This is the vertical direction y.
这个是垂直的，Y轨道。
And I will call this direction y, as I always do perpendicular to the surface.
照惯例把垂直方向，设为y方向。
So the consequence of my choosing this the direction in which y increases...
因此我的选择的结果,这个方向是y增长的方向。
So, the direction of the force is going to be that of x, y, z.
所以方向是。
There is no acceleration in the y direction, only in x direction when it starts moving, and that's why we split it.
在y方向上没有加速度，只有在x方向上移动时才有，这就是分解它的原因。
If you move in the y direction, it decreases at a constant rate.
如果你在y轴的方向上移动，函数值将匀速地减少。
现在看y轴方向。
the position of the highest point P we first ask ourselves the question from equation number four: 0 when is the velocity in the y direction zero?
为了找到最高点P的所在,我们首先得,解答我们刚才提出的问题：,究竟何时y方向上的速度减为？
Now what I want to find,actually, V is what are these two vectors, let's call them U and V, that correspond to moving a bit in the x direction or in the y direction?
现在我实际上想知道的是，这两个向量是什么，我们称它们为，U和，两者对应的是x方向和y方向一点位移？
Labels like Web 3.0, business driven architecture, and user centricity where empowerment of the end-user is a great subtle shift in this direction - or as Yahoo puts it - it's about y! Ou.
这些标签，诸如web3.0、业务驱动架构和给最终用户授权的用户中心(user Centricity)，就是这个方向上的一个极为隐蔽的转变，又或象yahoo给它贴上的：y ！ ou。
OK, so, if I have a point, at any point, I can slice the graph of my function by two planes, one that's going along the x, one along the y direction.
如果我有一个点，任意一点，我可以用2个平面把函数图像切开，一个是沿x方向，一个沿y方向。
in the minus y direction and there is a normal force in the plus y direction and these two exactly cancel each other.
,在y的负方向,那么也有一个常力,在y的正方向,这两个力互相抵消。
The velocity in the y direction started off plus 133, -133 but now it is minus 133.
在y轴方向的速度,从+133开始，而现在是。
For velocity in the x, y, or z direction.
对于x，y，或z三个方向的速度。
Let's first decompose this force as we did before, in the y direction.
还是照惯例,把力分解到y方向。
Transform.postScale(float sx, float sy, float sz): Scales the 3D object in the x, y, and z direction.
postScale(floatsx, floatsy, floatsz)：在x、y、z方向伸缩3D对象。
What is the acceleration in the y direction at time zero?
那y方向上,0时刻的加速度是多少？
x Well, we know how to compute partial w over partial x y or partial w over partial y, which measure how w changes if I move in the direction of the x axis or in the direction of the y axis.
我们知道如何计算∂w/∂,或者∂w/∂,这衡量了我在x轴,或者y轴方向上移动时，w的变化。
And this is the direction, positive direction, of y.
这个方向，是y的正方向。
Since there is no acceleration in the y direction, the normal force must be also mg cosine theta.
既然在y方向上,没有加速度，法向力也等于mgcosθ
In terms of forces and acceleration, I can write (now just in the y-direction).
关于力和加速度，我们可以写出以下公式(现在我们只要看纵轴方向就可以了)。
Suppose A itself is the unit vector in the y direction B Then A dot B is what?
假设A是y方向上的,单位矢量,则是z方向上的单位矢量。，and，B，is，the，unit，vector，in，the，z，direction。,那A点B是多少呢？
It comes to a halt in the y direction.
什么时刻它会在y方向停止前进。
There is almost no acceleration in the y direction.
在y方向几乎,没有加速度。
So if we visit those points, for instance here, then there is, mg of course, gravity, mg, if there is an object there in the y direction...
如果观察这些点，比如这个，那么那个，当然是等于重力，如果那里有个物体,在y方向。
这是y方向。
x but you better believe that there is one y So only when we deal with the y equations - does this acceleration come in-- not at all when we deal with the x direction.
方向上如果没有加速度，There，is，no，acceleration，in，the，x，direction,但是你最好相信,方向是有加速度的。，in，the，y，direction。,所以当我们处理y的方程时,需要考虑到加速度-,与处理x方向完全不同。
The x axis is positive in the right direction, while the y axis is positive in the downward direction (again, this is in contrast to its traditional upward direction).
x轴在右边为正值，但是y 轴在下方为正值（这与传统的上方为正值相反）。
This kind of control equipment for the control of the two computers can move in the direction of Zang Y roacl or galvanometer scanning laser beam.
这种控制设备用两台计算机控制的可以在臧Y方向移动的镜子或检流计扫描激光光束。
Then the direction of fracture zone can be determined by making both X, Y-component detection in data acquisition and coordinate revolution in data processing.
如果野外作了X、Y分量观测，室内又作了坐标旋转处理，那么，就可以进一步确定裂缝带的方向。
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