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Horological Meandering

correct

 

but if what I say is true and angular momentum is the important quantity, then you can calculate the moment of intertia of the two wheel sizes and the angular velocity corresponding to the frequency, multiply (which gives you angular momentum) and compare, the bigger value would be the "more stable" watch. Stricty speaking we are talking about vectorial quantities, and also we impose a forced osciallation through the spring, but to a first approximation it might be useful.

Best

Andreas

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