Ah, lovely! Who knew there would be so many different solutions? I'm
going to use Nicola's solution on this occasion, as it fits my thought
process best -- but thanks to Troy as your solutions showed me some
other things I didn't know.
All the best,
James
On Fri, Feb 26, 2010 at 9:42 PM,
In article
771da05a1002251718l55669a0co770b5a78bed84...@mail.gmail.com,
James Fisher jameshfis...@gmail.com wrote:
This isn't specifically a ConTeXt question, but via it I've run into a
seemingly simple problem in METAPOST that I just can't solve. I'm
trying to draw a parallelogram by
This isn't specifically a ConTeXt question, but via it I've run into a
seemingly simple problem in METAPOST that I just can't solve. I'm
trying to draw a parallelogram by specifying: (1) the length of sides
parallel to the x-axis; (2) the total height of the figure; (3) one of
the interior
James,
Try this:
z0 = origin;
z1 = (5,0);
z3 = 10*dir(87);
z2 = z3-z0+z1;
Troy Henderson
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James,
I apologize, but the previous information that I gave you was wrong.
Try this instead:
z0 = origin;
z1 = (5,0);
BL:=87;
r:=10/sind(BL);
z3 = r*dir(BL);
z2 = z3-z0+z1;
Troy
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Aha! That certainly works. I suspected I would have to fall back on
low-level trig :). Many thanks!
On Fri, Feb 26, 2010 at 1:44 AM, Troy Henderson thend...@gmail.com wrote:
James,
I apologize, but the previous information that I gave you was wrong.
Try this instead:
z0 = origin;
z1 =
James,
How about this.
BL:=87;
z0=origin;
z1=(5,0);
z4=dir(BL); % Intermediate point
z3=10/y4*z4;
z2=z3+z1;
or alternatively without having to define z4
BL:=87;
z0=origin;
z1=(5,0);
z3=10/(ypart dir(BL))*dir(BL);
z2=z3+z1;
