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3rd April 2002, 02:15 PM  #1 
diyAudio Member
Join Date: Apr 2002
Location: Amsterdam

effective mass
hello, you all,
It is great to have a forum like this!! Diytone arm building is something I am very interrested in. Only recently I came across the air bearing tangential tone arm by Poul Ladegaard and immediately I wanted to make something like it. From what I've been reading it seems that the effective mass of the arm is a very important factor to determine what kind of cartridge it is best for. But what is effective mass exactly? Is it the weight of the total arm? Or perhaps of the armtube? Or only the part of the tube in front of the bearing? I hope somebody can enlighten me on this. Peter 
3rd April 2002, 04:49 PM  #2 
diyAudio Member
Join Date: Feb 2002

It is the moment of inertia of the arm system. With the Ladegaard arm, the total mass of the arm assembly and the slider is your horizontal moving mass, and your vertical moving mass is more difficult to calculate.
It is related to the distribution of mass, m, times (the distance to the vertical pivot, d, squared)  in practice it is dominated by two things: the mass of the counter weight and the mass of the cartridge. A heavier counter weight closer to the pivot will reduce the vertical moving mass, a lighter CW further away will increase it (because of the square term in the d. The product d * m is a constant for a given tracking force and cartridge mass.) j 
3rd April 2002, 07:07 PM  #3 
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Join Date: Jan 2002
Location: Munich, Bavaria

Jeremy,
you get an A!
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Regards, Bernhard 
4th April 2002, 08:42 AM  #4 
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Join Date: Apr 2002
Location: Amsterdam

Thanks very much for your reply, Jeremy.
But I'm afraid I still don't get it: 1.What does distribution of mass mean? is it center of gravity? If so, it would mean that the effective mass is dependend on tracking force (which is the same thing as moving the center of gravity). 2.When I read about effective mass it is in gram and that is not what seems to come from your equation? 3. common specs for arms state effctive mass in the order of 10 gr. or even less so it can never be the total mass of the arm. Hope you can get me to understand all this! regards, Peter 
4th April 2002, 07:34 PM  #5  
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Join Date: Feb 2002

Quote:
Quote:
Quote:
The moment of rotational inertia applies to rotation about an axis. It represents the equivalent mass that would need to to be at the stylus location that, once in motion, would have the same opposition to a force applied to it. This must be why the dimension units drop out  you are specifying a specific distance from the pivot as the place where you are applying the force. (I am using college physics I learned in 1978 here, go easy on me!)  with a linear tracker, you see how the horizontal moving mass can be the total moving mass, right? Either you assume an infinite radius so the entire assembly turns out to be the same distance away, or you just ignore rotation altogether since you are not actually going about an axis to begin with. Special case.  with a pivoted arm, or in the vertical plane of a linear tracker, let's look instead at a playground seesaw. 5 kgs on one end, 1 kg on the other. Same distance for each from the fulcrum. Static condition: equivalent to 4 kgs on the heavy end. This is the gravitational force on this system. Set it in motion. The 5 kgs will have a moment of inertia of 5(d^2) and the 1 kg will have 1(d^2)  the moment of inertia adds, not subtracts, (I'm pretty sure.) Dynamic condition, now, the object is set in motion (and this is the same as what is in play when the groove excursion tries to set the arm in vertical motion.) Now let's balance it. 5kgs at d distance and 1 kg at distance 5d. The heavy end will have moment 5 (d^2) and the other will have 1(5d)^2 or 25(d^2). Moment will be 30 d^2 and that will be equivalent to a moving object of 30kg applied at a position d away from the pivot. Balance it with a smaller weight: 5 kgs on one end, .5 kgs at distance 10D. Now moment of inertia is 5(d^2)+100(d^2)=105(d^2) and will resist being moved like a weight of 105kg a distance d away from the pivot. But what you have with a tonearm is usually a heavy weight closer to the pivot: say 5kg at the heavy end and 50 kg a distance .1d away. Moment of inertia: 5(d^2)+50(.1d)^2=5.5(d^2) which is normalized at d to 5.5g. If I scratch my head I can may come up with the formulas that take into account the imbalance represented by the tracking force and gravity. Hopefully dice45 will be in here with his red pen soon and critique my efforts, I am working hard here! j 

4th April 2002, 09:03 PM  #6  
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Join Date: Jan 2002
Location: Munich, Bavaria

Jeremy,
had answered the question already, but Jason's server didn' t let me post it. I attach what i wrote below: Peter, Quote:
In reality mass is distributed continuously but not uniformly. Mass density can change. In a translationally moved mass, each tinytiny differntial mass element contributes equally to the masses intertia. But if If the mass is rotated, we must integrate over all differntial mass elements (dm): J = integral(r^2*dm) (limits from zero to total mass). This means that we have to build a sum over all mass elements dm miltiplied by the square of each mass element's radius. This sum is the total intertia momentum of this mass then and reference for this calculation is the center of inetria AKA center of gravity (consider: the moment of intertia is the rotatory equivalent to mass as far as +/ acceleration is concerned) We can conclude that a mass element contributes far more to the total momentum of inertia if is is rather distant to the center. Consequently, if we have a tonearm with counterweightadjusted tracking force, we have to move the counterweight a bit to change TF. So we slightly change the tonearm's total momentum of inertia. Maybe not as slightly as we want. If OTOH the TF is generated by a spring, inertia changes not at all. Quote:
See the "effective mass" as a translational approximation (small angle changes)f or the tonearm's rotation. The tonearm's momentum of inertia (a torque !) creates a force as soon as the tonearm is rotated (let's ignore for a moment the force is not constant): effective tonearm length multiplied by the momentum of inertia. If we translationally accelerate a mass of, say, 10 grams, this mass's intertia also creates a force, objecting to the acceleration. So if the tonearm shows the same amount of reactiing force against accelation, it is said to have an "effective mass" of 10 grams. This means, with a given cartridge, the same resonance frequency and resonance Qfactor occurs as if the cartridge would have a 10gram mass sitting piggyback on it (the mass of course guided by a virtual force field of the colour purple and prevented by this field from tilting or rotating) Quote:
hint: consult answers to QQ 1 and 2 ..
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Regards, Bernhard 

4th April 2002, 09:21 PM  #7 
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Join Date: Jan 2002
Location: Munich, Bavaria

Jeremy,
no objection from my side, good explanation. You seem to master your college physics very well, not as those bootscamp artists Jocko is snorting about. Now Peter sits caught in his skin, having to tackle the memory overload Peter, please work yourself thru both answer posts separately. Then crosscheck. Been there done that, with scientific posts i did not dig at once.
__________________
Regards, Bernhard 
5th April 2002, 08:10 AM  #8 
diyAudio Member
Join Date: Apr 2002
Location: Amsterdam

Jeremy, Bernhard,
Boy, I thought I was asking a simple question!! I will try to get to grips with your answers. It's years ago since I did this kind of math. So I expect it will be a while before you hear from me again. Thank you both for going through the trouble of writing these elaborate replies. I'll try to do you justice. Peter 
5th April 2002, 02:10 PM  #9 
diyAudio Member
Join Date: Feb 2002

I'm happy to help.
Hint for dice45's question: why is a flywheel made with as much mass as possible on the outer perimeter? 
5th April 2002, 02:52 PM  #10 
diyAudio Member
Join Date: Jan 2002
Location: Munich, Bavaria

Jeremy,
It's because of centrifugal force: if the flywheel would not be strengthened at the outer edge, the zentrifugal force would make it crack for higher rotational speeds ...ROTFLAMO Peter, don't get confused by me bullshitting Jeremy
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Regards, Bernhard 
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