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28th February 2012, 08:54 AM  #341 
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Join Date: Feb 2006
Location: Willy, VIC

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 
28th February 2012, 01:17 PM  #342 
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28th February 2012, 02:35 PM  #343  
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Quote:
"This tonearm, published in 1970, is the best known solution working almost tangential. The cartridge tip was mounted exactly below a pivotpoint. 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." __________________________________________________ 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 whitepaper. 

29th February 2012, 03:03 AM  #344 
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Location: Willamette Valley

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 Garrardtype 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. 
29th February 2012, 01:33 PM  #345  
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An Analysis of Skate force
Quote:
Quote:
We will perform a force balance on the arm by considering the longer segment of the arm as a free body. The coordinate 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. 

29th February 2012, 03:20 PM  #346  
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Quote:


29th February 2012, 03:24 PM  #347  
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Join Date: Feb 2006
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Quote:
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. 

29th February 2012, 05:14 PM  #348  
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Location: Gwynedd North Wales

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Cheers Steve
__________________
steve if it ain't broke, I ain't fixed it 

20th March 2012, 03:08 AM  #349 
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Join Date: Nov 2005

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. Last edited by directdriver; 20th March 2012 at 03:29 AM. 
20th March 2012, 09:20 AM  #350 
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Join Date: Mar 2011

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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