vault backup: 2025-07-04 17:08:59

工作OKRs/25.6-8 OKR.canvas

@@ -6,7 +6,7 @@
 		{"id":"82708a439812fdc7","type":"text","text":"# 7月已完成\n\n","x":-220,"y":134,"width":440,"height":560},
 		{"id":"505acb3e6b119076","type":"text","text":"# 6月已完成\n\n\nP1 结果对比\n- Herowind 带3.5气动与fast3.5对比 相同\n- Herowind 带4.0气动与fast4.0对比 相同\n- Herowind 带hrl气动与fast对比 需气动支持15MW\n- 叶根坐标系转换 \n\t- 叶尖变形量 - 变形向量 dot product 叶根坐标系方向\n\t- 叶片载荷输入量呢 载荷传递在blade mesh.force moment，mesh.orientation = coord_sys.n\n\nP1 Bladed交流问题汇总\n\nP1 模型线性化原理 done\n- Bladed 线性化理论手册 仔细阅读\n- multibody blade transform\n- fast线性化理论\n- 梳理Bladed线性化方法框架\n\n\nP1 编写线性化理论手册 done\nP1 上手Bladed \\ fast 线性化功能，研究OpenFAST线性化实现原理 done","x":-700,"y":134,"width":440,"height":560},
 		{"id":"30cb7486dc4e224c","type":"text","text":"# 8月已完成\n\n","x":260,"y":134,"width":440,"height":560},
-		{"id":"c18d25521d773705","type":"text","text":"# 计划\n这周要做的3~5件重要的事情，这些事情能有效推进实现OKR。\n\nP1 必须做。P2 应该做\n\n\nP1 柔性部件 叶片、塔架主动力惯性力算法 主线\n- 变形体动力学 简略看看ing\n- 柔性梁弯曲变形振动学习，主线 \n\t- 广义质量 刚度矩阵及含义\n- 如何静力学求解 \n\t- 基于本构方程 读孟的论文\n\t- normal mode shape 能否使用？\n\t- F = kx 外载与弹性势能相等\n\t\n- 梳理bladed动力学框架 this week\n\t- 子结构文献阅读\n\t- 叶片模型建模 done\n\nP1 工况点稳态载荷求解，F=kx\nP1 数值扰动+回归的线性化方法原理探究\n\nP2 如何优雅的存储、输出结果。\nP2 yaw 自由度再bug确认 已知原理了\n","x":-594,"y":-693,"width":450,"height":347}
+		{"id":"c18d25521d773705","type":"text","text":"# 计划\n这周要做的3~5件重要的事情，这些事情能有效推进实现OKR。\n\nP1 必须做。P2 应该做\n\n\nP1 柔性部件 叶片、塔架主动力惯性力算法 主线\n- 变形体动力学 简略看看ing\n- 柔性梁弯曲变形振动学习，主线 \n\t- 广义质量 刚度矩阵及含义\n- 如何静力学求解 \n\t- 基于本构方程 读孟的论文\n\t- normal mode shape 能否使用？\n\t- F = kx 外载与弹性势能相等\n\t\n- 梳理bladed动力学框架 this week\n\t- 子结构文献阅读\n\t- 叶片模型建模 done\n\nP1 工况点稳态变形量求解，F=kx\n- 文献调研，初步确定思路\nP1 数值扰动+回归的线性化方法原理探究\n\nP2 如何优雅的存储、输出结果。\nP2 yaw 自由度再bug确认 已知原理了\n","x":-597,"y":-693,"width":450,"height":347}
 	],
 	"edges":[]
 }

杂项/随礼清单.md

@@ -1,5 +1,5 @@
 
-|     | 事由  | 金额  |
+|     | 事项  | 金额  |
 | --- | --- | --- |
 | 乔延辉 | 结婚  | 888 |
 | 李小白 | 结婚  | 600 |

线性化求解器/参考文献/Kallesøe-Effect of steady deflections on the a/auto/Kallesøe-Effect of steady deflections on the a.md

@@ -63,7 +63,7 @@ $$
 
 where $\bar{\bf M}$ is the mass matrix, $\bar{\mathbf{F}}$ is a non-linear function that includes stiffness, damping, gyroscopic terms together with centrifugal force-based integral terms. The state vector $\bar{\mathbf{u}}=[u\left(t,s\right),\nu\left(t,s\right),\theta\left(t,s\right)]$ holds edgewise, flapwise and torsional deformations, respectively.  
 
-其中，$\bar{\bf M}$ 为质量矩阵，$\bar{\mathbf{F}}$ 为包含刚度、阻尼、陀螺惯性力以及基于离心力积分项的非线性函数。状态向量 $\bar{\mathbf{u}}=[u\left(t,s\right),\nu\left(t,s\right),\theta\left(t,s\right)]$ 分别表示摆振（edgewise）、挥舞（flapwise）和扭 角（torsional）变形。
+其中，$\bar{\bf M}$ 为质量矩阵，$\bar{\mathbf{F}}$ 为包含刚度、阻尼、陀螺惯性力以及基于离心力积分项的非线性函数。状态向量 $\bar{\mathbf{u}}=[u\left(t,s\right),\nu\left(t,s\right),\theta\left(t,s\right)]$ 分别表示摆振（edgewise）、挥舞（flapwise）和扭角（torsional）变形。
 
 Flapwise is defined as the direction normal to the rotor plane (positive downwind) and edgewise as in the rotor plane (positive towards leading edge) for a blade at zero pitch. When the blade pitches, the $(u,\,\nu)$ frame follows the blade. The position along the blades elastic axis is denoted $s$ , $t$ is the time, $\beta=\beta(t)$ is the global pitch of the blade, $\phi=\phi(t)$ is the azimuth angle of the rotor and the right hand side force function $\bar{\mathbf{f}}$ holds the effect of the aerodynamic forces $\mathbf{f}_{a e r o}$ and aerodynamic moment $M_{a e r o}$ on the blade. The dots denote time derivatives and the primes denote derivatives with respect to the longitudinal coordinate $s$ . As an example, the equation of motion for edgewise blade bending is given by  
 
@@ -102,7 +102,7 @@ where $s=R$ is the tip of the blade, $m=m(s)$ is the mass per length of the blad
 其中，$s=R$ 为叶片末端，$m=m(s)$ 为叶片单位长度的质量，$l_{c g}=l_{c g}(s)$ 为重心偏离弹性轴的偏移量，$E=E(s)$ 为杨氏模量，$I=I\left(s\right)$ 和 $I_{\eta}=I_{\eta}(s)$ 为惯性主矩，$w=$ $w(s,t)$ 为弹性轴上到位置 $s$ 的半径，$g$ 表示重力，$\tilde{\theta}=\tilde{\theta}\left(s\right)$ 为弦线与弹性主轴之间的夹角，且 $\tilde{\theta}=\tilde{\theta}\left(s\right)$ 为弦线与从弹性中心到重心之间的连线夹角，沿该连线测量 $l_{c g}$。当 $l_{c g}(R)\neq0$ 时，叶片末端的边界条件是风轮转速 $\dot{\phi}$ 和方位角 $\phi$ 的函数，因此随时间变化。这是因为叶片末端重心偏离弹性轴会导致重力和离心力引起的弯矩。大多数现代风电机组叶片在末端呈锥形，其中 $l_{c g}(s)\longrightarrow{\cal0}$ 且 $E I_{\xi}I_{\eta}\longrightarrow0$ 。因此，是否可以忽略这些方位角相关的边界条件取决于具体的叶片设计。在本工作中，叶片被构造成 $l_{c g}(R)=0$ 且 $E I_{\xi}I_{\eta}|_{s=R}\neq0$，从而使边界条件与方位角无关，并使方程 (5) 的所有右侧项变为零。
 # 3. STEADY–STATE AEROELASTIC MODEL  
 
-To determine the steady-state def lection for the blade, a non-linear steady-state aeroelastic model i.s derived. Steady-state conditions are def ined as uniform inf low, zero gravity, constant rotor speed and pitch an.gle $\ddot{\phi}=\dot{\beta}=0$ whereby all time derivatives in the structural equations of motion (1) become zero $\ddot{u}=\ddot{\nu}=\ddot{\theta}=\dot{u}=\dot{\nu}=\dot{\theta}=0$ . These uniform conditions remove the periodicity of the system. The steady-state aerodynamic model is based on blade element momentum (BEM) theory including Prendtl’s tip loss correction.26 The BEM theory computes a balance between the forces on the blade and the momentum change in the wind. The aerodynamic model is coupled to the structural model through the local wind speed and angle of attack and the structural model is coupled to the aerodynamic model through the aerodynamic forces acting on the blade.  
+To determine the steady-state def lection for the blade, a non-linear steady-state aeroelastic model i.s derived. Steady-state conditions are defined as uniform inf low, zero gravity, constant rotor speed and pitch angle $\ddot{\phi}=\dot{\beta}=0$ whereby all time derivatives in the structural equations of motion (1) become zero $\ddot{u}=\ddot{\nu}=\ddot{\theta}=\dot{u}=\dot{\nu}=\dot{\theta}=0$ . These uniform conditions remove the periodicity of the system. The steady-state aerodynamic model is based on blade element momentum (BEM) theory including Prendtl’s tip loss correction.26 The BEM theory computes a balance between the forces on the blade and the momentum change in the wind. The aerodynamic model is coupled to the structural model through the local wind speed and angle of attack and the structural model is coupled to the aerodynamic model through the aerodynamic forces acting on the blade.  
 为了确定叶片的稳态变形，推导了一个非线性稳态气弹耦合模型。稳态条件被定义为均匀来流、零重力、恒定风轮转速和变桨角度 $\ddot{\phi}=\dot{\beta}=0$，从而使运动方程(1)中的所有时间导数为零 $\ddot{u}=\ddot{\nu}=\ddot{\theta}=\dot{u}=\dot{\nu}=\dot{\theta}=0$。这些均匀条件消除了系统的周期性。稳态气动模型基于叶片元动量（BEM）理论，包括普兰德尔的梢流损失修正。BEM理论计算叶片上的力和风的动量变化之间的平衡。气动模型通过局部风速和迎角与结构模型耦合，结构模型通过作用在叶片上的气动力与气动模型耦合。
 # 3.1. Discretization of structural model  
 

线性化求解器/计算流程框架.canvas

@@ -6,11 +6,15 @@
 		{"id":"d3aa69200118cea0","type":"text","text":"小扰动求A, B, C, D","x":-200,"y":0,"width":250,"height":60},
 		{"id":"f8e0af85235be889","type":"text","text":"A上做特征值、模态","x":-200,"y":140,"width":250,"height":60},
 		{"id":"5818e7212360b063","type":"text","text":"可选是否MBC转换","x":160,"y":-140,"width":250,"height":60},
-		{"id":"03f26deb8603c7c3","x":-200,"y":280,"width":250,"height":60,"type":"text","text":"输出哪些量"},
-		{"id":"226774e95f4236f0","x":160,"y":140,"width":250,"height":60,"type":"text","text":"问题2 非线性，如何线性化"},
-		{"id":"65a392a60c82cf13","x":135,"y":-15,"width":300,"height":90,"type":"text","text":"问题1 动力学方程是二阶线性还是二阶非线性"},
-		{"id":"8c2eadcabf51301e","x":540,"y":-37,"width":250,"height":135,"type":"text","text":"非线性\n$$\n\\begin{array}{r}{\\dot{\\mathbf{x}}=f(t,\\mathbf{x},\\mathbf{u})}\\\\ {\\mathbf{y}=h(t,\\mathbf{x},\\mathbf{u})}\\end{array}\n$$"},
-		{"id":"e3f81d5e91896a13","x":540,"y":140,"width":250,"height":60,"type":"text","text":"小扰动+回归"}
+		{"id":"03f26deb8603c7c3","type":"text","text":"输出哪些量","x":-200,"y":280,"width":250,"height":60},
+		{"id":"226774e95f4236f0","type":"text","text":"问题2 非线性，如何线性化","x":160,"y":140,"width":250,"height":60},
+		{"id":"65a392a60c82cf13","type":"text","text":"问题1 动力学方程是二阶线性还是二阶非线性","x":135,"y":-15,"width":300,"height":90},
+		{"id":"8c2eadcabf51301e","type":"text","text":"非线性\n$$\n\\begin{array}{r}{\\dot{\\mathbf{x}}=f(t,\\mathbf{x},\\mathbf{u})}\\\\ {\\mathbf{y}=h(t,\\mathbf{x},\\mathbf{u})}\\end{array}\n$$","x":540,"y":-37,"width":250,"height":135},
+		{"id":"e3f81d5e91896a13","type":"text","text":"小扰动+回归","x":540,"y":140,"width":250,"height":60},
+		{"id":"4a884b5e3feb7e40","x":540,"y":-220,"width":250,"height":60,"type":"text","text":"气动力 = (K结构 + K钢化）q"},
+		{"id":"498cc5a9e1e188ac","x":880,"y":-220,"width":250,"height":60,"type":"text","text":"fast中两个K如何求？"},
+		{"id":"158be82aaa1bd0b4","x":880,"y":-140,"width":250,"height":60,"type":"text","text":"气动力如何求"},
+		{"id":"2a5628f97c424f0d","x":1220,"y":-220,"width":250,"height":60,"type":"text","text":"直接去掉$\\dot{q}、\\ddot{q}$项如何"}
 	],
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@@ -21,6 +25,9 @@
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-		{"id":"d8d1c9bccc6b3043","fromNode":"226774e95f4236f0","fromSide":"right","toNode":"e3f81d5e91896a13","toSide":"left"}
+		{"id":"d8d1c9bccc6b3043","fromNode":"226774e95f4236f0","fromSide":"right","toNode":"e3f81d5e91896a13","toSide":"left"},
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 }