3d 抛物线轨迹
我正在尝试使用 3d 空间中的弧进行一些计算，但有点迷失。假设我想在 3d 空间中制作一个弧线动画来连接 2 个 x,y,z 坐标（两个坐标的 az 值为 0，并且只是平面上的点）。我通过向弧发送起始 x、y、z 位置、旋转、速度和重力值来控制弧。如果我知道需要连接的 x、y、z 坐标，是否有一种方法可以计算将其从起始 x、y、z 坐标连接到结束坐标所需的旋转、速度和重力值？
谢谢。
编辑：谢谢汤姆10。为了澄清起见，我通过用粒子创建抛物线来制作“弧线”。我试图弄清楚如何（通过开始由一系列粒子形成的抛物线，以 x、y、z、速度、旋转和重力为起始点）确定它最终的位置（最后的 x、y、z 坐标） ）。因此，如果这是需要连接的两个坐标：
x1=240;
y1=140;
z1=0;
x2=300;
y2=200;
z2=0;
如何仅使用这些变量来计算该抛物线的旋转、速度和重力开始形成抛物线：
x1=240;
y1=140;
z1=0;
rotation;
velocity;
gravity;
我试图将角度保持为恒定值。
I'm trying to figure out some calculations using arcs in 3d space but am a bit lost. Lets say that I want to animate an arc in 3d space to connect 2 x,y,z coordinates (both coordinates have a z value of 0, and are just points on a plane). I'm controlling the arc by sending it a starting x,y,z position, a rotation, a velocity, and a gravity value. If I know both the x,y,z coordinates that need to be connected, is there a way to calculate what the necessary rotation, velocity, and gravity values to connect it from the starting x,y,z coordinate to the ending one?
Thanks.
EDIT: Thanks tom10. To clarify, I'm making "arcs" by creating a parabola with particles. I'm trying to figure out how to ( by starting a parabola formed by a series particles with an beginning x,y,z,velocity,rotation,and gravity) determine where it will in end(the last x,y,z coordinates). So if it if these are the two coordinates that need to be connected:
x1=240;
y1=140;
z1=0;
x2=300;
y2=200;
z2=0;
how can the rotation, velocity, and gravity of this parabola be calculated using only these variables start the formation of the parabola:
x1=240;
y1=140;
z1=0;
rotation;
velocity;
gravity;
I am trying to keep the angle a constant value.
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此链接描述了“击中”的弹道轨迹从 (0,0) 发射时距离 x 和高度 y 的目标，初始速度 v 为所需的发射角度 θ"，这就是您想要的，对吧？为了使变量变成正确的形式，请设置旋转角度（在 xy 平面中），以便指向正确的方向，即 atan(y/x)，从那时起，以匹配通常的术语2D问题，将你的z重写为y，并将到目标的水平距离（即sqrt(xx + yy)）重写为x，然后你可以直接使用链接中的公式。
This link describes the ballistic trajectory to "hit a target at range x and altitude y when fired from (0,0) and with initial velocity v the required angle(s) of launch θ", which is what you want, right? To get your variables into the right form, set the rotation angle (in the x-y plane) so you're pointing in the right direction, that is atan(y/x), and from then on out, to match the usual terminology for 2D problem, rewrite your z to y, and the horizontal distance to the target (which is sqrt(xx + yy)) as x, and then you can directly use the formula in link.
执行与 2D 中相同的操作。您只需通过旋转轴将图形转换为仿射空间，这样其中之一就为零；然后求解并撤消旋转。
Do the same as you'd do in 2D. You just have to convert your figures to an affine space by rotating the axis, so one of them becomes zero; then solve and undo the rotation.