Looks like there's a mistake in groff_out(5). It describes its arc-drawing command as:
*Draw arc from current position to (h1, v1)+(h2, v2) with center at (h1, > v1);* However, gropdf.pl tells a different story. To quote line #2791: <http://git.savannah.gnu.org/cgit/groff.git/tree/src/devices/gropdf/gropdf.pl?id=453a8aa7c8f8dd0c78160466301f81be8c40df2e#n2791> *Documentation is wrong. Groff does not use Dh1,Dv1 as centre of the > circle!* I noticed something odd when drawing an arc which wasn't an evenly-sized quarter-circle. For instance, the following Pic code: .PS > move down > A: box > arc at A.n > line right > .PE ... generates the following output: > … V84000 H99000 > Da 0 -36000 18000 54000 Which looks like this in gropdf and my own canvas code, respectively: [image: Inline images 3] I've marked the start (A) and end (Z) points in my output on the right. The circle at the top denotes where the centre-point is. Compare that with an ordinary arc (e.g., replace `arc at A.n` in the previous Pic code with `arc` instead): [image: Inline images 5] gropdf is also drawing each arc in 4 segments, instead of 1. The reasons why are unclear to me. Otherwise, I've succeeded in drawing every other (quartercircle-sized) arc using the following code: centreX = h1 + startX > centreY = v1 + startY > endX = centreX + h2 > endY = centreY + v2 > radius = sqrt(pow(endX - centreX, 2) + pow(endY - centreY, 2)) > startAngle = atan2(startY - centreY, startX - centreX) > endAngle = atan2(endY - centreY, endX - centreX) On 30 April 2017 at 20:18, John Gardner <gardnerjo...@gmail.com> wrote: > Sorry for the delayed response. Some pull-requests needed tending to > before I had a chance to go through all this. > > First, thank you all so much for your help and patience! It truly means a > lot. I've not yet got arcs drawing correctly, but I feel I'm coming close. > Branden's pointers on trigonometry cleared some cobwebs that hadn't been > lifted since high-school trigonometry lessons, making all this slightly > easier to follow. > > Turns out the HTML5 specification > <https://html.spec.whatwg.org/multipage/scripting.html#dom-context-2d-arc> > itself > offers a much clearer description of arc-drawing than the docs I linked to > earlier: > > >> *Consider an ellipse that has its origin at (x, y), that has a major-axis >> radius radiusX and a minor-axis radius radiusY, and that is rotated about >> its origin such that its semi-major axis is inclined rotation radians >> clockwise from the x-axis.* > > >> *If anticlockwise is false and endAngle-startAngle is equal to or greater >> than 2π, or, if anticlockwise is true and startAngle-endAngle is equal to >> or greater than 2π, then the arc is the whole circumference of this >> ellipse, and the point at startAngle along this circle's circumference, >> measured in radians clockwise from the ellipse's semi-major axis, acts as >> both the start point and the end point.* > > > > *Otherwise, the points at startAngle and endAngle along this circle's >> circumference, measured in radians clockwise from the ellipse's semi-major >> axis, are the start and end points respectively, and the arc is the path >> along the circumference of this ellipse from the start point to the end >> point, going anti-clockwise if anticlockwise is true, and clockwise >> otherwise. Since the points are on the ellipse, as opposed to being simply >> angles from zero, the arc can never cover an angle greater than 2π radians.* > > > It also explains how arcs and ellipses are drawn using the same > coordinates: > > *The arc() method is equivalent to the ellipse() method in the case where >> the two radii are equal. When the arc() method is invoked, it must act as >> if the ellipse() method had been invoked with the radiusX and radiusY >> arguments set to the value of the radius argument, the rotation argument >> set to zero, and the other arguments set to the same values as their >> identically named arguments on the arc() method.* > > > Going forward, I *think* I can work this out. > > I hope B-spline drawing doesn't turn out to be as difficult. =( > > Thank you all again! > > > > On 29 April 2017 at 01:38, Ralph Corderoy <ra...@inputplus.co.uk> wrote: > >> Hi John, >> >> > <https://cdn.rawgit.com/Alhadis/language-roff/a7c07744a9d44a >> df32a546646f7a8d57d52e6e58/preview-tty.html> >> > <https://cloud.githubusercontent.com/assets/2346707/ >> 25514878/732e09ea-2c23-11e7-97ea-9674d3845dd1.png> >> >> Gorgeous. :-) >> >> > Groff's output gives me these coordinates to go by: >> > >> > - startX, startY - Coordinates of the arc's starting point >> > - centreX, centreY - Coordinates of the arc's centre >> > - endX, endY - Coordinates of the arc's terminal point >> >> I'm assuming centreX and centreY are zero as it's just a simple >> translation to get there. And for the moment that startX and startY are >> in the x>0, y>0 quadrant. This gives a right-angled triangle (0, 0), >> (startX, 0), (startX, startY). >> >> > But the canvas arc method I'm working with requires all of these: >> > >> > - x - The x coordinate of the arc's centre. >> > - y - The y coordinate of the arc's centre. >> >> 0, 0. >> >> > - radius - The arc's radius. >> >> sqrt(startX**2, startY**2) as pic's arcs are always circular. >> >> > - startAngle - The angle at which the arc starts, measured clockwise >> > from the positive x axis and expressed in radians. >> >> That's the triangle's angle at its origin corner. That's the arc sine >> of the opposite edge's length, startY. But this assumes the radius, >> i.e. triangle's hypotenuse, is 1. When it's not, it's >> asin(startY/radius). >> >> This bit of Python might help. >> >> >>> from math import * >> >>> def d(r): return r / pi * 180 >> ... >> >>> d(asin(0.001)) >> 0.05729578906238321 >> >>> d(asin(0.999)) >> 87.43744126687686 >> >>> >> >>> d(asin(1)) >> 90.0 >> >>> d(asin(-1)) >> -90.0 >> >>> >> >>> d(asin(-0.001)) >> -0.05729578906238321 >> >>> d(asin(-0.999)) >> -87.43744126687686 >> >>> >> >>> d(asin(0.5)) >> 30.000000000000004 >> >>> d(asin(sqrt(0.75))) >> 60.0 >> >> You can see asin copes with [1, -1], giving [90, -90] in degrees. >> You'll need to check the signs of start{X,Y}, making π/2 adjustments to >> the radians you obtain to handle the other quadrants. >> >> Hope that helps. It's been many decades since I needed this. :-) >> >> -- >> Cheers, Ralph. >> https://plus.google.com/+RalphCorderoy >> >> >
