPlot[Lyles,Joined->True,PlotStyle->{Thick,Orange}];
Kerley = {{0.092,.0000},{1.092,-.001},{2.092,-.004},{3.092,-.009},{4.102,.0062},{5.102,.0453},{6.102,0.127},{7.102,.2619},{8.102,.4411},{9.102,.6065},{10.1,.7221},{11.1,.7833},{12.1,0.793},{13.1,.7395},{14.11,.6276},{15.1,.4993},{16.11,.3803},{17.11,.2825},{18.11,.2214},{19.11,.1985},{20.11,.1994},{21.11,.2018},{22.11,.1945},{23.11,0.202},{24.11,.2216},{25.12,.2297},{26.12,.2245},{27.12,.2132},{28.12,.1977},{29.12,.1619},{30.12,.1158},{31.12,.0767},{32.12,.0476},{33.12,0.025},{34.12,0.011},{35.13,.0029},{36.13,-.029},{37.13,-.099},{38.13,-.196},{39.13,-.296},{40.13,-0.4},{41.13,-.496},{42.13,-.556},{43.13,-0.56},{44.13,-.519},{45.14,-.449},{46.14,-0.36},{47.14,-.271},{48.14,-.199},{49.14,-.158},{50.14,-.133},{51.14,-.112},{52.14,-.082},{53.14,-.057},{54.14,-.037},{55.15,-.009},{56.15,.0208},{57.15,.0441},{58.15,.0636},{59.15,.0747},{60.15,.0808},{61.15,.0864},{62.15,.0817},{63.15,.0724},{64.15,.0634},{65.16,.0585},{66.16,.0446},{67.16,.0341},{68.16,.0344},{69.16,.0416},{70.16,.0474},{71.16,.0499},{72.16,.0543},{73.16,.0582},{74.16,.0549},{75.16,.0298},{76.17,-.005},{77.17,-.045},{78.17,-.096},{79.17,-0.15},{80.17,-.179},{81.17,-.173},{82.17,-0.15},{83.17,-.107},{84.17,-0.04},{85.18,.0481},{86.18,.1234},{87.18,.1736},{88.18,.1998},{89.18,.2197},{90.18,.2342},{91.18,.2376},{92.18,.2464},{93.18,.2747},{94.18,.3206},{95.19,.3751},{96.19,0.444},{97.19,.5322},{98.1,.5322}};
KerleyPlot=ListPlot[Kerley,Joined->True,PlotStyle->{Thick,Red}];

Now let’s include the rest of the data, and the special plots for our champions (thanks to https://community.wolfram.com/groups/-/m/t/253481 for labelling help):

Show[
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ListPlot[
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ListPlot[
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and compare to the NYTimes’s plot:

Unit Conversion

Now for the really stupid step: why did they mix English and metric units? Arghhhh!:)

Translating mph into meters/second:
mph = 5280/3600(ft/s)*12/1(in/ft)*2.54/1(cm/in)*0.01/1(m/cm)

meterspersecondConversion=1/(5280/3600*12/1*2.54/1*0.01/1) (*mph*)
dt=0.1; (* Speeds are taken on the tenth of a second *)

Rectangle Rules

What follows are almost the correct answers, which suggest that Thompson won the race: assuming equal times, he would have gone an additional 2.3 meters, versus 2.16 for Lyles.

Notice that our first time stamp occurs at about .01 second, and they’re all still in the blocks, essentially (at least they’re all tied at that point). So we don’t have the actual “left endpoint”, which would have a velocity of 0 (the gun goes off): each time series should have first element {0,0}.

Transpose[Lyles][[2]]
LylesSpeeds=Transpose[Lyles][[2]]*meterspersecondConversion (* Lyle's speeds in m/s *)
Total[LylesSpeeds]*dt (* Lyle's distance travelled beyond the average, in m *)
ThompsonSpeeds=Transpose[Thompson][[2]]*meterspersecondConversion;
Total[ThompsonSpeeds]*dt (* Thompson's distance travelled beyond the average *)
KerleySpeeds=Transpose[Kerley][[2]]*meterspersecondConversion;
Total[KerleySpeeds]*dt (* Kerley's distance travelled beyond the average *)

I say “almost” because we’re not accounting for differences in time intervals (they’re not actually all equal at 0.1 second intervals -- the last interval is a little shorter; and while Lyles and Thompson’s timeseries both show a final time of 9.79 to end, Kerley shows a time of 9.81).

LylesTimes=Transpose[Lyles][[1]]*dt;
LylesTimesIntervals=Rest[LylesTimes]-Most[LylesTimes]
ThompsonTimes=Transpose[Thompson][[1]]*dt;
ThompsonTimesIntervals=Rest[ThompsonTimes]-Most[ThompsonTimes];
KerleyTimes=Transpose[Kerley][[1]]*dt;
KerleyTimesIntervals=Rest[KerleyTimes]-Most[KerleyTimes];

Thompson wins!

So here are my estimates for 
								(∫_0^(Subscript[T, k])) (Subscript[α, k])(t)dt
which represents the little bit that their off-speeds contributed to their distance over the average. Whichever is larger represents the victor, because that runner will have gone a little farther in the same amount of time -- i.e. will have arrived first. And the winner is Thompson!

LylesLRR=Total[Most[LylesSpeeds]*LylesTimesIntervals]
ThompsonLRR=Total[Most[ThompsonSpeeds]*ThompsonTimesIntervals]
KerleyLRR=Total[Most[KerleySpeeds]*KerleyTimesIntervals]

That estimate for 
								(∫_0^(Subscript[T, k])) (Subscript[α, k])(t)dt
is a LRR (left rectangle rule), with a zero for the initial speed in each case. Let’s compute the RRR estimate, the right rectangle rule:

LylesRRR=Total[Rest[LylesSpeeds]*LylesTimesIntervals]
ThompsonRRR=Total[Rest[ThompsonSpeeds]*ThompsonTimesIntervals]
KerleyRRR=Total[Rest[KerleySpeeds]*KerleyTimesIntervals]

Then a trapezoidal rule is the average of those two:

LylesTrap=(LylesRRR+LylesLRR)/2
ThompsonTrap=(ThompsonRRR+ThompsonLRR)/2
KerleyTrap=(KerleyRRR+KerleyLRR)/2

Simpson’s Rule:

len=Length[LylesSpeeds];
oneFourTwoFourTwoStep=Table[3+(-1)^i,{i,0,len-3}];
AppendTo[oneFourTwoFourTwoStep,1];
PrependTo[oneFourTwoFourTwoStep,1]

LylesSimp=Total[oneFourTwoFourTwoStep*LylesSpeeds]*dt/3
ThompsonSimp=Total[oneFourTwoFourTwoStep*ThompsonSpeeds]*dt/3
KerleySimp=Total[oneFourTwoFourTwoStep*KerleySpeeds]*dt/3

Another approach:

Define the distance traveled, for runner k, as

	(Subscript[f, k])(T)≡(∫_0^T)(Subscript[r, k])(t)dt = (∫_0^T)(a(t) + (Subscript[α, k])(t))dt= (∫_0^T)a(t)dt + (∫_0^T)(Subscript[α, k])(t)dt

The problem is to solve for (Subscript[T, k]), the time at which (Subscript[f, k])(T)=100. 
Since we know when the first athlete purportedly crossed the line, at 9.784 seconds (the official time of the victory), we know that the victor will be the athlete whose (Subscript[f, k])(T) has the largest second term in this expression (all athletes will have the same value for the first term, the integral of the average speed over that interval):

					(∫_0^T)a(t)dt + (∫_0^T)(Subscript[α, k])(t)dt

We need to compute the terms		

			 (Subscript[w, k])(t)≡(∫_0^(  9.784))(Subscript[α, k])(t)dt

The winner will have gone the farthest (i.e., will have reached the finish line).

This explains why the linear transformation we did with the data, to make our graph look similar to (but not the same as) the NYTimes graph, won’t impact our results: the constant term will be the same for all runners, and the constant multiplier will only scale the results: it won’t change the order of the magnitudes (first is first!).