Is there official name for "Manhattan visibility" measure? I made up the measure (and its name) and I wonder if it is officially defined (and named!)? I need a name, because when I call it "Manhattan visibility" nobody will understand me unless I explain how it is measured.
Having two points in 2d space, those points are less distance if the angle is closer to 0, and far distance if the angle is close to 45°. Comparing two cases, if angle is the same, the second factor is radius -- the bigger, the more distant two points are.
And thus, I coined term visibility, because if the axis is the same, the points can "see" each other, if the radius (real distance) between is greater, the harder to "see" each other.
Please note, it is not Manhattan distance measure. Points (0,0) and (1,1) are closer than (0,0) and (0,10000), while with my measure the latter pair is closer than the former.
Update 1 to give some background, maybe this will help. I am writing the program which generates analog clocks (visually). Next I need to compute how 2 clocks are similar to each other. So let's say 12:00 and 24:00 are identical, 12:15 and 13:15 are pretty similar, less, but still similar are 12:15 and 16:15. 12:00 and 18:30 are completely different.
Update 2 (not verified yet)
This is the code I came up with to compute it:
var dx = Math.abs(a.x-b.x);
var dy = Math.abs(a.y-b.y);
var min = Math.min(dx,dy);
var max = Math.max(dx,dy);

if (max==0)  // same point
    return 0;
else
    return min*1.0/max+1-1.0/(1+dx*dx+dy*dy);

 A: The algorithm provided at the end of the question doesn't actually produce an ordering compatible with the requested order relation; in particular, under that code it's possible for two pairs of points at different angles to be equidistant; consider for instance $\langle0,0\rangle$ and $\langle1,2\rangle$ on one hand (where the computed value is $\frac12+1-\frac16=\frac43$) and $\langle0,0\rangle$ and $\langle2,y\rangle$ (with $y\gt2$ to be determined) on the other.  Then the computed value for this second pair of points is $\frac2y+1-\frac1{5+y^2}$ $=\frac{10+2y^2}{5y+y^3}+\frac{5y+y^3}{5y+y^3}-\frac{y}{5y+y^3} = \frac{y^3+2y^2+4y+10}{y^3+5y}$).  Then by setting this equal to $\frac43$ we get a cubic equation in $y$, $4y^3+20y = 3y^3+6y^2+12y+30$ or equivalently $y^3-6y^2+8y-30=0$; numerically solving this equation yields $y\approx 5.534$ as a solution.
Worse, there is no real function that satisfies the requested ordering.  By computing angles (or equivalently, something like tangents of angles - the $\frac{\mathrm{min}}{\mathrm{max}}$ component of the algorithm essentially computes a tangent) and then computing distances we transform the problem into needing an injective function $f:\mathbb{R}\times\mathbb{R}\mapsto\mathbb{R}$ such that $s\gt t$ implies $f(s, x)\gt f(t,y)$ for all $x,y$, and $f(x,s)\gt f(x,t)$ for all $x$.  But this ordering is equivalent to trying to fit $\mathbb{R}$ 'copies of' $\mathbb{R}$ side-by-side, and a little thought should convince you (though it's a bit trickier to prove!) that this can't be done in any sufficiently clean fashion.
