eci2lla altitude error?

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Derrick Early
Derrick Early 2023-11-7
评论： Les Beckham 2023-11-7
Ran in:
In the following example,
lla = eci2lla([-6.07 -1.28 0.66]*1e6,[2010 1 17 10 20 36])
lla = 1×3
1.0e+05 * 0.0001 -0.0008 -1.3940
How do you end up with a negative altitude?
The altitude should be approximately 312000 m.
  4 个评论
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Dyuman Joshi
Dyuman Joshi 2023-11-7
"The example should yield a positive altitude."
Why? Did you calculate the values by hand and compare?
Derrick Early
Derrick Early 2023-11-7
Oops. I made an error on computing the vector normal. I used
sqrt(sum([-6.07 -1.28 0.66]*1e6).^2)
Instead of
sqrt(sum(([-6.07 -1.28 0.66]*1e6).^2))

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采纳的回答

Les Beckham
Les Beckham 2023-11-7
编辑：Les Beckham 2023-11-7
Ran in:
lla = eci2lla([-6.07 -1.28 0.66]*1e6,[2010 1 17 10 20 36]);
lat = lla(1)
lat = 6.0574
lon = lla(2)
lon = -79.8476
So, this point is slightly above the Equator (by about 6 degrees)
dist = vecnorm([-6.07 -1.28 0.66]*1e6) % distance of this point from the center of the Earth
dist = 6.2385e+06
equatorialRadius = 6378e3;
dist - equatorialRadius
ans = -1.3950e+05
alt = lla(3)
alt = -1.3940e+05
So this point is beneath the surface of the Earth by about 140 kilometers (negative altitude).
  2 个评论
Derrick Early
Derrick Early 2023-11-7
Thank you! I messed up on computing the vector magnitude.
Les Beckham
Les Beckham 2023-11-7
You are quite welcome.

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