valleart
40
Gentlemen, Thank You ALL! Some Fun with Math...
Mkvc, Wim Bon and Daniel ? Thank you ALL very much for the insightful and helpful comments. You can be sure my next stop is the Chopard Service center on E63rd Street to start out fresh.
Ei8htohms ? You comment struck a nerve with me, so for fun, I did some calculations this morning to show how the moment of the inertia for a L.U.C. Microrotor compares to a typical, full size rotor found ubiquitously in conventional movements.
For simplicity , I model each rotor as a semicircular thin disk, with thickness, h, and diameter R. I made my Microrotor have a 15mm diameter and made it 4mm thick, while the larger rotor is 25mm in diameter and 1mm thick. Let?s assume both are made from copper, having density = 8.9 g/cm3, and ignore the thickness effects you get from electroplating the finished copper.
The Moment of Inertia for a semicircular cylindrical disk is then given by the expression:
I = ? (M)(R2), in which M = mass of the body. Since we are using copper for our rotors,
M = density x volume.
We need to first determine the volumes of each rotor:
Vmr = Volume of the Microrotor; Vlr = volume of the large rotor
Vmr = ?(pi)(R2)(h)
Vmr = ?(pi)(1.5cm)2(0.4cm)= 1.413 cm3
Likewise, Vlr = 0.981 cm3
Calculating the Moments we have:
Imr = 14.15 g-cm2
Ilr = 27.28 g-cm2
Or the larger diameter rotor will have an approximate 1.93:1 inertial advantage over the Microrotor. Practically speaking, the larger rotor will store and transfer more kinetic energy to the mainspring than the smaller one for a given angular velocity.
Conversely, the Microtor will be easier to rotate to begin with, assuming its mass and diameter are judiciously designed. And I know nothing of the parasitic frictional energy differences between a spinning L.U.C. rotor and a typical one either?perhaps someone else can comment..
What does it all mean to me? Not a darn thing! I love my Chopard Pro One exactly for what it is ? a superbly engineered mechanical watch.
Have a great weekend!
Best regards,
Frank