Fix lambda_infty notation

APRI0004-ADP/BEMT/bemt_prop_slides.tex

@@ -488,11 +488,11 @@ Assume
   The Blade Element Theory gave us previously
   $$dC_T = \dfrac{1}{2}\sigma c_l r^2 dr = \dfrac{\sigma c_{l_\alpha}}{2}(\theta r^2-\lambda r)dr$$
   Combining the two results gives a quadratic equation
-  $$\lambda^2 + \left(\dfrac{\sigma c_{l_\alpha}}{8}-\lambda_c\right)\lambda - \dfrac{\sigma c_{l_\alpha}}{8}\theta r = 0$$
+  $$\lambda^2 + \left(\dfrac{\sigma c_{l_\alpha}}{8}-\lambda_\infty\right)\lambda - \dfrac{\sigma c_{l_\alpha}}{8}\theta r = 0$$
   and solving for $\lambda$ leads to
-  $$ \lambda(r,\lambda_c) = \sqrt{\left(\dfrac{\sigma c_{l_\alpha}}{16} -
-  \dfrac{\lambda_c}{2}\right)^2 + \dfrac{\sigma c_{l_\alpha}}{8}\theta r} -
-  \left(\dfrac{\sigma c_{l_\alpha}}{16} - \dfrac{\lambda_c}{2}\right)$$
+  $$ \lambda(r,\lambda_\infty) = \sqrt{\left(\dfrac{\sigma c_{l_\alpha}}{16} -
+  \dfrac{\lambda_\infty}{2}\right)^2 + \dfrac{\sigma c_{l_\alpha}}{8}\theta r} -
+  \left(\dfrac{\sigma c_{l_\alpha}}{16} - \dfrac{\lambda_\infty}{2}\right)$$
 
 \end{frame}
 
@@ -610,14 +610,14 @@ Assume
   \textbf{Prandtl's tip-loss function}
   \begin{itemize}
     \item The correction factor can be incorporated in the BEMT\\[0.2cm]
-      $dC_T = 4 \textcolor{mLightBrown}{F} \lambda(\lambda-\lambda_c)rdr$
+      $dC_T = 4 \textcolor{mLightBrown}{F} \lambda(\lambda-\lambda_\infty)rdr$
       \vspace{0.35cm}
     \item In this case, the inflow velocity becomes\\[0.2cm]
-      $\lambda(r,\lambda_c) = \sqrt{\left(\dfrac{\sigma c_{l_\alpha}}{16
-        \textcolor{mLightBrown}{F}} - \dfrac{\lambda_c}{2}\right)^2 +
+      $\lambda(r,\lambda_\infty) = \sqrt{\left(\dfrac{\sigma c_{l_\alpha}}{16
+        \textcolor{mLightBrown}{F}} - \dfrac{\lambda_\infty}{2}\right)^2 +
       \dfrac{\sigma c_{l_\alpha}}{8 \textcolor{mLightBrown}{F}}\theta r} -
       \left(\dfrac{\sigma c_{l_\alpha}}{16 \textcolor{mLightBrown}{F}} -
-      \dfrac{\lambda_c}{2}\right)$
+      \dfrac{\lambda_\infty}{2}\right)$
       \vspace{0.35cm}
     \item Iterative process as $F = f(\phi) = f(\lambda)$!
   \end{itemize}