eci2lla altitude error?

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.

[Dyuman Joshi]

Though it is not specified in the documentation of eci2lla about this, but I presume that the altitude is measured from sea-level.

Going under it would give a negative value.

[Derrick Early]

The example should yield a positive altitude.

[Dyuman Joshi]

"The example should yield a positive altitude."

Why? Did you calculate the values by hand and compare?

[Derrick Early]

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))

Answer 1

[Les Beckham]

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).

[Derrick Early]

Thank you! I messed up on computing the vector magnitude.

[Les Beckham]

You are quite welcome.