Thanks very much for your reply, Jeremy.

But I'm afraid I still don't get it:

1.What does distribution of mass mean? is it center of gravity?

If so, it would mean that the effective mass is dependend on tracking force (which is the same thing as moving the center of gravity).

It's moment of inertia. Similar but not exactly the same (and calculated the same way, as a differential equation over the shape of an equally-distributed mass, but that's not really possible with a complex object like a tonearm!) Variation in tracking force is part of the picture but the difference between a 1.5g and 2g tracking force, say, is not dominant as I hope to demonstrate below.

2.When I read about effective mass it is in gram and that is not what seems to come from your equation?

Right, I gave you m x d^2 units and the spec is as you say, given in grams. If I'm remembering this right, for rotational moment of inertia, the units drop out as the mass is normalized to the distance from the pivot, of the stylus position. I'll show this below. It describes a force applied to an object at a distance d from an axis of motion. Since d is fixed the units drop out for a specific object and size.

3. common specs for arms state effctive mass in the order of 10 gr. or even less so it can never be the total mass of the arm.

Hope you can get me to understand all this!

regards, Peter

Trying my best!

The moment of rotational inertia applies to rotation about an axis. It represents the equivalent mass that would need to to be at the stylus location that, once in motion, would have the same opposition to a force applied to it. This must be why the dimension units drop out - you are specifying a specific distance from the pivot as the place where you are applying the force. (I am using college physics I learned in 1978 here, go easy on me!)

- with a linear tracker, you see how the horizontal moving mass can be the total moving mass, right? Either you assume an infinite radius so the entire assembly turns out to be the same distance away, or you just ignore rotation altogether since you are not actually going about an axis to begin with. Special case.

- with a pivoted arm, or in the vertical plane of a linear tracker, let's look instead at a playground seesaw.

5 kgs on one end, 1 kg on the other. Same distance for each from the fulcrum. Static condition: equivalent to 4 kgs on the heavy end. This is the gravitational force on this system.

Set it in motion. The 5 kgs will have a moment of inertia of 5(d^2) and the 1 kg will have 1(-d^2) - the moment of inertia adds, not subtracts, (I'm pretty sure.) Dynamic condition, now, the object is set in motion (and this is the same as what is in play when the groove excursion tries to set the arm in vertical motion.)

Now let's balance it. 5kgs at d distance and 1 kg at distance 5d. The heavy end will have moment 5 (d^2) and the other will have 1(5d)^2 or 25(d^2). Moment will be 30 d^2 and that will be equivalent to a moving object of 30kg applied at a position d away from the pivot.

Balance it with a smaller weight: 5 kgs on one end, .5 kgs at distance 10D. Now moment of inertia is 5(d^2)+100(d^2)=105(d^2) and will resist being moved like a weight of 105kg a distance d away from the pivot.

But what you have with a tonearm is usually a heavy weight closer to the pivot: say 5kg at the heavy end and 50 kg a distance .1d away. Moment of inertia: 5(d^2)+50(.1d)^2=5.5(d^2) which is normalized at d to 5.5g.

If I scratch my head I can may come up with the formulas that take into account the imbalance represented by the tracking force and gravity.

Hopefully dice45 will be in here with his red pen soon and critique my efforts, I am working hard here!

-j