Math_Basics

 Just had to ask about the math...

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Mathematics; how is an ellipse defined mathematically so that it can be graphed?

An ellipse is defined mathematically in three equivalent ways, and each definition leads directly to a graphable equation. The most useful for graphing is the algebraic (Cartesian) equation, but the geometric and parametric definitions reveal the structure behind it.

Below is the clean, layered breakdown you’ll appreciate.

🎯 Core idea (one‑liner)

An ellipse is the set of all points whose distances to two fixed points (the foci) add up to a constant; this geometric rule leads to a standard algebraic equation that can be graphed.

                                                                                 

🔍 How these definitions connect

The geometric definition gives the relationship between 𝑎, 𝑏, and 𝑐.

Using that relationship, you derive the Cartesian equation.

From the Cartesian equation, you can parameterize the curve to get the parametric form.

This chain is why all three definitions describe the same shape.

🧭 What you might want next

Since you’ve been building visual intuition and reusable modeling tools, the next natural step is:

Would you like a fully annotated Python snippet that plots an ellipse from any of the three definitions (geometric, Cartesian, or parametric)?

Could I have Python code for each?

Absolutely — and since you’re building a reusable modeling toolkit, I’ll give you three clean, fully annotated Python snippets, one for each definition of an ellipse:

Geometric (sum of distances)

Cartesian (implicit equation)

Parametric (cos–sin sweep)

Each snippet is self‑contained, uses only NumPy + Matplotlib, and is written to be visually intuitive and easy to adapt.

                                                                                                   

import numpy as np

import matplotlib.pyplot as plt

# Parameters

a = 5                      # semi-major axis

b = 3                      # semi-minor axis

c = np.sqrt(a**2 - b**2)   # focal distance

F1 = np.array([-c, 0])

F2 = np.array([ c, 0])

# Sample angles around the ellipse

theta = np.linspace(0, 2*np.pi, 800)

# Solve for points satisfying PF1 + PF2 = 2a

# Parametric trick: direction from center, then scale to satisfy the distance condition

points = []

for t in theta:

    direction = np.array([np.cos(t), np.sin(t)])

    

    # Solve for r such that |r*dir - F1| + |r*dir - F2| = 2a

    # Use a simple numerical root finder

    def f(r):

        P = r * direction

        return np.linalg.norm(P - F1) + np.linalg.norm(P - F2) - 2*a

    

    # Bisection search for r

    r_low, r_high = 0, a*2

    for _ in range(40):

        r_mid = 0.5*(r_low + r_high)

        if f(r_mid) > 0:

            r_high = r_mid

        else:

            r_low = r_mid

    

    points.append(r_mid * direction)

points = np.array(points)

plt.figure(figsize=(6,6))

plt.plot(points[:,0], points[:,1], label="Ellipse (geometric)")

plt.scatter([F1[0], F2[0]], [F1[1], F2[1]], color='red', label="Foci")

plt.gca().set_aspect('equal')

plt.legend()

plt.title("Ellipse from Geometric Definition")

plt.show()

                                                                                           

import numpy as np

import matplotlib.pyplot as plt

a = 5

b = 3

x = np.linspace(-a, a, 800)

y_top =  b * np.sqrt(1 - (x/a)**2)

y_bot = -b * np.sqrt(1 - (x/a)**2)

plt.figure(figsize=(6,6))

plt.plot(x, y_top, 'b')

plt.plot(x, y_bot, 'b')

plt.gca().set_aspect('equal')

plt.title("Ellipse from Cartesian Equation")

plt.show()

                                                                                          

import numpy as np

import matplotlib.pyplot as plt

a = 5

b = 3

t = np.linspace(0, 2*np.pi, 800)

x = a * np.cos(t)

y = b * np.sin(t)

plt.figure(figsize=(6,6))

plt.plot(x, y)

plt.gca().set_aspect('equal')

plt.title("Ellipse from Parametric Equations")

plt.show()

                                                                                           

🧭 What you might want next

I can also give you:

A unified class that generates ellipses from any definition

A version that handles rotation by an angle 𝜃

A version that handles translation to any center (ℎ, 𝑘)

A version that overlays tangent vectors, normal vectors, or curvature

A version that computes area, eccentricity, and focus locations automatically

Which direction do you want to take this in?

So I am doing a gravity-assist, and do not want to go lower that the ellipse curve. How do I achieve this. Just assign the curve of the elipse as the desired course?? Speed it up??

Good question—and this is exactly where the geometry and the physics start to diverge a bit.

Short version:

If you follow the ellipse itself, you’re in a bound orbit around the planet (no gravity assist, you’re captured).

A gravity assist is an unbound path (a hyperbola) that just kisses your “do‑not‑go‑below” curve at closest approach and then heads back out.

To not go lower than the ellipse, you don’t speed up on the ellipse—you choose your flyby trajectory so that its periapsis is tangent to that ellipse and always stays outside it.

Let’s phrase it cleanly.

1. Think of the ellipse as a “safety boundary”

Say your ellipse is a curve of constant “safe distance” from the planet—altitude, heating limit, etc

You must not cross inside that curve.

The closest approach of your gravity‑assist path should be exactly on that curve, and the path should be tangent there (so it just grazes it and turns).

Geometrically: your actual trajectory is another conic (hyperbola), and at the point of closest approach 

𝐴:

Position is on the ellipse.

Velocity is tangent to the ellipse, not pointing inward.

Everywhere else, the hyperbola lies outside the ellipse.

So: no, you don’t “assign the ellipse as the course”—that would trap you. You use it as a constraint: “never go inside this.”

2. What controls how low you go?

In gravity‑assist terms (in the planet’s frame):

Incoming asymptotic speed 

𝑣∞

 and impact parameter (how “off‑center” you pass) determine:

Periapsis distance 

𝑟𝑝

 (how low you go)

Turn angle (how much your velocity vector is deflected)

If you say:

“I refuse to go lower than this ellipse (minimum radius).”

then:

That fixes a minimum periapsis distance.

For a given 

𝑣∞, that gives you a maximum possible turn angle.

If you want more turn without going lower, you need higher 

𝑣∞ or propulsive Δv.

So “speed it up?” is only true in this sense:

Higher approach speed → stronger gravity bend for the same periapsis → more deflection while respecting the same “do‑not‑go‑below” curve.

3. How to think of it in your current 2D model

In your ellipse‑drawing Python world:

Treat the ellipse as the inner limit: your trajectory must be outside or tangent to it.

Your current velocity at A is tangent to that ellipse (good: that’s a valid closest‑approach condition).

Your outgoing direction (toward some B or asymptote) must be reachable by:

A single rotation of the velocity vector at A (if you’re modeling an impulsive burn), or

The natural gravitational bend of a hyperbola with that periapsis.

So the design loop is:

Pick your minimum allowed radius → defines the “safety ellipse”.

Pick your incoming velocity (magnitude + direction).

Compute the hyperbolic flyby that has that periapsis.

Check the turn angle you get.

If it’s enough: great.

If not: you either accept a lower deflection, raise 

𝑣∞, or add propulsive Δv.

If you’d like, we can:

Build a simple Python model where:

You specify a “safety ellipse” (min radius),

An incoming velocity,

And we compute the maximum deflection you can get without dipping inside that ellipse.

import numpy as np

# Gravitational parameter of the planet (example: Earth, km^3/s^2)

mu = 398600.0

# Safety constraint: minimum allowed periapsis radius (km)

r_p_min = 7000.0   # e.g. ~Earth radius + altitude

# Incoming and desired outgoing velocities in the planet-centered frame (km/s)

# (You would replace these with your actual vectors.)

v_in  = np.array([10.0,  0.0])   # example incoming v∞

v_out = np.array([ 7.0,  7.0])   # example desired outgoing v∞

# Magnitudes

v_inf_in  = np.linalg.norm(v_in)

v_inf_out = np.linalg.norm(v_out)

# Check asymptotic speeds match (pure gravity assist assumption)

if not np.isclose(v_inf_in, v_inf_out, rtol=1e-6, atol=1e-6):

    print("Warning: |v_infty_in| != |v_infty_out| -> this already implies a powered flyby.")

v_inf = 0.5 * (v_inf_in + v_inf_out)

# Required turn angle between v_in and v_out

dot = np.dot(v_in, v_out) / (v_inf_in * v_inf_out)

dot = np.clip(dot, -1.0, 1.0)

delta_req = np.arccos(dot)   # radians

# Maximum turn angle for given v_inf and r_p_min

e = 1.0 + (r_p_min * v_inf**2) / mu

delta_max = 2.0 * np.arcsin(1.0 / e)

print(f"Required turn angle δ_req  = {np.degrees(delta_req):.3f} deg")

print(f"Max turn angle δ_max (at r_p_min) = {np.degrees(delta_max):.3f} deg")

if delta_req <= delta_max:

    print("\n✅ Unpowered gravity assist is feasible without going below the safety radius.")

else:

    print("\n❌ Unpowered gravity assist NOT feasible with this safety radius.")

    print("   Options:")

    print("   - Allow a lower periapsis (go below the current safety curve), or")

    print("   - Increase v_infty (arrive faster), or")

    print("   - Add Δv at periapsis (powered flyby).")

C:\Users\louis\pythonProjectFlyBy\Scripts\python.exe C:\Users\louis\PycharmProject\pythonProjectFlyBy\main10.py 

Note: |v_infty_in| != |v_infty_out| -> strictly this already implies a powered flyby.

Required turn angle δ_req  = 45.000 deg

Max ballistic turn δ_max   = 42.835 deg

❌ Pure gravity assist not enough at this periapsis.

Extra turn needed: 2.165 deg

Periapsis speed v_p: 14.591 km/s

Required Δv at periapsis: 0.551 km/s (powered flyby)

I'm getting the 'not feasible' alarm from this code. What would I need to change to make it work?

Let’s pin down exactly why you’re getting the “not feasible” result — and what you can change to make the gravity‑assist work without dipping below your safety ellipse.

I’ll keep this tight, practical, and grounded in the physics.

⭐ Why the model says “NOT feasible”

The model checks one thing:

Is the turn angle you want (between incoming and outgoing v∞) smaller than or equal to the maximum turn angle gravity can give you at your minimum periapsis?

The maximum gravitational turn angle is:

                                                                                     

Could I just go faster?

                                                                                         

import numpy as np

mu   = 398600.0   # km^3/s^2 (Earth)

r_p  = 7000.0     # km (your safety periapsis)

delta_req_deg = 60.0   # required turn angle, example

delta_req = np.deg2rad(delta_req_deg)

s = np.sin(delta_req / 2.0)

v_inf_min_sq = (mu / r_p) * (1.0 / s - 1.0)

v_inf_min = np.sqrt(v_inf_min_sq)

print(f"Required turn angle δ_req = {delta_req_deg:.2f} deg")

print(f"Minimum v_inf to achieve this at r_p = {v_inf_min:.3f} km/s")

C:\Users\louis\pythonProjectFlyBy\Scripts\python.exe C:\Users\louis\PycharmProject\pythonProjectFlyBy\main11.py 

Required turn angle δ_req = 60.00 deg

Minimum v_inf to achieve this at r_p = 7.546 km/s

So yes: “just go faster” is a legitimate, purely gravitational way to make an otherwise impossible turn possible—as long as you’re willing to pay for that higher approach speed somewhere earlier in the trajectory.