	Truncating VSOP87
	Step 1 Step 2 Selecting the terms Questions The purpose of this page is to find a way to remove terms of VSOP87 in order to have a given geocentric maximal error. For this, we'll first consider that the position of Earth is considered as exactly known ; the incertitude on the Earth position will then be taken into account.

So we want to compute the position of the planets, and have a certain accuracy. We call α _Max the maximal angular error we tolerate on the position of a planet seen from Earth. The purpose is to find the number of terms to keep to get a precision of α_Max.

We introduce an intermediary variable : err_Max, defined as the error on heliocentric cartesian coordinates (given by VSOP87A). The angular geocentric error α_Max will be generated by heliocentric error err_Max.
Our purpose will be to find a relation between err_Max, and α_Max, and use it to filter the terms of VSOP87.

Step 1

In this first step, we will suppose that the position of the Earth is exactly known, and try to find a relation between err_Max and α_Max.

Let's see the extreme cases we can meet :

S : position of the Sun
E : position of the Earth
P : real position of the planet
P' : position of the planet affected of err_Max

Here, d1 and d2 represent the value of err_Max in two extreme configurations.
In case 1, the distance Earth - planet is minimal ; in case 2, it is maximal.

In both cases we have :

tg(α_Max) = d / EP

(formula 1)

then : d = EP * tg(α_Max)
And EP in case 2 > EP in case 1 ; this implies that : d2 > d1.

The same α_Max will be generated by a smaller err_Max on the position in case 1.
We will then consider case 1, where the distance between the Earth and the planet is minimal, because this is the worst case.

If we take into account the fact that the orbits are elliptic, let's see this situation for an outer planet :
(I didn't check if it was necessary to consider eccentricities here).

If we note :

a_E and a_P the semi major axis of the Earth and the planet ;

e_E and e_P the eccentricities of the Earth and the planet :

We have :

EP = SP - SE

SP = a_P(1 - e_P)

SE = a_E(1 + e_E)

Then :

EP = a_P(1 - e_P) - a_E(1 +e_E)

(formula 2.1)

A similar drawing for an inner planet would lead to :

EP = a_E(1 - e_E) - a_P(1 +e_P)

(formula 2.2)

For the Earth, the formula is :

ES = a_E(1 - e_E)

(formula 2.3)

Formula 1 and formulae 2 permit to calculate d from α_Max.

But d = PP', distance between the real position of the planet and the position affected by the error.
If we note :

, we have :

.
If we tolerate the same error on X, Y or Z, we have : ΔX = ΔY = ΔZ = err_Max.
Then :

As we have : d = EP * tg(α_Max), this leads to :

(formula 3)

err_Max can then be calculated using formulae 2 and 3 :

(formula 4.1), for an outer planet.

(formula 4.2), for an inner planet.

(formula 4.3), for the Earth.

In formulae 4, err_Max is expressed in the same unit as a_E and a_P.

α_Max is in radians.

Step 2

We will now take into account the incertitude on the position of the Earth, which poses a supplementary problem. Let's see the situation for an outer planet :

Here we introduce a new angle : α₁, wich indicates the incertitude on the Earth's position. α₁ is also the maximal angular error on the position of the Sun seen from Earth.

We call : α₀ the maximal angular error on the position of the planet seen from Earth.

α₂ is defined by the drawing.

With the notations of step 1, we have :

α₀ = α_Max(Planet)

α₁ = α_Max(Sun)

On figure 3, we see that, in the worst case, the incertitude of α₁ on the position of the Earth leads to an incertitude of α₀ + α₂ on the position of the planet as seen from Earth.

We call : α = α₀ + α₂. α is the total incertitude on the position of the planet.

We now calculate α₂ :

Then :

(formula 5)

A similar drawing for an inner planet leads to the same result.

The worst case (α maximal) happens when ES is maximal and EP is minimal.
So in formula 5, we should take ES = a_E(1 - e_E) and EP from formula 2.1 or 2.2.
(Notice that we'll never have at the same time ES maximal and EP minimal).

Selection of terms

So we are able to find the maximal geocentric angular error (α) from α_O and α₁.

We now have to choose α₁ (for the Earth) and α₀ for each planet such as :

α < α_Max for each planet, and

α₁ < α_Max (for the Earth).

Taking α_Max = 1 arc second, we choosed α₁ and the α₀s by successive attempts.
The method BuildVSOP87.calcAlphaMax() was written for this purpose.

This leads to the following result :

Planet	MER	VEN	EAR	MAR	JUP	SAT	URA	NEP
α₀	0.5	0.1	0.2	0.4	0.9	0.9	0.9	0.9
α	0.901	0.906	0.2	0.962	0.952	0.926	0.912	0.903

(Table 1)

Once we have these values, we can select the terms. We have :

, the contribution of each term to the sum is lower than

The condition to retain a term is then

In this formula, err_Max is expressed with formulae 4, in which α_Max are the value of α in table 1.

Choice of T_Max : as T varies, we must know the maximal values it can take. In VSOP87.doc in BDL FTP, we can read that the "intervals for which 1 arcsecond of precision is guaranteed". Anyway, I took these intervals as intervals of validity, and found T_Max :

Planet	T_Max
MER VEN EAR MAR	4
JUP SAT	2
URA NEP	6

Questions

As you'll see in the page Testing VSOP, this method does not give the expected results.
I listed here the points that could lead to this situation. Maybe you'll have better ideas.

The eccentricity varies with time. So in figure 2 and after, should we take this into account?

Is it right to admit the same error for X, Y, and Z? As planets vary much more in X and Y than in Z, maybe we should consider that norm(vector(PP')) = DeltaX2 + DeltaY2.

in figure 3, P and P' represent the planet seen from Earth. I supposed I could say that P and P' represent the planet as seen from E₀ and E₁. In fact, they are slightly different. There should be 4 points representing the planet. But à priori, the difference is very small, so I neglected that.

The choice of T_Max at the end is not really coherent.

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