[R] particle count probability

Thu Feb 21 11:58:37 CET 2019

OK. I got it.

Thanks.

Petr

> -----Original Message-----
> From: Jim Lemon <drjimlemon using gmail.com>
> Sent: Thursday, February 21, 2019 11:36 AM
> To: PIKAL Petr <petr.pikal using precheza.cz>
> Cc: Rolf Turner <r.turner using auckland.ac.nz>; r-help using r-project.org
> Subject: Re: [R] particle count probability
>
> Hi Petr,
> My second message was to show that if you take the limiting cases of "just
> inside" and "just outside" - which should have been:
>
> just inside the field:
> R0 = sqrt((x1+R1-x0)^2 + (y1+R1-y0)^2)
> just outside the field:
> R0 = sqrt((x2-R1-x0)^2 + (y2-R1-y0)^2)
>
> the two differences are equal along any radius, supporting the averaging
> strategy.
>
> Jim
>
> On Thu, Feb 21, 2019 at 7:53 PM PIKAL Petr <petr.pikal using precheza.cz> wrote:
> >
> > Hallo
> >
> > Thanks all for valuable suggestions. As always, people here are generous and
> clever. I will try to think through all your suggestions, including recommended
> literature.
> >
> > Jim. Standard practice in particle measurement is to count (and
> > mesure) only particles which are fully inside viewing area. So using
> > your equation I could compare probability for let say particles with
> > R1 = c(0.1, 1). But I probably misunderstand something. Having x0, y0
> > = 0 and x1 =10 and y1 = 0 I get
> >
> > > sqrt((10+c(0.1, 1)-0)^2 + (0+c(0.1,1)-0)^2)
> > [1] 10.10050 11.04536
> >
> > which gives in contrary higher value for bigger particle.
> >
> > OTOH, if I take your first reasoning I get quite satisfactory values.
> >
> > > 1-(10-c(0.1, 1))* (10-c(0.1,1))/(10^2)
> > [1] 0.0199 0.1900
> >
> > Cheers.
> > Petr
> >
> > > -----Original Message-----
> > > From: Jim Lemon <drjimlemon using gmail.com>
> > > Sent: Thursday, February 21, 2019 12:24 AM
> > > To: Rolf Turner <r.turner using auckland.ac.nz>
> > > Cc: PIKAL Petr <petr.pikal using precheza.cz>; r-help using r-project.org
> > > Subject: Re: [R] particle count probability
> > >
> > > 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.turner using 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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