iccsa-20-wind

commit 89a2e7dc78639177601746f92f7526c32e052e44
parent 26122bc797c140e8f5706d0dfd28b02e3b19e724
Author: Ivan Gankevich <i.gankevich@spbu.ru>
Date:   Sun, 15 Mar 2020 16:09:56 +0300

Spell check.

Diffstat:
main.tex
 | 
10
+++++-----

1 file changed, 5 insertions(+), 5 deletions(-)
diff --git a/main.tex b/main.tex
@@ -126,8 +126,8 @@ analysis of our programme.
 Air motion without turbulence can be decomposed into two components:
 translational motion~--- air particles travel in the same direction with
 constant velocity, circular motion~--- air particles travel on a circle.
-Translational motion describe sea breeze, that occures on the shore on the
-sunrise and after the sunset. Rotational motion describe storms suchs as
+Translational motion describe sea breeze, that occurs on the shore on the
+sunrise and after the sunset. Rotational motion describe storms such as
 typhoons and hurricane. Translational motion is a particular case of circular
 motion when the radius of the circle is infinite. Given the scale of circular
 motion relative to the scale of translational motion, and the size of a typical
@@ -292,11 +292,11 @@ following well-known formula:
 Here \(r\) and \(\theta\) are polar coordinates, \(R\) is cylinder radius and \(U\)
 is \(x\) component of velocity. Cylinder is placed at the origin. To prove that our
 solution on the boundary~\eqref{eq-solution-on-the-boundary} reduces to this solution
-we reduce it to cartesian form using polar coordinate identities
+we reduce it to Cartesian form using polar coordinate identities
 \begin{equation*}
     r = \sqrt{x^2+y^2}; \qquad \theta = \arccos{\frac{x}{\sqrt{x^2+y^2}}}.
 \end{equation*}
-Then in cartesian coordinates the solution is written as
+Then in Cartesian coordinates the solution is written as
 \begin{equation*}
     \phi(x,y) = U x \left(1 + \frac{R^2}{x^2+y^2}\right)
 \end{equation*}
@@ -317,7 +317,7 @@ quite surprisingly reduces to the same expression.
 To reduce solution near the boundary~\eqref{eq-solution-near-the-boundary} to
 the solution for potential flow around a cylinder, we let
 \(s=\Length{\vec{r}}^2/\Length{\vec{S}}^2\) (here \(\vec{r}\) is the radius
-vector in cartesian coordinates). Then the solution is written as
+vector in Cartesian coordinates). Then the solution is written as
 \begin{equation*}
     \vec\nabla\phi = \vec\upsilon + \frac{1}{s} \vec\upsilon_r
 \end{equation*}