如何解决物体卡在地板上
好的,我正在可汗学院使用 PJS 制作一个拖放程序,这是我的代码:
/* Just a simple drag and drop program */
// Todo:
// * Add realistic bouncing
// * Add particles when bouncing
// Changeable variables
var testShapeWidth = 45;
var testShapeHeight = 45;
var testShapeX = 200;
var testShapeY = 200;
var gravityStrength = 0.8;
// Unchangeable variables
var onGround = false;
var mouseDown = false;
var round2 = function (num) {
num = num / 2;
num = round(num);
num *= 2;
return num;
};
var checkIfClicked = function(x,y,w,h) {
var mx = mouseX;
var my = mouseY;
var rightSide = x + w;
var bottomSide = y + h;
if (mx > x && mx < rightSide && my > y && my < bottomSide) {
return true;
}
return false;
};
var draw = function() {
background(255,255,255);
fill(0);
rect(testShapeX,testShapeY,testShapeWidth,testShapeHeight);
testShapeY = round2(testShapeY);
// Check if on the ground or not and update the onGround variable
if (testShapeY + testShapeHeight < 400) { // If the y coorordinate of the bottom edge of the rectangle is less than 400...
onGround = false; // The rectangle is not on the ground
} else if (testShapeY + testShapeHeight >= 400) { // If the y coorordinate of the bottom edge of the rectangle is greater than or equal to 400...
gravityStrength = round2(gravityStrength * -1) / 2;
}
if (!onGround && !mouseDown) {
testShapeY += gravityStrength;
gravityStrength += 0.2;
}
if (mouseDown) {
testShapeX = mouseX - testShapeWidth / 2;
testShapeY = mouseY - testShapeHeight / 2;
gravityStrength = 0;
}
};
var mousepressed = function() {
if (checkIfClicked(testShapeX,testShapeHeight)) {
mouseDown = true;
}
};
var mouseReleased = function() {
mouseDown = false;
};
我刚刚通过替换这些行添加了弹跳:
else if (testShapeY + testShapeHeight >= 400) { // If the y coorordinate of the bottom edge of the rectangle is greater than or equal to 400...
testShapeY = 400 - testShapeHeight;
gravityStrength = 0;
}
这样:
else if (testShapeY + testShapeHeight >= 400) { // If the y coorordinate of the bottom edge of the rectangle is greater than or equal to 400...
gravityStrength = round2(gravityStrength * -1) / 2;
}
但现在有时矩形会卡在地板上。下面是一个例子:
一开始我以为是偶数和奇数的关系,这就是为什么我添加了round2函数(四舍五入到最接近的2)无济于事。我希望 stackoverflow 社区可以帮助我解决这个问题,因为我不能。您会找到指向我的项目 here 的链接。
解决方法
你是个幸运的人,因为我不必筛选所有这些代码来了解问题所在(否则我今晚会帮助其他人,真的,代码太多了)。你的问题是数学。您的解决方案很简单。
问题
这是怎么回事:当物体低到足以穿过地板时,您反转并除以 2 gravityStrength
。这正是出错的地方。
如果对象落在地平面下的像素数超过 gravityStrength / 2
,则它无法再次上升,因为添加新的 gravityStrength
时它的位置仍将位于地面之下。然后它会再次恢复它的方向和一半 gravityStrength
,确保它不会再从这个位置移动(除非你用手移动它)。它绝对卡住了。
解决办法
改变这个:
} else if (testShapeY + testShapeHeight >= 400) { // If the y coorordinate of the bottom edge of the rectangle is greater than or equal to 400...
gravityStrength = round2(gravityStrength * -1) / 2;
}
为此:
} else if (testShapeY + testShapeHeight >= 400) { // If the y coorordinate of the bottom edge of the rectangle is greater than or equal to 400...
testShapeY = 400; // now any speed will bounce. Lower this number to make sure that the object will "get stuck" after a while if otherwise it bounces forever
gravityStrength = round2(gravityStrength * -1) / 2;
}
这里的想法是,您的算法将保持相同的工作方式,但不会让对象下降到足够低的位置,gravityStrength
无法使其恢复弹跳。或者,您可以将 testShapeY = 400;
下移一行,并将 testShapeY
变量基于新的 gravityStrength
。那会很不错。
玩得开心!
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