effective mass - diyAudio
 effective mass
 User Name Stay logged in? Password
 Home Forums Rules Articles diyAudio Store Gallery Wiki Blogs Register Donations FAQ Calendar Search Today's Posts Mark Forums Read Search

 Analogue Source Turntables, Tonearms, Cartridges, Phono Stages, Tuners, Tape Recorders, etc.

 Please consider donating to help us continue to serve you. Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
 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!! Diy-tone 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 diyAudio Member   Join Date: Jan 2002 Location: Munich, Bavaria Jeremy, you get an A! __________________ Regards, Bernhard
 4th April 2002, 08:42 AM #4 diyAudio Member     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
diyAudio Member

Join Date: Feb 2002
Quote:
 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).
It's moment of inertia. Similar but not exactly the same (and calculated the same way, as a differential equation over the shape of an equally-distributed mass, but that's not really possible with a complex object like a tonearm!) Variation in tracking force is part of the picture but the difference between a 1.5g and 2g tracking force, say, is not dominant as I hope to demonstrate below.

Quote:
 2.When I read about effective mass it is in gram and that is not what seems to come from your equation?
Right, I gave you m x d^2 units and the spec is as you say, given in grams. If I'm remembering this right, for rotational moment of inertia, the units drop out as the mass is normalized to the distance from the pivot, of the stylus position. I'll show this below. It describes a force applied to an object at a distance d from an axis of motion. Since d is fixed the units drop out for a specific object and size.

Quote:
 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
Trying my best!

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

diyAudio Member

Join Date: Jan 2002
Location: Munich, Bavaria
Jeremy,

I attach what i wrote below:

Peter,
Quote:
 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).
i was glad Jeremy avoided differntial calculus here as this usually messes up explanations.

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 counterweight-adjusted 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:
 2.When I read about effective mass it is in gram and that is not what seems to come from your equation?
<sigh> most writers and reader find it difficult to mentally penetrate what is described above.
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 Q-factor occurs as if the cartridge would have a 10-gram mass sitting piggy-back 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:
 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.
The answer's elaboration is left as today's homework
hint: consult answers to QQ 1 and 2 ..
__________________
Regards,
Bernhard

 4th April 2002, 09:21 PM #7 diyAudio Member   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 cross-check. 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 __________________ Regards, Bernhard

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are Off Pingbacks are Off Refbacks are Off Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post IZHAKKATZ Analogue Source 18 3rd April 2016 10:50 AM planarboy Analogue Source 22 21st February 2009 07:49 AM dario54 Analogue Source 6 6th August 2008 07:40 PM nghiep Analogue Source 10 25th January 2008 04:07 PM stewartwen Analogue Source 4 11th April 2006 05:22 PM

 New To Site? Need Help?

All times are GMT. The time now is 03:49 AM.