No this is not specific to Kane's method. You, as the modeler, get to
choose whatever coordinates you want to describe a system including
redundant coordinates. But for efficiency and simplicity's sake you are
better off choosing a unique set of minimal coordinates, i.e. generalized
coordinates. Yo
To kind of expand on what Jason's saying a 3D pendulum can be completely
defined using just (x, y, z) and you can deduce the angles from these
coordinates. In your case the pendulum only has two degrees of freedom (x
and y for instance and z be calculated because the pendulum has a fixed
length
Thanks for the input!
I probably wasn't able to explain it properly, but in this pendulum system
the mass is actuated. Think of it as a differential drive robot with fans
instead of wheels (instead of the mass). The force acting on the x-axis of
the mass frame and the torque about the z-axis of
The system that you drew only has two degrees of freedom regardless of what
forces you apply to the system. It also isn't clear as to whether you
consider the mass a particle or a rigid body. The problem is very different
depending on that. If you want a conical pendulum that has forces applied
to
Hi,
I'm trying to represent a weird case using piecewise functions :
z = sympy.Piecewise((True, x>0), (False, True))
A piecewise function returning a boolean value... so far so good. The
problems comes when I want to use that as a condition inside another
Piecewise :
z2 = sympy.Piecewise((1,
Hello,
I try to perform calculations in GF(2^3); e.g. calculate (x^2+x)(x+1) = x^3
+ 2x^2 + x = x^3 + x as polynomial coefficients are modulo-2. Using the
irreducible polynomial x^3 + x + 1 I arrive at (x^2+x)(x+1) = 1.
I try to do the same in sympy
import sympy as sym
import sympy.polys as p
I would use ITE for this case. It's the boolean version of Piecewise.
Aaron Meurer
On Thu, Aug 18, 2016 at 9:51 AM, Vincent Noel wrote:
> Hi,
>
> I'm trying to represent a weird case using piecewise functions :
>
> z = sympy.Piecewise((True, x>0), (False, True))
>
> A piecewise function returnin
Of course, for Piecewise((True, x>0), (False, True)), you could just
replace it with x > 0. It seems ITE doesn't do this automatically but
it does if you call to_nnf() on it.
In [12]: ITE(x > 0, True, False).to_nnf()
Out[12]: x > 0
Aaron Meurer
On Thu, Aug 18, 2016 at 4:09 PM, Aaron Meurer wrot
On Thursday, August 18, 2016 at 9:30:23 PM UTC+3, clemens novak wrote:
>
> Hello,
>
> I try to perform calculations in GF(2^3); e.g. calculate (x^2+x)(x+1) =
> x^3 + 2x^2 + x = x^3 + x as polynomial coefficients are modulo-2. Using the
> irreducible polynomial x^3 + x + 1 I arrive at (x^2+x)(x+
