But 10+ year old industrial estate I work in, is missing completely .....(It's on Google Maps and even on StreetView)

This is what's great about free information -- you can add it yourself. Click on Edit above the map.

One point where Open Street Map shines is that it has actual roads and trails in such places as National Parks and forests...where the commercial maps have nothing but blank green areas.

Yep. For example, here is a place in the Alps in openstreetmap, and here is the same place in google maps, and here is the same thing in routes.tomtom.com. Only openstreetmap shows the hiking trail (as well as peaks with their elevations, and mountain huts). This is a really good thing for hikers, runners and mountain bikers.

You can also get topo maps based on OSM data from toposm.com, although this still seems pretty primitive and they only cover the US. Google's maps with contour shading are OK, but they don't let you print them through their web interface (although you can always print a screenshot), and they don't show contour lines.

What isn't so great about OSM is that driving directions from yournavigation.org are not usable at all. Also, the search functionality is (not surprisingly) inferior to the one in google maps -- if you don't put in exactly the right form of the name, it doesn't work.

Cool, thanks for checking that! As explained here, what's going on is that the expression is only a constant of the motion for v>0.

You missed the other factor of u. The expression u sqrt(u^2 + v^2) gives the x component of a vector with squared magnitude u^2 + v^2 that points in the direction of (u,v).

As often seems to be the case with these news articles about teenage prodigies, this has been overhyped. It turns out that what he did is not new and is not a complete solution to the problem.
Parker, Am J Phys 45 (1977) 606 has a summary of the preexisting results. The expression immediately after equation 23 is the constant of the motion that Ray rediscovered.

A reddit user has a nice simple derivation: http://redd.it/u74no (Note that there is an error because he claims to have proved it in general, but it's only valid when v (the vertical velocity) is positive.)

For more on the history of the problem:

Synge and Griffith, Principles of Mechanics, p.~154 http://archive.org/details/principlesofmech031468mbp

Whittaker, A treatise on the analytical dynamics of particles and rigid bodies, p.~229 http://archive.org/details/treatisanalytdyn00whitrich

According to Whittaker this was first done by D'Alembert in 1744.

Hmm...actually it only works for v>0, which is the reason for the goofy behavior I was seeing in this post. Notice how the numerical simulation shows Ray's expression as a perfect constant of the motion up until v becomes negative, and then it starts changing.

This is because if you simplify the total derivative from this post, you don't actually obtain zero, you get (v/|v|-1)/sqrt(u^2+v^2), which only vanishes for positive v.

So it looks like the expression shown on the big piece of paper in the photo is only for v>0. I don't know if there's some trivial modification that would generate a valid expression for the negative v case.

You can easily check this by differentiating his equation with respect to time, and then eliminating the derivatives of u and v using the expressions

du/dt = -b u sqrt(u^2 + v^2)
dv/dt = -b v sqrt(u^2 + v^2) - g

Awesome! I didn't want to grind out the calculus by hand, so I did it with maxima:

load (f90)$
 
r(u,v) := 1/(u**2)+v*sqrt(u*u+v*v) / (u*u) + asinh(abs(v/u)) ;
/* Ray's expression, multiplied by 2, with alpha=g=1 */
 
foo : diff(r(u,v),u)*u*sqrt(u*u+v*v) + diff(r(u,v),v)*(1+v*sqrt(u*u+v*v));
/* total derivative with respect to time */
 
f90('foo = foo);

Unfortunately maxima wasn't smart enough to simplify it show that it vanished identically, so I had it output it as fortran code, and then tested numerically that it vanished with random numbers as inputs:

double precision u,v,foo
u = 0.3776
v = 0.1209
foo = u*sqrt(v**2+u**2)*(-abs(v)/(u*abs(u)*sqrt(v**2/u**2+1))-2*v&
*sqrt(v**2+u**2)/u**3+v/(u*sqrt(v**2+u**2))-2/u**3)+(v*sqrt(v**2+&
u**2)+1)*(v/(abs(u)*sqrt(v**2/u**2+1)*abs(v))+sqrt(v**2+u**2)/u**&
2+v**2/(u**2*sqrt(v**2+u**2)))
write (*,*) foo
end

In fact, if you multiply his equation through by u^2, then set u=0 you are left with g+alpha v^2=0

Not quite, because v sqrt(v) isn't v^2, it's v^2 sign(v). As the projectile approaches terminal velocity, your approximations hold, but with the sign flipped relative to yours, and the expression you've found equals (with a sign flip) the acceleration, which is zero. When the projectile has small u but isn't close to terminal velocity, your approximations fail (u isn't small enough compared to v).

Similarly, setting v=0 you have g=const u^2

Setting v=0 isn't physically interesting in the same way that setting u=0 is. There are physical solutions of the equations of motion in which u is constant and zero -- vertical free fall. There aren't physical solutions of the equations of motion in which v is constant and zero.

Doing a reply-to-self because I checked my interpretation using a numerical simulation. I wrote some python 3 code, which does a reasonably realistic simulation of a baseball being hit for a home run. Slashdot's lameness filter wouldn't let me post it, so I put it here: http://ideone.com/yeP4y

The results:

u= 36.86184199300463 v= 25.810939635797073 Ray= 0.07075915491208162 KE+PE+heat= 147.825
u= 30.646253624059415 v= 12.467830176777555 Ray= 0.07075939744839914 KE+PE+heat= 147.82340481003814
u= 26.608846983666997 v= 1.6625489055858707 Ray= 0.07075957710355621 KE+PE+heat= 147.8224303518585
u= 23.559420165753 v= -7.761841618975968 Ray= 0.08597247439794412 KE+PE+heat= 147.82171310054588
u= 20.86163826256129 v= -16.094802395195508 Ray= 0.10413207421166563 KE+PE+heat= 147.82115230900214
range= 120.88936569485678 , vs 194.17117929504738 from theory without air resistance
u= 18.25141606403427 v= -23.242506129076933 Ray= 0.12066666645699123 KE+PE+heat= 147.8207286473949
u= 15.70673363979356 v= -29.088976584679852 Ray= 0.1353850869274781 KE+PE+heat= 147.8204307883206
u= 13.30143766684643 v= -33.65200048062784 Ray= 0.14867603720136566 KE+PE+heat= 147.8202356199746
u= 11.11267911406159 v= -37.07517115834146 Ray= 0.16096016949002218 KE+PE+heat= 147.8201144079141
u= 9.186200956690504 v= -39.564763699985484 Ray= 0.17255826567110216 KE+PE+heat= 147.82004160975018

The notation is that u and v are the x and y components of the velocity vector, "Ray" is the expression that Ray seems to be claiming is a constant of the motion, and the final column is the total energy, which should be conserved.

I tested my code two ways: (1) Energy is very nearly conserved. (2) If I turn off air friction, the range is very nearly as calculated by theory.

Let R be the expression that Ray says is a constant, under my interpretation of his variables. Then dR/dt appears to be very nearly zero early on in the simulation. However, later on it starts to drift upward. So I suspect that one of the following is true: (1) Ray is wrong; (2) my interpretation of his notation is wrong; or (3) my simulation doesn't use good enough numerical techniques to demonstrate with good precision that Ray is right.

Anyone who's got Runge-Kutta, etc., on the tip of their tongue want to try a better simulation of this?

That helps a little, but still doesn't really clarify completely what he did. I'll explain a little about what I know about the projectile problem and what I can figure out about what he might have accomplished here.

In the Principia, Newton poses three closely related problems. One is projectile motion under the influence of a frictional force that's proportional to velocity (book II, section I). Next he considers the case where the friction is proportional to the square of the velocity (book II, section II), and finally the case where it's of the form av+bv^2, where a and b are constants (book II, section III). Let's call these cases 1, 2, and 3.

Case 1 is pretty straightforward. The x and y motions are decoupled, and each of the motions is governed by a first-order, linear, inhomogeneous equation.

Case 2 is actually of more physical interest than case 1 for most real-world projectiles. For example, when you toss a baseball in air, its Reynolds number is about 10^4 or 10^5, and in that regime, a force proportional to v^2 is a pretty decent approximation. There is a well known closed-form solution for the one-dimensional subcase (I actually had a student a few years back who figured it out for herself, which was impressive), which is y=A ln[cosh(t sqrt(g/A))].

A hint is that this page has a photo of him holding up a large sheet of paper with his closed-form solution on it. The equation is clearly visible, and reads g^2/(2u^2)+(alpha g/2)[v sqrt(u^2+v^2) / u^2 + arsinh |v/u|] = const. The notation isn't explained, but clearly u and v are the components of some vector, probably the velocity vector. If so, then the constant alpha has to have units of inverse meters.

This makes me think that what he's solved is the full two-dimensional version of case 2. It can't be case 3, because besides g there is only the one constant alpha appearing in his equation. If you write down the equation of motion, a=F/m=(mg-bv^2)/m=g-(b/m)v^2, the constant that naturally occurs is b/m, which has units of inverse meters. It also makes sense that his solution has a hyperbolic trig function in it, since the y(t) for the one-dimensional version of case 2 has a hyperbolic trig function in it.

If my interpretation is right, then you should get a correct one-dimensional result from his equation when u=0. Unfortunately his equation blows up to infinity in that case, so I'm not sure how to extract any sane interpretation from it. By setting alpha=0, you should also get the case with zero friction. That does sort of make sense, since it says u is a constant, which it should be in that case.

It would be interesting to see if my interpretation is right by doing a numerical simulation and seeing if his expression really does seem to be a constant of the motion.

One thing to point out is that he may not have actually solved the full problem as set by Newton. He hasn't found the equation of the trajectory in closed form (which I think was what Newton was most interested in), and he also hasn't found the position in closed form as a function of time. (This is all assuming my interpretation is right.)