> Hi everybody,
> I am a programmer, but unfortunately I am not very good in Geometry
> and I have the following problem:
> I want to add a small circle at the end of a line. Whenever I drag
> the line, with my mouse or rotate it around, I want the circle to
> get updated appropriately and to be at the end of that line.
> Example:
> Start point End point
> and the circle
> |
> | |
> V V
> -------------------------------------o
> (something like that)
> I am writing that code in Java 1.1 ( I hope noone gets offended) and
> it's almost ready, except for that little smart calculation part,
> where I have to find out where exactly to draw the circle, based on
> the coordinates of the edge's start and end points, which are passed
> to me, everytime the edge (line) is moved or rotated. To make things
> even more interesting, in Java the function for drawing circles has
> the following prototype:
> drawOval(int x, int y, int width, int height)
> where:
> these coordinates specify a rectangle, in which my circle (ellipse)
> should fit.
I presume that the "little circle" (lc) should be concentric
to the endpoint (ep). Let's suppose you want the circle's
size to be "r" (think radius)
I don't see any geometry here. I think that all you have to do
is subtract (one offset from each of the x and the y of the
endpoint) and multiply (simply double the offset) to get the
lengths of the bounding box (square).
Here's a picture that might help:
x----------+
| /^^^^\ |
| / \ |
|( * )|
| \ / |
| \____/ |
+----------+
The "*" is the endpoint of your line, and the center
of my incredibly poor circle.
I'm guessing (with *no* knowledge of Java programming) that
the x would be the location of the point specified by the
bouding box (relative to the center of our circle (ep)).
If that's the case you want to subtract a constant about from
ep[x] and from ep[y] (the x and y co-ordinates of ep). Let's
call these offsets dx and dy respectively. Thus the upper
left of our bounding box is at the co-ordinates (dx, dy) for
any ep[x,y].
That's more easily expressed like:
corner(x) = ep(x) - offset
corner(y) = ep(y) - offset
If your function call calls for the co-ordinates of the
lower right corner you'd change both signs to "+", if
it requires the upper right you'd add the x and subtract
the y (assuming that you're using x for horizontal and
y for vertical and that these co-ordinates use upper left
as the origin, like every other computer video graphics
co-ordinate API I've ever seen), etc.
(note -- you want a square so the offset is constant --
i.e. you use the same value for both (unless you have
to adjust for some uneven aspect ratio).
So, now all we have to do is figure out the length of
the edges of the square. This will be a simple constant
(relative to the offset). We just double our offset and
use it for the lengths of both sites of our bounding
box. We know that the * in my diagram will be at the
center of the square (it is a by product of our method
of construction). Our offset is co-incidentally the
radius of our circle (unless this "drawOval" function
does something bizarre).
I'm presuming that the parameter to your drawOval
form a bounding box (a form into which the a
maximally sized circle will be inscribed). It
would be possible to create a function to "exscribe"
a circle *around* a rectangle --- but that would be
much harder to use by programmers since you'd have to
use the Pythagorean theorem to calculate the radius and
to figure out any interections betwee the circle that that
generated and things like the edge of the screen).
> I am really confused about how to choose that x,y point, that I
> should pass to the above function, based on the start,end point
> coordinates of the line.
I have no idea what the co-ordinates of the start point
have to do with this, unless you want to radius of the
circle to vary with the overall length of the line, or
something like that.
This is a pretty basic question, and your Java documentation
should include some examples and diagrams to explain this.
I still can't see why you expect there to be some mysterious
"geometry" in this --- unless it is simply an utter failure
to describe the problem (or to understand it on my part).
> Thank you. Please Help.
> Sincerely
> mich
> P.S.
> How can I join the newsgroup?
What newsgroup? This is a mailing list.
--
Jim Dennis (800) 938-4078 [EMAIL PROTECTED]
Proprietor, Starshine Technical Services: http://www.starshine.org
