From ffbea207f612cdf5fe5501d0fcd0b2e980e5b95e Mon Sep 17 00:00:00 2001 From: Jean-Baptiste Caillau Date: Thu, 12 Mar 2026 11:27:32 +0100 Subject: [PATCH] Fix formatting in balanced_field.md --- ext/Descriptions/balanced_field.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/ext/Descriptions/balanced_field.md b/ext/Descriptions/balanced_field.md index ca67aedd..8da31b5b 100644 --- a/ext/Descriptions/balanced_field.md +++ b/ext/Descriptions/balanced_field.md @@ -4,12 +4,12 @@ This implementation focuses on the **takeoff climb** phase with one engine out, ### Mathematical formulation The problem is to minimise the final range $r(t_f)$ to reach the screen height (usually 35 ft). -The state vector is $x(t) = [r(t), v(t), h(t), \gamma(t)]^ op$ and the control is the angle of attack $\alpha(t)$. +The state vector is $x(t) = (r(t), v(t), h(t), \gamma(t))$ and the control is the angle of attack $\alpha(t)$. ```math \begin{aligned} \min_{\alpha, t_f} \quad & r(t_f) \\ - ext{s.t.} \quad & \dot{r}(t) = v(t) \cos \gamma(t), \\ + \text{s.t.} \quad & \dot{r}(t) = v(t) \cos \gamma(t), \\ & \dot{v}(t) = \frac{T \cos \alpha(t) - D}{m} - g \sin \gamma(t), \\ & \dot{h}(t) = v(t) \sin \gamma(t), \\ & \dot{\gamma}(t) = \frac{T \sin \alpha(t) + L}{m v(t)} - \frac{g \cos \gamma(t)}{v(t)}, \\