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27th November 2002, 07:58 PM  #1 
diyAudio Member
Join Date: Aug 2002
Location: CA

Pinging DICE 45Diy tonearm question?
Dice 45, first I'd like to thank you for rousing up my interest in making a unipivot tonearm. I read some of the posts down below this one. I've used some ideas from the Altmann arm plus most of what you recommended. I'm 99.8% finished my 12"white oak arm, and have mounted a junkish cartridge just to be amazed at how stable and smoothly this thing appears to track.
My question to you is, by what method can the arm mass be determined so as to decide in what compliance range my "real"(not purchased yet) cart will be. It can't be as simple as weighing it, or choosing a "mid" compliance and go from there right? I think I am looking for the "rotational" mass as the arm tracks the groove, etc. Please keep in mind that I am just a techniciantype lacking the exposure to the physics behind the correct answer. Thanks for any tips and help! 
27th November 2002, 08:02 PM  #2 
diyAudio Member

I have nothing to add here technically but I'm very interested in how your tonearm look like. Do you have any pictures?
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28th November 2002, 01:50 AM  #3  
diyAudio Senior Member
Join Date: Aug 2002
Location: Belgium

FROM THE SOUND OF IT...
Hi,
Congrats. Since it is made out of wood which is medium weight as far as most tonearms go (unless you use balsa) it seems to me a low to medium compliance cartridge would work fine. Quote:
I can't remember the rough breakdown figures putting one type in one class or the other but Dice45 might. Ciao,
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Frank 

28th November 2002, 07:51 PM  #4 
diyAudio Member
Join Date: Jan 2002
Location: Munich, Bavaria

hard times,
thank you. your simple Q is hard to answer. Takes extended homework. My diagram is too small to post, have to redraw it with CAD help to a usable size. Too tired this evening, i come back later with details. Possible procedures: 1a) You have a soft helical spring with a known spring constant. You apply the string to the tonearm at the radius the stylus would be. Then you move it up and down with different frequency and find the systems natural resonance frequency. From that frequency and the known spring constant you calculate the rotational inertia in reference to the tonearm's horizontal pivot axis. And jconvert that inertia to a virtual mass located at stylus radius. 1b) You also can assume with sufficient precision this as a translational problem and use the formula for a mass hanging on a spring: f= (1/2*pi)*sqrt(c/m) with f being frequency, c being spring constant and m being the effective mass. Solve that equation for m. 2) You can take an unknown cartridge an mount it into a tonearm of known effective mass. You measure the frequency (use a test record). Solve the equation above for c. c is the cartridge compliance. Then mount this cartridge into your new tonearm and measure the frequency, from that and from the formula above solved for m you can calculate your new tonearms effective mass. 1b) and 2) is easier if you have that durned diagram, giving a family of curves with c as parameter, effective mass as Yaxis and frequnecy as Xaxis. 3) make a 3D model of your tonearm and integrate about all infinitesimal mass points, each having a densitiy attached and calculate the total inertia in reference to the horizontal pivot. Like 1a). This gives you the "effective mass" with highest precision. Better use a highend CAD for that and tell me when you are ready, i marry your girl and get old with her in the meantime i just mentioned it for completeness' sake.  Expect horizontal and vertical effective mass to be almost the same for a pivoted tonearm. I see when i can bring the diagram.
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Regards, Bernhard 
28th November 2002, 08:05 PM  #5 
diyAudio Member
Join Date: Feb 2002
Location: Belgium

In my case I guess I have it done in CAD before I find a cartridge and an arm
Even if you take into account that I'm now off installing W2K again. 
22nd January 2003, 03:40 AM  #6 
diyAudio Moderator Emeritus
Join Date: Jan 2003
Location: Near London. UK

Effective mass
The "effective mass" of the arm has little to do with the actual mass of the arm. It's all about inertia, which is proportional to the square of distance of masses from the pivot. Thus, a heavier counterweight actually reduces mass because its counterbalancing force (the moment) is proportional only to the distance. That's why arms have big, heavy counterweights close to their pivot.
To return to your question. Because effective mass is determined by the square of distance, only the headshell mass is really significant. As a huge generalisation, unless your arm is a spindly little thing with holes drilled everywhere, it would probably prefer to see a moving coil cartridge. Onepiece arms like the Rega are lower mass than you might expect because the lack of joints (particularly at the headshell) greatly reduces effective mass. 
22nd December 2003, 06:06 PM  #7 
diyAudio Member
Join Date: Oct 2003
Location: Tejas

Ping EC8010
Ping EC8010
I want to figure the lowest amout of actual mass in a 12" tonearm to acheive 4.7 grams effective mass. Can a small shifting weight about the center axis be used to alter effective mass? The cartridge weighs 6.3 grams. I have calculated that with my compliance of 25cm/dyne, to get the resonance to 10 hz, I need 10 grams effective mass. The SME SeriesIII improved has a resonance of 10Hz with the same cartridge. It has an effective mass of 6.1 grams. I wonder if it can be made lighter, and still keep the same effective lateral mass, and how. Are there any formulas for this. To figure the effective mass, I used this formula Resonance 10 = 1000 * 6.28 sqrt (cartridge mass +effective mass) HELP 
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