http://imagej.273.s1.nabble.com/Measure-of-the-angles-of-a-segmented-line-tp5001041p5001380.html
> On 11. Jan 2013, at 17:09, Michael Schmid wrote:
>
> Hi Norbert,
>
> for calculating angles from x & y, usually the best method is the atan2 function. It is not sensitive to rounding errors and takes care of all special cases like 0°, 90°, 180° etc.
> This can be also used for the angle between two vectors:
>
> dotprod = (x[i+1]-x[i])*(x[i-1]-x[i])+(y[i+1]-y[i])*(y[i-1]-y[i]);
> crossprod = (x[i+1]-x[i])*(y[i-1]-y[i])-(y[i+1]-y[i])*(x[i-1]-x[i]);
> angle = (180/PI)*atan2(crossprod, dotprod);
>
> I you don't care about positive or negative angles, take the absolute value, abs(angle).
>
>
> Michael
> ________________________________________________________________
> On Jan 11, 2013, at 16:04, Norbert Vischer wrote:
>
>> I was using Wayne's macro below and experienced NaN values. This happened on straight lines, when
>> the argument of acos() was outside range -1 .. 1 due to rounding errors.
>> An improved version is here:
>>
>> getSelectionCoordinates(x, y);
>> for (i=1; i<x.length-1; i++) {
>> dotprod = (x[i+1]-x[i])*(x[i-1]-x[i])+(y[i+1]-y[i])*(y[i-1]-y[i]);
>> len1 = sqrt((x[i-1]-x[i])*(x[i-1]-x[i])+(y[i-1]-y[i])*(y[i-1]-y[i]));
>> len2 = sqrt((x[i+1]-x[i])*(x[i+1]-x[i])+(y[i+1]-y[i])*(y[i+1]-y[i]));
>> val = minOf(1, dotprod/(len1*len2));
>> val = maxOf(-1, val);
>> angle = (180/PI)*acos(val);
>> print(i, angle);
>> }
>>
>> Norbert Vischer
>>
>>
>>> On 5. Dec 2012, at 19:11, Rasband, Wayne (NIH/NIMH) [E] wrote:
>>>
>>> On Dec 5, 2012, at 8:23 AM, Eric Denarier wrote:
>>>
>>>> Hi all,
>>>> Is there a macro/plugin to measure all the angles formed by a segmented
>>>> line selection ?
>>>
>>> Here is a macro that does this:
>>>
>>> getSelectionCoordinates(x, y);
>>> for (i=1; i<x.length-1; i++) {
>>> dotprod = (x[i+1]-x[i])*(x[i-1]-x[i])+(y[i+1]-y[i])*(y[i-1]-y[i]);
>>> len1 = sqrt((x[i-1]-x[i])*(x[i-1]-x[i])+(y[i-1]-y[i])*(y[i-1]-y[i]));
>>> len2 = sqrt((x[i+1]-x[i])*(x[i+1]-x[i])+(y[i+1]-y[i])*(y[i+1]-y[i]));
>>> angle = (180/PI)*acos(dotprod/(len1*len2));
>>> print(i, angle);
>>> }
>>>
>>> I found the code for calculating an angle given 3 points using the dot product at
>>>
>>>
http://stackoverflow.com/questions/6719563/>>>
>>> -wayne
>