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Old 06-02-2019, 01:35 AM   #1
ncc1701d
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Default Can I get the Vector components xyz given only this information?

hello,
I have these 3 dates
and I know the position X Y Z of moon Amalthea relative to Jupiter
and then I know the Velocity Vectors V for each one.
I need to get the VX VY VZ components for each of the 3 date-times.
Do I have enough information to get those from what you see?
I believe I need the instantaneous values.
I am aware of VX = dx/dt, VY = dy/dt, VZ = dz/dt
and I know VX + VY + VZ = V
I am interested in the math behind finding those components given what I have.
If you know the procedure can you tell me what it is? or shed some light on how I might solve the problem if its possible?
thank you.

1979 MAR 05 00:00:00
-178932.619 km -28063.045 km -17448.755 km V = -29.804114 km/sec

1979 MAR 05 00:02:00
-178337.419 km -30878.143 km -18771.071 km V = -29.231286 km/sec

1979 MAR 05 00:04:00
-177688.033 km -33683.859 km -20087.682 km V = 28.649604 km/sec

thank you.
I do have the answers if someone knew how to do it. I can check to see if your right.

Last edited by ncc1701d; 06-02-2019 at 01:45 AM.
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Old 06-02-2019, 01:48 AM   #2
Quick_Nick
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If I understand correctly, you have XYZ position information and a speed, and you want XYZ velocity information.

This can get really complicated if you care about a perfect answer.

However I would say simply knowing that the orbit is nearly circular and nearly equatorial is enough information to have a pretty good idea of the velocity vector.
You need to know what coordinate frame those XYZ values are in though.

Edit: I'm thinking you may actually care about a more accurate answer given your three times are less than an orbit apart.
I suspect you can in fact evaluate those three points together to get some orbital parameters out.

Last edited by Quick_Nick; 06-02-2019 at 01:55 AM.
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Old 06-03-2019, 01:16 PM   #3
Marijn
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I think this article describes the required math. It's not easy. https://science.larouchepac.com/gaus...erProblem.html
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Old 06-06-2019, 02:32 PM   #4
BrianJ
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Hi,
thanks for the interesting problem to think about :-)

I can get the semimajor axis (a) and the standard gravitational parameter (mu) from the vis-viva equation;

v^2 = mu*(2/r - 1/a)

But then I get stuck!

Then I thought maybe it was an energy-conservation problem (Potential + Kinetic = Constant) but I'm not getting anywhere with that.

Since the times of the data points are given, I'm pretty sure that is necessary information. Needs some kind of insight regarding Kepler's 2nd law (radius sweeps out equal areas in equal times) maybe?

I think that the shape of an ellipse is completely determined by three points (pos vectors), so theoretically you should be able to find the equation of the ellipse and the tangent at a given point - but the math for doing that in 3D is looking really gnarly, so I'm not even thinking about that :-)

Anyway, I'm having fun playing with this one.
Cheers,
Brian

P.S. Your first two Velocity values are negative - typo?
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Old 06-06-2019, 06:08 PM   #5
ncc1701d
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the checkmarked answer I found here might provide the answer.
https://stackoverflow.com/questions/...elocity-vector

Although not sure its that simple yet. I am still researching.
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Old 07-12-2019, 02:35 PM   #6
MontBlanc2012
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Quote:
Originally Posted by BrianJ View Post
 Hi,
thanks for the interesting problem to think about :-)

I can get the semimajor axis (a) and the standard gravitational parameter (mu) from the vis-viva equation;

v^2 = mu*(2/r - 1/a)

But then I get stuck!
I'm a bit suspicious of the orbital speeds given in the opening post, but here is the general procedure for solving this problem for Keplerian orbits where the points are separated by less than half an orbit:

Let's suppose, as above, that you are given two (3-vector) points on an Keplerian orbit \mathbf{R}_i and \mathbf{R}_f and that you know the (scalar) orbital speeds at those points, V_i and V_f respectively.

Now calculate:

\begin{aligned}r_i &= \sqrt{\mathbf{R}_i.\mathbf{R}_i} \\r_f &= \sqrt{\mathbf{R}_f.\mathbf{R}_f} \\\theta &=\cos^{-1}\left(\frac{\mathbf{R}_i.\mathbf{R}_f}{r_i\,r_f}\right) \\A &= r_i + r_f \\B &= 2\,\sqrt{r_i\,r_f}\,\cos\left(\frac{\theta}{2}\right)\end{aligned}

In addition, calculate the semi-major axis at either the first or last point using:

V_i^2 = \mu\,\left(\frac{2}{r_i}-\frac{1}{a}\right)

or

V_f^2 = \mu\,\left(\frac{2}{r_f}-\frac{1}{a}\right)

Then (and here's the magic!), solve for the two roots of the quadratic equation:

A-B\,X = 2\,a\,(1 - X^2)

Call these roots X_1 and X_2. These two numbers encode information about the two Keplerian arcs that join the points \mathbf{R}_i and \mathbf{R}_f with semi-major axis a.

Then, the radial and transverse components of the velocity vector at the \mathbf{R}_i are given by:

\left(v_{i,r},v_{i,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X}}\,\left(\sqrt{\frac{r_f}{r_i}}\,\cos\left(\frac{\theta}{2}\right)- X,\sqrt{\frac{r_f}{r_i}}\,\sin\left(\frac{\theta}{2}\right)\right)

and the radial and transverse components of the velocity vector at the \mathbf{R}_f are given by:

\left(v_{f,r},v_{f,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X}}\,\left(X - \sqrt{\frac{r_i}{r_f}}\,\cos\left(\frac{\theta}{2}\right),\sqrt{\frac{r_i}{r_f}}\,\sin\left(\frac{\theta}{2}\right)\right)

As mentioned, this gives two possible solutions - corresponding to the two possible Keplerian arcs that pass through the initial and final points with set-major-axis a. For each solution, we calculate the eccentricity in the usual way. In practice, for problems involving nearly circular orbits, one value of X will give a high eccentricity; and the other will give one close to zero. Obviously, we choose the solutions corresponding to the low orbital eccentricity - thereby removing the degeneracy.

(N.B., I have a proof of the above but it's a little long-winded - too long for this post. But I can share if anyone is sufficiently interested)

---------- Post added at 02:35 PM ---------- Previous post was at 01:23 PM ----------

Let's take a simple worked example:

Suppose that we choose units such that \mu=1 with:

\begin{aligned}
\mathbf{R}_i &= \left(-0.9803802, 0.241571, 0\right) \\
\mathbf{R}_f &= \left(-0.9972285, -0.159303, 0\right) \\
\\
V_i &= 0.9903428 \\
V_f &= 0.9901760
\end{aligned}

Find the velocity vectors at \mathbf{R}_i and \mathbf{R}_f.

Using the above algorithm we calculate:

\begin{aligned}
r_i &= \sqrt{\mathbf{R}_i.\mathbf{R}_i} = 1.009704\\
r_f &= \sqrt{\mathbf{R}_f.\mathbf{R}_f} = 1.009872 \\
\theta &=\cos^{-1}\left(\frac{\mathbf{R}_i.\mathbf{R}_f}{r_i\,r_f}\right) = 0.4000000\\
A &= r_i + r_f = 2.019576\\
B &= 2\,\sqrt{r_i\,r_f}\,\cos\left(\frac{\theta}{2}\right) = 1.979319
\end{aligned}

We next calculate the semi-major axis using

V_i^2 = \mu\,\left(\frac{2}{r_i}-\frac{1}{a}\right)

or

V_f^2 = \mu\,\left(\frac{2}{r_f}-\frac{1}{a}\right)

and in both cases we calculate the same semi-major axis a = 1.00000. (You may find this value a little suspicious - but it's just a result of me setting up a test problem with simple parameters.)

Next, we solve the quadratic:

A-B\,X = 2\,a\,(1 - X^2)

The two roots of this are X_1 = 0.00999118 and X_2 = 0.9796683.

With X = X_1 = 0.00999118, the radial and transverse components at \mathbf{R}_i are:

\left(v_{i,r},v_{i,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X}}\,\left(\sqrt{\frac{r_f}{r_i}}\,\cos\left(\frac{\theta}{2}\right)- X,\sqrt{\frac{r_f}{r_i}}\,\sin\left(\frac{\theta}{2}\right)\right) = \left(0.9702056, 0.1986958\right)

and the radial and transverse components at \mathbf{R}_f are:


\left(v_{f,r},v_{f,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X}}\,\left(X - \sqrt{\frac{r_i}{r_f}}\,\cos\left(\frac{\theta}{2}\right),\sqrt{\frac{r_i}{r_f}}\,\sin\left(\frac{\theta}{2}\right)\right) = \left(-0.9700421, 0.1986627\right)

And with X = X_2 = 0.9796683, the radial and transverse components at \mathbf{R}_i are:

\left(v_{i,r},v_{i,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X}}\,\left(\sqrt{\frac{r_f}{r_i}}\,\cos\left(\frac{\theta}{2}\right)- X,\sqrt{\frac{r_f}{r_i}}\,\sin\left(\frac{\theta}{2}\right)\right) = \left(0.002392613, 0.9903399\right)

and the radial and transverse components at \mathbf{R}_f are:


\left(v_{f,r},v_{f,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X}}\,\left(X - \sqrt{\frac{r_i}{r_f}}\,\cos\left(\frac{\theta}{2}\right),\sqrt{\frac{r_i}{r_f}}\,\sin\left(\frac{\theta}{2}\right)\right) = \left(-0.001577536, 0.9901747\right)

Based on this, if we go through the straightforward exercise of calculating the eccentricity with X=X_1 we find that the orbital eccentricity is 0.9796683; and with X=X_2 we find that the orbital eccentricity is 0.010000. Since I know that the orbit is nearly circular (i.e., so that the eccentricity is close to zero), the appropriate solution to use is the one corresponding to X=X_2.

So, finally, the radial and transverse components of the velocity vector at \mathbf{R}_i are (0.002392613, 0.9903399); and the radial and transverse components of the velocity vector at \mathbf{R}_f are (-0.001577536, 0.9901747).

Last edited by MontBlanc2012; 07-12-2019 at 02:10 PM.
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Old 07-12-2019, 08:49 PM   #7
BrianJ
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Hi,
wow! Thanks so much, I would never have got those later steps. Magic indeed
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Old 07-13-2019, 03:38 AM   #8
MontBlanc2012
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Actually, it occurs to me that the solution to ncc1701d's original problem is actually quite trivial - given that we are given the time of flight on each of the arc segments (120 seconds each). This means that the information about speeds is superfluous and one just needs to use a standard Lambert Solver to solve the problem.

However, we can also use the maths set out above to do the same job of solving the Lambert Problem, - so, in outline, here's the solution:

In SI units, the first and second Jupiter-centric positions of Amalthea are:

\begin{aligned}
R_i &= \left(-178932619.,-28063045.,-17448755.\right)\\
R_f &= \left(-178337419.,-30878143.,-18771071.\right)
\end{aligned}

We also know that for Jupiter, the accepted value (in SI units) of the gravitational parameter is \mu = 1.26687\times 10^{17}. And we know the time of flight is 120 seconds.

As before calculate:

\begin{aligned}
r_i &= \sqrt{\mathbf{R}_i.\mathbf{R}_i} = 181958444.95\\
r_f &= \sqrt{\mathbf{R}_f.\mathbf{R}_f} = 181961665.85\\
\theta &=\cos^{-1}\left(\frac{\mathbf{R}_i.\mathbf{R}_f}{r_i\,r_f}\right) = 0.01740310807\\
A &= r_i + r_f = 363920110.79\\
B &= 2\,\sqrt{r_i\,r_f}\,\cos\left(\frac{\theta}{2}\right) = 363906333.39
\end{aligned}

Now set X and Y as functions of an as yet unknown semi-major axis a:

\begin{aligned}
X_+(a) &= \frac{B+\sqrt{16\,a^2 - 8\,a\,A + B^2}}{4\,a} \\
X_-(a) &= \frac{B-\sqrt{16\,a^2 - 8\,a\,A + B^2}}{4\,a}
\end{aligned}

where X_+ and X_- are the two roots of the equation A - B\,X =2\,a\,(1-X^2).

As it turns out, for elliptical orbits, the transfer time from the initial to final points can be calculated as:

\Delta T =120 = 2\,\sqrt{\frac{a^3}{\mu}}\,\left(\cos^{-1}X_+(a) + X_-(a)\,\sqrt{1-X_+(a)^2}\right)

So, now we have an expression that is a function of a only. And since we know that the orbit of Amalthea is almost circular, we use a standard Newton root-finding algorithm with an initial guess of 180,000 km. From this, we deduce that:

a = 181997386.85

and hence

X_+ = 0.99996215721

and

X_- = -0.0002051286

Then, the radial and transverse components of the velocity vector at the \mathbf{R}_i are given by:

\left(v_{i,r},v_{i,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X_+}}\,\left(\sqrt{\frac{r_f}{r_i}}\,\cos\left(\frac{\theta}{2}\right)- X_+,\sqrt{\frac{r_f}{r_i}}\,\sin\left(\frac{\theta}{2}\right)\right) = (26.79327, 26389.15)

and the radial and transverse components of the velocity vector at the \mathbf{R}_f are given by:

\left(v_{f,r},v_{f,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X_+}}\,\left(X_+ - \sqrt{\frac{r_i}{r_f}}\,\cos\left(\frac{\theta}{2}\right),\sqrt{\frac{r_i}{r_f}}\,\sin\left(\frac{\theta}{2}\right)\right) = (26.88698, 26388.68)

And from this it is straightforward to convert back to the x-y-z coordinates of the original coordinate system.

N.B. In case you were wondering what the orbital eccentricity is, we get this from the expression:

e = \sqrt{X_-^2 + \frac{(r_1-r_2)^2}{A^2 - B^2}\,(1 - X_-^2)} = 0.001037615

i.e., very nearly a circular orbit.

Yay!


---------- Post added at 03:38 AM ---------- Previous post was at 02:15 AM ----------

And finally, just a quick comment on ncc1701d's summation of velocity vector components (V = VX + VY + VZ). Initially, I thought this was a typo and ncc1701d really meant V^2 = VX^2 + VY^2 + VZ^2 - which is, by far and away the more conventional way of 'adding' velocity components.

As it turns, out, I was wrong: ncc1701d actually did mean the straight arithmetic sum of the velocity components. How do I know this? Well, I can calculate the x-y-z representation of the radial and transverse unit vectors at that point. Although I'm not going to bother showing the intermediate calculations, in x-y-z coordinates the radial unit vector is:

\hat{\mathbf{r}} = \left(-0.9833708, -0.1542278, -0.09589418\right)

and the transverse unit vector is:

\hat{\mathbf{t}} = \left(0.1804087, -0.8901994, -0.4183273\right)

So, the velocity vector at the first point in ncc1701d's list can be written as:

\mathbf{V}_i = 26.79327\,\hat{\mathbf{r}} + 26389.15\,\hat{\mathbf{t}} = \left(4734.483, -23495.74, -11041.87\right)

So, at that point, VX = 4734.483; VY = -23495.74; and VZ = -11041.87. And the straight arithmetic sum of those is -29803.12 which is suspiciously close to the value given by ncc1701d of -29804.11.

Last edited by MontBlanc2012; 07-17-2019 at 03:43 AM.
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Old 07-14-2019, 03:26 PM   #9
BrianJ
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Thanks for the further insight.
I was on the wrong track to start with, since I had assumed the OP actually meant V^2 = VX^2 + VY^2 + VZ^2.
Note to self: always read the question carefully!
Cheers,
Brian
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Old 07-16-2019, 07:35 AM   #10
ncc1701d
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thank you everyone for all the feedback!
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