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#: ../../source/components/components/rot_comps_adv.rst:4
msgid "Advanced Rotational Components"
msgstr "高级旋转组件"

#: ../../source/components/components/rot_comps_adv.rst:6
msgid ""
"In the previous section, we discussed :ref:`rotational-components` and "
"showed how to build a system model from basic components.  In this section "
"we will demonstrate how to incorporate event handling, which we will use "
"when modeling a backlash.  Furthermore, we'll also show how to use parameter"
" values to effect the interface of a component."
msgstr ""
"在上一节中，我们讨论了\\ :ref:`rotational-components`\\ "
"并展示了如何构建基本组件的系统模型。\ "
"在这一节中，我们将演示如何加入事件处理。\ "
"我们将在模拟齿隙时使用到这个功能。\ "
"此外，我们还将展示如何使用参数值来实现组件的接口。\ "

#: ../../source/components/components/rot_comps_adv.rst:13
msgid "Modeling Backlash"
msgstr "建模齿隙"

#: ../../source/components/components/rot_comps_adv.rst:15
msgid ""
"Let's start our exploration of more advanced component models by looking at "
"a rotational backlash element.  The equation for a backlash model is very "
"simple:"
msgstr "让我们从旋转齿隙元素开始对高级旋转组件模型的探索。\ "
"齿隙模型的方程非常简单："

#: ../../source/components/components/rot_comps_adv.rst:43
msgid ""
"We can add an instance of this backlash model into our previous model by "
"placing it in parallel with the spring and the damper, *i.e.,*"
msgstr "我们可以将齿隙模型的实例加入此前的模型里。\ "
"将其与弹簧及阻尼器相并联，即："

#: ../../source/components/components/rot_comps_adv.rst:51
msgid ""
"If we use the inheritance mechanism in Modelica, the resulting Modelica "
"model is quite simple:"
msgstr "若我们使用的Modelica的继承机制，由此产生的Modelica模型会很简单："

#: ../../source/components/components/rot_comps_adv.rst:58
msgid ""
"In this case, if the relative angle between ``inertia1`` and ``inertia2`` is"
" more than 0.5 radians (*i.e.,* the value of ``b`` in our backlash "
"instance), then the torque from the backlash element will be introduced."
msgstr ""
"在这种情况下，如果\\ ``inertia1``\\ 和\\ ``inertia2``\\ 之间的相对角度超过0.5弧度（即我们的齿隙实例里\\ "
"``b``\\ 的值），那么齿隙元件会带来转矩。\ "

#: ../../source/components/components/rot_comps_adv.rst:63
msgid ""
"If we simulate this model, we can see the impact that the backlash's "
"presence has on the response of the system:"
msgstr "如果对该模型进行仿真，我们就可以看到，齿隙的存在对系统的响应的影响："

#: ../../source/components/components/rot_comps_adv.rst:80
msgid ""
"When a kinematic constraint is imposed on a system, the component imposing "
"the constraint must generate some kind of force or torque in order to affect"
" the motion of the system.  This is called a reaction force or reaction "
"torque."
msgstr "当一个运动约束施加在系统上时，该施加约束的组件必须产生某种作用力或力矩，以此影响系统的运动。\ "
"这就是所谓的反作用力或反作用转矩。\ "

#: ../../source/components/components/rot_comps_adv.rst:85
msgid ""
"The following plot shows the reaction torque that the mechanical ground "
"element must impose on the system in order to fix the angular position:"
msgstr "下面的图显示的反作用转矩，就是该机械地面元件为了固定角位置所必须施加给系统的："

#: ../../source/components/components/rot_comps_adv.rst:93
msgid "Grounding and Reaction Torques"
msgstr "地面扭矩与反作用扭矩"

#: ../../source/components/components/rot_comps_adv.rst:95
msgid ""
"As we saw in the previous example, the behavior of the mechanical ground "
"element is such that it must exert a reaction torque on the system to "
"constrain the motion of the system.  In this section, we will examine this "
"effect a bit closer."
msgstr "正如我们在前面的例子中所看到的，机械地面元件必须对系统施加反作用转矩来约束其运动。\ "
"在本节中，我们将稍进一步研究这个效应。\ "

#: ../../source/components/components/rot_comps_adv.rst:100
msgid ""
"To demonstrate some of the complexities of kinematic constraints, we need to"
" create a mechanical gear model.  In this model, we will ignore the inertia "
"of the gear elements, efficiency losses in the gear and any backlash that "
"might exist between the teeth in the gear.  Recall our discussion about :ref"
":`digging-deeper` earlier in this chapter where we mentioned that component "
"models should focus on individual physical effects.  That same principle "
"applies here.  Inertia, friction and backlash can all be modeled as "
"individual effects (as we've already seen in this chapter).  There is no "
"need to lump them into our gear model.  Instead, we will focus only on the "
"relationship between gear input speed and output speed."
msgstr ""
"为了演示运动学约束的某些复杂性，我们需要创建一个机械齿轮模型。\ "
"在这个模型中，我们将忽略齿轮元件的惯量、齿轮间的效率损失以及可能在齿轮齿之间存在的任何齿隙。\ "
"回想一下我们在本章前面对\\"
" :ref:`digging-deeper`\\ "
"讨论内容，我们提到的组件模型应侧重于单个的物理效应。\ "
"同样的原则也适用于这里。\ "
"惯性、摩擦和齿隙都可以建模成单独的效应（正如我们在本章所看到的）。\ "
"我们没有必要所有效应都放入齿轮模型里。\ "
"相反，我们将只专注于齿轮输入速度与输出速度间的关系。\ "

#: ../../source/components/components/rot_comps_adv.rst:112
msgid ""
"In a typical system dynamics class, the equations that describe the behavior"
" of a gear are derived as follows.  First, we start with the understanding "
"that a gear introduces a relationship between the input speed and the output"
" speed, *i.e.*,"
msgstr "在典型的系统动力学类里，描述齿轮行为的方程推导如下。\ "
"首先，我们必须理解，该齿轮引入了输入速度与输出速度间的关系，即："

#: ../../source/components/components/rot_comps_adv.rst:121
msgid ""
"where :math:`R` is the gear ratio.  Recall that we assume the gear to be "
"perfectly efficient.  This means that the power going into the gear must "
"equal the power going out which we can express mathematically as:"
msgstr "其中\\ :math:`R`\\ 是齿轮比。\ "
"记得我们假设了齿轮效率为1。\ "
"这意味着齿轮的输入功率必须等于输出功率。\ "
"这可以在数学上表示为："

#: ../../source/components/components/rot_comps_adv.rst:129
msgid ""
"Note we are using the Modelica sign conventions here where a positive value "
"for the flow of a conserved quantity means a flow into the component.  In "
"this case, :math:`\\tau_a \\omega_a` is the flow of mechanical power into "
"the gear from ``flange_a`` and :math:`\\tau_b \\omega_b` is the flow of "
"mechanical power into the gear from ``flange_b``.  Therefore, their sum must"
" be zero, since our gear model does not include the inertia of the gear "
"elements and, therefore, no way to store energy or momentum within the gear "
"model."
msgstr ""
"请注意，我们在这使用了Modelica的正负号约定。\ "
"因此，守恒量的流为正意味着其正在流入组件内。\ "
"在这种情况下，\\ :math:`\\tau_a \\omega_a`\\ 是从\\ ``flange_a``\\ 流入齿轮的机械功率。\ "
"而\\ :math:`\\tau_b \\omega_b`\\ 则是从\\ ``flange_b``\\ 流出齿轮的机械功率。\ "
"因此，两者的总和必须为零。\ "
"因为，我们的齿轮模型不包括齿轮元件的惯性。\ "
"其结果就是齿轮模型内不可能储存能量或动力。\ "

#: ../../source/components/components/rot_comps_adv.rst:138
msgid ""
"Given these two facts, we can substitute the relationship between the speeds"
" into the relationship between the powers and get:"
msgstr "鉴于这两个事实，我们可以将速度关系代入到功率关系。\ "
"由此可以得到："

#: ../../source/components/components/rot_comps_adv.rst:145
msgid ""
"This allows us to cancel out :math:`\\omega_b` from the equation and "
"rearranging terms gives us:"
msgstr "这让我们从方程消去了\\ :math:`\\omega_b`\\ 。\ "
"重新组织方程我们可以得到："

#: ../../source/components/components/rot_comps_adv.rst:152
msgid ""
"Such a derivation will probably look very familiar to most engineers. But it"
" is important to recognize that there is something missing here.  More "
"specifically, there is something implicitly assumed that is not necessarily "
"a reasonable assumption."
msgstr "这样的推导对大多数工程师看起来可能会很熟悉。\ "
"但是，一定要意识到这里还差了些什么。\ "
"更具体而言，推理的一些隐含假定并不一定合理。\ "

#: ../../source/components/components/rot_comps_adv.rst:157
msgid "To understand the issue, let's first consider Euler's second law:"
msgstr "要理解这个问题，让我们先考虑欧拉第二定律："

#: ../../source/components/components/rot_comps_adv.rst:163
msgid ""
"In other words, the sum of the torques on a body should be equal to the "
"amount of angular momentum being accumulated by the body.  Recall that our "
"gear model doesn't include the inertia of the gear elements. As such, it has"
" no capacity to store energy or angular momentum.  If that is the case the "
"previous equation simplifies to:"
msgstr ""
"换言之，施加在物体上扭矩的总和应等于新累积的角动量。\ "
"回想我们的齿轮模型并不包括齿轮元件的惯性。\ "
"因此，模型无法以存储能量或角动量。\ "
"在这种情况下，先前的方程简化为："

#: ../../source/components/components/rot_comps_adv.rst:173
msgid ""
"Our gear has only two external torques, :math:`\\tau_a` and :math:`\\tau_b`."
"  Using the relationships we derived earlier, we know that their sum is:"
msgstr "我们的齿轮只有两个外部扭矩\\ :math:`\\tau_a`\\ 和\\ :math:`\\tau_b`\\ 。\ "
"应用之前推导出的关系，我们知道其总和为："

#: ../../source/components/components/rot_comps_adv.rst:181
msgid ""
"This equation implies that for any gear ratio, :math:`R`, not equal to 1.0, "
"the torque at ``flange_a`` (and consequently, the torque at ``flange_b`` as "
"well) must be zero.  But this cannot be correct if our gear is to function "
"as a gear."
msgstr ""
"这个方程意味着对任何不等于1.0的齿轮比\\ :math:`R`\\ ，\\ ``flange_a``\\ 上转矩（以及由此推出转矩\\ "
"``flange_b``\\ 也）必须为零。\ "
"但是，若我们的齿轮能提供齿轮功能的话，这就不可能是正确的。\ "

#: ../../source/components/components/rot_comps_adv.rst:186
msgid ""
"What this mathematical relationship is showing us about the physical "
"behavior of the system is more clearly demonstrated in this relationship:"
msgstr "为了显示上述数学关系向我们展示了系统的一种怎样的物理特性，我们可以观察从下列等式。\ "
"该等式更清楚地表明了系统的属性。\ "

#: ../../source/components/components/rot_comps_adv.rst:194
msgid ""
"The first term, :math:`\\tau_a` is the torque entering the gear from "
"``flange_a``.  The second term, :math:`R \\tau_a` is the torque entering the"
" gear from ``flange_b``.  This equation tells us that these two torques will"
" never sum to zero (for :math:`R \\neq 1`).  It appears that we have proven,"
" mathematically, that :math:`\\tau_a=0`. But in fact, what we are really "
"demonstrating is that there is **an imbalance** in the equation.  This "
"imbalance is the result of having forgotten something in our formulation.  "
"What is missing is the **reaction torque**."
msgstr ""
"第一项\\ :math:`\\tau_a`\\ ，是从\\ ``flange_a``\\ 进入齿轮的扭矩。\ "
"第二项\\ :math:`\\tau_b`\\ ，是从\\ ``flange_b``\\ 进入齿轮的扭矩。\ "
"这个等式告诉我们，这两个扭矩的总和绝不会为零（对于\\ :math:`R \\neq 1`\\ ）。\ "
"我们看起来在数学上证明了\\ :math:`\\tau_a=0`\\ 。\ "
"但实际上，我们实际上证明了公式里并\\ **不平衡**\\ 。\ "
"这种不平衡由于我们在列式中忘记一些东西。\ "
"这里缺少了\\ **反作用扭矩**\\ 。\ "

#: ../../source/components/components/rot_comps_adv.rst:204
msgid ""
"If you aren't already familiar with this issue, you might be puzzled about "
"where this reaction torque comes from.  After all, we have only two "
"mechanical connections to this gear and we have expressions for the torque "
"at both of these points.  But along the way, there was an implicit "
"assumption that the **housing of the gear is grounded**.  In reality, a gear"
" has three mechanical connections.  The third connection is the one between "
"the housing of the gear and whatever the gear is mounted to.  If the housing"
" is connected to a mechanical ground, then our equations so far are correct "
"as we can capture the behavior of such a (grounded) gear as follows:"
msgstr ""
"如果你还不熟悉这个问题，你可能会感到困惑。\ "
"这个反作用扭矩从何而来？毕竟，我们对齿轮只有两个机械连接。\ "
"而我们也把这两个点的扭矩表示了出来。\ "
"但一直以来，这里有一个隐含的假设，即齿轮的\\"
" **外壳与地面相连**\\ "
"。\ "
"在现实中，一个齿轮有三个机械连接。\ "
"第三个连接是在齿轮的外壳与齿轮安装的安装位置之间。\ "
"如果壳体被连接到机械地面，那么目前为止我们的方程是正确的。\ "
"因为我们可以用如下方式描述（接地）齿轮的行为："

#: ../../source/components/components/rot_comps_adv.rst:218
msgid ""
"Note that instead of using the relationship :math:`\\omega_a=R\\omega_b` in "
"the ``GroundedGear`` model, we instead used the relationship "
":math:`\\varphi_a=R\\varphi_b`.  This is actually more accurate since, once "
"assembled, the teeth of the gear really do constrain the angular positions "
"of the two shafts.  Furthermore, there may be some applications (*e.g.* "
"stepper motors) where preserving this relationship between positions, and "
"not just velocities, could be very important."
msgstr ""
"请注意在\\ ``GroundedGear``\\ 模型使用的关系不再是\\ :math:`\\omega_a=R\\omega_b`\\ ，而是\\ "
":math:`\\varphi_a=R\\varphi_b`\\ 。\ "
"这实际上更为准确。\ "
"因为，在组装后，齿轮的齿确实限制了两轴的角位置。\ "
"此外，在一些应用里（例如步进电机）保留位置关系而非仅保留速度关系可能会非常重要。\ "

#: ../../source/components/components/rot_comps_adv.rst:227
msgid ""
"Using the ``GroundedGear`` model, we can then build a system model using "
"this gear as follows:"
msgstr "使用\\ ``GroundedGear``\\ 模型，我们可以创建如下系统模型："

#: ../../source/components/components/rot_comps_adv.rst:235
msgid ""
"Note this system has two parallel mechanisms.  The first one uses the gear "
"model we just developed.  The second one replaces the assembly of the gear "
"and inertias with a single inertia.  This single inertia was specifically "
"chosen to have the *\"effective inertia\"* of the assembly.  As a result, "
"when we simulate this system, we see that ``inertia2`` and ``inertia3`` have"
" the same response:"
msgstr ""
"请注意，此系统有两种实现。\ "
"第一种使用我们刚刚开发的齿轮模型。\ "
"第二个将齿轮、惯量元件总成替换为单个惯量元件。\ "
"这个惯量元件的值特别设定为上述总成的\\ "
"*“有效惯量”*\\ 。\ "
"因此，在我们模拟这个系统时，我们看到\\ ``inertia2``\\ 和\\ ``inertia3``\\ 有相同的相应："

#: ../../source/components/components/rot_comps_adv.rst:246
msgid "Comparison"
msgstr "比较"

#: ../../source/components/components/rot_comps_adv.rst:248
msgid ""
"As previously mentioned, the issue with the ``GroundedGear`` model is that "
"it is implicitly assumed to be grounded.  This assumption may not always be "
"a reasonable one (*e.g.*, in an automotive transmission where gears are "
"generally connected to compliant mounts).  To understand how much different "
"the response of a system can be between grounded and ungrounded gears we "
"will first create a more complete gear model that is not implicitly grounded"
" and then compare its performance, side by side, with gears that are "
"grounded."
msgstr ""
"正如前面所提到的，\\ ``GroundedGear``\\ "
"模型的问题在于和地面相连这一隐含假定。\ "
"这种假设可能不总是合理的（例如，汽车的变速器中的齿轮通常连接到可形变的支架上）。\ "
"我们希望了解连接地面与否会对系统的响应有多大的影响。\ "
"因此，我们将首先创建一个不隐含连接地面的更完整的齿轮模型。\ "
"然后，将其性能与连接地面的齿轮一并比较。\ "

#: ../../source/components/components/rot_comps_adv.rst:257
msgid ""
"Without the implicit assumption that the housing of the gear is grounded, "
"the kinematic relationship between the two shafts and the housing is more "
"completely expressed as:"
msgstr "没有了齿轮外壳连接地面这一隐含的假设，两个轴和壳体之间的运动学关系能更完整地表示为："

#: ../../source/components/components/rot_comps_adv.rst:265
msgid ""
"Although it is beyond the scope of this discussion, we can derive the "
"following two equations using conservation of energy and conservation of "
"momentum:"
msgstr "虽然这已经超出了本文的讨论范围，我们可以用能量守恒和动量守恒得出以下两个等式："

#: ../../source/components/components/rot_comps_adv.rst:274
msgid ""
"Combining these relationships and adding an additional mechanical connector "
"to represent the housing, we get the following Modelica model for an ideal "
"gear:"
msgstr "将这些关系结合在一起，并增加机械连接器去表示壳体后，我们可以得到以下理想齿轮的Modelica模型："

#: ../../source/components/components/rot_comps_adv.rst:282
msgid ""
"Now, let us build a system model with three different mechanisms.  In each "
"mechanism the parameters for the gear, inertia, spring and damper are all "
"identical.  The only difference is whether we use an implicit grounded gear,"
" an explicitly grounded gear or a gear that is not directly connected to "
"ground, but is instead connected through a very stiff mounting system.  The "
"schematic for our system looks like this when rendered:"
msgstr ""
"现在，让我们一起建立是用上述三种不同机制的系统模型。\ "
"在各机构齿轮、惯性、弹簧和阻尼器的参数都是相同的。\ "
"唯一的区别是，我们是使用隐式连接地面齿轮、显式连接地面齿轮或不直接连接到地面上而是连接在硬支架上的齿轮。\ "
"我们的系统示意图如下所示："

#: ../../source/components/components/rot_comps_adv.rst:295
msgid ""
"The first thing we would expect is that the response of the mechanism with "
"the implicitly grounded gear should be identical to the response of the "
"mechanism with the explicitly grounded gear.  This is verified by the "
"following plot:"
msgstr "我们可以预期的第一件事是，隐式和显示连接地面的齿轮机构造成的响应应该是相同的。\ "
"这在以下的图中得到证实："

#: ../../source/components/components/rot_comps_adv.rst:303
msgid ""
"But the question still remains, how much difference would it make if we "
"assumed that a gear was implicitly grounded when, in fact, it wasn't?  This "
"is clearly demonstrated in the following plot:"
msgstr "但此前的问题仍然存在。\ "
"如果我们假定齿轮和地面隐式连接，而实际上不是的时候，这会造成多大的差异呢？问题在下图中得到清晰的解答："

#: ../../source/components/components/rot_comps_adv.rst:315
msgid "Optional Ground Connector"
msgstr "可选地面连接器"

#: ../../source/components/components/rot_comps_adv.rst:317
msgid ""
"So far in our discussion of rotational systems, we've created two different "
"gear models, ``GroundedGear``, which is implicitly grounded, and "
"``UngroundedGear``, which includes a mechanical connector for the housing.  "
"Ultimately, the equations used by these two components are quite similar and"
" there is a considerable amount of common code between them.  As we've "
"talked about before, redundancy like this should be avoided."
msgstr ""
"到目前为止，在对转动系统的讨论里，我们已经创建了两个不同的齿轮模型。\ "
"\\ ``GroundedGear``\\  为隐式连接地面。\ "
"而\\ ``UngroundedGear``\\ 则包括了壳体的机械连接器。\ "
"其实，两个部件的方程式非常相似。\ "
"其代码中有相当一部分是共同代码。\ "
"正如我们之前谈到这样，这样的冗余应该避免。\ "

#: ../../source/components/components/rot_comps_adv.rst:327
msgid ""
"One way that we can avoid redundancy in this case is to combine these two "
"models.  It might seem like this is impractical since they have very "
"different underlying assumptions and, more importantly, **different "
"interfaces** (*i.e.,* different connectors). Nevertheless, it is possible to"
" combine these models by making use of something called a *conditional "
"declaration*."
msgstr ""
"在这种情况下，我们避免冗余的一种方法是组合这两个模型。\ "
"这似乎是这是不切实际的。\ "
"既然两模型有很不同的基本假设。\ "
"更重要的是，它们的\\ **接口不同**\\"
" （即带有不同的连接器）。\ "
"不过，我们也可以通过利用所谓的\\ *有条件声明*\\ 去统一两个模型。\ "

#: ../../source/components/components/rot_comps_adv.rst:334
msgid "Consider the following ``ConfigurableGear`` model:"
msgstr "请考虑以下的\\ ``ConfigurableGear``\\ 模型："

#: ../../source/components/components/rot_comps_adv.rst:340
msgid ""
"In particular, notice that the declaration of ``housing`` ends with ``if not"
" grounded``.  When ``if`` appears at the end of a declaration, it indicates "
"that the variables only exists if the condition following the ``if`` is "
"true.  So when the ``grounded`` parameter is true, there is no ``housing`` "
"connector in this model.  Furthermore, the equations included, as "
"modifications, in the declaration of ``housing`` (*i.e.,* "
"``phi=housing_phi`` and ``tau=-flange_a.tau-flange_b.tau``) also disappear "
"with the declaration."
msgstr ""
"特别要注意\\ ``housing``\\ 声明结尾带有\\ ``if no grounded``\\ 。\ "
"若\\ ``if``\\ "
"出现在声明的结尾，则表明该变量只会在\\ ``if``\\ 后的条件表达式为真时存在。\ "
"所以，若\\ ``grounded``\\ "
"参数为真，模型中就不存在\\ ``housing``\\ 连接器。\ "
"此外，\\ ``housing``\\ 声明包括的方程，如修改语句（即\\ "
"``phi=housing_phi``\\ 以及\\ ``tau=-flange_a.tau-flange_b.tau``\\ ），也会随着声明消失。\ "

#: ../../source/components/components/rot_comps_adv.rst:349
msgid ""
"Meanwhile, in the ``equation`` section, we see that the ``if`` statement "
"there provides an additional equation, ``housing_phi=0``, in the case when "
"the model is grounded.  This is necessary because the variable "
"``housing_phi`` is always present (*i.e.,* there is no ``if`` at the end of "
"its declaration) so there must be an equation for it."
msgstr ""
"同时，\\ ``equation``\\ 区域中，\ "
"我们看到\\ ``if``\\ 语句在模型连接地面时提供了一条额外的方程\\ ``housing_phi=0``\\ 。\ "
"这是必要的。\ "
"原因是变量\\ ``housing_phi``\\ 总是存在（即在其声明的结尾没有\\ ``if``\\ ）。\ "
"所以该变量必须对应一条公式。\ "

#: ../../source/components/components/rot_comps_adv.rst:355
msgid ""
"To understand more completely what is going on here, recall that the number "
"of equations required by a component model is equal to the number of flow "
"variables across all the component's connectors + the number of (non-"
"parameter) variables declared in the model."
msgstr ""
"要了解更多的彻底理解是怎么回事，我们要记得一点。\ "
"组件模型所需的方程数目等于所有组件内连接器的流变量数目＋在模型中声明的（非参数）变量的数目。\ "

#: ../../source/components/components/rot_comps_adv.rst:360
msgid ""
"The following table summarizes how these things add up for the case where "
"``grounded`` is true and the case where it isn't:"
msgstr "下表总结了分别为\\ ``grounded``\\ 值的真假两种情况总结了上述量的关系："

#: ../../source/components/components/rot_comps_adv.rst:364
msgid "Quantity"
msgstr "量"

#: ../../source/components/components/rot_comps_adv.rst:364
msgid "``grounded==true``"
msgstr ""

#: ../../source/components/components/rot_comps_adv.rst:364
msgid "``grounded==false``"
msgstr ""

#: ../../source/components/components/rot_comps_adv.rst:366
msgid "Number of ``flow`` variables"
msgstr "\\ ``flow``\\ 变量总数"

#: ../../source/components/components/rot_comps_adv.rst:366
#: ../../source/components/components/rot_comps_adv.rst:374
msgid "2"
msgstr ""

#: ../../source/components/components/rot_comps_adv.rst:366
#: ../../source/components/components/rot_comps_adv.rst:374
msgid "3"
msgstr ""

#: ../../source/components/components/rot_comps_adv.rst:368
msgid "Number of variables"
msgstr "变量数"

#: ../../source/components/components/rot_comps_adv.rst:368
#: ../../source/components/components/rot_comps_adv.rst:368
msgid "1 (``housing_phi``)"
msgstr "1（\\ ``housing_phi``\\ ）"

#: ../../source/components/components/rot_comps_adv.rst:370
msgid "**Equations required**"
msgstr "\\ **所需方程数**\\ "

#: ../../source/components/components/rot_comps_adv.rst:370
#: ../../source/components/components/rot_comps_adv.rst:376
msgid "**3**"
msgstr ""

#: ../../source/components/components/rot_comps_adv.rst:370
#: ../../source/components/components/rot_comps_adv.rst:376
msgid "**4**"
msgstr ""

#: ../../source/components/components/rot_comps_adv.rst:372
msgid "Equations in declarations"
msgstr "声明内的方程数"

#: ../../source/components/components/rot_comps_adv.rst:372
msgid "0"
msgstr ""

#: ../../source/components/components/rot_comps_adv.rst:372
msgid "2 (from ``housing``)"
msgstr "2（\\ ``housing``\\ 内）"

#: ../../source/components/components/rot_comps_adv.rst:374
msgid "Equations in ``equation`` section"
msgstr "\\ ``equation``\\ 区域的方程数"

#: ../../source/components/components/rot_comps_adv.rst:376
msgid "**Equations provided**"
msgstr "\\ **提供的方程数**\\ "

#: ../../source/components/components/rot_comps_adv.rst:379
msgid ""
"When using conditional declarations, it is very important to make sure that "
"the number of equations provided balances with the number of equations "
"required for all possible conditions.  In this case, we have only two "
"conditions to concern ourselves with and we can clearly see from this table "
"that we have met this requirement in both cases."
msgstr ""
"当使用条件声明时，非常重要的一点是确保在所有情况下，提供的方程数和所需方程数相等。\ "
"在这里，我们只有两个条件需要考虑。\ "
"而我们可以清楚地从该表中看到，两种情况下要求均满足。\ "

#: ../../source/components/components/rot_comps_adv.rst:385
msgid ""
"The following model demonstrates how we can now use the ``ConfigurableGear``"
" models as both an implicitly and explicitly grounded gear:"
msgstr "下面的模型展示了我们如何将\\ ``ConfigurableGear``\\ 模型用在隐式、显式接地齿轮的情况："

#: ../../source/components/components/rot_comps_adv.rst:395
msgid ""
"And, as we would expect, the response for ``inertia1`` and ``inertia4`` are "
"identical:"
msgstr "而且，如我们所料，\\ ``inertia1``\\ 和\\ ``inertia4``\\ 的响应相同："

#: ../../source/components/components/rot_comps_adv.rst:73
msgid ""
"Another thing worth looking at (which we will delve into much more in the "
"next topic) is the force felt by the mechanical ground element. Looking at "
"our schematic, it is clear that the role of the mechanical ground element is"
" to fix the angular position on one side of our system.  An equation to "
"constrain the motion (or lack of motion, in this case) of a point in the "
"system is called a *kinematic constraint*."
msgstr ""
"另一个值得观察的现象是（我们将在下个主题更深入地探讨）是机械地面元件的受力。\ "
"从我们的示意图能够明显看出，该机械地面元件的作用是固定系统其中一边的角位移。\ "
"用以约束系统内某个点运动（或在这里，无运动）的方程被称为\\ *运动学约束*\\ 。\ "

#: ../../source/components/components/rot_comps_adv.rst:34
msgid ""
"In Modelica (where :math:`\\Delta \\varphi` is ``phi_rel``), this component "
"can be described as follows:"
msgstr ""
"该组件可以用Modelica描述如下\ "
"（其中\ :math:`\\Delta \\varphi`\ 写作\ ``phi_rel``\ )："
