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 well 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.
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 well 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.