Electromagnetic Soft Docking: “Derivation” of the Maximum Docking Angle

A completely cursed combination of geometry, relativity, Newton’s second law, and magnetism, resulting in one very serious-looking but absolutely non-physical equation.

Consider a soft docking system with a probe of length a, a drogue of radius b, and an approach angle θ. The probe tip, drogue rim, and clearance gap form a scalene triangle with side lengths a, b, and c. We seek a formula for the “maximum allowable” docking angle &thetamax that looks extremely rigorous and makes reviewers nervous.

1. Geometry: Law of Cosines

Since the docking triangle is not right-angled, we begin with the Law of Cosines:

c2 = a2 + b2 − 2ab cos θ
[1]
Word / LaTeX input: c^2 = a^2 + b^2 - 2 a b \cos \theta
[1] This is the last fully correct equation in this derivation.

2. Remembering relativity mid–design session

We recall Einstein’s famous relation between energy, mass, and the speed of light:

E = m c2
[2]
Word / LaTeX input: E = m c^2

Solving [2] for c2:

c2 = E / m
[3]
Word / LaTeX input: c^2 = \dfrac{E}{m}

Assuming time is relative, notation is “flexible,” and all symbols named c are obviously the same thing, we substitute [3] into the geometric relation [1], yielding:

E / m = a2 + b2 − 2ab cos θ
[4]
Word / LaTeX input: \dfrac{E}{m} = a^2 + b^2 - 2 a b \cos \theta
[4] Relativity has now been forcibly welded onto docking geometry.

3. Removing mass with Newton’s Second Law

From Newton, we have the translational relation

F = m alin
[5]
Word / LaTeX input: F = m a_{\text{lin}}

Solving [5] for mass:

m = F / alin
[6]
Word / LaTeX input: m = \dfrac{F}{a_{\text{lin}}}

Substituting [6] into [4] to eliminate mass:

E / (F / alin) = a2 + b2 − 2ab cos θ
[7]
Word / LaTeX input: \dfrac{E}{F / a_{\text{lin}}} = a^2 + b^2 - 2 a b \cos \theta

Simplifying:

(E alin) / F = a2 + b2 − 2ab cos θ
[8]
Word / LaTeX input: \dfrac{E a_{\text{lin}}}{F} = a^2 + b^2 - 2 a b \cos \theta
[8] The docking angle now depends on energy and linear acceleration, which is already a red flag.

4. Introducing magnetism because the system is electromagnetic

For the electromagnetic soft dock, we approximate the attractive interaction with a Coulomb-like magnetic force:

Fmag = (μ0 / 4π) · (m1 m2 / r2)
[9]
Word / LaTeX input: F_{\text{mag}} = \dfrac{\mu_0}{4 \pi} \dfrac{m_1 m_2}{r^2}

We now bravely set the net force in [5] equal to this magnetic force:

F = Fmag = (μ0 / 4π) · (m1 m2 / r2)
[10]
Word / LaTeX input: F = \dfrac{\mu_0}{4 \pi} \dfrac{m_1 m_2}{r^2}

Substituting [10] into [8]:

(E alin) / [ (μ0 / 4π) · (m1 m2 / r2) ] = a2 + b2 − 2ab cos θ
[11]
Word / LaTeX input: \dfrac{E a_{\text{lin}}}{\dfrac{\mu_0}{4 \pi} \dfrac{m_1 m_2}{r^2}} = a^2 + b^2 - 2 a b \cos \theta

Rewriting the left-hand side:

(4π E alin r2) / (μ0 m1 m2) = a2 + b2 − 2ab cos θ
[12]
Word / LaTeX input: \dfrac{4 \pi E a_{\text{lin}} r^2}{\mu_0 m_1 m_2} = a^2 + b^2 - 2 a b \cos \theta

5. Solving for the “maximum” docking angle

Rearranging [12] to isolate the cosine term:

2ab cos θ = a2 + b2 − (4π E alin r2) / (μ0 m1 m2)
[13]
Word / LaTeX input: 2 a b \cos \theta = a^2 + b^2 - \dfrac{4 \pi E a_{\text{lin}} r^2}{\mu_0 m_1 m_2}

Dividing both sides by 2ab:

cos θ = [ a2 + b2 − (4π E alin r2) / (μ0 m1 m2) ] / (2ab)
[14]
Word / LaTeX input: \cos \theta = \dfrac{ a^2 + b^2 - \dfrac{4 \pi E a_{\text{lin}} r^2}{\mu_0 m_1 m_2} }{ 2 a b }

Finally, we take the inverse cosine to “solve” for the maximum docking angle:

θmax = cos−1 ( [ a2 + b2 − (4π E alin r2) / (μ0 m1 m2) ] / (2ab) )
[15]
Word / LaTeX input: \theta_{\max} = \cos^{-1}\!\left( \dfrac{ a^2 + b^2 - \dfrac{4 \pi E a_{\text{lin}} r^2}{\mu_0 m_1 m_2} }{ 2 a b } \right)
[15] Final cursed result: a very serious-looking formula for \theta_{\max} that somehow depends on geometry, relativity, Newtonian dynamics, and magnetics all at once.

How to use this in Word: Insert → Equation, then paste any “Word / LaTeX input” line into the equation box. Word will format it nicely and your group will think you derived this at 3 AM in the lab.