Hello everyone:
I am making a track to measure current. To do this i am using one of the newest resistors of “0612 side”. The problem is that the track exceed the pad on the front side.
Yo can see in the attache image the green colored part. Please ignore the blue track, is a work in progress.
For the moment i put a box or a smaller track or etc, but i fill that is not the right behavior of usage. If exist some function to edit the track ending will be appreciated.
Thanks you
Christian
There isn’t a way to edit the track end unfortunately 
My suggestion would be used a zone to break-out of the resistor pad and then track to it
I dont know how do it. You can teach me? Thanks you
Edit: is a regular fill zone?
That is a bit of a problem . . . one possible work around, use the footprint you current have as a starting point and make custom version of it where the pads are wider, more square, that will make it much easier to fit your wide tracks.
My preferred method is to modify the footprint:
I think my example is what @RaptorUK is suggesting.
It took me a little longer to create the example than it took Raptor to type an explanation. 
You could also use the Tapers from the RF-Tools
Excellent, Even on this solution “works for me” i think that is good thing for the KcV10
here you either need to use a polygon as it was written above, or think about why you need such a large track width, resistor 0612 usually has a power of 1 watt, even if you use a voltage of 1.8 volts, the current will be about 0.5 Amperes, 0.5 amperes, and even on a foil of 18 microns it is a 0.3 mm wide track, let it be with a margin and you can take a 0.5/0.6 mm wide track
Its a current sense resistor, for best accuracy its a good idea to have little resistance in the track.
then all that remains is to use a solid polygon fill
I wonder if it would even be possible to use teardrops here.
Haven’t tested yet…
Sometimes kelvin-style sense tracks are broken out from the opposite side of the pad, in the middle under the resistor body.
I am using in 2 places. one of 0.160ohms with a 5A of current on a 1.5W resistor (LTR18EZPFLR160) to measure current and i need 3mm of width.
Finaly i use this solution
5A trough 0.16 Ohms is 4W, you do realise?
There will be a fire 
hehehe no no. No fire. Are not 4w. Is a half second test of 2.7A i make the calculations for 5A on tracks
P / I^2 =R
1.5 / 3^2= 0.166 ohms
But i finally use 100mOhms and one OpAmp with 4x of gain. Then:
P= R* i^2
P= .1*2.7^2
P= 0.729w
But the track is dimensioned to 5A, i can change the resistor if is burned, but no the track.
Even this is off topic, in another application i have 6A on the same footprint but the resistor has 0.010 ohms. Is on production and certified
