[amsat-bb] Re: AO-51

Stephen Melachrinos melachri at verizon.net
Mon Mar 1 16:09:55 PST 2010

```Drew -

But remember that this is three-dimensional; your example is 2-D.

I'm pretty sure that "39 meters radial" is referencing the instantaneous radius of the orbit; that means that at the closest approach, the predicted altitude of one spacecraft was 39 meters higher than the other. Pythagoras tells us that if the hypotenuse of a right triangle is 893 meters, and the height (=radial distance) is 39 meters, then the length of the base of the triangle is just over 892 meters. View this as the radius of a circle, centered on one satellite but 39 meters either above or below it, and the other satellite flies by, in the plane of and tangent to the circle.

AO-51 has an inclination of about 100 degrees; Formosat-3D has an inclination of (I believe) 72 degrees. Without going through the calculations I'm not sure of the actual trajectories, but my modification of your visualization is the following:

The two satellites are each "driving" along interstate highways (at 17,500 mph). The two highways do cross each other (at some angle which is determined by the inclinations and trajectories as mentioned above), but one highway is elevated 39 meters (about 128 feet) above the other. When the two vehicles get to their closest point, there's 892 meters between them horizontally, and 39 meters between them vertically.

Now I'm not making light of how uncomfortable this approach is. Even though space is big (that's why they call it space), this is definitely uncomfortably close. The reason is that our ability to predict exactly where a spacecraft is at a precise time in the future is not perfect. The uncertainties are small from the perspective of antenna pointing angles on the ground, but they're large compared to the dimensions of this example. Instead of viewing the orbits as roadways, where you know that a car is exactly in a lane that's ten feet wide, a better view is that these are tubes whose internal radius is a hundred meters or so; the vehicle can be anywhere within the tube's cross-section. (Those are known as radial and cross-track errors.) And its position lengthwise in the tube (the "in-track" position) at any precise moment in time is only known to a hundred meters or so. So if two orbits are only separated by 39 meters, their tubes actually intersect for a large portion of their cross-sections. And when you add in the in-track uncertainties, that 893 meter separation gets eaten up really quick.

Steve
W3HF

Mar 1, 2010 03:40:54 PM, glasbrenner at mindspring.com wrote:

> It's all predictive until they smack together, but the estimates this
> morning said 893m overall, but only 39m radially. I'm visualizing that
> as one car merging on a highway at 17,500 mph, with one 3000 feet in
> front of, and 120 feet left or right of the other. Scale the 17,500 mph
> to 70 mph, and the distances in the analogy would be 12 feet in front
> of, and 6 inches abreast. Just to put the velocity and distances in
> perspective!

> 73, Drew KO4MA

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