Hero System (Champions) STUN / BODY dice (Display total pips and display count of dice modified by the number of one's and six's)

Has anyone seen any further work making the Hero System Stun / Body count work. Example I roll 5d6 'damage' I get roll 4, 1, 5, 6, 1 This would show I have rolled 17 'Stun' which is the total of the pips on the dice I count the number of dice and modify the result, subtracting 1 for each '1' rolled and adding 1 for each '6' rolled. 5 -1 +1 -1 = 4 'Body'    it would be great boon if this function could be added

Yes, it would!

These dice mechanics are not likely something Roll20 will implement without a thread first being made in the Suggestions & Ideas forum . I'm going to move this thread over to the API forum; there are many existant dice scripts, some of which could probably be adapted for your use with some patience and guidance. (Please note that access to the API is available only within games in which the Creator has an active Pro subscription.)

This would be pretty straight forward to write for the API.  I have several similar scripts already.  Let me know if you'd like one modified for this purpose. =D

Incidentally, you could use a rollable table to simulate this.  Basically, each die has the following possible values: 0,2,3,4,5,7  You'd just need to setup the table with those values and a weight of 1 each.  Assuming you named the table hero , you'd just call it like this: [[ 5t[hero] ]] or /roll 5t[hero]

Looking into it some more, It looks like the mechanic actually leads to these possible values: 0,1,1,1,1,2, so a rollable table would have 3 values: 0 at weight 1, 1 at weight 4, 2 at weight 1  Still totally doable with a rollable table, I believe.

They still need to sum of the pips on the original dice to calculate stun.  One throw of the dice generates (in his example above) 17 stun (the sum of the dice) and 4 body.  For each throw the dice need to be interpreted twice using two different methods.  Squarely in the realm of the API

Ah, back to "post a request in API" and someone  will likely help you.