Singular

The fastes way I know is:
- create a ring with elimiantion ordering for everything but x,v:
- use modStd to compute a Groebner basis
[code]ring r=0,(c1,c2,c3,c4,c5,x,v),(dp(5),dp);
ideal I=....;
LIB"modstd.lib";
ideal J=modStd(I);
[/code]
The first polynomial of the result is in x,v only.
But is is not trivial: computing the probably right solution took several hours
and checking it some days (and 100 GB RAM) on a fast machine.

The result is one polynomial, 20 kB

Hi,
I'm trying to obtain the planar projection of a curve by eliminating 5 variables in a system with 7 unknowns and 6 equations:

[code]
ring r=0,(x,v,c1,c2,c3,c4,c5),lp;
ideal I=(-16*c1*c2*c3*c4*c5+16*v,16*c1*c2*c3*c4-16*(-c1*c2*c3-(c1*c2-(-c1-c2)*c3)*c4)*c5+40*v,-16*c1*c2*c3-16*(c1*c2-(-c1-c2)*c3)*c4-16*(c1*c2-(-c1-c2)*c3-(-c1-c2-c3)*c4)*c5+25*v,16*c1*c2-16*(-c1-c2)*c3-16*(-c1-c2-c3)*c4-16*(-c1-c2-c3-c4)*c5-25,-16*c1-16*c2-16*c3-16*c4-16*c5-40,(c1-c3)*(c2-c4)-x*(c1-c4)*(c2-c3));
eliminate(I,c1*c2*c3*c4*c5);
[/code]

but it seems that I'm being far too naive. I know that the answer should have degree 30 in v and 16 in x, so it should be manageable; and I also remember that I had computed it in 2013, I just forgot how :(

Note that all equations are symmetric in c1...c5, except for the last one which says that x is the cross ratio of c1...c4.

Many thanks in advance! Laurent