Why is the price of new tonearms so high

Next a look at the effect of "warp riding". Assume a standard "taco warp" so the warp frequency is 7 rad.s^-1. This gives a maximal velocity of .007 m.s^-1 in the vertical plane for each mm of vertical warp and a maximal acceleration of .049 m.s^-2 (again for each mm of vertical warp).

The product of this acceleration and the effective inertial mass of the arm / cart combination gives a maximal VTF variation when riding the warp. If the inertial mass were 25 g (say 15g arm plus 10g cartridge) and the warp were 5mm high, the maximal variation would be around 6.2 mN. A similar calculation allows a maximal warp tolerance to be derived for any arm / cart combination as a function of VTF.

The important point is that it has nothing to do with the balance of the arm but is stictly related to the moment of inertia (and the mass of the cart).

Mark Kelly

If it were as you stated, in consequence the derivation in VTF would be worse with increased effective length (= increased inertia) and increased effective mass (= increased inertia). Thus a super lightweight short (9") tonearm would be best in conjunction with a low mass body cartridge.
What brings up the Black Widow w/MM again.

However the sonic results do show us a different picture.
The derivation in VTF with a dynamically balanced tonearm is less than with the same tonearm in static balanced mode.
As all dynamic balanced tonearms can be used in static balanced mode too, this is easy to illustrate in demo. The static balanced mode to some does sound more "livelike" due to more alternations in VTF. The dynamically balanced mode often is mistaken for being too "remote - less lively".
But it is due to more constant VTF.

So same -static- inertia, same effective mass, same effective length.
The whole static spring-mass-system is the same in both modes - but we face different behavior.

I do not think we have yet reached the verdict with the model as described by Quiddity.
Some dynamic aspects has to be taken into consideration too - aside from the pure static model.

Completely wrong.

Your first paragraph makes no sense: the effective mass of an arm is simply the moment of inertia divided by the square of the effective length.

In the second para you present a supposition which I have already shown to be wrong but you do not support it with evidence.


Mark Kelly

Mark,
Are you implying that in all situations where the effective mass or "inertial mass" (not sure which one you mean, or if you mean to say they are the same) is 25g, and the effective length is identical (let's just say that the combination of total mass at the effective length is identical and the effective length is identical), that the maximal variation due to warp riding (which you put at 6.2nM) will be identical, regardless of compliance of cart, and regardless of what force mechanism is keeping the stylus on the record?

T-Bone

Yes, that is what I am saying.

It follows from a simple torque balance on the arm according to D'Alembert's principle. Rather than boring everyone by converting forces to torques and computing moments of inertia, I used the concept of equivalent mass.

BTW the figure given only applies to the example given, obviously different constraints will result in different figures.

Note that this is a force variation, the way the cart responds to the force variation will depend on the compliance. Also note the assumption of equal inertia is not equivalent to an assumption of equal structure but the differences are so small as to be immaterial to the argument.

Mark Kelly