I take that back (can't edit otherwise I would) and assume that by "offset" you mean the angle of the pivoting head vs the rest of the tonearm.
If so, it should be obvious by inspection that no pivoting tangential tonearm can have constant offset. The nearest you can get is to lengthen the pivoting head so it becomes a pivoting wand, like the Schroeder
If so, it should be obvious by inspection that no pivoting tangential tonearm can have constant offset. The nearest you can get is to lengthen the pivoting head so it becomes a pivoting wand, like the Schroeder
AS long as head cantilever and Diamond to pivot line are not perfectly aligned, you will get an centipede force. Hence distortion.
I have to believe your phrase "consistent offset" is to mean consistent tangency or consistently close to tangency. In the Thales website, Michel Huber stated the specs for the Garrard Zero 100:
"This tonearm, published in 1970, is the best known solution working almost tangential. The cartridge tip was mounted exactly below a pivot-point. This idea made it possible to reduce the tracking error to +0.025 / -0.018°"
The Thales Simplicity tonearm is able to "reduce the tracking error to 0.008° which is three times less than all solution published so far."
An externally hosted image should be here but it was not working when we last tested it.
__________________________________________________
The other published specs that I am aware of is from the Slovakian inventor Marek Bundzel.
And this is his DIY tonearm that looks like a variant of the Garrard Zero 100:
And the error is +0.009° / -0.040° and you can download the specs from his white-paper.
offset
Please understand: I just a mechanic. Maybe I cannot love music as much as an engineer, or a physicist, or (of course not) a mathematician. Nevertheless, I view a bit, a schosche maybe, of skating force to be useful in getting that wee little stylus from here to there. So my question is, suppose the offset of a Garrard-type arm was about the same at both extremes of its traverse of the record. I do get that the angle must increase in the middle due to the arc. I conceive of the two arms as converging at a finite pivot about 24 inches from the stylus. The VE calculator doesn't go that far so I'll use 13 degrees. I have collected most of the bits for this experiment; now I'll just have to do it.
It won't be pretty.
Please understand: I just a mechanic. Maybe I cannot love music as much as an engineer, or a physicist, or (of course not) a mathematician. Nevertheless, I view a bit, a schosche maybe, of skating force to be useful in getting that wee little stylus from here to there. So my question is, suppose the offset of a Garrard-type arm was about the same at both extremes of its traverse of the record. I do get that the angle must increase in the middle due to the arc. I conceive of the two arms as converging at a finite pivot about 24 inches from the stylus. The VE calculator doesn't go that far so I'll use 13 degrees. I have collected most of the bits for this experiment; now I'll just have to do it.
It won't be pretty.
An Analysis of Skate force
It looks as if I was wrong. After thinking about this some more I decided to do a force balance on a split plane arm design using some simple assumptions and based on the same geometry as used in my posts #219 and #222.
We will perform a force balance on the arm by considering the longer segment of the arm as a free body. The co-ordinate system is established by making the X axis the axial length of this body , Y the horizontal axis orthogonal to this and Z the vertical. We need only consider three degrees of freedom, namely X and Y translation and Z axis rotation, the other three being controlled by the vertical bearing and counterweight.
All forces in any direction must balance for the arm to resist translation, similarly all torques around any pivot must balance for the arm to resist rotation.
Since the object of the design is to achieve zero tracking error, we posit that all the frictional reaction force (Ff) acting at point C due to groove contact is in the X direction and normalise this to one force unit. (This force is usually of the order of 10mN).
Any skate force (Fs) will act in the Y direction though point C.
The torque around the pivot P must be zero, otherwise the arm PD would rotate. Thus the force acting at pivot D (Fd) must be axial to the segment DP. The angle PDE is known for any arm position so we can resolve this into an X vector Fd*cosPDE and a Y vector Fd*sinPDE.
If the guide at point E is frictionless the force it exerts on the arm at this point (Fe) must be normal to the tangent of the curve, otherwise the force would move the pivot along the curve. Again the angle θ between this normal force and the body is known for any arm position so we can resolve this into an X vector Fe*cosθ and a Y vector Fe*sinθ.
Force balance in the X direction requires that:
1) Fe*cosθ + Fd*cosPDE + Ff = 0
Since Ff = 1 we have
2) Fe*cosθ + Fd*cosPDE +1 = 0
Force balance in the Y direction requires that:
3) Fe*sinθ + Fd*sinPDE + Fs = 0
Torque balance around point D requires that:
4) Fe*sinθ*ED = Fs*DC which rearranges to
4a) Fs = Fe*sinθ*ED/DC
Substituting eqn 4a into eqn 3 gives
5) Fe*sinθ + Fd*sinPDE + Fe*sinθ*ED/DC = 0 which rearranges to
5a) Fe*sinθ(1+ED/DC) + Fd*sinPDE = 0
Sinx = cosx/tanx so from eqn 2 we write
6) Fd*sinPDE/tanPDE + Fe*sinθ/tanθ + 1 = 0 which rearranges to
6a) Fd*sinPDE = - tanPDE( Fe*sinθ/tanθ + 1) which rearranges to
6b) Fd*sinPDE = -Fe*sinθ*tanPDE/tanθ - tanPDE
Substituting eqn 6b into eqn5a we get
7) Fe*sinθ(1+ED/DC) - Fe*sinθ*tanPDE/tanθ - tanPDE =0 which rearranges to
Fe*sinθ(1+ED/DC - tanPDE/tanθ) = tanPDE which rearranges to
Fe*sinθ = tanPDE/(1+ED/DC - tanPDE/tanθ)
all of which are known quantities for any arm position.
Once we know Fe * sinθ we can easily calculate Fs by substitution of the known value into eqn 4a.
As a reality check if the angle between the tangent to the curve at point E and the free body is 90 degrees then all the frictional force will be taken by this point and all other forces should fall to zero, so substituting a value of zero for sinθ should give Fe = 1, Fs = Fd +0 which it does so we shouldn't be too far from the truth.
For plausible values of the segments PD, DE and DC the range of skate force is from about 50% of Ff at the outer groove to under 20* at the inner groove. For reference, the skate force on a standard arm is around 45% of Ff.
We can therefore conclude that skate compensation is a requirement for this type of arm. The variation of skate force with position can make this difficult to achieve.
It looks as if I was wrong. After thinking about this some more I decided to do a force balance on a split plane arm design using some simple assumptions and based on the same geometry as used in my posts #219 and #222.
We will perform a force balance on the arm by considering the longer segment of the arm as a free body. The co-ordinate system is established by making the X axis the axial length of this body , Y the horizontal axis orthogonal to this and Z the vertical. We need only consider three degrees of freedom, namely X and Y translation and Z axis rotation, the other three being controlled by the vertical bearing and counterweight.
All forces in any direction must balance for the arm to resist translation, similarly all torques around any pivot must balance for the arm to resist rotation.
Since the object of the design is to achieve zero tracking error, we posit that all the frictional reaction force (Ff) acting at point C due to groove contact is in the X direction and normalise this to one force unit. (This force is usually of the order of 10mN).
Any skate force (Fs) will act in the Y direction though point C.
The torque around the pivot P must be zero, otherwise the arm PD would rotate. Thus the force acting at pivot D (Fd) must be axial to the segment DP. The angle PDE is known for any arm position so we can resolve this into an X vector Fd*cosPDE and a Y vector Fd*sinPDE.
If the guide at point E is frictionless the force it exerts on the arm at this point (Fe) must be normal to the tangent of the curve, otherwise the force would move the pivot along the curve. Again the angle θ between this normal force and the body is known for any arm position so we can resolve this into an X vector Fe*cosθ and a Y vector Fe*sinθ.
Force balance in the X direction requires that:
1) Fe*cosθ + Fd*cosPDE + Ff = 0
Since Ff = 1 we have
2) Fe*cosθ + Fd*cosPDE +1 = 0
Force balance in the Y direction requires that:
3) Fe*sinθ + Fd*sinPDE + Fs = 0
Torque balance around point D requires that:
4) Fe*sinθ*ED = Fs*DC which rearranges to
4a) Fs = Fe*sinθ*ED/DC
Substituting eqn 4a into eqn 3 gives
5) Fe*sinθ + Fd*sinPDE + Fe*sinθ*ED/DC = 0 which rearranges to
5a) Fe*sinθ(1+ED/DC) + Fd*sinPDE = 0
Sinx = cosx/tanx so from eqn 2 we write
6) Fd*sinPDE/tanPDE + Fe*sinθ/tanθ + 1 = 0 which rearranges to
6a) Fd*sinPDE = - tanPDE( Fe*sinθ/tanθ + 1) which rearranges to
6b) Fd*sinPDE = -Fe*sinθ*tanPDE/tanθ - tanPDE
Substituting eqn 6b into eqn5a we get
7) Fe*sinθ(1+ED/DC) - Fe*sinθ*tanPDE/tanθ - tanPDE =0 which rearranges to
Fe*sinθ(1+ED/DC - tanPDE/tanθ) = tanPDE which rearranges to
Fe*sinθ = tanPDE/(1+ED/DC - tanPDE/tanθ)
all of which are known quantities for any arm position.
Once we know Fe * sinθ we can easily calculate Fs by substitution of the known value into eqn 4a.
As a reality check if the angle between the tangent to the curve at point E and the free body is 90 degrees then all the frictional force will be taken by this point and all other forces should fall to zero, so substituting a value of zero for sinθ should give Fe = 1, Fs = Fd +0 which it does so we shouldn't be too far from the truth.
For plausible values of the segments PD, DE and DC the range of skate force is from about 50% of Ff at the outer groove to under 20* at the inner groove. For reference, the skate force on a standard arm is around 45% of Ff.
We can therefore conclude that skate compensation is a requirement for this type of arm. The variation of skate force with position can make this difficult to achieve.
Last edited:
So my original speculation was right then, that, as Conrad Hoffman mentioned, if the pivot and the cartridge cantilever do not form a straight line there will be skating force?
Unless some form of compensation is included, yes. As mentioned, the variable geometry can make this difficult to achieve.
Note that the analysis says that if it were possible to make the curve at E so that the segment ED was always normal to the tangent, this would remove the need for antiskate. The only such curve is a circle (or section thereof) which would take us back to Birch geometry. No free lunch.
Last edited:
string driven tonearm
From a post here. I cannot read Japanese. We need a translator here!
Looks like the pivoting headshell is being rotated by the strings to correct for tangency. Clever.
Clearly the above tonearm works in the same concept as this mysterious arm.
From a post here. I cannot read Japanese. We need a translator here!
Looks like the pivoting headshell is being rotated by the strings to correct for tangency. Clever.
An externally hosted image should be here but it was not working when we last tested it.
An externally hosted image should be here but it was not working when we last tested it.
An externally hosted image should be here but it was not working when we last tested it.
An externally hosted image should be here but it was not working when we last tested it.
An externally hosted image should be here but it was not working when we last tested it.
An externally hosted image should be here but it was not working when we last tested it.
An externally hosted image should be here but it was not working when we last tested it.
Clearly the above tonearm works in the same concept as this mysterious arm.
An externally hosted image should be here but it was not working when we last tested it.
Last edited:
Interesting and fine solution. The only problem I can see there is a skating force, gradually decreasing to 0 in the middle of record, and than changing direction to the opposite, and slowly increasing up to the record end. As a result, to construct proper compensation would be close to Mission Impossible. Good news is that stylus wear will be symmetrically equal.
Based on one of the drawings on page 82, it has a rather unusual alignment, that it uses the middle of the record as starting point, instead of the spindle. And at the middle of the record is not even tangent! Again, since I can't read Japanese, I don't know if I am interpreting this right.
The alignment reminds me of the one for RS Labs headshell which is was designed to lessen skating force or "needle talk" and has no guiding mechanism. But at least the RS arm uses the middle of the record at tangency.
Based on the drawings, this tonearm has rather strange geometry... I have to say I prefer the other string tonearm because at least that one is geometrically a variation of the Van Eps concept or Garrard Zero 100. I don't think it's too hard to readjust the geometry on this Japanese arm to the one used in the Thales Simplicity.
Perhaps, Mark Kelly can chime in on this one?
An externally hosted image should be here but it was not working when we last tested it.
The Japanese arm reminds me of this patent from 1985 by Andrew M. Wolff
An externally hosted image should be here but it was not working when we last tested it.
An externally hosted image should be here but it was not working when we last tested it.
And still it continues... possible data collection?
I wonder if someone has one of Dr. Frickert's Adjust+ measuring software setup so that some sort of meaningful data could be collected or compared. I know calculations will result in what the theoretical numbers should be, just curious what real world numbers are.
Also the last tonearm picture posted could be simplified , with just a pulley at the point directly above the stylus, and attaching the ends of the string to a stationary portion on the tonearm mount and the other end to a rotating portion of the tonearm mount.
Again, perhaps somebody (Frank?) might have some data that he would be willing to share.
I wonder if someone has one of Dr. Frickert's Adjust+ measuring software setup so that some sort of meaningful data could be collected or compared. I know calculations will result in what the theoretical numbers should be, just curious what real world numbers are.
Also the last tonearm picture posted could be simplified , with just a pulley at the point directly above the stylus, and attaching the ends of the string to a stationary portion on the tonearm mount and the other end to a rotating portion of the tonearm mount.
Again, perhaps somebody (Frank?) might have some data that he would be willing to share.
Last edited:
I can't read Japanese so I can only go from the drawing supplied. Either the drawing is in error or the arm cannot maintain tangency across the disc. Assuming the arm is 225mm long, the drawing shows a tracking error of 2.5 degrees at the centre position. This is simple geometry : taking the outer groove at 150mm and the inner at 60 mm, the centre point is at a radius of 105mm. The length O' - O'' must be 225 - SQRT(225^2 - 45^2) which is 4.54mm. The angular error is arcsin (4.54 / 105) which is 2.5 degrees.
Nanook, there is a simple way of directly measuring the real world angular error in any of these designs. Attach a long, thin, straight indicator rod to the headshell so that it is directly above the stylus position and at a right angle to the cartridge axis. I use a length of carbon fibre rod chosen for straightness.
If the arm maintains tangency the rod will cross the centre of the spindle as the arm scans across the disc. Any deviation from tangency will show as an offset between the rod and the spindle centre. The quotient of the offset and the groove radius is the angular error in radians.
On my design the maximal error is less than 0.1mm so the max angular error is less than .0015 radians or 0.1 degrees. I say less than because at present I'm not set up to measure more accurately than that.
I believe 0.1mm / 0.1 degrees is the practical limit of cartridge alignment - you'd need a very specialised mounting jig to get better alignment than that.
Last edited:
Thanks for the math, Mark. Given the drawings supplied, I don't think the arm can maintain tangency either..... until we get the Japanese text translated. I just noticed on page 81, the effective length is given as 270mm.
An externally hosted image should be here but it was not working when we last tested it.
