OpenEuphoria

Re: submarine commander project

• Posted by David Money <dmoney290 at ao?.c?m> Dec 10, 2007
• 509 views

don cole wrote:
> 
> Can any of you geometry people out there help me?
> 
> }}}
<eucode>
> function p2xy(atom r, atom a)   -- polar to xy
>    return {-r*sin(a),-r*cos(a)}
> end function
> </eucode>
{{{

> all measurements in pixels.
> 
> r=distance from center
> x=clock in degrees 3600 = center to top
> center=vector {400,273 } 
> fin=cen+p2xy(r,x/1800*PI)  --the outmost position of the radius
> drawLine(Pix,cen[1],cen[2],floor(fin[1]),floor(fin[2])) --one line of the
> sweep.
>   
> 
> This formula gives me the position of the final dot of the radius. Works good.
> 
> What I would like to do is the opposite  to find the clock degrees and last
> radius position 
> given the center and the vector.
> 
> For example let's say center={400,273}.
> 
> A dot is out there at {210,120}.
> 
> What would be the clock degrees to that dot?
> 3600=12 o'clock
> 1800=6 o'clock
> 2700=9 o'clock etc...
> And how many pixels from the center to that dot?
> 
> Don Cole

I would include the conversions in the functions, so you could use your
0~3600 angle directly.  Something like this:

<code>
constant PI=3.1415926535893238 --or include misc.e

function p2xy(atom r,atom sec)
--r is distance, sec is angle (seconds after hour, measured from 12 o'clock)
    atom a
    a=sec*PI/1800 --convert to radians
    return {r*sin(a),-r*cos(a)}
end function

function vector(sequence pt1,sequence pt2)
--distance and angle from pt1 to pt2
    integer dx,dy
    atom r,a
    dx=floor(pt2[1]-pt1[1])
    dy=floor(pt2[2]-pt1[2])
    r=sqrt(dx*dx+dy*dy)
    if r=0 then
	return {0,0}
    end if
    if dy<0 then
	if dx<0 then
	    a=2*PI-arctan(dx/dy)
	else
	    a=-arctan(dx/dy)
	end if
    elsif dy>0 then
	a=PI-arctan(dx/dy)
    else --dy=0 (one point above the other); can't use arctan formula
	if dx<0 then
	    a=3*PI/2
	else
	    a=PI/2
	end if
    end if
    return {r,1800*a/PI} --convert radians to decidegrees
end function
</code>

  The arctan function just works for -pi/2 to pi/2, so you have to figure
out which quadrant you are in, and offset it.  Also, you have to avoid
dividing by zero.  If r=0, the points are the same; the function will return
0 for the angle, too, but it's really undefined.
  You had two minus signs in your function--was that a typo?  That would work
for conventional angle measurements (counterclockwise), but it would make
your clock run backwards.
--David