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 at 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 at gmail.com> Sent: Thursday, February 21, 2019 12:24 AM To: Rolf Turner <r.turner at auckland.ac.nz> Cc: PIKAL Petr <petr.pikal at precheza.cz>; r-help at 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 at 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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