AgentR wrote:Plantagenet wrote:satellite up there to within a centimeter at all times.
Please provide a link to an official claiming such a degree of accuracy.
No problemo.
From the US patent #6016117 for satellite positioning:
"Specifically, the ephemeris for a given satellite comprises:
a) The square-root of the semi-major orbital axis, which is half the greatest dimension of the elliptical orbit. The square-root ranges from 0 to about 8192 square-root-meters, corresponding to a semi-major axis a of about 67 megameters, with
a resolution corresponding to about 20 millimeters for the typical orbit that has a semi-major axis of 26560 kilometers. Transmission of the square-root rather than the semi-major axis itself allows calculation of the mean motion n0, which is the average orbital angular velocity, without using a square-root operation:
n0 =(μ/a3)0.5 =μ0.5 /(a0.5)3
where μ is the earth's gravitational parameter, commonly the WGS-84 value of 398600.5 kilometers-cubed-per-second-squared. The value of n0 is typically 146 microradians-per-second, corresponding to about 3874 meters-per-second.
b) The ephemeris reference time of the week T0, which is the time from the beginning of the current GPS week to a time near the middle of the time interval for which the ephemeris is valid. The reference time T0 is an integer multiple of 16 seconds. Time T0 is commonly subtracted from the given time of interest T to get ephemeris time t, that is, time referred to the ephemeris's epoch:
t=T-T0
c) The mean-motion difference ∆n, which is a correction to the mean motion no as computed above:
n=n0 ∆n
The range of ∆n corresponds to about ±311 millimeters-per-second with
a resolution of about 9.5 micrometers-per-second for a typical orbit.
d) The reference-time mean anomaly M0, which is the angle of the satellite at the reference time T0 from perigee which would result from uniform angular motion. Perigee is the point in the orbit closest to the earth. The
resolution of M0 corresponds to about 39 millimeters for a typical orbit. M0 is used to find the satellite's mean anomaly M at ephemeris time t:
M=M0 nt
e) The orbital eccentricity e, which is the fraction of the semi-major axis by which the earth is distant from the orbital center. The dimensionless fraction e ranges from 0 to 0.03 with a resolution corresponding to about 3 millimeters for a typical orbit. Kepler's equation relates mean anomaly M to eccentric anomaly E, which is the actual at the reference time T0 about the orbital center from perigee to the satellite:
M=E-e sin E
Kepler's equation is commonly solved iteratively to get eccentric anomaly E from mean anomaly M. True anomaly v, which is the actual angle about the earth's center from perigee to the satellite, is commonly calculated from eccentric anomaly E by any of various formulas, for example: ##EQU1##
f) The argument of perigee ω, which is the angle about the earth's center from the ascending node to perigee. The ascending node is the intersection of the orbit with the plane of the equator at which the satellite passes from the southern to the northern hemisphere.
The resolution of ω corresponds to about 39 millimeters for a typical orbit. The argument of perigee ω is commonly added to the true anomaly v to get the uncorrected argument of latitude Φ, which is the angle about the earth's center from the ascending node to the satellite:
Φ=ω+v
g) The second-harmonic correction coefficients Cuc and Cus to the argument of latitude Φ. The ranges of Cuc and Cus correspond to about ±1621 meters with
a resolution of about 49 millimeters for a typical orbit. They multiply the cosine and sine respectively of twice the uncorrected argument of latitude Φ to correct the argument of latitude u:
u=Φ+Cus cos 2Φ+Cus sin 2Φ
h) The second-harmonic correction coefficients Crc and Crs to the radius r. The ranges of Crc and Crs are about ±1024 meters with
a resolution of about 31 millimeters. They multiply the cosine and sine respectively of twice the uncorrected argument of latitude Φ to correct the radius r:
r=a(1-e cos E)+Crc cos 2Φ+Crs sin 2Φ
The argument of latitude u and the radius r are commonly combined to calculate the satellite's position in the orbital plane (x',y'):
x'=r cos u
y'=r sin u
i) The orbital inclination i0 at the reference time, which is the angle that the orbital plane makes with the plane of the equator, typically about 55 degrees. The
resolution of i0 corresponds to about 39 millimeters for a typical orbit.j) The rate of orbital inclination i. The range of i corresponds to ±155 millimeters-per-second with
a resolution of about 9.5 micrometers-per-second for a typical orbit.k) The second-harmonic correction coefficients Cic and Cis to the inclination i. The ranges of Cic and Cis correspond to about ±1621 meters with
a resolution of about 49 millimeters for a typical orbit. They multiply the cosine and sine respectively of twice the uncorrected argument of latitude Φ to correct the inclination i:
i=i0 +it+Cuc cos 2Φ+Cus sin 2Φ
l) The right ascension of the ascending node at the reference time Ω0, which is the angle measured eastward from the vernal equinox to the ascending node.
The resolution of Ω0 corresponds to about 39 millimeters for a typical orbit.l) The rate of right ascension Ω. The range corresponds to about ±80 meters-per-second with
a resolution of about 9.5 micrometers-per-second for a typical orbit. Ω0 and Ω are commonly combined with Ωe, the WGS-84 earth's rotation rate, to calculate the ascending node's longitude:
Ω=Ω0 Ωt-Ωe T
The ascending node's longitude Ω and the inclination i are commonly combined with the satellite's position in the orbital plane to get the satellite's position in earth-centered earth-fixed coordinates (x, y, z)."
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This 9-year-old patent describes how satellite positions in 3d space can be determined to within 30-40 mm, and the velocity is known to within a few micrometers per second (thats velocity to within a millionth of meter per second).
When combined, these factors allow the determination of a satellite's position at all times on centimeter scales.
(---NOTE: the patent I quote above is 9 years old and PUBLIC....much better resolution can be obtained by the secret military satellite tracking systems---they
are secret, but since they are better then the public systems, one might guess they are accurate to the millimeter or even sub millimeter scale.)
The bottom line is that the US (and Russia) knew EXACTLY where both these satellites were in their orbits at all times. The US didn't move the Iridium satellite because it wasn't in danger of colliding with the Russian satellite based on the centimeter-scale positions calculated for their conventional orbits.
Its impossible for them to have collided accidently. Clearly the Russians activated their military satellite and moved it to collide with the US Iridium satellite.