[Math] How to find the start and end point of an Arc? Using center xy, radius and start/end angle values.

geometry

I'm defining an arc by calling a function like this: arc(x, y, radius*2, radius*2, start, end);

x: center x.
y: center y.
radius*2: width.
radius*2: height.
start: start angle in radians or degrees.
end: end angle in radians or degrees.

And now I need to "close" the arc by drawing two lines, both from the center of the arc, to the start and end point of the arc.

How do I find the start and end points?

Best Answer

Given start angles and end angles you get

sX = offsetX + radius * cos(startAngle * PI/180)
sY = offsetY + radius * sin(startAngle * PI/180)
eX = offsetX + radius * cos(endAngle * PI/180)
eY = offsetY + radius * sin(endAngle * PI/180)

Multiplying an angle by PI/180 converts degrees to radians.

Related Solutions

Geometry – How to Find Center of an Arc Given Start Point, End Point, Radius, and Arc Direction

Don't write down too many equations to be solved, but produce the desired center ${\bf c}=(a,b)$ in a forward movement instead. Let ${\bf z}_i=(x_i,y_i)$ $\ (i=0,1)$ be the two given points, put $\epsilon:=1$ if the arc should go from ${\bf z}_0$ to ${\bf z}_1$ counterclockwise, and put $\epsilon:=-1$ otherwise.

Next, let $d:=|{\bf z_1}-{\bf z_0}|=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$ be the distance and ${\bf m}:=\bigl({x_0+x_1 \over2}, {y_0+y_1\over2}\bigr)$ be the midpoint of ${\bf z_0}$ and ${\bf z_1}$. Then

$${\bf n}:=(u,v):=\Bigl({x_1-x_0\over d},{y_1-y_0\over d}\Bigr)$$

is the unit normal in the direction ${\bf z_1}-{\bf z}_0$, and ${\bf n}^*:=(-v,u)$ is the unit vector you get by rotating ${\bf n}$ counterclockwise by $90^\circ$.

Given $r>0$ the center ${\bf c}$ has a distance $h:=\sqrt{r^2 -d^2/4}$ from ${\bf m}$, and the given $\epsilon$ together with ${\bf n}^*$ tell us in which direction we should go. In vectorial notation the center is given by

$${\bf c}\ =\ {\bf m}+\epsilon\ h\ {\bf n}^*\ ,$$

so that coordinate-wise we get

$$a={x_0+x_1 \over2}-\epsilon\ h\ v, \qquad b={y_0+y_1 \over2}+\epsilon\ h\ u\ .$$

[Math] Formula for calculating the center of an arc

Had an epiphany just as I hit submit, would this work?

$cP \{ Px +$ $\cos (eA - sA) \over 2$ $\times r, Py +$ $-\sin (eA - sA) \over 2$ $\times r\}$

SOLVED:

Using the piece-wise function:

$ cP( Px +$ $\cos($ $sA + eA \over 2$ $ + n) \times r, Py +$ $-\sin($ $sA + eA \over 2$ $ + n) \times r) = \begin{cases} n = 0, & \text{if }eA - sA \text{ is } >= 0 \\ n = \pi, & \text{if }eA - sA \text{ is } < 0 \end{cases} $

For you computer science-y types here's some pseudo-code:

double e = GetEndAngle();
double s = GetStartAngle();
double d = e - s;

double x = 0.0;
double y = 0.0;
double offset = 0.0;
if(d < 0.0) {
    offset = PI;
}
x = (GetPosition().GetX() + std::cos(((s + e) / 2.0) + offset) * GetRadius());
y = (GetPosition().GetY() + -std::sin(((s + e) / 2.0) + offset) * GetRadius());

Best Answer

Related Solutions

Geometry – How to Find Center of an Arc Given Start Point, End Point, Radius, and Arc Direction

[Math] Formula for calculating the center of an arc

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