javascript - 物体卡在地板上
问题描述
好的,我正在可汗学院使用 PJS 制作拖放程序,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, testShapeY, testShapeWidth, testShapeHeight)) {
mouseDown = true;
}
};
var mouseReleased = function() {
mouseDown = false;
};
我只是通过替换这些行来添加弹跳:
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;
}
有了这个:
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;
}
起初我认为它与偶数和奇数有关,这就是为什么我添加了 round2 函数(它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。您可以在此处找到我的项目的链接。
解决方案
你是一个幸运的人,因为我不必筛选所有这些代码来知道问题所在(否则我今晚会帮助其他人,真的,这是太多的代码)。你的问题是数学。您的解决方案很简单。
问题
这是发生了什么:当物体足够低可以穿过地板时,你反转并除以 2 gravityStrength
。这正是事情出错的地方。
如果对象下落的像素多于gravityStrength / 2
地平面下的像素,则它不能再次上升,因为添加新对象时它的位置gravityStrength
仍将在地下。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。
解决方案
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;
}
为了这:
} 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,但不会让对象下降到足够低以至于无法gravityStrength
将其带回弹跳。或者,您可以将testShapeY = 400;
一行下移,并将testShapeY
变量基于新的gravityStrength
. 那会很不错。
玩得开心!