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Old 28th February 2012, 08:54 AM   #341
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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
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Old 28th February 2012, 01:17 PM   #342
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Quote:
Originally Posted by Mark Kelly View Post
angle of the pivoting head vs the rest of the tonearm.
AS long as head cantilever and Diamond to pivot line are not perfectly aligned, you will get an centipede force. Hence distortion.
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Old 28th February 2012, 02:35 PM   #343
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Quote:
Originally Posted by phivates View Post
I'd like to know if it(Garrard Zero 100) maintains a consistent offset, which the B-J definitely does not, or a progressively decreasing offset, a la Van Eps or Warden.
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."

Click the image to open in full size.

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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:

Click the image to open in full size.

And the error is +0.009° / -0.040° and you can download the specs from his white-paper.

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Old 29th February 2012, 03:03 AM   #344
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Default 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.
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Old 29th February 2012, 01:33 PM   #345
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Default An Analysis of Skate force

Quote:
Originally Posted by Mark Kelly View Post
If Frank has done what I think he's done, there will be no skating force.
Quote:
Originally Posted by berlinta View Post
At first it may appear as if there ought to be a skating force present - and there arguably can be, but the arm/cartridge doesn't exhibit a tendency to move inwards when properly set up. The cantilever is NOT subjected to any side thrust other than what results from bearing friction and the minute influence of the wiring.

Frank
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 by Mark Kelly; 29th February 2012 at 01:39 PM.
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Old 29th February 2012, 03:20 PM   #346
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Quote:
Originally Posted by Mark Kelly View Post
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.
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?
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Old 29th February 2012, 03:24 PM   #347
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Quote:
Originally Posted by directdriver View Post
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 by Mark Kelly; 29th February 2012 at 03:33 PM.
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Old 29th February 2012, 05:14 PM   #348
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Quote:
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.
Magnetic antiskate?

Cheers Steve
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Old 20th March 2012, 03:08 AM   #349
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Default 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.

Click the image to open in full size.

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Clearly the above tonearm works in the same concept as this mysterious arm.

Click the image to open in full size.

Last edited by directdriver; 20th March 2012 at 03:29 AM.
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Old 20th March 2012, 09:20 AM   #350
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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.
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