This question came to mind while I was mowing the lawn. I had glanced down at my wrist to check the time and, to my horror, I saw that, instead of my Traser "beater", I was wearing a vintage 1929 Bulova cut-corner watch with no shock protection on the balance jewels. Not only that but, because of the unbalanced mower blade and general crappy state of the machine, the vibration was actually blurring the watch with about a 1/8 inch amplitude! I just

__had__to do the calcs - and here are the results . .

Machinery Vibration and the Strength of Materials can both get pretty complicated, especially if you don't know the steel alloy used to make a balance staff, so I simplified the problem and assumed that any breakage would be to a balance staff pivot by a shearing force, i.e. applied at right angles to to the staff axis (centerline). Then I assumed that the pivots were good fits in the jewels and that the oil would cushion any tendency for the staff to "rattle" under vibration. I also assumed that the watch itself being moving in a sinusoidal manner which eased the calculation a lot. The mower runs at about 600 rpm which equals a vibration frequency of 10 Hz. The first job was to calculate the peak force on the balance . .

For that, it was necessary to weigh the balance. Presenting it to the 0-500g kitchen scales proved to be a waste of time, as one might expect! The needle didn't even budge. Fortunately, I still had the beam-balance scales that I use for re-loading gun cartridges. The balance weighed in at

**1.3 grains**(there are 7000 grains to 1 lb). Next I measured the balance pivot diameter which was 0.09mm or

**90um**.

Now, if we knew the peak acceleration, we could calculate the force on the pivots. Fortunately, that is a common enough calculation in the world of machinery protection and, these days, there are calculators like this available on the 'net. Using that calculator and entering 10Hz and 3mm, the result was

**0.6 G's**. Didn't seem a lot but, just to be sure, the next step was to calculate the stress on the pivots.

As a worst case, I assumed that the watch was vertical. That way, it is necessary to add 1 G (the constant force of gravity) to the acceleration figure obtained previously, which gives a total acceleration of 1.6 G's. Then, the formula for Stress is simply Stress = Force/Area. The force is simply 1.6 x the weight (in Newtons). The balance weight of 1.3 grains converts to 826uN (micro-newtons), and so the force is 826uN x 1.6 =

**1.322mN**(milli-Newtons).

*(The use of Newtons in metric engineering calculations is commonly done to distinguish between force and mass.)*With a pivot diameter of only 0.09mm, the area will be very small.

*(Small areas give high stresses - consider the case of the fat lady tottering along on stiletto heels!)*. However,we will be doubling it because the force is assumed to be shared equally between the two pivots. The pivot cross-sectional area (from πD

^{2}/4) came out to 6.362nm

^{2}, twice that being

**12.72nm**. Therefore, the Stress is 1.322mN/12.72nm

^{2}^{2}= 104 kPa or

**15 psi**.

**15 psi**doesn't sound much to me. To make any judgement though, we need to know what level of Stress steel can withstand. As stated above, it's not easy to find an exact value beyond which a pivot would suddenly snap. However, a lot of Googling suggested that somewhere around 200,000 kPa

*(Shear Yield Stress)*would be a reasonable guess which is

**29,000 psi**. From that, it's obvious that you can mow the grass with a lawnmower

__way__worse than mine while wearing even the most delicate wristwatch from your collection.

cheers,

**xpatUSA**
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