Distance between 2 joints?

[fred22]

What is the best way to measure the distance between 2 joints. My first thought is to take the world space positions of the 2 joints and create a distanceDimension node get the measurement from the attribute it creates.... But a floor I don't want to keep this node in my scene I want it gone but without a naming flag on the node I can't create multiple distanceDimension nodes without trouble. Thanks in advance.

[empleh]

You can script it, although it might be slow.

$distBetween=sqrt((p1x-p2x)(p1x-p2x)+(p1y-p2y)(p1y-p2y)+(p1z-p2z)*(p1z-p2z));

I believe this is correct. If it isn't I'm sure someone will be happy to let me know :-), hopefully.

Does this help?
ww

[fred22]

I came across this post by TNT in a search. using the distanceBetween node I'm about to try it this morning I'll post when I find out what happens.

You can create a distanceBetween node:

createNode distanceBetween;

this creates distanceBetween1. You can now enter the coordinates of two points in the attribute editor or you can use an expression that does this for you:

//
float $Tmp1[] = getAttr object1.translate;
vector $point1 = <>;
float $Tmp2[] = getAttr object2.translate;
vector $point2 = <>;

eval ("setAttr distanceBetween1.point1 " + $point1);
eval ("setAttr distanceBetween1.point2 " + $point2);

float $distance = getAttr distanceBetween1.distance;
//

SOLUTION 3:
You can use the distance tool (modify->measure->distance_tool)
This creates two locators. You can point-constrain them to the two objects and use this line:

float $distance = getAttr distanceDimension1.distance;

hope this helps

[joojaa]

In reply to:
You can script it, although it might be slow.

$distBetween=sqrt((p1x-p2x)(p1x-p2x)+(p1y-p2y)(p1y-p2y)+(p1z-p2z)*(p1z-p2z));

if you put that in a expression then it should be pretty fast. Altough measure distance is a bit better! Hou often do you need to evaluate this once in a fortnight or each frame?

[tbriggs]

get worldspace positions regardless of position in hierarchy with xform:

$tmp1= xform -q -ws -t joint1;//returns a float array
$tmp2= xform -q -ws -t joint2;// ditto

$dist = sqrt(pow($tmp2[0] - $tmp1[0]),2)+pow($tmp2[1] - $tmp1[1]),2) +pow($tmp2[2] - $tmp1[2]),2));//thank you Pythagoras

[dwalden74]

Huh? Nobody

[joojaa]

vector times 1/2, 1/3 etc...

Yeah mag works provided that you get the world postions of the joints... But the measure node is better if this needs to be querried!

Actualy the mag you describe is exactly the same function as sqrt((p1x-p2x)(p1x-p2x)+(p1y-p2y)(p1y-p2y)+(p1z-p2z)*(p1z-p2z)); Just shorter to write (and slightly faster to calculate since its compiled in but as its so short its not much more optimal)

[fred22]

wouldn't you just divide the distance by 1/3 and add it to 0 or take it from the total distance?