Why is the price of new tonearms so high

Well it comes down in several aspects to the static vs dynamic model....
Aber es hat weder mit dem Hammer zu tun, noch damit, wo er hängt (im Zweifel immer an der Wand...).

I don't get your last comment (unless Mark's comment that effective mass = moment of inertia divided by square of effective length is wrong). If Mark's equation is right, the two could be different and still result in an identical third (effective mass) value.

No, the equation isn't wrong; I believe that Mark was using it to make a point about how little variation there is in the deflection of a dynamic VTF spring, and thus a correspondingly extremely small change in the VTF at the end of the tonearm as a result of this change in spring deflection. For this calculation, the tonearm length is of course relevant.

The misconception that I understood from Dertonarm's comments is that given two tonearms of different effective length, but identical effective mass . . . that they will somehow present different forces against the cartridge when tracking vertical undulations and warps:

Only if the moving mass is homogenous distributed in the whole moving corpus - which is not the case in a tonearm with mounted cartridge.
Brings up again the picture of the Micro Seiki and other turntables which increased their moment of inertia by moving most of the mass towards the outer rim.

You are correct in that the effective mass at one end of a tonearm is the result of its static mass, length, and distribution of mass along the length. But the 'effective mass' specification takes all that into account, and if you change two of these factors (i.e. length and static mass) to acheive the same effective mass at the cartridge . . . the cartridge doesn't care one iota.

To use the platter analogy . . . you can either increase the rotational mass by adding a BUNCH of weight close to the center, or a whole lot less weight at the edge. But which ever way you do it (practical considerations aside), the platter can have the EXACT same rotational mass either way. As long as the "effective mass" at the circumference of the platter is the same, they will exhibit the same inertial characteristics.

Kirkus

You make a good point so I'll answer that first. The answer should make it plain that everything Dertonarm has said in response is wrong.

I chose the example I used because the frequency was far enough away from practical cart/arm resonances for that factor to be safely ignored. As the frequency of the warp increases two things happen. The first is that the maximal acceleration (per mm of warp) increases with the square of frequency so the effect becomes more and more pronounced.

The second is that the phase of the arm's response to the warp becomes important. The easiest way to analyse this is to convert to electrical analogy. We can use either a force current or a force voltage analogy, the first is more elegant mathematically so I'll use that. In this analogy the moment of inertia of the arm (with the cartridge attached) becomes an inductance. The rotational compliance of the cartridge (which is the actual compliance times the square of the effective length) becomes a capacitance.

These two form an LC low pass filter and it is obvious by inspection that the product is identical to the usual product used in resonance calculations (effective mass x compliance) so the f0 of the filter is the same as the resonant frequency.

The other thing we need is to know the Q of this filter, which is determined by the hysteresis loss in the cartridge suspension. Since this will also affect the high frequency response of the cartridge way may assume that it is low enough that the Q of the filter is quite high. Perhaps JCarr can chime in here with some accurate values, if not we’ll just assume a range of values greater than 5.

The phase angle of the response is given by :

Tan^-1(Q(2.f/f0 + SQRT(4-1/Q^2)) - Tan^-1(Q(2.f/f0 + SQRT(4+1/Q^2))

From which is may be seen that if Q is greater 5 and f/f0 is less than 0.5 then the phase error is less than 7.5 degrees.

Similarly the amplitude response may be calculated from the formula

1/(1 – (f/f0)^2 +j.f/Q.f0)

From which it can be seen that there is some amplitude peaking: about 30% for Q = 5 and f/f0 = 0.5, reducing to a few percent when f/f0 is less than 0.2. If we take 10% as an acceptable error then as long as the warp frequency is less than 0.3 times the resonant frequency of the arm / cart combination the calculation I gave in my post above is “good enough for Jazz”. Using your range of warp frequencies this is equivalent to arm cart resonance being around 10.

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.

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.