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Rasa Bhattarai

Hello @grant,

Using the method you stated for the calculation, I tried using these steps [Taking distance between user and screen = 60cm] :
1. For points (2 and 3) -> (x1, y1) = (1037.115, 367.1904) at time(t = 3.35) and (x2, y2) = (684.5436, 716.046) at time (t = 4.45), I used pythagoras theorem as you stated and found the value as 495.990 px which changed to 13.12 cm during conversion. This gave the value of theta = arctan(13.12/60) = 12.35 degrees [Assuming a standard distance of 60cm]
2. Now, from my understanding 12.35 degrees is the saccade amplitude (angular distance eye traveled during this fixation movement). So is the saccade velocity = 12.35degrees/(4.45-3.45)seconds = 12.35 degrees/seconds ?

Is my calculation wrong somewhere ? Usually according to many papers saccade velocity begins from 100 degree/seconds. Would you help me point out my mistake and clear my confusion?

Thank you sir again.