# Orientation and Rotation

G4CKT

## Homework Statement

I was given a question to transform the LA system to the CI system. (Local astronomic and Conventional Inertial)

## Homework Equations

r^LA=P1*R2(90-Ф)*R3(Λ)*R2(-xp)*R1(-yp)*R3(GAST)*N*P*r^CL

## The Attempt at a Solution

I attempted this on matlab but I used the wrong rotation and got the question wrong so I was wondering if anyone knew how to convert LA -> CI

## Answers and Replies

Homework Helper
I see "LA" on the left side of the equation but I see no "CI". Was the "CL" on the right supposed to be "CI"?

If so, first divide both sides by all the numbers on the right to get
$$r^{CI}= \frac{r^{LA}}{P1*R2(90-Ф)*R3(Λ)*R2(-xp)*R1(-yp)*R3(GAST)*N*P}$$

Now, take the logarithm of both sides (it doesn't matter what base)
$$CI log(r)= LA log(r)- log(P1*R2(90-Ф)*R3(Λ)*R2(-xp)*R1(-yp)*R3(GAST)*N*P)$$
Finally, divide both sides by log(r) to get
$$CI= LA- \frac{log(P1*R2(90-Ф)*R3(Λ)*R2(-xp)*R1(-yp)*R3(GAST)*N*P)}{log(r)}$$

G4CKT
I see "LA" on the left side of the equation but I see no "CI". Was the "CL" on the right supposed to be "CI"?

If so, first divide both sides by all the numbers on the right to get
$$r^{CI}= \frac{r^{LA}}{P1*R2(90-Ф)*R3(Λ)*R2(-xp)*R1(-yp)*R3(GAST)*N*P}$$

Now, take the logarithm of both sides (it doesn't matter what base)
$$CI log(r)= LA log(r)- log(P1*R2(90-Ф)*R3(Λ)*R2(-xp)*R1(-yp)*R3(GAST)*N*P)$$
Finally, divide both sides by log(r) to get
$$CI= LA- \frac{log(P1*R2(90-Ф)*R3(Λ)*R2(-xp)*R1(-yp)*R3(GAST)*N*P)}{log(r)}$$

Yes, that is correct I was confusing myself, sorry xD.

I have another question in order to find the CI using right ascension and declination. How would I be able to compute that?

Thanks!!!!

Maybe I should be in the astronomy section xD

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