Neutrino decoupling

Hansen's problem is a problem in planar surveying, named after the astronomer Peter Andreas Hansen (1795–1874), who worked on the geodetic survey of Denmark. There are two known points A and B, and two unknown points P₁ and P₂. From P₁ and P₂ an observer measures the angles made by the lines of sight to each of the other three points. The problem is to find the positions of P₁ and P₂. See figure; the angles measured are (α₁, β₁, α₂, β₂).

Since it involves observations of angles made at unknown points, the problem is an example of resection (as opposed to intersection).

Solution method overview

Define the following angles: γ = P₁AP₂, δ = P₁BP₂, φ = P₂AB, ψ = P₁BA. As a first step we will solve for φ and ψ. The sum of these two unknown angles is equal to the sum of β₁ and β₂, yielding the following equation:

$\varphi +\psi ={\beta }_1+{\beta }_2$

A second equation can be found more laboriously, as follows. The law of sines yields

$\frac{AB}{P_{2}B}=\frac{\sin {\alpha }_2}{\sin \varphi }$ and

$\frac{P_{2}B}{P_1{P}_2}=\frac{\sin {\beta }_1}{\sin \delta }$

combining these together we get

$\frac{AB}{P_1{P}_2}=\frac{\sin {\alpha }_2\sin {\beta }_1}{\sin \varphi \sin \delta }$

An entirely analogous reasoning on the other side yields

$\frac{AB}{P_1{P}_2}=\frac{\sin {\alpha }_1\sin {\beta }_2}{\sin \psi \sin \gamma }$

Setting these two equal gives

$\frac{\sin \varphi }{\sin \psi }=\frac{\sin \gamma \sin {\alpha }_2\sin {\beta }_1}{\sin \delta \sin {\alpha }_1\sin {\beta }_2}=k$

Using a known trigonometric identity this ratio of sines can be expressed as the tangent of an angle difference:

$\tan \frac{\varphi -\psi }{2}=\frac{k-1}{k+1}\tan \frac{\varphi +\psi }{2}$

This is the second equation we need. Once we solve the two equations for the two unknowns $\varphi$ and $\psi$, we can use either of the two expressions above for $\frac{AB}{P_1{P}_2}$ to find P₁P₂ since AB is known. We can then find all the other segments using the law of sines.

Solution algorithm

We are given four angles (α₁, β₁, α₂, β₂) and the distance AB. The calculation proceeds as follows:

• Calculate $\gamma =\pi -{\alpha }_1-{\beta }_1-{\beta }_2,\delta =\pi -{\alpha }_2-{\beta }_1-{\beta }_2$

• Calculate $k=\frac{\sin \gamma \sin {\alpha }_2\sin {\beta }_1}{\sin \delta \sin {\alpha }_1\sin {\beta }_2}$

• Let $s={\beta }_1+{\beta }_2$, $d=2\arctan [\frac{k-1}{k+1}\tan (s/2)]$ and then $\varphi =(s+d)/2$, $\psi =(s-d)/2.$

• Calculate

$P_1{P}_2=AB\frac{\sin \varphi \sin \delta }{\sin {\alpha }_2\sin {\beta }_1},$

or equivalently,

$P_1{P}_2=AB\frac{\sin \psi \sin \gamma }{\sin {\alpha }_1\sin {\beta }_2}.$

If one of these fractions has a denominator close to zero, use the other one.

References

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