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Robin Clark 2010-08-30 08:46:29 +01:00
parent d2ab642231
commit 4ffb457ef3
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26
component_failure_modes_definition/component_failure_modes_definition.tex
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@ -339,23 +339,6 @@ Thus if the failure modes of a component $F$ are unitary~state, we can say $F \
An example of a component with an obvious set of ``unitary~state'' failure modes is the electrical resistor. An example of a component with an obvious set of ``unitary~state'' failure modes is the electrical resistor.
Electrical resistors can fail by going OPEN or SHORTED. Electrical resistors can fail by going OPEN or SHORTED.
%% CUNT
%% CUNT For a given resistor R we can apply the
%% CUNT the function $fm$ to find its set of failure modes thus $ fm(R) = \{R_{SHORTED}, R_{OPEN}\} $.
%% CUNT A resistor cannot fail with both conditions open and short active at the same time! The conditions
%% CUNT OPEN and SHORT are thus mutually exclusive.
%% CUNT Because of this, the failure mode set $F=fm(R)$ is `unitary~state'.
%% CUNT
%% CUNT
%% CUNT Thus because both fault modes cannot be active at the same time, the intersection of $ R_{SHORTED} $ and $ R_{OPEN} $ cannot exist.
%% CUNT
%% CUNT The intersection of these is therefore the empty set, $$ R_{SHORTED} \cap R_{OPEN} \eq \emptyset $$,
%% CUNT therefore
%% CUNT $ fm(R) \in \mathcal{U} $.
%% CUNT
%% CUNT
For a given resistor R we can apply the For a given resistor R we can apply the
the function $fm$ to find its set of failure modes thus $ fm(R) = \{R_{SHORTED}, R_{OPEN}\} $. the function $fm$ to find its set of failure modes thus $ fm(R) = \{R_{SHORTED}, R_{OPEN}\} $.
@ -366,7 +349,6 @@ Because of this, the failure mode set $F=fm(R)$ is `unitary~state'.
Thus because both fault modes cannot be active at the same time, the intersection of $ R_{SHORTED} $ and $ R_{OPEN} $ cannot exist. Thus because both fault modes cannot be active at the same time, the intersection of $ R_{SHORTED} $ and $ R_{OPEN} $ cannot exist.
%%CUNT The intersection of these is therefore the empty set, $$ R_{SHORTED} \cap R_{OPEN} \eq \emptyset $$,
The intersection of these is therefore the empty set, $$ R_{SHORTED} \cap R_{OPEN} = \emptyset $$, The intersection of these is therefore the empty set, $$ R_{SHORTED} \cap R_{OPEN} = \emptyset $$,
therefore therefore
$ fm(R) \in \mathcal{U} $. $ fm(R) \in \mathcal{U} $.
@ -578,12 +560,6 @@ components $C_j$ are in
\end{itemize} \end{itemize}
%} %}
%% CUNT
%% CUNT \begin{equation}
%% CUNT |{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1} \frac{|{SU}|!}{k!(|{SU}| - k)!}}
%% CUNT - {\sum{j \in J} {|FM({C_{j})}| \choose 2}}}
%% CUNT \label{eqn:correctedccps}
%% CUNT \end{equation}
\begin{equation} \begin{equation}
|{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1} \frac{|{SU}|!}{k!(|{SU}| - k)!}} |{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1} \frac{|{SU}|!}{k!(|{SU}| - k)!}}
- {\sum_{j \in J} {|FM({C_{j})}| \choose 2}} - {\sum_{j \in J} {|FM({C_{j})}| \choose 2}}
@ -667,7 +643,7 @@ Thus the statistical sample space $\Omega$ for a component or derived~component
$$ \Omega(C) = \{OK, failure\_mode_{1},failure\_mode_{2},failure\_mode_{3}, \ldots ,failure\_mode_{N}\} $$ $$ \Omega(C) = \{OK, failure\_mode_{1},failure\_mode_{2},failure\_mode_{3}, \ldots ,failure\_mode_{N}\} $$
The failure mode set $F$ for a given component or derived~component $C$ The failure mode set $F$ for a given component or derived~component $C$
is therefore is therefore
$$ F = \Omega(C) \backslash OK $$ $$ F = \Omega(C) \backslash \{OK\} $$
The $OK$ statistical case is the largest in probability, and is therefore The $OK$ statistical case is the largest in probability, and is therefore
of interest when analysing systems from a statistical perspective. of interest when analysing systems from a statistical perspective.