Вот код, который я только что построил, чтобы решить эту проблему, приближаясь к нему с точки зрения центра круга, с объявлениями и примерами значений:
CGPoint arcCenter = CGPointMake(30,20);
float arcLengthRad = M_PI_4; // Whatever, the full span of the arc in radians
float radius = 10;
float arcCenterRad = M_PI_2; // the angle of the center of the arc, in radians
float arcP1hyp = 1/cos(arcLengthRad/2) * radius;
float arcP1x = arcCenter.x + cosf(arcCenterRad)*arcP1hyp;
float arcP1y = arcCenter.y + sinf(arcCenterRad)*arcP1hyp;
float arcP2tx = arcCenter.x + cosf(arcCenterRad+(arcLengthRad/2))*radius;
float arcP2ty = arcCenter.y + sinf(arcCenterRad+(arcLengthRad/2))*radius;
float arcP2x = (arcP1x - arcP2tx)*-1 + arcP2tx;
float arcP2y = (arcP1y - arcP2ty)*-1 + arcP2ty;
CGContextAddArcToPoint(context,
arcP1x,
arcP1y,
arcP2x,
arcP2y,
radius);
Таким образом, приведенный выше код должен создавать небольшую 45-градуснуюугловая дуга в верхней части круга.
Отредактировано: В ответ на полученный комментарий приведенный выше супер-краткий код показан ниже с комментариями и завернут вметод (плюс небольшая корректировка для вычисления arcP2)
/*
EOTContext:addArcWithCenter:arcLength:radius:arcMiddlePointAngle:
Use this method for building a circle with breaks at certain points,
for example to use other CGContext methods to draw notches in the
circle, or protruding points like gear teeth.
This method builds up the values to use in CGContextAddArcToPoint(),
which are the x and y coordinates of two points. First added to
the current point in context, form two lines that are the tangents of
the entry and exit angles of the arc.
This method's arguments define the length of the arc in radians, and
the position of start and end using the angle centerpoint of the arc.
This is useful when drawing a certain defined amount of gear teeth,
rotating around the circle.
It is beyond this method's scope to maintain or calculate the
centerpoint relative to an arbitrary current point in the context, because this
is primarily used for drawing a gear/notch circle.
*/
-(void)EOTContext:(CGContext*)context
addArcWithCenter:(CGPoint)arcCenter
arcLength:(CGFloat)arcLengthRad
radius:(CGFloat)radius
arcMiddlePointAngle:(CGFloat)arcCenterRad {
/*
Calculate the hypotenuse of the larger, outer circle where the
points of the tangent lines would rest upon (imagine wrapping
the drawn circle in a bounding regular convex polygon of tangent
lines, then wrap that polygon in an outer circle)
*/
float arcP1hyp = 1/cos(arcLengthRad/2) * radius;
// Build first tangent point
CGPoint arcP1 = (CGPoint){
arcCenter.x + cosf(arcCenterRad)*arcP1hyp,
arcCenter.y + sinf(arcCenterRad)*arcP1hyp
};
// Build the final endpoint of the arc
CGPoint arcP2final = (CGPoint){
arcCenter.x + cosf(arcCenterRad+(arcLengthRad/2))*radius,
arcCenter.y + sinf(arcCenterRad+(arcLengthRad/2))*radius
};
// Build second tangent point using the first tangent point and the final point of the arc.
// This point is resting on the bounding outer circle like arcP1 is.
// This would also work using the final point itself, using the simple assignment of arcP2 = arcP2final;
// or of course simply omitting arcP2 altogether.
CGPoint arcP2 = (CGPoint){
(arcP2final.x - arcP1.x) + arcP2final.x,
(arcP2final.y - arcP1.y) + arcP2final.y
};
// The following adds an arc of a circle to the current path, using a radius and tangent points.
CGContextAddArcToPoint(context,
arcP1.x,
arcP1.y,
arcP2.x,
arcP2.y,
radius);
}