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Orientation and Rotation

  • Thread starter G4CKT
  • Start date
  • #1
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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

  • #2
HallsofIvy
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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
[tex]r^{CI}= \frac{r^{LA}}{P1*R2(90-Ф)*R3(Λ)*R2(-xp)*R1(-yp)*R3(GAST)*N*P}[/tex]

Now, take the logarithm of both sides (it doesn't matter what base)
[tex]CI log(r)= LA log(r)- log(P1*R2(90-Ф)*R3(Λ)*R2(-xp)*R1(-yp)*R3(GAST)*N*P)[/tex]
Finally, divide both sides by log(r) to get
[tex]CI= LA- \frac{log(P1*R2(90-Ф)*R3(Λ)*R2(-xp)*R1(-yp)*R3(GAST)*N*P)}{log(r)}[/tex]
 
  • #3
12
0
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
[tex]r^{CI}= \frac{r^{LA}}{P1*R2(90-Ф)*R3(Λ)*R2(-xp)*R1(-yp)*R3(GAST)*N*P}[/tex]

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

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