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The LVLH frame is widely used in rockets during launch. The space shuttle crew used it in its attitude readings beginning very early after launch, meaning it was very reliable as soon as the rocket was in motion.

This frame is built from the rocket's position and velocity vectors such that:

  • z is -r
  • y is -r x v
  • x is y x z

But what frame are the r and v vectors resolved in? In an earth-centered inertial frame (ECI), a rocket sitting in the pad has a sizeable eastward velocity even before launch.

In an earth-centered earth-fixed frame (ECEF), the rocket on the pad has zero velocity (as the rocket is then moving eastward as fast as the ground).

This means different velocity vectors, which means different LVLH frames.

I've tried both approaches, and the one that works is the one I least expected. My rocket's attitude and my orbital inclination (given by r x v) are all good if my r and v vectors are defined in ECEF.

The attitude and inclination both break down if r and v are defined in ECI.

Can someone clarify for me which frame the r and v vectors should be defined in to properly build an LVLH frame?

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    $\begingroup$ Check out ntrs.nasa.gov/api/citations/19740026178/downloads/… for all the coordinate systems. $\endgroup$ – Organic Marble Jan 26 at 22:34
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    $\begingroup$ Re The LVLH frame is widely used in rockets during launch. I disagree. It is widely used once a vehicle is on orbit. Other frames are much more useful during launch. $\endgroup$ – David Hammen Jan 27 at 0:26
  • $\begingroup$ @DavidHammen it's even called "local orbital" in the shuttle book. $\endgroup$ – Organic Marble Jan 27 at 1:25
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    $\begingroup$ @OrganicMarble Get a more modern reference. Nobody uses Mean of 50 anymore. The posted answer is correct. $\endgroup$ – David Hammen Jan 27 at 1:51
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    $\begingroup$ @OrganicMarble Sorry about that. $\endgroup$ – David Hammen Jan 27 at 1:52
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The LVLH can be realized using the position and velocity vectors in the ECI frame. Other treatments are possible, but they would involve transformation of the vectors to the right frame. Conversion from ECI seems the most straightfoward, and aligned with the definition you provide above!

Assume the following notation for the LVLH axes vectors, expressed in the ECI frame (notice the "descending" listing):

$$\begin{align} & \mathbf{o}_{3I}: \text{LVLH Z-axis - Aligned with the spacecraft geocentric position vector}\\ & \mathbf{o}_{2I}: \text{LVLH Y-axis - Aligned with the negative orbit-normal}\\ & \mathbf{o}_{1I}: \text{LVLH X-axis - Completes the right-hand triad}\\ \end{align}$$

Assume the additional notation: $$\begin{align} \mathbf {r}_{I}: \text {Spacecraft position vector, expressed in the ECI frame}\\ \mathbf{v}_{I} \equiv \mathbf{\dot r}_{I}: \text {Spacecraft velocity vector, expressed in the ECI frame} \end{align}$$

The representation of the LVLH frame axes vectors, expresed in ECI frame, is given by:

$$\begin{align} & \mathbf{o}_{3I} = \mathbf{-{r}}_{I} / \lvert \lvert \mathbf{{r}}_{I} \rvert \rvert\\ & \mathbf{o}_{2I} = \mathbf{-({r}}_{I}\times \mathbf {v}_{I}) / \lvert \lvert \mathbf{{r}}_{I}\times \mathbf{{v}}_{I} \rvert \rvert\\ & \mathbf{o}_{1I} = \mathbf {o}_{2I} \times \mathbf {o}_{3I}\\ \end{align}$$

Finally, the rotation matrix from the LVLH frame (denoted as O) to the ECI frame (denoted as I) is:

$${A}_{IO} = \left[ \begin{array}{ccc} \mathbf{o}_{1I}&\mathbf{o}_{2I}&\mathbf{o}_{3I} \end{array} \right] $$

A graph of the LVLH is shown below. This is in line with the use of the "Geocentric Radius Vector" in the definition of the Orbital Frame as provided by the source from @OrganicMarble (page 17 of the attached document in the comments, also provided here in case the comment is deleted).

enter image description here

Source: Markley, Crassidis: Fundamentals of Attitude Determination and Control, page 36

An alternative method to convert from ECI to the similar (but not identical for all sources) RSW frame is provided in page 169, Section 3.4.3 from Vallado's "Fundamentals of Astrodynamics and Applications". That method perfoms three principal rotations using the orbits keplerian elements, with some hints to the fact that those angles are not necessarily independent, but rather more intuitive.

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    $\begingroup$ +1 but posting an image of a page of text as the actual answer is inadequate because some people use text readers rather than vision, the answer is not searchable as text, and the image may be deleted as it may violate the copyright of the publisher. Can you add some text and equations using MathJax that summarizes what it is on the page that is central, in order to make your post an actual stand-alone answer? Thanks! $\endgroup$ – uhoh Jan 27 at 1:03
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    $\begingroup$ Yes @uhoh I understand, I will type the equations using MathJax and update both my answer and comment. $\endgroup$ – Manny Jan 27 at 1:07
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    $\begingroup$ Thanks to you, I was able to copy the exact bolding on the subscripts! Question updated @uhoh! $\endgroup$ – Manny Jan 27 at 2:13
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    $\begingroup$ Looks great, thanks! :-) $\endgroup$ – uhoh Jan 27 at 2:17

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