How to change effective mass of a SME V.


I have a SME V. Its effective mass is quoted as 10-11. I wish to alter this to around 15-16. If I add mass to the headshell (and know how much) and crank the counterweight back how can I accurately re-calculate the ffective mass? The cartridge is 9.1 Gm, compliance 6. I know that the arm resonant frequency will change as well.

Regards Geoff
clarkie
I am very glad to read your question. I had a similar one in a thread that I started. Unfortunately I did not get a satisfactory answer. So I put my thinking cap on...

Here is my basic theory.

When your tone arm is horizontally balanced the effective mass of the counter weight end is equal to the effective mass of the cartridge end. The problem is you do not know the effective mass of either.

Small note before proceeding, my definition of effective mass is the mass that would be measured at the tip of the tone arm if the arm itself had a mass of zero and a weight were hung a the tip of it.

What do we know? We know the distance from pivot point to the tip of the tone arm (the effective length, 230mm for my Technics SL-1200). We also know or can easily determine the distance from the pivot point to the center of mass of the balance weight. And finally we know the mass of the balance weight.

Definition of variables

Mass of Balance Weight end of the tone are = Mbal
Distance from pivot point to center of balance weight =Dbal
Mass of tone arm = Mtarm (unknown)
Effective Lenght of tone-arm = Dtarm

In the horizontal balanced position the equation is:

(Mbarm x Dbarm) - (Mtm x Dtm)= 0

We now have one equation two unknowns

For the second equation I propose to move the balance weight back by a small easily measurble increment. I have a technics SL-1200, which has a very adjustable and user friendly tone arm. The pitch of the balance weight is 10mm and there are 40 graduations on weight, thus by turning the weight by 2 graduations it moves back 0.5mm.

I not familiar with SME thus you might have to use a vernier caliper to accurately measure your change in balance weight position.

Once the weight is moved back the cartridge tip of the tone arm will rise by a few millimeters (10 to 20) before finding it new point of equilibrium (or not at all, if hits its end stop or the point of equilibrium has shifted to the vertical position, in which case you must move the balance weight back by a smaller increment).

Now the difficult part, you must measure the change in height of the tip of the tone arm. With the change height and some high school trigonometry you can determine the angle the tone arm has shifted (actually not really required) and the new vertical distance between the pivot point and the various other points.

Now you have:
Mbal = no change
Dbal = initial position
Dbal1= new position
Mtarm = no change
Dtarm = initial position
Dtarm1 = new position

Mbal X dbal - Mtarm x dtarm = 0
Mbal x dbal1 - Mtarm x dtarm1 = 0

Mbal x dbal - Mtarm x dtarm = Mbal x dbal1 - Mtarm x Dtarm1

Mtarm = Mbal x[(dbal1 - dbal)/(dtarm - dtarm1)]

Now for some caveats!
1- In theory this great but in practice it requires the precise and accurate measurement of small distance (0.5mm) of parts that are not fixed but balanced. So measuring technic is everything.
2- futher to the point above small measuring errors can have a large impact on the end result.
3- I have ingnored the mass of the tone arm on balance weight end, my feeling is that its mass is small compared to the balance weight and distance is small from the pivot thus the change in distance should (this has not been verified mathimatically) have negligible effect on the final result.

I am very curious to hear feed-back from the forum. I must go to sleep now but I will share the results of my measurements tomorrow.
Holly mo!
Nick this sound very involved indeed.
I think a reasonable and quick approximation is to take off the counterweight and cart, then place the arm on a micro-scale. I did this with a 9c (Pro-Ject) arm and is was the exact weight (mass) as stated it should have been, 11g.

Go figure,

It's actually very easy.

You know the effective mass of the SME V.

You know its effective length. Call this Xel.

You know the mass of the extra headhsell weight.

You can take the counterweight off and weigh it (in grams). Call this M.

You can measure the counterweight's distance from the pivot before the additional weight and after it. Call these Xb and Xa.

The counterweight's additional contribution is then given by

M . (Xa^2 - Xb^2 / Xel^2)

The new effective mass of the arm is the sum of this number, the original effective mass and the additional headshell weight.

Mark Kelly

BTW for a first approximation you can simply assume that the counterweight's contribution will be approximately 10% of the added headshell weight's.

If you wish to add say 5 g overall, you would add a headshell weight of about 4.5 g and expect the counterweight movement to get you (most of) the rest.

Mark Kelly

God I hate that I can't edit

The bracket is in the wrong place, the equation is

M . (Xa^2 - Xb^2) / Xel^2

Mark Kelly