计算饼图中的旋转角度

Question

我想围绕其中心点旋转图像。我面临的问题是我需要获取在触摸移动事件中计算的角度（我不想使用多点触摸）。我当前使用下面的代码

- (void)touchesMoved:(NSSet *)touches withEvent:(UIEvent *)event{

NSArray *allTouches = [touches allObjects]; 
gestureStartPoint = gestureMovedPoint;//i am getting the gestureStartPoint on touch began event

gestureMovedPoint = [[allTouches objectAtIndex:0] locationInView:[self superview]];

NSLog(@"gestureMovedPoint = %@",NSStringFromCGPoint(gestureMovedPoint));

}

CGFloat previousAngle = [self angleBetweenPoints:gestureStartPoint second11:gestureMovedPoint]; // atan2(gestureMovedPoint.y - gestureStartPoint.y, gestureMovedPoint.x - gestureStartPoint.x) * 180 / M_PI;
CGFloat currentAngle =atan2(self.transform.b, self.transform.a);//atan2(gestureMovedPoint.y - gestureStartPoint.y,gestureMovedPoint.x - gestureStartPoint.x) * 180 / M_PI;



CGFloat angleToRotate = currentAngle - previousAngle;



float xpoint = (((atan2((gestureMovedPoint.x - gestureStartPoint.x) , (gestureMovedPoint.y - gestureStartPoint.y)))*180)/M_PI);

    CGAffineTransform transform = CGAffineTransformMakeRotation(angleToRotate-100);



self.transform = transform;

请帮助我找到解决方案，因为我被困在这里，需要在截止日期后尽快完成此申请。 提前致谢

Answer 1

很高兴我记得三角函数

-(void)degreesToRotateObjectWithPosition:(CGPoint)objPos andTouchPoint:(CGPoint)touchPoint{

   float dX = touchPoint.x-objPos.x;        // distance along X
   float dY = touchPoint.y-objPos.y;        // distance along Y
   float radians = atan2(dY, dX);          // tan = opp / adj

   //Now we have to convert radians to degrees:
   float degrees = radians*M_PI/360;

   return degrees;
}

一旦你有了好的方法，只需在触摸事件方法中执行此操作即可。 （我忘了它叫什么...）

CGAffineTransform current = view.transform;

[view setTransform:CGAffineTransformRotate(current,
                 [self degreesTorotateObjectWithPosition:view.frame.origin 
                                  andTouchPoint:[touch locationInView:parentView]]

//Note: parentView = The view that your object to rotate is sitting in.

这几乎是您需要的所有代码。数学是正确的，但我不确定 setTransform 的东西。我在学校用浏览器写这个。从这里你应该能够弄清楚。

祝你好运，

奥鲁姆·阿奎拉

Answer 2

必须思考这一点。但我更喜欢通过两次触摸来旋转视图。会简单很多。

Answer 3

我确实在如何获得触摸驱动的旋转方面遇到了一些困难，更重要的是因为我希望 100% 理解我正在使用的代码。因此，在多次失败的尝试之后，我得出了这样的结论：

- (CGFloat) pointToAngleFromCenter: (CGPoint) point {
    // transform point to a self.center'ed origin based coordinate system
    point.x = point.x - self.center.x ;
    // ditto for y, but compensate for y going downwards to y going upwards
    point.y = self.center.y - point.y ;
    return ::atan2(point.y, point.x) ;
}

如果有人对这种方法有更好的名称，我会洗耳恭听。

它的作用是获取父视图坐标中的一个点，相对于视图中心（位于父视图坐标中）重新映射它，并计算该重新映射点与轴 [0X] 之间的角度。为此，它将 y 标准化为正常的数学坐标（y 在其值增加时上升，而不是下降），因此 self.center.y - point.y 而不是相反。

最后，在touchesMoved中：

- (void) touchesMoved: (NSSet *) touches withEvent: (UIEvent *) event {
    UITouch * touch = [touches anyObject] ;

    CGFloat currA = [self pointToAngleFromCenter:[touch locationInView:self.superview]] ;
    CGFloat prevA = [self pointToAngleFromCenter:[touch previousLocationInView:self.superview]] ;
    // the current value of the rotation angle
    CGFloat tranA = ::atan2(self.transform.b, self.transform.a) ;
    // the angle difference between last touch and the current touch
    CGFloat diffA = currA - prevA ;
    // the new angle resulting from applying the difference
    CGFloat angle = tranA - diffA ;

    CGAffineTransform t = ::CGAffineTransformMakeRotation(angle) ;

    self.transform = t ;
    [self setNeedsDisplay] ;
}