Hi! Is anyone that manually insert load data into KUKA starting from SOLIDWORKS centre of mass?
The centre of mass X, Y and Z are clear
A, B and C?
Jx, Jy and Jz?
Hi! Is anyone that manually insert load data into KUKA starting from SOLIDWORKS centre of mass?
The centre of mass X, Y and Z are clear
A, B and C?
Jx, Jy and Jz?
https://www.cati.com/blog/2017/03/d…-in-solidworks/
You could change units to kg and m to get inertia values already in proper units so you that don't need any recalculating. or if you like to do it the hard way, apply what you learned in school:
1g = 10^-3kg
1mm = 10^-3m
(1mm)^2 = 10^-6 m^2
and therefore scale factor is:
1g*1mm^2 = 10^-9 kg*m^2
so all the values for inertia expressed in g*mm^2 will need decimal point moved 9 places to get kg*m^2 (power of metric system, try that in your head but using imperial units).
Kuka robot expects data of inertia at the center of gravity of mass or load. So if you use principal inertia values, you get:
Px = 0.463 kgm^2
Py = 1.053 kgm^2
Px = 1.11 kgm^2
Then you need to compute angles ABC from matrix Ix,Iy,Iz
When principal inertia axes are aligned with flange axes, this would be identity matrix (ones across main diagonal and all other values are zero) and ABC values would be all zero degrees.
Example of that calculation is found on the forum when computing base from three points. it is the same math... so in this case you should be getting about
A=174.14 deg
B=66.92 deg
C=-91.38 deg
The other option is to leave values A,B, C as zeroes so no calculation of angles is needed. But that means that inertia values Px, Py, Pz would need to be recalculated. Fortunately for you SolidWorks already offers those values as Lxx, Lyy and Lzz. That is inertia values about CM but calculated using axes parallel to robot flange (not principal axes).
Then you get:
A = 0 deg
B = 0 deg
C = 0 deg
Jx = Lxx = 0.964 kgm^2
Jy = Lyy = 1.110 kgm^2
Jz = Lzz = 0.553 kgm^2
Hi! Thanks a lot for you detailed explanation... I think this will help many other around here. Thanks again...