3 ms·
I understand the division of units, what I’d like to see is an illustration of the signal’s cone at the astronomical distances in the article.
by semireg 2y ago
I understand the division of units, what I’d like to see is an illustration of the signal’s cone at the astronomical distances in the article.
- marcosdumay 2y agoWell, the area taken by a cone with angle "a" at the distance "r" is pi * r^2 * sin(a)^2. For an arcsec, the square sin is ~85e-9. At the distance of the Moon, that means the signal occupies ~400km^2. At the distance of the Sun, ~5 billions of km^2 (5Mm^2). The signal intensity is inversely proportional to that area.
- semireg 2y agoFor example, if the screw/servo controlling one axis of the dish is turned by a minimum adjustment, how many hundreds or thousands of miles does it throw the cone’s target at the distance of the probe? What’s the remaining margin for error at these scales?
- mikewarot 2y agoThe angle is about 9x10^-5 radians. If you were to take a meter stick (or a yard stick) and stick a piece of paper under one end, that's the precision within which it can point the dish. (While the earth is spinning, and going around the sun!) The actual width of the beam is about 5 times that angle.
- deleted 2y ago[deleted]