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Re: Cycloid grids

Posted by Sinead Roberts on Nov 18, 2009; 8:22am
URL: http://imagej.273.s1.nabble.com/Cycloid-grids-tp3688866p3688869.html

That is great, thanks very much for your help- there is just one more  
thing; when you say you define the height of the cycloid as a fraction  
of the image width- this is not necessarily the width of the image,  
but the fraction of whichever is smaller out of the width/height isn't  
it? Or, that is what mine seems to be doing!

Thanks very much.

Sinead


On 17 Nov 2009, at 16:17, Kischell, Eric R. wrote:

> Sinead,
>> how do you calculate the length of the individual cycloid arcs?
> The length of the cycloid per test point is displayed in the Results
> Table (l/p).
> Also,
> Length of Cycloid per Point = 2.0 * cycloidHeight (in physical units)
> for Sequence C
> Length of Cycloid per Point = cycloidHeight (in physical units) for
> Sequence A
>
>> what is the 'image width' they are referring to?
> We define the height of the cycloid  as a fraction/ percentage of the
> image width.
> The image width is the width of the current image (ImagePlus).
>
> keesh
> Work Life Plan:
> To improve the quality of life for all mankind through better pattern
> recognition techniques.
> Home Life Plan:
> Married, but happy. Three bright beacons light my way.
> "Life goes by so fast, that if you don't stop and look around, you  
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>
> -----Original Message-----
> From: ImageJ Interest Group [mailto:[hidden email]] On Behalf Of
> sinead roberts
> Sent: Monday, November 16, 2009 12:29 PM
> To: [hidden email]
> Subject: Cycloid grids
>
> Dear all,
>
> I am trying to use the cycloid arc plug-in and I was wondering; how do
> you calculate the length of the individual cycloid arcs please? I know
> you input the percent of the image width you wish the height of the
> cycloid to be, but what is the 'image width' they are referring to? It
> cannot be of the whole on screen image, but is it of the 'box' area
> each cycloid is contained within?
>
> I am very sorry if this is a silly question,
> Thank you in advance for your help,
>
> Sinead Roberts