Effective mass is the summation of the mass at each point of the tone arm times the distance from that point to the pivot. In other words, it is an integral of the mathematical function which describes the tone arm, the headshell, the cart, and the counter weight. Since that mathematical function is unknown to us, it is very difficult if not impossible to calculate the effective mass accurately. Perhaps other members with better math skills can come up with a way to estimate that.
One way to find out how much more weights are needed is by trial and error. You can incrementally add some weight to the headshell, make all the VTF adjustment necessary, and play a test record to see if the resonant frequency falls within the ideal range. 
Dear Clarkie: My advise is that you mount that cartridge and hear it, in this first " step " don't worry about tonearm effective mass, this factor is important but not the only that define the tonearm/cartridge quality level performance.
Btw,I see at your system ( sorry is out of the thread ) and IMHO you can improve in an important way your analog rig performance if you take your AS TT out of those " terrible " own footers. You can see at my system in that same TT, something similar that you can do will help you.
Regards and enjoy the music, Raul. 
Won't that cause tracking problems on the slightest warped records? What I'm talking about is the stylus shank bending more having to move the extra mass even though it's balanced perfectly.Just a thought. 
No, the low compliance cartridges REQUIRED more mass to work effectively. Low compliance means that it takes more mass to bend the stylus a given amount. When you have insufficient mass the stylus will not properly follow the groove. Imagine a very stiff shock absorber that is intended for a heavy auto. Put it on a light auto and it will hop from bump to bump. The tracking weight will be the same but if the arm has insufficient mass it will be easily deflected, the more massive arm will absorb a warp [ for example] more easily as it will allow the cantilever to flex to accommodate it. So the more massive arm works better precisely because it allows the cantilever to move more, the lighter arm will allow the arm to move rather than the cantilever. 
I always learn something new here.No dealer ever brought this to my attention either.I remember how popular the B&O linear tracking turntables were.I didn't like their cartridge stylus combination.They had aftermarket mods to use your own cartridge. I figured the arm was to light and would resonate too much.I guess I had some sense,but not the proper knowledge about it. Thanks for setting me straight. 
It's complicated enough, during the "Compliance races" of the 70s some cartridges had such a high compliance that they would have worked best with an arm of NEGATIVE mass. If you had ever seen a Transcripters Vestigial arm you would not have believed it, held together with thread. High rigidity was out of the question. The Technics linear tracking tables worked quite well, linear arms cam be very good but expensive to implement well at the highest level. 
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 SL1200). 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 tonearm = 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 SL1200, 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 feedback 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 microscale. I did this with a 9c (ProJect) 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

I don't want to beat a dead horse, but for good measure. After some careful consideration the calculation I proposed above is wrong.
It appears as though it could be okay in practice, at least with my technics tonearm. That is when one pull the balance weight back by .5mm the tone arm move into a new inclined equilibrium.
In theory, this should not occur unless the pivot point is displaced to the new center of mass. I assume that the fact that in practice a new equilibrium point is found is due to friction in the pivot point and that the change of the center of mass is minimal. 