Fix errors

bryanwweber · Jun 16, 2024 · a85f9f9 · a85f9f9
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Showing 8 changed files with 16 additions and 7 deletions.
diff --git a/classical-orbital-elements/scripts/orbital_elements_and_the_state_vector.py b/classical-orbital-elements/scripts/orbital_elements_and_the_state_vector.py
@@ -51,6 +51,7 @@
  warnings.simplefilter("ignore")
  from myst_nb import glue
 
+ np.set_printoptions(legacy="1.25")
  glue("orbital-elements-radius", r, display=False)
  glue("orbital-elements-velocity", v, display=False)
  glue("orbital-elements-v_r", v_r, display=False)

diff --git a/interplanetary-maneuvers/planetary-arrival-flyby.md b/interplanetary-maneuvers/planetary-arrival-flyby.md
@@ -250,6 +250,7 @@ h_1 = np.sqrt(mu_sun * R_Neptune * (1 - e_1))
 from functools import partial
 from myst_nb import glue as myst_glue
 glue = partial(myst_glue, display=False)
+np.set_printoptions(legacy="1.25")
 glue("interplanetary-flyby-e_1", e_1)
 glue("interplanetary-flyby-h_1", h_1)
 ```

diff --git a/orbital-maneuvers/comparison-of-bielliptic-and-hohmann.md b/orbital-maneuvers/comparison-of-bielliptic-and-hohmann.md
@@ -41,6 +41,7 @@ v_3 = np.sqrt(mu / r_3) # km/s
 
 ```{code-cell} ipython3
 :tags: [remove-cell]
+np.set_printoptions(legacy="1.25")
 glue("moon_circle_velocity", v_1)
 glue("leo_circle_velocity", v_3)
 ```

diff --git a/orbital-maneuvers/plane-change-example.md b/orbital-maneuvers/plane-change-example.md
@@ -66,6 +66,7 @@ Delta_v_1 = Delta_v_1_LEO + Delta_v_1_GEO
 
 ```{code-cell} ipython3
 :tags: [remove-cell]
+np.set_printoptions(legacy="1.25")
 from functools import partial
 from myst_nb import glue as myst_glue
 glue = partial(myst_glue, display=False)

diff --git a/orbital-maneuvers/single-impulse-example.md b/orbital-maneuvers/single-impulse-example.md
@@ -29,6 +29,8 @@ from matplotlib.patches import Circle, Arc
 import numpy as np
 from myst_nb import glue
 
+np.set_printoptions(legacy="1.25")
+
 R_E = 6378 # km
 orbit_radius = 1000 # km
 mu = 398_600 # km**2/s**3

diff --git a/time-since-periapsis-and-keplers-equation/elliptical-orbit-example.md b/time-since-periapsis-and-keplers-equation/elliptical-orbit-example.md
@@ -35,7 +35,6 @@ e = \frac{r_a - r_p}{r_a + r_p}
 :::
 
 ```{code-cell} ipython3
-# %matplotlib notebook
 import numpy as np
 from scipy.optimize import newton
 
@@ -51,6 +50,7 @@ e = (r_a - r_p)/(r_a + r_p)
 :tags: [remove-cell]
 from functools import partial
 from myst_nb import glue as myst_glue
+np.set_printoptions(legacy="1.25")
 glue = partial(myst_glue, display=False)
 glue("ellipse-time-since-periapsis-e", e)
 ```
@@ -149,7 +149,7 @@ def kepler(E, M_e, e):
 
 def d_kepler_d_E(E, M_e, e):
  """The derivative of Kepler's equation, to be used in a Newton solver.
- 
+
  Note that the argument M_e is unused, but must be present so the function
  arguments are consistent with the kepler function.
  """

diff --git a/time-since-periapsis-and-keplers-equation/elliptical-orbit-time-in-shadow.md b/time-since-periapsis-and-keplers-equation/elliptical-orbit-time-in-shadow.md
@@ -29,6 +29,7 @@ from matplotlib.patches import Ellipse, Circle, Arc, Rectangle
 import numpy as np
 
 glue = partial(myst_glue, display=False)
+np.set_printoptions(legacy="1.25")
 
 mu = 3.986004418E5 # km**3/s**2
 R_E = 6378 # km

diff --git a/time-since-periapsis-and-keplers-equation/hyperbolic-trajectory-example.md b/time-since-periapsis-and-keplers-equation/hyperbolic-trajectory-example.md
@@ -41,6 +41,8 @@ e = h**2 / (r_p * mu) - 1
 :tags: [remove-cell]
 from functools import partial
 from myst_nb import glue as myst_glue
+
+np.set_printoptions(legacy="1.25")
 glue = partial(myst_glue, display=False)
 
 glue("hyperbolic-time-since-perigee-h", h)
@@ -117,7 +119,7 @@ def kepler(F, M_h, e):
 
 def d_kepler_d_F(F, M_h, e):
  """The derivative of Kepler's equation, to be used in a Newton solver.
- 
+
  Note that the argument M_h is unused, but must be present so the function
  arguments are consistent with the kepler function.
  """
@@ -129,7 +131,7 @@ F_2 = newton(func=kepler, fprime=d_kepler_d_F, x0=np.pi, args=(M_h2, e))
 With this value for $F$, we can calculate the value for $\nu$. To avoid quadrant ambiguity problems, we will use Eq. {eq}`eq:eccentric-anomaly-true-anomaly-hyperbola`.
 
 ```{code-cell} ipython3
-sqrt_e = np.sqrt((e + 1) / (e - 1)) 
+sqrt_e = np.sqrt((e + 1) / (e - 1))
 nu_2 = (2 * np.arctan(sqrt_e * np.tanh(F_2 / 2))) % (2 * np.pi)
 ```
 
@@ -181,12 +183,12 @@ function kepler
  v_p = 15; % km/s
  h = r_p * v_p;
  e = h^2 / (mu * r_p) - 1;
- 
+
  nu_1 = deg2rad(100);
  F_1 = 2 * atanh(sqrt((e - 1) / (e + 1)) * tan(nu_1 / 2));
  M_h1 = e * sinh(F_1) - F_1;
  t_1 = h^3 / mu^2 * 1 / (e^2 - 1)^(3 / 2) * M_h1;
- 
+
  t_2 = t_1 + 3 * 3600;
  M_h2 = mu^2 / h^3 * (e^2 - 1)^(3 / 2) * t_2;
 
@@ -198,7 +200,7 @@ function kepler
  t2 = 2 * atan(sqrt((e + 1) / (e - 1)) * tanh(F_2 / 2));
  nu_2 = mod(t2, 2 * pi);
  disp(rad2deg(nu_2))
- 
+
  r_2 = h^2 / mu * 1 / (1 + e * cos(nu_2));
  v_perp = h / r_2;
  v_r = mu / h * e * sin(nu_2);