# Re: Debating the athletic merits of Ultimate

On Dec 3, 2:43 pm, "Mike Gerics" <mger...@xxxxxxxxx> wrote:
So i was having a discussion with a friend online about playing
Ultimate, and this friend challenged me to justify calling Ultimate a
"sport".

Lets take it from the science angle. Is it possible to calculate the
work that a person does while playing at a tournament?
How about total energy expended? Total Kinetic Energy? Seems
unlikely but here goes...

assumptions about the average, competitive ultimate tournament...
with a roster of 20 (i'll low ball every number to show the minimum
results)

25 meters per cut/clear (this prolly depends on the type of offense
being run, but immediately omits high value deep cuts)
4 cut/clears per point (a low balance between ladders and hell points)
8 points played per game (also questionable, gehret and gibson played
what, 46 consecutive points to close out nationals in 06?)
4 games per day (reasonable, imo, sometimes 5)
2 days (except pres day, vegas, and centex)

now the tricky part. assume that standing still requires
no work or energy (which in terms of reality is not the case, as
simply remaining alert is super trying)
avg body mass is 70 kg (F = 70 x 9.8 = 686)
avg applied force to run is... 45 degrees of vertical at best (further
than this and you're prolly falling over or attempting to layout)
and now... how many newtons does it take to maintain a high velocity?
let's say a vertical force applied by my legs to the ground for each
stride is 686 newtons (maybe low balling it... but...)
this is rendered even less when we look at the value with respect to
forward motion and the work of running. Since we can't
push ourselves directly forward, ever, considering we walk on two legs
on a ground perpendicular to gravity

lets say constant velocity is maintained by a 686 newton force applied
at 30 degrees of the vertical. total work would
therefore push you forward with Work = 686 Newtons x (total distance
in meters above) x cos (60)

my dad the physics professor would rip me a new one for such loose
figures, but lets see what we get.

2.2 x 10^6 joules, or roughly just under TWICE the amount of work it
would require to
pull a 2000 kg roller coaster car up a 60 meter tall hill at any angle
(since force is applied along a parallel plane to the angle
regardless,
the roller coaster's chain)

so this method is horribly flawed in hindsight, but it does show that
even
with low assumed values, it all adds up. It doesn't take into account
the fact that
people change directions and vary their speeds when cutting. maybe a
better way to calculate
work would have been to use work = change in kinetic energy,
so go from the ke of standing still to the ke of running at full
speed,

unsure of how to factor residual hangovers/morning vape sessions into
this equation.
now who wants to calculate the energy it takes to get to tulsa?

-taco
.