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Jaeger-LeCoultre

Some maths

 

A simple question re: pushers. Why are non-screw chrono pushers not activated at depth? If there's enough "drag" to cause deflection of the stem then what about the static pressure inadvertantly depressing a pusher.

So, what is the force on the end of a pusher?

(now this is all done in my head, so I beg forgiveness if it is stupidly wrong)

Pressure=Force / Area.

50m=5 atm=5 kgf/cm^2 approx.

Area of pusher=(assume 4mm dia pusher) pi x r^2

using 3 for pi (I like simple numbers!) gives us: 3 x (0.2)^2=0.12

So back to P=F/A:

5=F/0.12

0.6=F

So at 50m there is roughly 0.6 kg (~6 N) on the end of the pusher.  And 50m is pretty deep.

(Out of interest, what is the force on the dial? if we assume 40mm dial: 60kg (~600N).... dare you stand on the dial?

I will do some delving in my old fluids texts for dynamic pressure to work out the sideways force on a crown. but the above gives the magnitudes of forces involved.

Something to dwell on.

Please comment on any of the maths or assumptions, all feedback is great.

 

This message has been edited by BDLJ on 2008-12-04 16:19:07

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