Thanks ... And I did some tests this morning:
Plotted the *bisector (3rd, 1st quadrant) and *sin using
pd 'xrange _1 1'
pd 'yrange _1.1 1.1'
and
pd 'aspect xx' with these values: xx= 0.866 | 1.0 | 1.155
Results:
aspect: 0.866 | 1.0 | 1.155 'xrange _1 1'
-------------------------------------------------------------
vertical/horizontal unit length: 0.81 | 0.92 | 1.05
height/width of grid frame: 0.87 | 1.0 | 1.15
'aspect 1.0' does not produce an angle of 45 deg of tangent versus
horizontal at the root as expected.
Try it with a different xrange and have a look how distorted it becomes:
aspect: 0.866 | 1.0 | 1.155 'xrange _2 2'
-------------------------------------------------------------
vertical/horizontal unit length: 1.57 | 1.83 | 2.13
height/width of grid frame: 0.87 | 1.0 | 1.15
This clearly shows (at least to me) that the aspect value is
definitely governing the rectangular grid frame, and somehow (on the
way) influencing the units ratio.
I think that's a bug (as it should at least be the other way around).
Therefore I still plead for a separate pf option of e.g. 'compression
ratio' or 'units ratio' if possible.
-M
At 2022-03-21 21:43, you wrote:
On Mon, 21 Mar 2022, Martin Kreuzer wrote:
[I'm writing under the influence of a (so long) mild attack of big
C, so beg your pardon if reads a bit incoherently in places.]
Sad to hear that. Hope you get better soon.
When plotting a unit circle (using pd commands and options) I got
disappointed again as it looked like a ballon from last weeks birthday party.
When I run the following:
plot r.2p1 * 128 %~ i.128
I get something that looks stretched out and disappointing. But the
aspect ratio follows the window size. If, instead, I run this:
'aspect 1' plot r.2p1 * 128 %~ i.129
The aspect ratio remains fixed regardless of the window size, and I
get something that looks like a circle.
On to the issue of the hexagon:
pd'reset'
pd'aspect 1'
pd r.2p1*6%~i.7
pd'show'
If I do this, I get a slightly distorted hexagon. However, if I
inscribe it in a circle, it looks correct:
pd r.2p1*128%~i.129
pd'show'
So I think it is a bug in plot. The aspect ratio is being computed
relative to the max/min bounds of the function being plotted, not in
absolute terms.
So where does 0.866 come from? We can find the max/min bounds of
the points in our hexagon, and thereby find the (incorrect) aspect
ratio used by plot:
+. p=. r.2p1*6%~i.6
1 0
0.5 0.866025
_0.5 0.866025
_1 1.22465e_16
_0.5 _0.866025
0.5 _0.866025
(>./ - <./)+.p
2 1.73205
%/(>./ - <./)+.p
1.1547
plot is using ~1.1547 as its aspect ratio, so we need the reciprocal,
which is ~0.866. 0.866 is also visible in p, as it really comes from
1 o. o. %3. And yes, it really is -:%:3.
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