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Phantom Degrees of Freedom (by Thovthe)
SolveSpace tells me I have 2 Degrees of Freedom (DoF) in the g008 group in AquariumStand.slvs but I don't think I do. Seems to be the middle box on the side farther along the x-axis in that sketch group. I've tried many changes and redrawing the offending area without deleting the group. Is this a known issue or am I just missing the unconstrained DoF? If it is indeed a bug and you want more data I can supply a file from before it was broken too.
(no subject) (by ruevs)
This is very interesting. The culprit is the "bottom middle" rectangle (compression stand sketch). See the attached screen shot.
It is constrained differently from the other five - and a bit unconventionally, but I do not se why there should be 2 degrees of freedom left. This is worth investigating closer.
It is constrained differently from the other five - and a bit unconventionally, but I do not se why there should be 2 degrees of freedom left. This is worth investigating closer.
(no subject) (by ruevs)
Here is another screen shot with the problematic "rectangle" selected - for context.
(no subject) (by Jonathan Westhues)
I'm thinking the point-on-line constraints are on lines perpendicular to the workplane, so the one equation actually consumes two DOF. The reported DOF count fails to capture that.
Equations of that form are bad for the solver, but I guess it happened to work here. If we wrote two equations when the line is perpendicular then that would fix the count and improve convergence, though that could break if the line position is also getting solved for. If we wrote two equations only when the line is perpendicular and also from a previous group then I think that could only help, though that's not very elegant and not a full solution.
A general equation-level solution would be better. That's tricky though. I don't believe the linearized equations contain sufficient information, since the partials approach zero as the solution converges. (For example, consider f(x1,x2) = x1^2 + x2^2 = 0, which implies x1 = x2 = 0. ∂f/∂xi = 2*xi = 0 when converged. So the linearized equation constrains 0 DOF, even as the full equation constrains 2 DOF.)
Equations of that form are bad for the solver, but I guess it happened to work here. If we wrote two equations when the line is perpendicular then that would fix the count and improve convergence, though that could break if the line position is also getting solved for. If we wrote two equations only when the line is perpendicular and also from a previous group then I think that could only help, though that's not very elegant and not a full solution.
A general equation-level solution would be better. That's tricky though. I don't believe the linearized equations contain sufficient information, since the partials approach zero as the solution converges. (For example, consider f(x1,x2) = x1^2 + x2^2 = 0, which implies x1 = x2 = 0. ∂f/∂xi = 2*xi = 0 when converged. So the linearized equation constrains 0 DOF, even as the full equation constrains 2 DOF.)
This might be interesting. (by Thovthe)
In an earlier version of this file there appears to be no issue. I'm not sure if that way that it is constrained is how it was originally (I fiddled with it trying to fix the issue) but I am pretty sure that it spontaneously changed to having more stated degrees of freedom without me messing around with that sketch.
Earlier version attached.
Earlier version attached.
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