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Euclidean distance between articulatory movement points

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mahesh bvm
mahesh bvm 2017 年 8 月 18 日
編集済み: Image Analyst 2017 年 8 月 19 日
I have data from two sensors that collect the articulatory movements in 3 dimensions (x, y, z). Assume that I have used channel 1(sensor 1) for Upper Lip and Channel 2 (sensor 2) for lower lip where each channel/sensor has recorded equal number of samples with a sampling rate of 200Hz. Now, how to find the euclidean distance for 'z' coordinate between upper lip and lower lip.

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Image Analyst
Image Analyst 2017 年 8 月 18 日
Try this:
distances = sqrt((xUpper-xLower).^2 + (yUpper-yLower).^2 + (zUpper-zLower).^2);
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mahesh bvm
mahesh bvm 2017 年 8 月 19 日
Thank you very much for the quick reply. It works. However, as I had mentioned in the question itself I was looking the difference between the upper and lower lip only in Z dimension so I have discarded and x and y dimensions. so the final script, if i am not wrong is... distance = sqrt(zUpper-zLower).^2;
Kindly Correct me if am wrong.
Image Analyst
Image Analyst 2017 年 8 月 19 日
編集済み: Image Analyst 2017 年 8 月 19 日
Yes, you can get the distance along each dimension separately if you want.
It would be
zdistance = abs(zUpper - zLower);
No sqrt needed.

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