Why do nearsighted people see better with their glasses *rotated*? If you are nearsighted (like me), you may have noticed that if you tilt your glasses, you can see distant objects more clear than with normally-positioned glasses. If you already see completely clear, you can distance your glasses a little more from your eyes and then do it. To do so, rotate the temples while keeping the nosepads fixed on your nose, as is shown in the figures.  
As I said, starting with your glasses farther than normal from your eyes, you can observe the effect for near objects too. (By distant, I mean more than 10 meters and by near I mean where you can't see clear without glasses)
Note that if you rotate more than enough, it will distort the light completely. Start from a small $\theta$ and increase it until you see blurry, distant objects more clear. (You should be able to observe this at $\theta\approx20^\circ $ or maybe a little more)
When looking at distant objects, light rays that encounter lenses are parallel, and it seems the effect happens because of oblique incidence of light with lenses:

The optical effect of oblique incidence for convex lenses is called coma, and is shown here (from Wikipedia):

I am looking for an explanation of how this effect for concave lenses (that are used for nearsightedness) causes to see better.
One last point: It seems they use plano-concave or convexo-concave lenses (yellowed lenses below) for glasses instead of biconcave ones.

 A: Your basic premise that one can see better with the glasses tilted is false.  If the lenses have the right correction for your eyes, then tilting them will make things worse.
The reason this does work often is that the lenses are not at the right correction.  In young nearsighted (myopic) people, the myopia usually gets worse with age.  The glasses may have had the right correction 2 years ago, but meanwhile the eyes have gotten a little more myopic and a stronger correction is required.
Tilting the lenses makes light pass thru them in a way that effectively makes the lens seem stronger at that angle.  If distant objects are a little blurry due to more myopia than the lenses are correcting, then tilting the lenses will make horizontal edges sharper.  It does nothing for vertical edges.  But, overall the image will appear sharper.
A: When you place two lenses one behind the other, the effective focal length is a function of their distance. This is the principle behind a zoom lens - with the same pieces of glass, a zoom lens achieves a range of distances. 
What you are describing is the simplest case of a zoom lens - just two elements. The first element is the cornea + lens in the eye - for a myopic person, this has a focal length that is slightly too short for the distance to the retina, which is why close-up things are in focus but distant objects are fuzzy. To correct this, you add a second lens with a negative focal length - a diverging lens.
The task then is to compute the apparent focal length of the combination, and show that it depends on the distance between the lenses.
From http://en.wikipedia.org/wiki/Lens_(optics)#Compound_lenses you can see
1) when two lenses touch, their compound focal length is found by
$$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$$
2) when there is a distance $d$ between the lenses, this complicates to
$$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}-\frac{d}{f_1 f_2}$$
Rearranging things and taking the focal distance from the second element (the eye - this is the focal length of interest) you get the "Back focal length" or BFL:
$$\mbox{BFL} = \frac{f_2 (d - f_1) } { d - (f_1 +f_2) }$$
For a typical eye, $f_2$ is about 1.7 cm. If you have mild myopia your prescription might be -2 diopters, or $f_2=-0.5m$ since $diopter = 1/f$. When you move the glasses away from your eye, the focal length of the compounds system actually gets shorter - this means for a myopic person that they will work "less well":

Now when you tilt your glasses, you "compress" the vertical dimension a little bit - in effect, you increase the radius of curvature in the vertical direction (although not in the horizontal). The net result is that you make the lens stronger in one direction - it becomes a slightly astigmatic lens (cylindrical), with greater strength vertically.
On average then, this makes a stronger lens. And if your eyes have a slight cylindrical component of aberration that is not fully corrected by your glasses, this will really help.
In my case, I have strong astigmatism (although no myopia). I find that slightly rotating my glasses (which changes the cylindrical axis direction) can help with focus - as will pressing against the side of my eye (which distorts the eye ball and has the same effect).
In short - what you are doing when you move one lens relative to the other is change the optical properties. Tilting adds a cylindrical component, while moving them apart changes the compound focal length.
Whether this helps you depends on the gap between your prescription and your aberration.
Interesting reading: http://www.oculist.net/downaton502/prof/ebook/duanes/pages/v1/v1c033.html
A: This is a real effect, but it doesn't have anything to do with coma or any of the optical aberrations.  It is caused by the fact that the effective focal length shortens as you tilt a lens.  When your eyesight gets worse, you need a stronger focal length lens in your eyeglasses, and tilting the lenses has this effect.  The problem with doing this all of the time is that it introduces distortions such as the coma you've pointed out in your question.  
This presentation and this journal paper show the the effective focal length of a tilted lens from ray tracing simulations and from theory respectively.  From the paper, the focal length for the tangential focal point (up and down as you look through your glasses) and the sagittal focal point (left and right as you look forward) are given from the paper by
$$
\begin{align}
f_{tan}&=f_0\frac{n-1}{n}\frac{\cos\theta\sqrt{n^2-\sin^2\theta}}
{\sqrt{n^2-\sin^2\theta}-\cos\theta}\\
f_{sag}&=f_0(n-1)\frac{1}{\sqrt{n^2-\sin^2\theta}-\cos\theta}
\end{align}
$$
where $n$ is the index of refraction of the material and $f_0$ is the original focal length. I've plotted $f/f_0$ for both of these in the figure below for an index of refraction of 1.5.  I believe that this is special to a lens which has equal radii of curvature on both sides (bi-convex or bi-concave), but the results for other types of lenses will have similar outcomes.

A: My theory is that this effect is not seen by most myopes. Rather, it indicates that you and I have an uncorrected higher-order aberration in the eye, i.e. coma, which is corrected by inducing the opposite amount of coma in the spectacle lens by rotation.
I have noticed that rotating my left spectacle lens about the vertical (not horizontal) axis by $10 ^\circ$ in a particular direction gives me much crisper vision. I've also noticed that when I look at a distant circular light source (like a traffic light) with the lens unrotated, it looks somewhat like this image of uncorrected coma. The image is much better with the lens rotated.
Note that the sphere and cylinder prescriptions in the (unrotated) lens were confirmed yesterday by an optometrist as being the best I could get -- but simply by yawing the lens I can see much better.
I'd appreciate it if someone familiar with optics could confirm whether or not coma can actually be corrected in this way. Everything I can find online about higher-order aberrations in the eye suggests that they can only be corrected by refractive surgery or contact lenses. In addition, could a lens be constructed to perform the same correction without rotation?
A: In the image you have posted, the "rotated lens" results to be a "worse lens" (very poor converging ray concentration) but a stronger one, with a nearer focal plane (if you only consider upper 1/3 of rays, you can try to see a blurred convergence point 30% nearer), wich may feel more suitable if your graduation has got worse.
