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Log Polar Transform

asked 2012-09-10 14:26:50 -0500

Sid gravatar image

I have been going through the latest OpenCV implementation of the Log Polar Transform using the Blind Spot Model. I am a bit confused as to how the cortical image size (sectors x rings) is being calculated. The inputs I provide are width & height of original image, center point, rings, blindspot size and I ask OpenCV to internally calculate the sectors.

Given this, my cortical image size turns out to be 120 x 134 when I give the following values as inputs : width = 800, height = 600, center(400,300), rings R = 120, blindspot size ro0 = 1, romax = 0.5*min(width,height). This means that my no of sectors is being internally calculated to be 134 to keep the aspect ratio of the log-polar template = 1.

If I manually calculate S = 2pi/(a-1) where a = exp(ln(romax/ro0)/R) I however get S = 130

I am not sure why I am getting this discrepancy of 4 rows. Any suggestions would be helpful.


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answered 2012-09-12 00:57:53 -0500

mrgloom gravatar image

maybe a rounding/interpolation error about 3%?(I read about it somewhere). I don't know how can it be done more precise, but interested in more precise LogPolarTransform

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Asked: 2012-09-10 14:26:50 -0500

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Last updated: Sep 12 '12