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Omega

Yes that is correct!

 

Here is how I think about it. Both the sub-seconds and chronograph seconds hands have 1 minute to complete a single rotation. The 'distance' they need to travel is the circumference of circle they 'trace' with the tip of the their hands. That formula is circumference=( 2 * pi * hand length) ,  ignoring any counterbalance on the hands. Since distance=rate * time, you can compute the rate that the hands have to travel (in mm/tick) each tick, based on the bpm of the movement. The sub-seconds has to travel a much shorter distance in the same amount of time, and that comes across as a smoother motion to the eye. 

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