Am 23.10.2011 16:12, schrieb Paul Menzel:
Am Freitag, den 21.10.2011, 18:08 +0200 schrieb Peter Rolf:
Am 21.10.2011 17:26, schrieb Paul Menzel:
Am Freitag, den 21.10.2011, 17:02 +0200 schrieb Peter Rolf:
I agree, this is confusing on the first sight. But scaling is not meant as 'scaling to' a dimension. In fact is is just a simple multiplication. The reason why it seems to work this way with 'fullsquare' and such predefined paths is, that they have a 'neutral' size/scale (bounding box size of filled path is (1pt,1pt)).
So how can I find out what the dimension of the path of a function is? Not scaling it, it also looked pretty small, so I am guessing (1pt,1pt).
I guess in this case its size is (10,log(10)) + pen size when drawn.
That was it kind of. Probably the most of my confusion resulting from the use of the inverse function.
So all in all the width is 1 – the x-coordinates seem to be scaled to the width of 1 – and the height is 10, as ten is the highest value. Therefore the width has to be multiplied with 10cm to get it to 10cm and since the height it 10 already multiplying it with 2cm gets it to 20cm.
I see. I must admit that I haven't looked deeper into function (never used it). If I want to test the bounding box of a graphic, I normally use a small debugging macro (last code line of nearly all of my graphics). It simply draws a small red frame at the bounding box borders. You can use the 'Measuring Tool' in Acrobat (also Reader?) or even better the 'Object Inspector' to get the exact size. minlinewidth:= .143pt; def wirepen= pencircle scaled minlinewidth enddef ; def drawbb = drawoptions() ; interim linejoin:= mitered; draw (llcorner currentpicture) shifted ( .5minlinewidth, .5minlinewidth) -- (lrcorner currentpicture) shifted (-.5minlinewidth, .5minlinewidth) -- (urcorner currentpicture) shifted (-.5minlinewidth,-.5minlinewidth) -- (ulcorner currentpicture) shifted ( .5minlinewidth,-.5minlinewidth) -- cycle withpen wirepen withcolor red; % draw boundingbox currentpicture withpen wirepen withcolor red ; % was not neutral! enddef ;
Multiplying such a path with (x,y) gives an object with size (1*x,1*y). In general: if the bounding box of an object has the size (a,b) and you scale it with (x,y), the resulting object has a size of (ax,by). That's all the magic.
but if you use numbers with a unit than it should not be multiplied but expanded to that value, should not it? Otherwise I am unsure how multiplication works with a unit.
1pt is the base unit in MP (used if no dimension is given; probably stored as 65536sp (scaled point) units).
In the manual I only found bp for »big point«, which equal PostScript points equaling 1/72 of an inch.
It can't be helped, I always mix this up. :-D It's 'bp' for MP and for Adobe products, and 'pt' for TeX. Ask me in a few weeks and I can't give the correct answer.
Now if you use pure numerics for scale, such as 'xscaled 2', this is interpreted as '2pt' (or 2*65536sp). If you use any dimension, it is also converted into scaled points. All the same for MP.
If I'm right this should all be the same (untested)
xscaled 2 xscaled 2pt
This gives the same result.
xscaled (2*65536sp)
I could not test the above because the numbers got too big and MetaPost complained.
My fault. In the meantime I have checked it and 'sp' is not a predefined constant in MP (couldn't find it in the manual). Also never used it in MP.
1 = 1pt = 65536sp
[…]
Thanks,
Paul
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