How Do You Convert ENU to ECEF Coordinates?

  • Thread starter Thread starter rjjansky
  • Start date Start date
  • Tags Tags
    Formula
rjjansky
Messages
2
Reaction score
0
Dear Readers,

It has been many (30+) years since I have had to work this kind of math and it is kickin' my butt. So, if you can help a software weenie with a need for a conversion formula for ENU to ECEF (or at least an explanation of things more elementary than what I have found thus far) it would be a great help. Not quite sure why it seems I can find plenty of discussions of going from ECEF to ENU but I must not be using the correct word set in my searches. Remember, please explain things as if I were a four year old.

Thanks in advance for what ever you can provide!

Cheers...
 
Physics news on Phys.org
What are ENU and ECEF?
 
Sorry. They are different Earth axes conventions. ENU is East North Up while ECEF is Earth Centered Earth Fixed. There is also NED which is North East Down. At any rate, ECEF is Cartesian and ENU is not so those bits of data that I received in ENU need to be converted to ECEF in order to work through other formulas to estimate a new position. Being away from linear algebra and matrix math these many years has made it tough to get traction on solutions.
 
Since I am not familiar with the conventions by name, it would be helpful if you gave a complete definition for each.
 
The first item google gives me for "enu to ecef" is
http://www.navipedia.net/index.php/Transformations_between_ECEF_and_ENU_coordinates
Did you already look at the top google results for this topic? If so, what was confusing to you?

jason
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
Back
Top