lray - horiz*v - verti*u = whatever*(LightSource -
R), a system of three linear equations for three unknowns:
horiz, verti, whatever?
u and explicitly multiply all the coordinates by
u. Is there a better way to do this in Asymptote?
interp(a,b,c);
See the https://asymptote.sourceforge.io/gallery/dimension.asy example for an example of how deferred drawing is used to accomodate both user and true-size (PostScript) coordinates.
(i) use the default size(0,0) for the entire picture and do all of the scaling by hand, just like in MetaPost;
(ii) draw to a separate picture pic and add(pic.fit());
(iii) use frames.
:: is a macro for tension atleast 1:
size(100);
pair z0=(0,0);
pair z1=(1,0.25);
pair z2=(2,0);
draw(z0{up}::z1{right}::z2{down});
currentpen=red;
whatever unknown. Such a facility could certainly be added (perhaps using the notation
?= since = means assignment). However, the most common uses of whatever in MetaPost are covered by functions like extension in math.asy: pair extension(pair P, pair Q, pair p, pair q);this returns the intersection point of the extensions of the line segments
PQ and pq. We find using routines like extension more explicit and less confusing to new users. But we
could be persuaded to add something similar if someone can justify the
need. In the meantime, one can always use the explicit built-in linear
solver solve (see https://asymptote.sourceforge.io/doc/solve.html), which uses LU decomposition.
lray - horiz*v - verti*u = whatever*(LightSource -
R), a system of three linear equations for three unknowns:
horiz, verti, whatever?
horiz*v+verti*u spans a plane, you could use real intersect(vector P, vector Q, vector n, vector Z);to find the intersection time for the line
lray-whatever*(LightSource - R) and then extract the three desired values from there. (You'll still
need to use the built-in explicit linear solver to solve a 2x2 system to get
horiz and verti.)
u and explicitly multiply all the coordinates by
u. Is there a better way to do this in Asymptote?
1cm: unitsize(1cm); draw(unitsquare);One can also specify different x and y unit sizes:
unitsize(x=1cm,y=2cm); draw(unitsquare);Another way is to draw your fixed size object to a frame and add it to currentpicture like this:
path p=(0,0)--(1,0); frame object; draw(object,scale(100)*p); add(object); add(object,(0,-10));To understand the difference between frames and pictures, try this:
size(300,300); path p=(0,0)--(1,0); picture object; draw(object,scale(100)*p); add(object); add(object,(0,-10)); // Adds truesize object to currentpicture
fill((0,0)--(100,100)--(200,0)--cycle);
pair center(picture pic=currentpicture) {return 0.5*(pic.min()+pic.max());}
real height=100;
real width=100;
pair delta=0.5(width,height);
pair c=center();
clip(box(c-delta,c+delta));
However, drawing in PostScript coordinates is often inconvenient.
Here's the Asymptote way of doing the same thing, using deferred drawing:
size(200,100);
fill((0,0)--(1,1)--(2,0)--cycle);
void clip(picture pic=currentpicture, real width, real height)
{
pic.clip(new void (frame f, transform) {
pair center=0.5(min(f)+max(f));
pair delta=0.5(width,height);
clip(f,box(center-delta,center+delta));
});
}
clip(100,100);
See also the discussion of tilings in the documentation:
https://asymptote.sourceforge.io/doc/Pens.html. Asymptote - 2024-10-07