Why is the price of new tonearms so high

Kirkus

I composed the above offline before you posted again, it is in response to your post from last night (my time).

Since the issue of efective mass is causing confusion, I can take it out of the argument by reverting to rotational units. If we assume the arm is 225mm long then we have an angular deviation of 4.4 mrad per mm of warp so a 5mm warp will be 22mrad. At the frequency given this is a maximal acceleration of 1.1 rad.s^-2. If the moment of inertia of the arm and cartridge combination is 1.26 x 10^-3 kg.m^2 the maximal torque transmitted to the arm is 1.38 x mNm which is equivalent to a force of 6.2 mN acting at a distance of 225mm. This is exactly equivalent to the previous calculation.

Mark Kelly

BTW I meant to write "the phase and amplitude of the response becomes important" in the post above.

DIVIDED BY

Rotational compliance is linear compliance DIVIDED BY the square of the effective length. Sorry I wrote it the wrong way around.

The point to note is that all this is dependent on the moment of inertia of the arm and the compliance of the cartridge. It has nothing to do with the method by which VTF is applied.

Absolutely agreed - and we're on the same page as far as the proper electrical model for the cartridge resonance equations.

If the moment of inertia of the arm and cartridge combination is 1.26 x 10^-3 kg.m^2 the maximal torque transmitted to the arm is 1.38 x mNm which is equivalent to a force of 6.2 mN acting at a distance of 225mm.

Sure, absolutely - and calculating the rotational torque of the tonearm is necessary for evaluating the particulars of the downforce spring itself . . . but that's about all.

I guess some of the assertions got jumbled around in various different posts . . . but my point is mainly that to calculate deviation of VTF across a warp . . . if you have an accurate figure for effective mass from the cartridge's point of view, then neither the rotational torque nor the length of the tonearm matters - the tonearm can be viewed simply as a pure mass that moves only on two axes.

Kirkus

We are indeed "singing from sheet".

The point that seems to be lost on the other participants is that this is a dynamic analysis. This follows directly from D'Alemberts principle.

Mark Kelly

Hi,
excellent maths no doubt.
Let's look at some practical part of it all then.

1) Is the **effective** mass of an arm increased if the counter weight it moved further away from the pivot bearing?
(I think so, because the mass / moment is increased)

2) If a spring is used for VTF, the CW is further back from the fulcrum as the spring provides the down force and the CW only the arm balance.
What is the effect, if assumption of 1) is correct?

3) If static down force is used, the CW is closer to the pivot/fulcrum, the mass moment should therefore be reduced. What is the effect as compared to 2)?

Inertia has of course ONLY an effect if acceleration is present, and with any tonearm particularly riding a 'taco warp'.
Incidentally, I find the type of warp lifting a smaller area from the start-wax (~ 1" into the record) more common and more radical in vertical acceleration, plus the one that actually pushed in the (still soft?) start-wax for ~ 1/8" causing VERY nasty lateral acceleration.

This leads me to think that lower accelerated mass (effetive mass?), plus higher compliance be the better solution to this kind of problem.
But I still can not make the connection to the 'dynamic VTF' being of any advantage, since (given 1)'s assumption is right) the accelerated mass of the same arm with only using static VTF be somewhat lower.

Note: SME quotes, the V arm's effective mass 10 - 11 gr. (depending where the CW is positioned...)
Axel