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Re: Why not post the details here so everyone can read them? (nt)

Posted by Auxarcs (A) on March 22, 2002 at 19:11:18

In Reply to: Why not post the details here so everyone can read them? (nt) posted by John Elison on March 22, 2002 at 08:57:06:

Well, I'm new here, and I wasn't sure if everyone (or _anyone_) was interested in the "tech side" of things.

At any rate, this is a fairly simple way to *estimate* (i.e., there are more elaborate methods that will give more accurate results)effective mass that should be within in the means of most DIY type hobbyists. Remember, this is but one approach to the issue.

To keep things simple and easy, secure a cartridge of known weight (mass) to your headshell. Next, "zero balance" the tonearm. Next measure the distance from the pivot to the center of the counterweight. Then measure the distance from the pivot to the stylus tip (or, you may glean this from the tonearm's spec list). Next, remove and weigh the headshell and counterweight (_seperately_, of course!).

Now, to calculate the effective mass of the arm:

Mass of cart+headshell X (distance to the pivot-squared)
plus
Mass of counterweight X (distance to the pivot-squared)
equals
the moment of inertia (in g cm-squared), which, when divided by the square of the distance to the pivot, yeilds the effective mass.

As an example, if we have an arm with a 20cm stylus-to-pivot dimension and a cartridge headshell mass of 10g, along with a counterweight weighing 100g, 2cm from the pivot (nice, round numbers), we have this:

10g X 20cm-squared + 100g X 2cm-squared
*or*
10 x 400 = 4000 plus 100 X 4 = 400

so, 4000 + 400 = 4400

Therefore, the moment of enertia is: 4400g cm-squared

The moment of inertia (4400g cm-squared) is then divided by 20cm-squared (the square of the distance to the pivot), giving an effective mass of 11 grams.

I hope this all makes sense and I apologize for not having the appropriate notation available on my keyboard.

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