Re: Image transform

Posted by Liu, Dongfang (NIH/NIAID) [F] on
URL: http://imagej.273.s1.nabble.com/Image-transform-tp3703413p3703420.html

Dear Gabriel,

Thank you for your help.
Yes, the system is a miroscope. Could you tell what other optics I should consider to calibrate a image? Could you give me more detail information about this?

Best regards
df

> ----------
> From: Gabriel Landini
> Reply To: List IMAGEJ
> Sent: Friday, March 3, 2006 10:53 AM
> To: List IMAGEJ
> Subject: Re: Image transform
>
> On Friday 03 March 2006 15:40, Liu, Dongfang (NIH/NIAID) [E] wrote:
> > I am using a non-calibrated image. I don't know how to calibrate a image
> > and what is function of this calibration? You mean the spatial calibration?
>
> Yes. The result is the coordinates in your image.
> If you have a calibrated image (you do this by going to Image>Properties and
> set the width and height of you pixels.)
>
> > In our system, we use the 10.35 microns pixel size of CCD and
> > 100*objective. So I think of the pixel size of image is 100nm/pixel.
>
> And not other optics in between? Is it a microscope? if so, it is very likely
> that there are other correcting lenses in between.
>
> I can't tell if your calculation is correct. You need to put a known sized
> object under your imaging system and see how many pixels it spans. Then you
> estimate the size of your pixels dividing the known size of your object by
> the number of pixels it occupies.
>
> The XM and YM coordinates are relative to the image coordinates. Since the
> resolution of the image is given in pixels, there is no advantage in
> calculating the coordinates in calibrated units. The result is not any more
> accurate than in pixel units.
>
> Cheers,
>
> Gabriel
>
>