无限旋转图像

Question

我尝试旋转其超级视图中的图像视图，以使该图像视图在旋转时始终与超级视图的边界相交而不交叉，并适当调整其大小。我该如何实施？图片视图应该能够绕360˚旋转。

在这里，我使用基于三角形公式的计算，同时考虑了初始图像视图的对角线。

也许我应该考虑旋转后的图像视图的新边框（它的x和y坐标为负，并且变换后的帧尺寸也变大）。

到目前为止没有成功，我的图像视图缩小得太快和太大了。因此，据我了解，我的目标是为CGAffineTransformScale获得适当的比例因子。也许还有其他方法可以做到这一点。

// set initial values

    _planImageView.layer.affineTransform = CGAffineTransformScale(CGAffineTransformIdentity, 1, 1);

    _degrees = 0;

    _initialWidth = _planImageView.frame.size.width;
    _initialHeight = _planImageView.frame.size.height;
    _initialAngle = MathUtils::radiansToDegrees(atan((_initialWidth / 2) / (_initialHeight / 2)));

// rotation routine

- (void)rotatePlanWithDegrees:(double)degrees
{
    double deltaDegrees = degrees - _degrees;
    _initialAngle -= deltaDegrees;
    double newAngle = _initialAngle;
    double newWidth = (_initialWidth / 2) * tan(MathUtils::degreesToRadians(newAngle)) * 2;
    double newHeight = newWidth * (_initialHeight / _initialWidth);

    NSLog(@"DEG %f DELTA %f A %f W %f H %f", degrees, deltaDegrees, newAngle, newWidth, newHeight);

    double currentScale = newWidth / _initialWidth;

    _planImageView.layer.affineTransform = CGAffineTransformScale(CGAffineTransformIdentity, currentScale, currentScale);
    _planImageView.layer.affineTransform = CGAffineTransformRotate(_planImageView.layer.affineTransform, (CGFloat) MathUtils::degreesToRadians(degrees));

    _degrees = degrees;

    self->_planImageView.center = _center;

//    NSLog(@"%@", NSStringFromCGRect(_planImageView.frame));
}

编辑

多亏了答案，我改写了例行程序，现在可以了！

- (void)rotatePlanWithDegrees:(double)degrees
{
    double newWidth =
            _initialWidth  * abs(cos(MathUtils::degreesToRadians(degrees))) +
            _initialHeight * abs(sin(MathUtils::degreesToRadians(degrees)));
    double newHeight =
            _initialWidth  * abs(sin(MathUtils::degreesToRadians(degrees))) +
            _initialHeight * abs(cos(MathUtils::degreesToRadians(degrees)));

    CGFloat scale = (CGFloat) MIN(
            self.planImageScrollView.frame.size.width / newWidth,
            self.planImageScrollView.frame.size.height / newHeight);

    CGAffineTransform rotationTransform = CGAffineTransformMakeRotation((CGFloat) MathUtils::degreesToRadians(degrees));
    CGAffineTransform scaleTransform  = CGAffineTransformMakeScale(scale, scale);
    _planImageView.layer.affineTransform = CGAffineTransformConcat(rotationTransform, scaleTransform);

    self->_planImageView.center = _center;
}

Answer 1

旋转矩形W x H时，边界框的尺寸为W' = W |cos Θ| + H |sin Θ|，H' = W |sin Θ| + H |cos Θ|。

如果您需要将其适合在W" x H"矩形中，则比例因子是W"/W'和H"/H'中的最小值。