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[edit] December 7

[edit] Efficient point mass forces

Suppose I have a set of points in Euclidean 2-space, with each point A represented on a computer by a vector (A_x, A_y). The points repel each other by an inverse exponent, i.e. the force of repulsion varies with 1 / rⁿ, with n an arbitrary integer. For a particular point, A, I want to calculate the resultant force from all the other points. Assume all points are identical in all respects except their location.

What is the most computationally efficient way to calculate the resultant force on the point A from the other points? Once I have calculated the resultant force I will apply it and adjust the position of point A appropriately, then repeat the process on the next point, etc. What I need is a way to calculate the corresponding force in each dimension with as few computations as possible. Any ideas? Maelin (Talk | Contribs) 05:12, 7 December 2006 (UTC)

I'd say:

Use two floats/doubles/whatever to store the total force on x and y for each point other than the one considered, calulate and store dx and dy (algebraic distances between the two points), add them and store it too. Then take the root of the sum of the squares to get r, calculate 1/r^n, multiply it by ~~dx/(dx+dy)~~dx/r and put it in x, multiply it by ~~dy/(dx+dy)~~dy/r and put it in y. Then continue. I know this is rather simplistic, but I don't think there's a simpler algorithm. yandman 10:33, 7 December 2006 (UTC)
How are you thinking of "adjusting" the position of point A? I suppose you were thinking of Euler's method? If you want to avoid divergence problems, it might be better to use retrograde techniques, and if precision is what you want, a fEhlberg or RK4 method might be a better idea. yandman 10:41, 7 December 2006 (UTC)

Comment point 1, shouldn't that be dx/sqrt(dx²+dy²) not dx/(dx+dy) (I assumed you just simplifed the notation.

I can't see any real way to make it easier - there are n points, n-1 forces to sum. n(n-1) point point interactions overall.

You could attempt to speed r^n using r,r^2,(r^2)^2=r^4 etc eg if n=7 x=r^2 x2=x^2=x^4 there fore x^7=x2 times x times r. Thats 4 multiplies instead of 6 - it gets better as n increses depending on n. Did you mean speed it on a computational machine - or just the simplest method?83.100.174.147 11:20, 7 December 2006 (UTC)

Note if you store the Particle(n) to Particle(m) distance in memory when calculating it for particle(n) - you don't need to recalculate it for particle m reduced the number of calculations by 1/2 = 1/2n(n-1) particle particle calculation + 1/2n(n-1) particle particle calculation lookup from a stored memory. (Where you thinking interms of computers?)83.100.174.147 11:24, 7 December 2006 (UTC)

WHoops, I meant to put dx/r. Corrected. yandman 13:12, 7 December 2006 (UTC)

I would strongly suggest asking this at the science reference desk. There are important computational results that the typical mathematician would know nothing about. Among these are Leslie Greengard's 1987 Yale doctoral dissertation, The Rapid Evaluation of Potential Fields in Particle Systems, which won an ACM Distinguished Dissertation award. Sophisticated methods are hugely important in computational astronomy, physics, and chemistry. You might want to read these brief summaries, and consult this longer list of resources. Our mathematically-oriented article on the n-body problem offers no hint of this rich computational literature. --KSmrq^T 15:17, 7 December 2006 (UTC)

Out of curiosity, what two-dimensional system gives you n > 1? Melchoir 16:09, 7 December 2006 (UTC)

A very small steel ball on a very thin table with very small magnets just underneath it is the usual excuse for asking students to do this, I think. yandman 16:30, 7 December 2006 (UTC)

[edit] Integration-related question

Had a math test today, didn't go all that well but what the heck. I found one particular question rather amusing but sadly I wasn't able to solve it. Here goes:

The figure shows an area A that is limited by the functions f(x) = x + sin x and

g(x) = x. Calculate a so that the line x = a divides the area in two equal parts.

My approach was this: if f and g intersect in P_1(0:0) and P_2(w:z), then S(0->a)f(x)dx - S(0->a)g(x)dx = S(a->w)f(x)dx - S(a->w)g(x)dx Is this the right approach? I didn't get any further than that so I'd be glad if someone could provide an answer to the question. I tried to get the coordinates for P_2 by adding the two formulas but it didn't make much sense to me - isn't that how you're supposed to do to find out where two equations would intersect? Anyway if someone could provide an answer it'd be great, of course feel free to rename any variables you like. Thanks in advance. Jack Daw 10:05, 7 December 2006 (UTC)

Good work. All you then had to do was merge the integrals, you do S(0->a)(f(x)-g(x))dx = S(a->w)(f(x)-g(x))dx, which is S(0->a)(sin x)dx = S(a->w)(sin x)dx, integrate ( [-cosx](0->a) = [-cosx](a->w) ), you know that ~~w=Pi/2~~w=Pi, you use arccos, and robert's your father's brother. yandman 10:21, 7 December 2006 (UTC)

Yeah I figured that it was possible to merge, wasn't sure though. But how did you get pi/2 as the x-coordinate for the second intersection? Jack Daw 10:46, 7 December 2006 (UTC)

Corrected. Please forgive me, Euler... yandman 10:48, 7 December 2006 (UTC)

Using Cavalieri's principle, you can see that the answer is the same for f(x) = sin x and g(x) = 0, which should be obvious.--80.136.173.216 10:55, 7 December 2006 (UTC)

It would be kind to either say a little more, or offer a link. Let's try something simpler still. Recall that the area of a triangle is half the product of base and height. Notice especially that it does not depend on the horizontal position of the top vertex with respect to the base; the area is invariant under horizontal shearing. We immediately deduce that the area of any region is invariant over shearing in any direction. (This is a limited form of Cavalieri principle, but all we need.) Now observe that f(x) = x+sin x and g(x) = x are sheared forms of F(x) = sin x and G(x) = 0. I'm going to assume that the figure, which we are not shown here, limits x to the interval [0,π]. Thus the total area is

$\int_0^{\pi} \sin x \, dx ,$

which is exactly 2. However, we will not need to integrate. We wish to split this with a vertical line, x = a, to divide the area exactly in half. But this is trivial, because the region is symmetric. Furthermore, the shearing does not affect the line, so we can obtain our answer with no calculus. --KSmrq^T 16:04, 7 December 2006 (UTC)

That was a bit too complicated for me. I don't doubt that your answer is wrong, however I doubt this is the way we were supposed to solve the problem. We haven't been taught the math you use to solve the problem, that's why. The figure doesn't add any extra information, the problem was part of the test where you weren't allowed to use a calculator so the figure was merely for the sake of showing how f and g intersect. So, do you know a method involving calculus that would solve the problem? Jack Daw 18:49, 7 December 2006 (UTC)

The moral of the story is: either you do what they expect you to and integrate, either you think a bit and are called KSmrq. yandman 16:18, 7 December 2006 (UTC)

My apologies; I was trying for a simpler answer, to explain the invocation of Cavalieri's principle. If you prefer to use integration, more explicit calculations are involved but much is alike. The area under f(x) = x+sin x is given by the obvious integral; similarly for the area under g(x) = x. The area between the two curves is either the difference of these area integrals, or the integral of the difference function, f(x)−g(x) = sin x. Taking the latter route, we are looking at

$\int_0^{x_0} \sin x \, dx$

for the total area, where x₀ is the x value of the intersection, and

$\int_0^a \sin x \, dx$

for the split. To find the total area, we must know that the curves intersect at x₀ = π. This comes from knowing where sin x equals zero, for then x+sin x will equal x. We have already computed the first integral in this case, and found that the total area is 2. The second integral is no more difficult, and gives an equation to solve for 1. The equation involves cos a, and the solution requires basic knowledge of its behavior. --KSmrq^T 18:30, 8 December 2006 (UTC)

First of all, when I wrote "I don't doubt that your answer is wrong" I meant RIGHT. Hope you got that... Second... How do you know that the second intersection is at x = pi? arcsin 0 = 2 * pi * n, is it not? So how does that become pi? Jack Daw 03:32, 9 December 2006 (UTC)

Yes, I assumed that was what you meant. :-D

No, sin x is zero at integer multiples of π, not 2π. It has period 2π, which is not the same thing. --KSmrq^T 17:11, 9 December 2006 (UTC)

Sine function

Aha OK now I get it. Thank you. Too bad I didn't nail this during the test, gave a lot of points. Oh well... Jack Daw 02:15, 10 December 2006 (UTC)

[edit] Probability question

Note: I am not a student, and this is not a homework question.

Say I have a 10-sided die. What are the chances of getting, say, a 3 at least nine times out of 10?

Please explain how you arrived at your answer.

Thanks -- 192.206.151.130 15:43, 7 December 2006 (UTC)

The simplest way to consider this is to divide the number of "good" outcomes by the total possible outcomes. Here the total is 10^10, i.e. 10 billion. The number of "good" outcomes is:

1 for the outcome of getting 3's every time.
10 for each "order" of 9 3's (say, 5 3's, followed by another number, followed by another 4 3's etc...). You multiply this by 9 because the "other number" could be a 1,2,4,5...10.

Which gives 91/10 billion. Not much, huh?

There are simpler but more abstract ways of calculating it, though. 0.1^10 (the chance of getting straight 3's) + (0.1^9)*0.9*10 (the chance of getting 9 3's on 9 rolls multiplied by the chance of not getting a 3 on the last roll multiplied by the number of orders possible).

Basically, don't take the bet.... :-) yandman 16:05, 7 December 2006 (UTC)

You mean 9.1/1 billion, surely? Melchoir 16:07, 7 December 2006 (UTC)

Yeah, I get 91/(10 billion), which is 9.1x10^-9 -sthomson 16:20, 7 December 2006 (UTC)

Time for me to get some coffee. Yeah, it was the 10 billion I wrote 4 lines above... Corrected. yandman 16:24, 7 December 2006 (UTC)

So throwing 9 or more 3s in 10 throws with a fair dice is stupendously unlikely - you would probably have to roll 10 dice once a second, 24x7, for over 3 years before you saw this result. If this is a practical problem, say someone has defeated your battalion of orcs against overwhelming odds, then you can reasonably assume that something dodgy is going on. Gandalf61 17:15, 7 December 2006 (UTC)

The ten-sided die itself is a little problem. No regular polygon available ... you have to turn to a bar with ten angles, something like a Grecian temple column. Don't stay under it when someone throws it :-)--DLL ^{.. T} 19:02, 8 December 2006 (UTC)

There are perfectly good ten-sided dice, they're just not platonic solids. Confusing Manifestation 13:25, 9 December 2006 (UTC)

(edit conflict) I think you meant no regular polyhedron. However, the five Platonic solids are the tetrahedron (4 faces), the cube (6 faces), the octahedron (8 faces), the dodecahedron (12 faces), and the icosahedron (20 faces). An obvious adaptation suggests itself.

But there is no reason that fair dice must be regular polyhedra. Remarkably, Commons has a collection of examples. A natural choice is the bipyramid, which guarantees equal faces without limiting their number (except to be even); but the pentagonal antiprism dual, the pentagonal trapezohedron, remains fair while rolling better. From the dice article:

The full geometric set of "uniform fair dice" (with all congruent sides) are:

Platonic solids: 5 regular polyhedra: (4, 6, 8, 12, 20 sides)
Catalan solids: 13 Archimedean duals: (12, 24, 30, 48, 60, 120 sides)
Bipyramids: infinite set of prism duals, triangle faces: (6, 8, 10, 12, ... sides)
Trapezohedrons: infinite set of antiprism duals, kite faces: (6, 8, 10, 12, ... sides)

Rolling-pin style dice are usually made so that all the faces they may actually land on are congruent, so they are equally fair.

Nor must we restrict to flat faces; we might take a dodecahedron and glue halves of a sphere to two opposite faces to prevent them being stable resting bases, thus reducing the 12 fair faces to 10. See this collection for more creative inspiration. --KSmrq^T 16:52, 9 December 2006 (UTC)

[edit] Average number of workdays

Thanks for the probability answer. Here's another question:

Say you live in a country with a five-day workweek. In addition, there are nine weekday holidays on which people don't work.

What is the average number of workdays in any 10-day period ending in a weekday?

Thanks -- 192.206.151.130 17:05, 7 December 2006 (UTC)

Let f(x) be the number of workdays in the 10-day period that ends on day x, and let W be the set of all workdays. Then the average number of workdays is $\frac{\sum_{x \in W} f(x)}{|W|}$ .

So the first question to answer is how to calculate $\sum_{x \in W} f(x)$ . At this point, though, I'm not sure when you say "10-day period" whether you mean "10 workdays" or "10 days, including weekends". Which you mean would affect the outcome. Also, the cardinality of W (ie the number of workdays in the year) needs to be defined. And finally, do holidays count as workdays, or do they count as extra "weekends"? Dugwiki 17:34, 7 December 2006 (UTC)

Isn't it easier to simply count them in your example? I did that some time ago and the average I found was 22 - not counting February. Jack Daw 19:38, 7 December 2006 (UTC)

Assume that "weekday" means: any day of the week except Saturday and Sunday. In a random period of 10 consecutive calendar days of which the last is a weekday, there are on the average:

1.6 Mondays

1.6 Tuesdays

1.6 Wednesdays

1.4 Thursdays

1.2 Fridays

1.2 Saturdays

1.4 Sundays

This can be established by straightforward counting for each possible final weekday. Together this gives us on the average 7.4 weekdays. If some of the holidays are bound to a specific day of the week (say the fourth Thursday in November), then we need to know the exact distribution. If the weekday holidays are uniformly distributed over the weekdays, then we need to know the number of weekdays in a year. Taking 365.2425 days for the average length of a year, the average number of weekdays is 260.8875. If 9 of these are holidays, the remaining 251.8875 days are workdays. So a fraction of 251.8875/260.8875 of the weekdays are workdays. Under the assumption of uniform distribution, the average number of workdays out of these 7.4 weekdays is then 251.8875/260.8875 × 7.4 = 82843/11595 = 7.14471755... workdays. --Lambiam ^Talk 06:38, 8 December 2006 (UTC)

Here is a little table, to support computing the weekday probabilities:

1	2	3	4	5	6	7	8	9	X
M	T	W	R	F	S	U	M	T	W
T	W	R	F	S	U	M	T	W	R
W	R	F	S	U	M	T	W	R	F
S	U	M	T	W	R	F	S	U	M
U	M	T	W	R	F	S	U	M	T

We count 8 Mondays (M), 8 Tuesdays (T), 8 Wednesdays (W), 7 Thursdays (R), 6 Fridays (F), 6 Saturdays (S), 7 Sundays (U). Thus from a total of 50 days, we find that 37 are weekdays. But in practical terms, of the nine Canadian national holidays, one is coerced to Friday and three are coerced to Monday. Of course, every country is different, but it is quite likely that Easter will be celebrated on a Sunday (if at all), and Good Friday on a Friday, to give two obvious examples. --KSmrq^T 19:31, 8 December 2006 (UTC)

[edit] Why not invent a symbol for 1/0

I am thinking about 1/0 = w

SO 2/0 = 2 * (1/0) = 2w

and (3/0) * (4/0) = 3w * 4w = 12 w²

7 + 5/0 = 7 + 5w

(42/0) * 0 = 42w * 0 = 42

My question is this. I'm sure this has been thought up before by other people before me, so why was it rejected. Is there a fatal flaw in representing 1/0 as w?

202.168.50.40 21:13, 7 December 2006 (UTC)

See Division_by_zero. Friday (talk) 21:19, 7 December 2006 (UTC)

The reason is that division is defined to be the inverse operation of multiplication. This means that the value of

a / b

is the solution x of the equation

b x = a

whenever such a value exists and is unique. Otherwise the value is left undefined.

For b = 0, the equation bx = a can be rewritten as 0x = a or simply 0 = a. Thus, in this case, the equation bx = a has no solution if a is not equal to 0, and has any x as a solution if a equals 0. In either case, there is no unique value, so

a / b

is undefined. Conversely, in a field, the expression

a / b

is always defined if b is not equal to zero.

I think you misunderstood.

$\frac{2}{0} \ne \frac{1}{0}$ because $2 w \ne 1 w$

202.168.50.40 21:35, 7 December 2006 (UTC)

Division by zero would break other rules of math, hence the contradictions explained in Division_by_zero#Fallacies_based_on_division_by_zero. Friday (talk) 21:38, 7 December 2006 (UTC)

Argh! yes you are right. Replacing (1/0) with w ( or 0/0 with 0w ) does not resolve the paradox, hence it is rejected as not useful. Thank you very much. 202.168.50.40 23:26, 7 December 2006 (UTC)

There's a recent Slashdot story about someone claiming to have done something very similar. BungaDunga 05:03, 9 December 2006 (UTC)

[edit] whilst on zero

Seeing as the subject has come up can someone direct me to a page covering 0⁰. Is it undefined...87.102.36.136 22:30, 7 December 2006 (UTC)

See Exponent#Exponents_one_and_zero. Friday (talk) 22:32, 7 December 2006 (UTC)

More directly, see Empty product#0 raised to the 0th power Dugwiki 22:40, 7 December 2006 (UTC)

Taking the nth root of zero is zero - it is only in this case that

lim n to infinity | 0^1/n

that the result can be considered zero.? The page doesn't seem to consider the case x^y where x is alreadly zero and y approaches zero. Just wondering.87.102.36.136 23:01, 7 December 2006 (UTC)

Limits only need to be considered when defining powers for non-integers. 0 is an integer, so 0^0 has nothing to do with limits. It is just the product of no factors, which is 1. True, when we later try to extend the definition to any nonnegative real, we get the inconvenience that the function x^yM/ is defined but discontinuous at (0, 0), but this is beside the point. -- Meni Rosenfeld (talk) 11:04, 8 December 2006 (UTC)

I noticed that for the graph z=x^y that the planes x=0 and y=0 are the boundaries of 'massive' discontinuities. Wondering if anyone has a link to an image of such a graph, (just real values obviously) just to satisfy my curiousity..87.102.37.17 11:10, 8 December 2006 (UTC)

There is much more of a discontinuity with respect to x than with respect to y, because you can a positive number to any power but negative numbers only to integers and specific roots (in the real domain), hence when x > 0 you get a nice solid shape but when x < 0 you get a very sparse-looking graph. Try plotting it in gnuplot:

gnuplot > splot [x = -1:1][y = -1:1] x**y

And you'll see what I mean.137.99.174.5 23:33, 8 December 2006 (UTC)

Actually Meni's answer elides a subtlety. It is possible to make a distinction between the integer or natural number zero, and the real number zero. Since this distinction almost never makes a difference, we rarely bother.

But it does arguably make a difference when considering the expression 0⁰. If both 0s are considered to be the natural number 0, then the arguments for defining the value of the expression to be 1 are quite convincing. Even if the base is taken to be the real number (I'll write it as 0.0 to make the distinction) it seems clear that 0.0⁰ should be the real number 1.0 (as possibly distinct from the natural number 1).

However if the exponent is taken to be a real number, in the expression 0.0^0.0, most of these arguments lose their force. I personally think that 0.0^0.0 should be considered undefined. --Trovatore 22:16, 10 December 2006 (UTC)

[edit] Probability w/ coins

Say I have a coin with height (or thickness) h and bases w/ radius r. What are the odds that if I flip this coin, it will land on its side (as opposed to the bases)?? --Tuvwxyz (T) (C) 22:39, 7 December 2006 (UTC)

Obviously this needs to be highly idealized. The coin has to be a perfect cylinder. You have to assume that the coin will have no bounce, otherwise it's way complicated. Reduce it to "assume the coin will land on its side if it hits the ground within a certain range of angles", and then it will be whatever that range is divided by π. Or something rather like that. For a real world number, I guess you just have to start flipping. --jpgordon^∇∆∇∆ 23:38, 7 December 2006 (UTC)

When you consider that the answer is effected by all sorts of factors such as the motion and angular momentum of the flipping coin and the physics of how it bounces when it lands, I doubt there's a simple answer besides "almost zero" for standard coins.

A simpler related problem to consider might be to consider a coin that is in a random orientation relative to a plane below it. Then what would be the probability that the coin's center of gravity, when projected onto the plane, lies over the coin's rim?

Assume the coin is circular. Then let's look at a planar cross section perpendicular to the ground that includes the coin's center of gravity. To land on its edge at the moment of impact, the center of gravity would have to lie over the rim when looking along that cross section would be the ratio of the angles formed by tracing edges from the center of the coin's rectangle in the plane to its corners. This would have to be true of all possible such cross sections; if the center of gravity projects beyond the coin's rim on any cross section, then the coin would likely tip over onto its base along that angle.

I'm going to bow out at this point to the experts on how to calculate that probability, since it's getting a bit over my head. But I'm thinking it might involve integrating the angular ratios over all possible cross sections and orientations for the coin. Dugwiki 00:00, 8 December 2006 (UTC)

Well, it won't always be "almost zero". For instance, if a "coin" has a radius of one inch and and a height of 1 foot, the probability would be "almost 100%". That was the entire point of me supplying variables for the radius and height. --Tuvwxyz (T) (C) 00:27, 8 December 2006 (UTC)

That's why I said it was almost zero "for standard coins". You'll have a hard time finding a coin that's shaped like a piston. :) Dugwiki 17:04, 11 December 2006 (UTC)

Out of a full circle (of 2π for the measure) of possible orientations of the coin in a perpendicular cross section, two orientations will let the coin land, on first impact, exactly flush on its edge. With a tolerance of ±arctan(h/(2r)), the centre of gravity will be over a spot inside the rim. The measure of these orientations is then 2 × 2 × arctan(h/(2r)) – where one factor 2 is for the two exactly-flush orientations, and the other for the two-sided tolerance. Expressed as a fraction, this results in 4×arctan(h/(2r))/(2π). For example, if h = 1 and r = 1/2, we get 4×arctan(1)/(2π) = 1/2. Of course, as already noted above, this has little to do with the eventual result; if h<<r, the chance of the coin coming to rest on its side is typically much smaller. Also, in actual coin flipping not all orientations are equally likely. --Lambiam ^Talk 07:07, 8 December 2006 (UTC)

If considering 'coin dropping' and nor coin flipping (the coin is not allowed to rotate) then the probaility of it landing on the 'edge' will be the solid angle subtended by the edge divided by the total solid angle (4pi), the angle the rim makes to the plane of symmetry of the coin is a=arctan(h/2r),

The solid angle is

2 $\int_{0}^{a} 2pi\cos {a}\;da$

=4π sin(arctan(h/2r) so the probability is sin(arctan[h/2r]) (in the case h=1 r=0.5) p=1/sqrt(2) ~ 0.7 </math>

If the coin is spinning in any direction but along the axis this method will not be right.87.102.37.17 10:28, 8 December 2006 (UTC)

I do not know the odds of landing on edge, but Persi Diaconis has evidence that coin flipping is biased so that the face that is up when it is flipped is slightly more likely to be the face that is up afterwards. (He himself can flip a coin so it comes up heads every time.) Fun to know; hell to analyze! --KSmrq^T 19:46, 8 December 2006 (UTC)

I think the problem is that not only must the coin be on its edge at some point, it has to have just the right energy to stay there. I'll look at the two-dimensional problem and assume with each bounce, the coin's total mechanical energy decreases by a factor of f > 1. (This might not be a good assumption, but it'll produce a reasonable scalaing law.) Let gravity be g = 1. Then the coin settles on its edge if and only if:

After some bounce, its energy is between r and sqrt(r^2 + h^2/4) or about r + h^2/8r.
At that point in time, its orientation is within ±arctan(h/(2r)).

As Lambiam calculates, the latter probability is 4×arctan(h/(2r))/(2π) or about h/3r. But the former probability is even smaller: a random geometric progression must happen to intersect with a band of width h²/8r at a height of r, which happens with probability log ([r+dr]/r)/log f or about dr/r log f = h²/8r² log f. So my final answer is

$\frac{1}{24\log f}\left(\frac{h}{r}\right)^3$

for small h/r. Not very good. Melchoir 20:33, 8 December 2006 (UTC)

And then there is research. For example, we might read what Justin Smith has to say about "Constructing a fair 3 sided coin". Or, we might consult the article by Daniel B. Murray and Scott W. Teare, "Probability of a tossed coin landing on edge", in Physical Review E, 48(4), October 1993, pp. 2547–2552. The abstract states:

An experiment is reported in which an object which can rest in multiple stable configurations is dropped with randomized initial conditions from a height onto a flat surface. The effect of varying the object's shape on the probability of landing in the less stable configuration is measured. A dynamical model of the experiment is introduced and solved by numerical simulations. Results of the experiments and simulations are in good agreement, confirming that the model incorporates the essential features of the dynamics of the tossing experiment. Extrapolations based on the model suggest that the probability of an American nickel landing on edge is approximately 1 in 6000 tosses.

This is what references are really about among professionals: knowing what has been done before so we can build on the work of our predecessors, acknowledging and benefiting from their successes and their failures. This concept bewilders crank contributors, but it works. To quote the old chestnut by Sir Isaac Newton:

If I have seen further it is by standing on ye shoulders of Giants. (Letter to Robert Hooke, February 5, 1676)

Still, we must balance this with a lovely quotation attributed to Carl Sagan:

When you make the finding yourself — even if you’re the last person on Earth to see the light — you’ll never forget it.

Apropos of quotations and references, I always hesitate when I find an enticing line without a definite reference, even if it is widely repeated. Sometimes the quotation is wrong, sometimes the attribution. Hence, "attributed". Any help on this one? --KSmrq^T 19:23, 9 December 2006 (UTC)

Excellent reference; we ought to have an article on this problem! I see that Murray and Teare's Figure 5, p.2551 is a log-log plot with a slope of 3 and an intercept that varies with the coefficient of restitution. I wonder if there's another model as simple as mine that also predicts that result? Melchoir 23:08, 9 December 2006 (UTC)

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