Re: [Groff] Problems with arcs and angles

[John Gardner]

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 <address@hidden> wrote:

> Hi John,
>
> > <https://cdn.rawgit.com/Alhadis/language-roff/
> a7c07744a9d44adf32a546646f7a8d57d52e6e58/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
>
>