Azimuth shift and vertical bearing orientation
The azimuth shift when riding warps depends on the orientation of the vertical pivot relative to the axis perpendicular to the cartridge body and the offset angle, and has no relation to whether the arm is straight, ‘S’ shape, or ‘J’ shape.
I took a spare 3009/S2 Improved (SN 197057) that I have and made some measurements in order to run a few calculations. To model this, one of the measurements I had to make is the angle by which the vertical pivot (the knife edge) deviates from the "effective length" line that runs through the stylus point and the pivot. Because the 3009 is a "J" shape arm, this angle must be measured separately, since it is not the same as the offset angle.
I'll call this angle T (tau), and on this 3009 it appears to be 'exactly' 15deg. The total angle between the cartridge axis and the perpendicular of the knife edge axis, which I will call B (beta), is
B = T + Offset Angle = 15deg + 22.6deg = 37.6deg.
The common base for tying everything together is the geometric plane that the arm is in. This 'tonearm plane' tilts or rotates about the arm's vertical pivot, i.e., the knife edge bearings. The location and orientation of the cartridge in this plane determines whether and by how much the cartridge azimuth will rotate, or shift, because of the knife edge angle when the cartridge rides up and over a warp.
For the analysis, let's consider a hypothetical 1/4 inch (6.35mm) warp. We first have to project the effective length of the arm to the effective length acting to rotate the tonearm plane about the knife edge axis.
Projected arm length = (231.2mm) x (Cos 15deg) = 223.32mm
The 1/4 inch (6.35mm) warp acts on this projected arm length to rotate, or tilt, the tonearm plane.
Tonearm plane tilt = Arcsin(6.35mm/223.32mm) = 1.63deg
Sounds reasonable. Now we need to translate that into azimuth shift experienced by the cartridge. The amount of this 1.63deg tonearm plane rotation that gets translated into cartridge azimuth shift is proportional to the Sine of the angle B:
Azimuth shift = (1.63deg) x (Sin B)
= (1.63deg) x (Sin 37.6deg)
= 0.99deg
So, in this 3009, the cartridge riding up and over a 1/4 inch warp also rotates counterclockwise for an azimuth change or shift of 0.99deg. Let's do a quick reality check on the math model. In arms where the vertical pivot bearing axis is perpendicular to the cartridge axis, the angle B is zero, and so the azimuth shift on a warp is (1.63deg) x (0) = zero. A classic and clear visual example of this is a Dual 1019.
So, how audible would a 0.99deg shift in azimuth be? Well, that's one of those questions where, if you ask 3 people, you get 4 answers. I would think at this point that that the direct audible effects of the warp (flutter effects due to stylus scrubbing or skipping due to the cartridge body bottoming out) would be of more concern than the azimuth shift.
For comparison, I looked at a SME M2. I don't have one of these but the dimensions are in the VE database. The M2 is a straight tube design with an offset angle at the 'headshell'. This is easier to analyze because, in this case, the angle T is zero and the angle B is simply equal to the offset angle. For a 1/4 inch warp:
Effective Length = 233.62mm
Offset Angle = 23.62deg
Tonearm Plane Tilt = Arcsin(6.35mm/233.62mm)
= 1.56deg
Azimuth Shift = (Plane Tilt) x (Sin B)
= (1.56deg) x (Sin 23.62deg)
= 0.62deg, which is about 1/3 lower than the 3009.
Consider this:
The warp that's causing the azimuth shift in the cartridge is also causing an azimuth tilt in the record surface itself:
Let's say there's a 1/4 inch warp at the outer edge of the record. The record label is still sitting flat on the platter. (If the label is warped too, then the record is most certainly toast.) For a 1/4 inch rise over the distance of the recorded area (5.75 - 2.375 = 3.375 inches), the record surface itself has a tilt of:
Arcsin(0.25 inch/3.375 inches) = 4.25deg
Sooo, the azimuth tilt in the warped record surface itself is far greater than the azimuth shift caused by a 3009 or M2 vertical knife edge being "sub-optimally" oriented. Because the azimuth shift is in the same direction as the azimuth tilt in the record surface, the relative azimuth shift between the cartridge and the record surface deceases, rather than getting worse, i.e., the combined effects subtract instead of adding together.
Now, get ready to throw rocks at me:
The calculated data contradicts the widely held belief that the azimuth shift due to vertical bearing mis-orientation is a design fault. In practice, the effect of the azimuth shift is a good thing, as it helps compensate for some of the azimuth tilt introduced by the warp in the record surface itself. This would also hold true for other arms that have a vertical pivot orientation similar to the 3009/M2.