Okay, suppose the viewing field is circular and we consider two
particles as in the attached image.
Probability of being within the field:
R0 > sqrt((x1+R1-x0)^2 + (y1+R1-y0)^2)
Probability of being outside the field:
R0 < sqrt((x2-R1-x0)^2 + (y2-R1-y0)^2)
Since these are the limiting cases, it looks like the averaging I
suggested will work.
Jim
On Thu, Feb 21, 2019 at 9:23 AM Rolf Turner <r.tur...@auckland.ac.nz> wrote:
>
> On 2/21/19 12:16 AM, PIKAL Petr wrote:
> > Dear all
> >
> > Sorry, this is probably the most off-topic mail I have ever sent to
> > this help list. However maybe somebody could point me to right
> > direction or give some advice.
> >
> > In microscopy particle counting you have finite viewing field and
> > some particles could be partly outside of this field. My
> > problem/question is:
> >
> > Do bigger particles have also bigger probability that they will be
> > partly outside this viewing field than smaller ones?
> >
> > Saying it differently, although there is equal count of bigger
> > (white) and smaller (black) particles in enclosed picture (8), due to
> > the fact that more bigger particles are on the edge I count more
> > small particles (6) than big (4).
> >
> > Is it possible to evaluate this feature exactly i.e. calculate some
> > bias towards smaller particles based on particle size distribution,
> > mean particle size and/or image magnification?
>
> This is fundamentally a stereology problem (or so it seems to me) and as
> such twists my head. Stereology is tricky and can be full of apparent
> paradoxes.
>
> "Generally speaking" it surely must be the case that larger particles
> have a larger probability of intersecting the complement of the window,
> but to say something solid, some assumptions would have to be made. I'm
> not sure what.
>
> To take a simple case: If the particles are discs whose centres are
> uniformly distributed on the window W which is an (a x b) rectangle,
> the probability that a particle, whose radius is R, intersects the
> complement of W is
>
> 1 - (a-R)(b-R)/ab
>
> for R <= min{a,b}, and is 1 otherwise. I think! (I could be muddling
> things up, as I so often do; check my reasoning.)
>
> This is an increasing function of R for R in [0,min{a,b}].
>
> I hope this helps a bit.
>
> Should you wish to learn more about stereology, may I recommend:
>
> > @Book{baddvede05,
> > author = {A. Baddeley and E.B. Vedel Jensen},
> > title = {Stereology for Statisticians},
> > publisher = {Chapman and Hall/CRC},
> > year = 2005,
> > address = {Boca Raton},
> > note = {{ISBN} 1-58488-405-3}
> > }
>
> cheers,
>
> Rolf
>
> --
> Honorary Research Fellow
> Department of Statistics
> University of Auckland
> Phone: +64-9-373-7599 ext. 88276
>
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