WinRT中如何获取一指触摸总旋转角度？

Question

我正在尝试按照Jerry's tutorial进行拨号控制。因为方法 GetAngle 是基于触摸点的位置，所以计算出的角度总是等于或小于 360°。但我需要的是，如果我将结旋转两个半圈，我想得到 900° (360°*2 + 180°)。

我知道e.Cumulative.Rotation（e是ManipulationDeltaRoutedEventArgs）可以给我总旋转角度，但是好像只能用两根手指触发旋转。

我是不是漏掉了一些简单的解决方案？？

Answer 1

我在这上面花了一段时间，结果证明这是一个很好玩的程序 (thanks to Jerry for great blog post)。最后，我得到了一些不同的解决方案 - 下面是一些。

如果你需要计算整个圆圈，那么你需要记住角度从360度变为0的那一刻，并记住有一个完整的圆圈。在我的例子中，我记得最后一个象限，这有助于我处理这个问题。主要代码如下所示：

private Quadrants currentQuadrant = Quadrants.I;
private Quadrants lastQuadrant = Quadrants.I;
private double lastAngle = 0;

public enum Quadrants { I, IV, III, II } // counterclockwise quadrants
private double CheckAngle(Point touchPoint, Size bounds)
{
    // find point position relative to bounds
    double valX = touchPoint.X - (bounds.Width / 2d);
    double valY = (bounds.Height / 2d) - touchPoint.Y;
    // determine the quadrant and save it
    currentQuadrant = (valX >= 0) ?
        (valY >= 0) ? Quadrants.I : Quadrants.IV :
        (valY >= 0) ? Quadrants.II : Quadrants.III;

    double quadrantAngle = Math.Atan(valX / valY) * (180 / Math.PI); // calculate angle within quadrant
    return currentQuadrant == Quadrants.II ? 360 + quadrantAngle :
            currentQuadrant != Quadrants.I ? 180 + quadrantAngle : quadrantAngle;
}

private void Grid_ManipulationDelta(object sender, ManipulationDeltaRoutedEventArgs e)
{
    double currentAngle = CheckAngle(e.Position, this.RenderSize);

    if (Math.Abs(lastAngle - currentAngle) < 1) return; // don't update UI always - performance

    if (currentQuadrant == Quadrants.I && lastQuadrant == Quadrants.II) // check for full circle
    { spinNumber++; RaiseProperty("SpinNumber"); } 
    else if ((currentQuadrant > lastQuadrant + 1)) return; // check if proper movement

    // update quadrants and Angle
    lastQuadrant = currentQuadrant;
    Angle = lastAngle = currentAngle;
}

private void Grid_ManipulationCompleted(object sender, ManipulationCompletedRoutedEventArgs e)
{
    // reset all things
    Angle = lastAngle = spinNumber =0;
    currentQuadrant = lastQuadrant = Quadrants.I;
}

和一些 XAML 部分：

<Grid>
    <Grid ManipulationMode="All" ManipulationDelta="Grid_ManipulationDelta" ManipulationCompleted="Grid_ManipulationCompleted">
        <Ellipse Width="300" Height="300" Fill="Red"/>
        <Path Fill="Green" Data="M 0 0 L 50 0 L 100 150 L 0 150 L 50 0 Z" HorizontalAlignment="Center" VerticalAlignment="Center" RenderTransformOrigin="0.5,0.5">
            <Path.RenderTransform>
                <RotateTransform Angle="{Binding Angle}"/>
            </Path.RenderTransform>
        </Path>
    </Grid>
    <TextBlock Text="{Binding AngleTxt}" VerticalAlignment="Bottom" Margin="0,0,0,50" HorizontalAlignment="Center"/>
    <TextBlock Text="{Binding SpinNumber}" VerticalAlignment="Bottom" Margin="0,0,0,25" HorizontalAlignment="Center"/>
</Grid>

请注意，如果您从第一象限开始顺时针移动手指，这将起作用。它还需要更多改进，比如处理逆时针移动时的情况，尤其是从 0 度到 360 度的变化。

WinRT中如何获取一指触摸总旋转角度？

How to get the total rotation angle by one finger touch in WinRT?

xaml

windows-8

windows-runtime

windows-phone-8.1

windows-8.1