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Correct wrong sentence about mechanical impedance

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Thomas Dehaeze
2020-04-07 16:19:15 +02:00
parent ca42338fe3
commit 48b2d36418
2 changed files with 25 additions and 25 deletions
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org/uncertainty_payload.org
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@@ -337,10 +337,10 @@ The obtained dynamics from $F$ to $x$ for the three isolation platform are shown
* Generalization to arbitrary dynamics
<<sec:arbitrary_dynamics>>
** Introduction
Let's now consider a general payload described by its *impedance* $G^\prime(s) = \frac{x}{F^\prime}$ as shown in Figure [[fig:general_payload_impedance]].
Let's now consider a general payload described by its *impedance* $G^\prime(s) = \frac{F^\prime}{x}$ as shown in Figure [[fig:general_payload_impedance]].
#+begin_note
Note here that we use the term /impedance/, however, the mechanical impedance is usually defined as the ratio of the velocity over the force $\dot{x}/F^\prime$. We should refer to /resistance/ instead of /impedance/.
Note here that we use the term /impedance/, however, the mechanical impedance is usually defined as the ratio of the force over the velocity $F^\prime/\dot{x}$. We should refer to /resistance/ instead of /impedance/.
#+end_note
#+begin_src latex :file general_payload_impedance.pdf
@@ -455,12 +455,11 @@ In order to verify that the formula is correct, let's take the same mass-spring-
By eliminating $x^\prime$ of the equations, we obtain:
#+begin_important
#+name: eq:impedance_mass_spring_damper
\begin{equation}
G^\prime(s) = \frac{F^\prime}{x} = \frac{m^\prime s^2 (c^\prime s + k)}{m^\prime s^2 + c^\prime s + k^\prime}
G^\prime(s) = \frac{F^\prime}{x} = \frac{m^\prime s^2 (c^\prime s + k)}{m^\prime s^2 + c^\prime s + k^\prime} \label{eq:impedance_mass_spring_damper}
\end{equation}
The impedance of a 1dof mass-spring-damper system is described by Eq. [[eq:impedance_mass_spring_damper]].
The impedance of a 1dof mass-spring-damper system is described by Eq. eqref:eq:impedance_mass_spring_damper.
#+end_important
And we obtain