Is there an easy way to read jld files in Maple?  Apparently JLD (Julia Data) and JLD2 files are binary formats primarily designed for saving and loading Julia variables, preserving types.

In the screen shot below taken from a 3D animation I want to vastly shrink the area I have stippled with spots (it's actually white on the worksheet).  It's outside the cube defining the 3D animation and inside the square that defines the part of the worksheet used for an image.

I have created an animation of a flexible octahedron (following Bricard) with Maple.   It shows the six vertices and twelve edges.  How do I add the faces (triangles)?  Can the animation show their lines of intersection?

Based on information I learned here last week, I tried the following.  Summary: I enter three functions f,g,h. f Is a function of x and t, g is a function of y and t, h of z and t.  I want to plot a 3D curve. I know an intial value.  The curve looks good but quite very early saying "cannot evaluate the solution further right of 3.4774922, probably a singularity."  HELP!

f := 4.61376*x+1.320192000*cos(t)^2*x-.320192000*sin(t)*cos(t)^2*x+4.304656800*cos(t)^2*x^2+17.21862720*cos(t)*x^2+12.9477168*sin(t)^2*x^2-25.8954336*sin(t)*x^2+7.147008*sin(t)^3*x-21.441024*sin(t)^2*x+9.68025600*sin(t)*x+3.181248*sin(t)*x^3-1.3251456*sin(t)*cos(t)^2+2.6502912*sin(t)^2*cos(t)+0.6625728*sin(t)^2*cos(t)^2+5.2807680*cos(t)*x-5.3005824*sin(t)*cos(t)+0.9443529*x^4-3.181248*x^3+147.0230064*cos(t)^2+67.10736960*sin(t)^2+23.693094*x^2+6.8757264*cos(t)^4-61.58338560*sin(t)^3-11.047968*sin(t)+15.3958464*sin(t)^4+148.045536*cos(t)+55.0058112*cos(t)^3-5.2807680*sin(t)*cos(t)*x+38.16801;

g := 0.9443529*y^4-42.29607960*y^2-3.437942400*y^3-1.3314096*sin(t)^2*y^2+2.662819200*sin(t)*y^2-17.5101696*sin(t)^3*y+52.530508800*sin(t)^2*y+25.45344*cos(t)^2*y+3.437942400*sin(t)*y^3-11.85948*cos(t)^2*y^2-47.43792*cos(t)*y^2-138.4682688*sin(t)*y+101.81376*cos(t)*y-101.81376*sin(t)*cos(t)*y-25.45344*sin(t)*cos(t)^2*y+632.4697764-91.635840*sin(t)*cos(t)^2+183.27168*sin(t)^2*cos(t)+45.8179200*sin(t)^2*cos(t)^2-366.5433600*sin(t)*cos(t)+942.16392*cos(t)^2+247.7287584*sin(t)^2+39.20400*cos(t)^4-61.5833856*sin(t)^3-372.2907456*sin(t)+15.3958464*sin(t)^4+1259.59968*cos(t)+313.63200*cos(t)^3+103.4479296*y; 
h := z-2*sin(t)+2;
eqs := {f, g, h};
eqs2 := subs(x = x(t), y = y(t), z = z(t), eqs);
x0 := -0.896661124832438;
y0 := -1.7877356703982348;
z0 := -2;
ics := {x(Pi) = x0, y(Pi) = y0, z(Pi) = z0};
odes := diff~(eqs2, t);
res := dsolve(odes union ics), numeric);
plots:-odeplot(res, [x(t), y(t), z(t)], Pi .. 2*Pi); p1 := %;

A good graph is produced of maybe 10% of the answer.  Then:

Warning, cannot evaluate the solution further right of 3.4774922, probably a singularity

For example, 

f(x,t) = t^2 x^2 + t x + 2x - 1, g(y,t) = t^2 y^3 + t y^2 + 2y - 1, h(z,t) = 2t^2 z^3 + t z + 3z - t^2, 0 < t < 4.

I have an initial point on the curve corresponding to t=0.  The answer is a curve in space, or maybe several curves.  The real example that I care about is much more complex than this and has trig functions of t.