Detect 360 degrees turn algorithm

问题

I'm successfully detecting 0-360 degrees rotation (roll) of phone around an axis, but now I am having hard times designing an effective algorithm to detect one full turn. My working but I think not elegant and as effective algorithm as I would like is:

private boolean detectRoll;
private boolean[] checkpointsR = new boolean[4];
private boolean fullRollTurn;

public void detectingRoll() {
    setDetectRoll(true);
    checkpointsR[0] = true;
    for (int i = 1; i < 4; i++) {
        if (roll > 90 * i && roll < 90 * (i + 1)
            && checkpointsR[i - 1] == true) {
            checkpointsR[i] = true;
        }
    }

    if (areAllTrue(checkpointsR) && roll > 0 && roll < 45) {
        fullRollTurn = true;
        // reset rollCheckpoints
        for (int i = 1; i < 4; i++) {
            checkpointsR[i] = false;
        }
    }
}

public static boolean areAllTrue(boolean[] array) {
    for (boolean b : array)
        if (!b)
            return false;
    return true;
}

public void setDetectRoll(boolean detectRoll) {
    this.detectRoll = detectRoll;
}

Any help would be really appreciated.

回答1:

Well, your code only detects turns that result from increasing roll. In the other direction, ie a turn resulting from roll falling,

    if (checkpointsR[i - 1] == true) 
    {
        checkpointsR[i] = true; 
    }

won't ever trigger.

While that's easily fixable, if the only sensor input you've got is roll, the fixed-checkpoint is approach always going to have problems. The best way to visualize it is that your checkpoints are little red points on a circle, and the rotation of the phone corresponds to an ant crawling around the edge of the circle. Suppose that when the ant passes a point, it turns green, and that this represents the checkpoint being set to true. If the ant starts between two checkpoints A and B, it can "fool" them by crawling through B, all the way round to A, and then turning around and heading the other way again. All the checkpoints will be green, but the ant won't have completed a full circle.

The way to resolve this is to change two things: first, give the checkpoints three states: unvisited, clockwise and anticlockwise. Second, start the ant on top of a checkpoint.

Here're the new rules:

Every point starts as unvisited (a red point).
If the ant passes a checkpoint in a clockwise direction, set it's state to clockwise (a little green arrow pointing clockwise from the point)
If the ant passes a checkpoint in the anticlockwise direction, set it's state to anticlockwise (a little green arrow pointing on the anticlockwise direction from the point)
A turn has been completed if all the checkpoints are clockwise or if they're all anticlockwise.

This can be achieved with three checkpoints, and placed wherever you want on the circle as long as the first one is under the ant's initial position. I'd recommend {initial, inital + 120, initial + 240} though for symmetry's sake.

(The ant thought-experiment works with two checkpoints, but when you've got two checkpoints the problem is there are a pair a regions with a non-unique checkpoint between them, which confuses the detection of which checkpoint the ant actually passed through when the ant goes from one region to the other)

来源：https://stackoverflow.com/questions/18847164/detect-360-degrees-turn-algorithm

标签

java

android

algorithm

rotation

accelerometer