hi,
im looking for a way to compute a arccos in pd ? i fail to find an object directly computing it, did i miss it ? is there a way to compute it in a more indirect way (e.g. with a "limited developpment" really free translation of the french "developpement limite") ?
Hallo, Jerome Etienne hat gesagt: // Jerome Etienne wrote:
im looking for a way to compute a arccos in pd ? i fail to find an object directly computing it, did i miss it ? is there a way to compute it in a more indirect way (e.g. with a "limited developpment" really free translation of the french "developpement limite") ?
Everytime I read sin, cos, exp, sqrt or similar I must think of expr, expr~ and relatives. ;)
Try [expr acos($f1)]
ciao
Frank Barknecht _ ______footils.org__
On Wed, 10 Dec 2003, Jerome Etienne wrote:
im looking for a way to compute a arccos in pd ? i fail to find an object directly computing it, did i miss it ? is there a way to compute it in a more indirect way (e.g. with a "limited developpment" really free translation of the french "developpement limite") ?
sin(t)^2 + cos(t)^2 = 1 so sin(arccos(x)) = sqrt(1-x^2) so tan(arccos(x)) = sin(arccos(x))/cos(arccos(x)) = sqrt(1-x^2)/x so arccos(x) = arctan(tan(arccos(x))) = arctan(sqrt(1-x^2)/x)
and pd has [atan].
Using this trigonometric identity is easier and faster than computing partial taylor series.
Mathieu Bouchard http://artengine.ca/matju
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Frank Barknecht -
Jerome Etienne -
Mathieu Bouchard