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Effective Mass and Moment of Inertia (again)

I worked through a couple of examples in response to one of the questions in the original thread and I thought the results illuminating, so here goes.

Example 1: Bearing 40mm across, centre balanced, weight 60g. Armtube 180mm long, constant weight for length, total weight 15g. Headshell 40mm long, centre balanced, 7 g. Counterweight 150g, assumed as point mass, positioned to attain static balance (x = 21.27mm). Stylus position at x = 225mm.

Total I cm = 637594 g.mm 2 , effective mass = 12.5 g.

Now lets's substitute a 200 g counterweight, moved further in to maintain balance (x = 15.95mm).

Total I cm = 6210628 g.mm 2 , effective mass = 12.25 g. This means that a 33% greater CW mass reduces effective mass (and I cm ) by 2%.

Now let's take the effective mass back to where it was by adding mass to the armtube. Armtube mass for equal I cm becomes 19.75g so the 33% greater CW mass allows us to increase the armtube mass by about the same amount. BTW with good engineering this would allow us to nearly double torsional rigidity.

Just to satisfy the the original question, let's swap the masses of the headshell and armtube, so the total mass is the same. The I cm increases to 907460 g.mm 2 so the effective mass increases to 17.9 g, an increase of 43 percent. A light headshell will do much more than a heavy counterweight.

If anyone wants me to run through the actual maths I'd be only too happy, I love maths. The basic formula is

I cm = ∫ x1 x2 x 2 dM

All the masses and dimensions used are measured from bits of tonearms I had around.

Mark Kelly


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Topic - Effective Mass and Moment of Inertia (again) - Mark Kelly 03:57:40 11/18/05 (43)


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