Track Math Quiz

1. Yep it's Rupp-certified. There's an interview somewhere where Salazar talks about traveling in Europe, and often they just do a workout on a soccer field. Often they don't even measure it.

2. Each rectangle is about 314.78m. I think it would have been better if you just told us the length of one straight or of the turn radius.

3. Gonna assume you meant 4 x 4 rectangles? 1 rectangle wouldn't take 3:44. 4 rectangles in 3:44 is on pace for 4:44 for 1600m. By a performance calculator it might "convert" to 1600s in 4:55, so 4 x 1600 in that.

HardLoper wrote:

2. Each rectangle is about 314.78m. I think it would have been better if you just told us the length of one straight or of the turn radius.

Except that the infield also includes what is inside the turns. I didn't say you just run along the straights and then cross.

By Rupp-certified I meant as fast as Rupp could do it. But you didn't run the biggest possible rectangle. Rupp would have done your workout in about 3:08.

i get 323.6363m, but more from trial'n'error

you do need calculus to work out the maximum perimeter of a right angle traingle

the trialn'error method is a 45 degree triangle

the calculus approach shown here :

http://tinyurl.com/o63vtdz

( by Wal C from 7y ago )

I'm gonna guess the largest infield rectangle is ~375.255m.

4 rectangles is ~1501.021m.

Assuming same pace, this makes 3:58.77 for 1600m.

Assuming equivalence tables (Purdy), it's 4:01.00 for 1600m.

not quite, v3, but you are on the "right" track

373m I measured.

Bad Wigins wrote:not quite, v3, but you are on the "right" track

forgot to add all the little "bits" : 323.636 + 2*(36.5/root 2) = 375.25m

ok now here's a hard one.

The meet finishes up but your coach tells you to continue working out on the track. Unfortunately an elite female javelin thrower in a bad mood is hurling spears at you from the kerb in the middle of the backstretch.

You know that her PR is 65m, therefore as long as you are across the track you are safe. But at some point on the turns near the backstretch you will be in range of the javelins. Where precisely is that point in terms of meters along the turn from the backstretch?

and a medium one:

after your workout, your coach brings out a lawnmower, marks out the largest possible rectangular area of the infield and makes you mow it. How big an area was it? How fast could you or Rupp run around it 4 times?

Bad Wigins wrote:

ok now here's a hard one.

The meet finishes up but your coach tells you to continue working out on the track. Unfortunately an elite female javelin thrower in a bad mood is hurling spears at you from the kerb in the middle of the backstretch.

You know that her PR is 65m, therefore as long as you are across the track you are safe. But at some point on the turns near the backstretch you will be in range of the javelins. Where precisely is that point in terms of meters along the turn from the backstretch?

If the thrower is on the inside curb, 90.6m into the first curve you will be in range.

Medium one first:Area is 7648.612 m^24 laps around this area is 1466.4mRupp could run this in 3:27.72Me? Harder to say. Maybe with proper training 4:45 to 5:25

Bad Wigins wrote:

...

and a medium one:

after your workout, your coach brings out a lawnmower, marks out the largest possible rectangular area of the infield and makes you mow it. How big an area was it? How fast could you or Rupp run around it 4 times?

The hard one:24.06m along the turn from the backstretch(I agree with "think" -- well 90.61m -- but he answered from the other direction).

Bad Wigins wrote:

ok now here's a hard one.

The meet finishes up but your coach tells you to continue working out on the track. Unfortunately an elite female javelin thrower in a bad mood is hurling spears at you from the kerb in the middle of the backstretch.

You know that her PR is 65m, therefore as long as you are across the track you are safe. But at some point on the turns near the backstretch you will be in range of the javelins. Where precisely is that point in terms of meters along the turn from the backstretch?

Extra credit to think for remembering the kerb is actually 398.1 meters around. But rekrunner seems to have used good honest algebra, not Euclidean hocus-pocus.

One more easy one. Your coach flies you back to corporate headquarters and the next day unveils a new high-tech track built vertically like a hamster wheel. However fast you run on the track, a motor will spin the wheel just as fast so you can run in place.

Disregarding inconvenient things like drag, elevation and friction, and assuming your top speed is 8.4 m/s and a gravitational constant of -9.8 m/s^2, what is the largest possible track radius that will allow you to suddenly stop running and do a backwards loop on the turning track without falling? Assume that the track will not decelerate until it has made at least one loop.

Wow I'm embarrassed for doing the first problem wrong. Anyways.

a = v^2/r

a = 9.8 m/s^2

v = 8.4 m/s

r = (8.4 m/s)^2/(9.8 m/s^2) = 7.2 m

Hopefully didn't screw that one up

Random thought... 84.39 is exactly 1/500th of a marathon (the accepted metric measurement)

Disregarding those things, and a few others, I find 7.2m also.

Bad Wigins wrote:

Extra credit to think for remembering the kerb is actually 398.1 meters around. But rekrunner seems to have used good honest algebra, not Euclidean hocus-pocus.

One more easy one. Your coach flies you back to corporate headquarters and the next day unveils a new high-tech track built vertically like a hamster wheel. However fast you run on the track, a motor will spin the wheel just as fast so you can run in place.

Disregarding inconvenient things like drag, elevation and friction, and assuming your top speed is 8.4 m/s and a gravitational constant of -9.8 m/s^2, what is the largest possible track radius that will allow you to suddenly stop running and do a backwards loop on the turning track without falling? Assume that the track will not decelerate until it has made at least one loop.

Bad Wigins wrote:Extra credit to think for remembering the kerb is actually 398.1 meters around. But rekrunner seems to have used good honest algebra, not Euclidean hocus-pocus

i'd really like to see how someone solved it with algebra

i coudn't find any easy similar triangles to compare & using algebraic brute force involved solving for x & y :

65^2 = (42.195 + x)^2 + y^2

36.5^2 = x^2 + (36.5 - y)^2

trying to solve that manually is virtually impossible ( unless someone spots a fancy substitution )

i solved it by trig & using cosine rule ( honestly, i hadn't looked at/used that in 35y+ ! ) on relevant triangle, which was pretty quick'n'easy for 24.06m eventually

if anyone can find a neat substitution to solve for x & y above, that woud be impressive...

the brute force algebra

let s = 42.2, r = 36.5, j = 65. Look for x and y such that

x^2 + y^2 = j^2 and

(x-s)^2 + (y-r)^2 = r^2, x > s

2nd => (x^2 - 2sx + s^2 + y^2 - 2ry = 0

s^2 + j^2 = 2sx + 2ry

x = (s^2 + j^2 - 2ry)/2s

= 71.2 - .865y (kill it before it grows!)

1st => (71.2 - .865y - s)^2 + (y-r)^2 = r^2

(29.0 -.865y)^2 + y^2 - 73 = 0

...

y^2 - 70.5y + 481 = 0

solve for y and x, compute arctan and arc length etc etc.

HardLoper wrote:

Random thought... 84.39 is exactly 1/500th of a marathon (the accepted metric measurement)

If that's an accident then it's the coolest factoid I've heard in a while.

Bad Wigins wrote:

the brute force algebra

let s = 42.2, r = 36.5, j = 65. Look for x and y such that

x^2 + y^2 = j^2 and

(x-s)^2 + (y-r)^2 = r^2, x > s

2nd => (x^2 - 2sx + s^2 + y^2 - 2ry = 0

s^2 + j^2 = 2sx + 2ry

x = (s^2 + j^2 - 2ry)/2s

= 71.2 - .865y (kill it before it grows!)

1st => (71.2 - .865y - s)^2 + (y-r)^2 = r^2

(29.0 -.865y)^2 + y^2 - 73 = 0

...

y^2 - 70.5y + 481 = 0

solve for y and x, compute arctan and arc length etc etc.

I was missing the x^2+y^2 substitution. I cheated using an online solver after about 2min of looking at it.