particle count probability

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particle count probability

PIKAL Petr
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?

Best regards
Petr Pikal
Osobn? ?daje: Informace o zpracov?n? a ochran? osobn?ch ?daj? obchodn?ch partner? PRECHEZA a.s. jsou zve?ejn?ny na: https://www.precheza.cz/zasady-ochrany-osobnich-udaju/ | Information about processing and protection of business partner's personal data are available on website: https://www.precheza.cz/en/personal-data-protection-principles/
D?v?rnost: Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a podl?haj? tomuto pr?vn? z?vazn?mu prohl??en? o vylou?en? odpov?dnosti: https://www.precheza.cz/01-dovetek/ | This email and any documents attached to it may be confidential and are subject to the legally binding disclaimer: https://www.precheza.cz/en/01-disclaimer/


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Re: particle count probability

Roger Koenker-3

Somewhere, buried in the vast literature on the Wicksell problem, there is probably an answer, or at least a hint.

> On Feb 20, 2019, at 11:16 AM, PIKAL Petr <[hidden email]> 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?
>
> Best regards
> Petr Pikal
> Osobn? ?daje: Informace o zpracov?n? a ochran? osobn?ch ?daj? obchodn?ch partner? PRECHEZA a.s. jsou zve?ejn?ny na: https://www.precheza.cz/zasady-ochrany-osobnich-udaju/ | Information about processing and protection of business partner's personal data are available on website: https://www.precheza.cz/en/personal-data-protection-principles/
> D?v?rnost: Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a podl?haj? tomuto pr?vn? z?vazn?mu prohl??en? o vylou?en? odpov?dnosti: https://www.precheza.cz/01-dovetek/ | This email and any documents attached to it may be confidential and are subject to the legally binding disclaimer: https://www.precheza.cz/en/01-disclaimer/
>
> <particle.pdf>______________________________________________
> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

______________________________________________
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Re: particle count probability

Jim Lemon-4
Hi Petr,
This is off the top of my head, but I assume that the shape of the
particle is not considered in counting. Assume particles are uniformly
distributed in the viewing field. If all particles entirely within the
field are counted, large particles will be undercounted. If all
particles within or intruding on the field are counted, large
particles will be overcounted. I do not have an analytic proof of
this, but the average of the two counts may provide a true count.

Hi Roger,
The Wicksell problem looks fascinating, although I don't think it
solves Petr's problem. I will study it further.

Jim

On Wed, Feb 20, 2019 at 10:34 PM Roger Koenker <[hidden email]> wrote:

>
>
> Somewhere, buried in the vast literature on the Wicksell problem, there is probably an answer, or at least a hint.
>
> > On Feb 20, 2019, at 11:16 AM, PIKAL Petr <[hidden email]> 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?
> >
> > Best regards
> > Petr Pikal
> > Osobn? ?daje: Informace o zpracov?n? a ochran? osobn?ch ?daj? obchodn?ch partner? PRECHEZA a.s. jsou zve?ejn?ny na: https://www.precheza.cz/zasady-ochrany-osobnich-udaju/ | Information about processing and protection of business partner's personal data are available on website: https://www.precheza.cz/en/personal-data-protection-principles/
> > D?v?rnost: Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a podl?haj? tomuto pr?vn? z?vazn?mu prohl??en? o vylou?en? odpov?dnosti: https://www.precheza.cz/01-dovetek/ | This email and any documents attached to it may be confidential and are subject to the legally binding disclaimer: https://www.precheza.cz/en/01-disclaimer/
> >
> > <particle.pdf>______________________________________________
> > [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

______________________________________________
[hidden email] mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: particle count probability

Rolf Turner
In reply to this post by PIKAL Petr
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

______________________________________________
[hidden email] mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: particle count probability

Jim Lemon-4
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 <[hidden email]> 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
>
> ______________________________________________
> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

______________________________________________
[hidden email] mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

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Re: particle count probability

PIKAL Petr
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 <[hidden email]>
> Sent: Thursday, February 21, 2019 12:24 AM
> To: Rolf Turner <[hidden email]>
> Cc: PIKAL Petr <[hidden email]>; [hidden email]
> 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 <[hidden email]>
> 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
> >
> > ______________________________________________
> > [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
Osobní údaje: Informace o zpracování a ochraně osobních údajů obchodních partnerů PRECHEZA a.s. jsou zveřejněny na: https://www.precheza.cz/zasady-ochrany-osobnich-udaju/ | Information about processing and protection of business partner’s personal data are available on website: https://www.precheza.cz/en/personal-data-protection-principles/
Důvěrnost: Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a podléhají tomuto právně závaznému prohláąení o vyloučení odpovědnosti: https://www.precheza.cz/01-dovetek/ | This email and any documents attached to it may be confidential and are subject to the legally binding disclaimer: https://www.precheza.cz/en/01-disclaimer/

______________________________________________
[hidden email] mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: particle count probability

Jim Lemon-4
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 <[hidden email]> 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 <[hidden email]>
> > Sent: Thursday, February 21, 2019 12:24 AM
> > To: Rolf Turner <[hidden email]>
> > Cc: PIKAL Petr <[hidden email]>; [hidden email]
> > 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 <[hidden email]>
> > 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
> > >
> > > ______________________________________________
> > > [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide
> > > http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible code.
> Osobní údaje: Informace o zpracování a ochraně osobních údajů obchodních partnerů PRECHEZA a.s. jsou zveřejněny na: https://www.precheza.cz/zasady-ochrany-osobnich-udaju/ | Information about processing and protection of business partner’s personal data are available on website: https://www.precheza.cz/en/personal-data-protection-principles/
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Re: particle count probability

PIKAL Petr
OK. I got it.

Thanks.

Petr

> -----Original Message-----
> From: Jim Lemon <[hidden email]>
> Sent: Thursday, February 21, 2019 11:36 AM
> To: PIKAL Petr <[hidden email]>
> Cc: Rolf Turner <[hidden email]>; [hidden email]
> 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 <[hidden email]> 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 <[hidden email]>
> > > Sent: Thursday, February 21, 2019 12:24 AM
> > > To: Rolf Turner <[hidden email]>
> > > Cc: PIKAL Petr <[hidden email]>; [hidden email]
> > > 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
> > > <[hidden email]>
> > > 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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> > > > http://www.R-project.org/posting-guide.html
> > > > and provide commented, minimal, self-contained, reproducible code.
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