I'll look at it later -- I'm still trying to work it out.
I'll look at it later -- I'm still trying to work it out.
rekrunner wrote:
I'll look at it later -- I'm still trying to work it out.
You're wasting your time. The general solution has been known for decades and is famously difficult.
I posted the specific answer already.
This one's tricky but easy. After the marathon comes the 4x100 relay. The athletes line up in lanes 2 through 5.
But when the gun goes off, each takes the shortest route (i.e. negligible extra distance due to lateral motion) to a different random opponent's lane on the first turn and hands off to the wrong 2nd leg. The 2nd legs run the backstretch properly, but the 3rd legs shuffle their lane assignments the same way the 1st legs did. The 4th legs finish normally.
Of course they are all DQ'd. But which baton took the longest journey around the track?
1) what is the chance that all four batons travelled the proper 400 meters?
2) what is the probability that the baton starting in lane 3 travelled more than a few centimeters less distance than the baton starting in lane 5?
3) what is the average expected distance travelled for each baton (mathematical expectation over time)?
Assume a standard 400m track with 1-turn staggers for lanes 2-5 of 3.52m, 7.35m, 11.2m and 15.0m.
As "ventolin" recently found out, just before he forced me to look up "Sisyphusian", I'm not so interested in the final answer, as in the derivation. I can easily write a script in 10 minutes to iteratively search all solutions with a predetermined precision.Seems like you should know the point of all these math problems is learning how to build the right models to solve them. That is unless you actually want to score 100 ideal marathon runners running 250 out and back laps on the 84.39m straights of an (8-lane?) track.
Bad Wigins wrote:
rekrunner wrote:I'll look at it later -- I'm still trying to work it out.
You're wasting your time. The general solution has been known for decades and is famously difficult.
I posted the specific answer already.
Of course. I'm just letting you know it's surely going to take you a lot longer than any of the problems on this thread, besides not being about track.
There is a theorem named after it.
I think I managed to get as far as your "solution". It's possible with pure algebra, and some geometry, to get to 4th order equations. Unless I missed something in your link, seems that you and everyone else, no matter what technique they used, got stuck with having to solve a quartic equation, which, not unlike using the quadratic formula, becomes trivial when you know the formulas (and have the patience):http://upload.wikimedia.org/wikipedia/commons/9/99/Quartic_Formula.svgOf course these formulas are only possible to derive with a firm grasp of calculus to find critical points, meaning we've strayed somewhat beyond algebra.Without resorting to trigonometry, imaginary numbers, or rules of logs, I managed to derive a deceptively simple relation for the height of intersecting ladders with the height of the two ladders on the passage walls:1 = h/r + h/lIt really doesn't get much simpler than that, does it?Yet, in our specific case, based on what little we know, this equation becomes:1 = 8/sqrt(400-b^2) + 8/(sqrt(900-b^2)All that remains now is to solve for "b".Yet all of my efforts to eliminate messy square roots from the denominator ended up in 8th order equations, with very large coefficients. Since there were no "odd" powers, we can look at it as a 4th order equation in "b^2", something solvable by using the quartic formula for finding the roots, which I will leave as an exercise for the interested reader.I thought I might have better luck solving for r and l, but still ended up with pseudo 4th order equations, and and very large coefficients. If anyone is still reading, and is interested, just ask, and I'll post my 4th order equations for r^2, l^2, and b^2. If not, maybe I'll do it anyway, in a few days, just to bump the thread.
ventolin^3 wrote:
rekrunner wrote:That is deceptively hard. The math gets messy quicklyventolin^3 wrote:
this was always "simplest/hardest" problem i ever saw :
"2 ladders 20 & 30 feet long, lean in opposite directions across a passageway. They cross at a point 8 feet above the floor. How wide is the passage ?"
solve it & i'll buy you serious single-malt at confluence...
i did say so
fortunately, i have seen a solution for all the mess :
http://www.physicsforums.com/showthread.php?t=119674
OK math fanatics, the OP "mcguirkthejerk" never came back with an answer:
http://www.letsrun.com/forum/flat_read.php?thread=3913652
My summary:
"What's the most ping pong balls (circumference of ~4.9634in.) that can fit into a larger sphere (circumference of ~96in), without crushing or cutting them?"
Or if you want to make it track related, use shotputs, and a 3D omni-directional track purpose built by Nike for Galen's workouts.
Sphere stacking is an old problem too but it all boils down to the hexagonal shape formed by alternating rows with rows on top of them resting in the dimples between them. (I cheated a bit and found this was recently proven the most efficient stacking method)
Each sphere rests on top of 3 others. I haven't fully checked, but my gut says the arrangement puts the centers of the spheres in a regular tetrahedron. If the radius of each sphere is r, then that tetrahedron has sides of length 2r. The height of each face is
i = r/tan(pi/6) =rsqrt3
and the distance from one vertex to the center of each face is
rsqrt3 - rtan(pi/6) = rsqrt3 - r/sqrt3 = 2r/sqrt3
so the height of the TH is
sqrt((2r)^2 - (2r/sqrt3)^2) = sqrt(4r^2 - 4r^2/3)
= sqrt (8r^2/3) = 2rsqrt(2/3)
the height of the 2 stacks is simply the rest of the vertical distance from the centers of the spheres to the top or bottom
2rsqrt(2/3) + 2r
subtracting the radius of the upper sphere gives the height of the bottom of the 2nd layer above the bottom of the 1st layer
2rsqrt(2/3) + 2r - 2r = 2rsqrt(2/3)
This difference, plus the height of the 2nd and a 3rd layer, gives the height of 3 layers
2r + 2rsqrt(2/3) + 2rsqrt(2/3)
the bottom of the 3rd layer is of course 2x2rsqrt(2/3) above the bottom of the first. In general, for a number of layers n,
h(stack) = 2r + 2(n-1)rsqrt(2/3)
stacking them in a sphere could use 2 approaches, density, or calculating or writing a function for the width of the sphere at each layer. Density is easier. I know the actual maximum density but I haven't derived it. All it should require is to is set a constant width and length for the layers - stack them in a box, basically - for which there is a known constant number of spheres in each layer, allowing the volume of the spheres to be a 1st order function of n, and the volume of the box to be h times some coefficient. Then find lim(n->inf) of the volume of the spheres over the volume of the box.
I'll do that later if noone else does, and if I'm not in error so far.
The wikipedia links give the formulas for ball placement and height. The numbers look similar to your formulas (but I didn't check).
I'd be interested in seeing if there is a mathematical approach better than brute force.
I've more or less got it.
Consider a box of length 9sqrt(3)/2 and width 9r.
Arrange the first layer of spheres in alternating rows along the length such that there is a half sphere at the end of the 1st and 3rd rows, at the beginning of the 2nd and 4th rows, and the 5th row consists of 4 half spheres with a quarter-sphere at the end. Draw a picture if it's confusing. There is a little bit of space left over after the 1st and 5th rows, but that converges out as I'll explain later.
The 2nd layer can be constructed the same way, with the row of half spheres and quarter sphere at the top, and the other half spheres on opposite sides from the first row. And so on ad infinitum. Then the 3rd like the 1st, and so on ad infinitum.
Let m be the number of layers beyond 1. Then the volume of the spheres
U(m) = (4.5x4 + .5x2 + .25)x4pir^3/3 = 20.5x4pir^3/3 = 27pir^3
from my earlier post, the volume of the box containing the spheres is the area of the box times h.
V(m) = (2r + 2mrsqrt(2/3)) x (9rsqrt3)x9r/2
Then the density
D(m) = U(m)/V(m) = 27pir^3/(((2r + 2mrsqrt(2/3))x81r^2sqrt3)/2)
= pirm/((2r + 2mrsqrt(2/3)x(3sqrt3)/2)
as m --> inf this converges to
pim/(msqrt(2/3)x3sqrt3 = pi/(3sqrt2) ~ 0.74
Now, what about the little extra space in each layer above the top or bottom rows? Well, for a given length of the rows and number of layers, that extra space will remain constant if you increase the number of rows and thus the width of box, so the ratio of that space to the volume of the box for a given height converges to zero when the number of rows ---> inf. In general the density of this stacking method converges to pi/3sqrt2, though in this example it would be slightly lower.
There may be irregular arrangements that fill a sphere more densely than that, but I doubt the OP of the other thread was asking about that. It was probably some jerk asking about homework.
So, for the track question - after practice your ever-sadistic coach makes you clean up after the shot putters. You must put them in a cubic box 30cm on a side. How many will fit in the box?
But we are not putting balls in a box of infinite height. We are putting them in a finite sphere, and the extra space between the sphere boundary, and our cluster of ping pong balls becomes a very important limitation.There are many ways you can arrange the different layers to achieve this proven maximum density, and the choices you make result in different placement of the balls, and different clipping. FCC and HCP are two special cases.My brute force simulations show that the clipping is a function of:- Which layers you chose- The placement of the initial ball relative to the larger sphere
Bad Wigins wrote:
I've more or less got it.
...
pim/(msqrt(2/3)x3sqrt3 = pi/(3sqrt2) ~ 0.74
The best answer I could find, using maximum packing density resulted in an additional 13.44% wasted "extra space" (or 86.56% of theoretical upper bound).
McJerk wrote:The larger sphere has a circumference of ~96in.
Any precision greater than a two-digit approximation is meaningless.
A sphere of circumference ~96in has radius ~31in (for computation, 30.573). A ping-pong ball of circumference 4.9634in has diameter 1.5807in, so approximately 19 balls (19.341) can fit in a row across the sphere.
The maximum density could be derived for a sphere the same way I did it using a box - put the balls in a small cluster, consider the smallest sphere around it and then find the limit of density as the radius of that cluster and of the sphere tend to infinity. Any error in finite spheres comes from that margin of uncounted extra space at the edge of the cluster. That space can be shown to be less than one sphere diameter thick no matter what method is used, so it introduces uncertainty no greater than the 31 +/- 1 in or 31 +/- .5in or whatever introduced by the question itself.
Sure, it's interesting to consider what the answer would be if it were 96.000 in instead, but this thread has already moved way over the heads of the average track athlete. I will have to wait a few days and start another one.
Irish gymnast shows you can have sex in the "anti-sex" cardboard beds in the Olympic village (video)
2024 College Track & Field Open Coaching Positions Discussion
Per sources, Colorado expected to hire NAU assistant coach Jarred Cornfield as head xc coach
Olympic village has opened and Dutch beach volleyball player who raped a 12-year-old isn't in it