tag:blogger.com,1999:blog-1592992209402300549.post5695080546221792134..comments2024-03-28T08:06:43.198-04:00Comments on The Silicon Graybeard: Not Exactly Ham Radio, Just Amazing Radio MonitoringSiGraybeardhttp://www.blogger.com/profile/00280583031339062059noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-1592992209402300549.post-32439746011051683202020-10-11T11:01:08.070-04:002020-10-11T11:01:08.070-04:00Every path loss equation I've ever seen has th...Every path loss equation I've ever seen has that direct dependence on frequency, i.e., loss goes up as frequency goes up. In reality, all radio frequencies drop off with distance as 1/(distance squared) so there's no dependence on frequency. The wavelength effect is due to the model of an antenna aperture that's embedded in the assumptions. I use these relationships because it's easy to add and subtract antenna gains in dB for the systems engineering.<br /><br /><a href="https://www.electronics-notes.com/articles/antennas-propagation/propagation-overview/free-space-path-loss.php" rel="nofollow">This discussion </a>may help clear it up, I hope. <br />SiGraybeardhttps://www.blogger.com/profile/00280583031339062059noreply@blogger.comtag:blogger.com,1999:blog-1592992209402300549.post-48695163267698469322020-10-11T10:47:22.109-04:002020-10-11T10:47:22.109-04:00I get about a 3.5 degree beam width (at -3 dB). C...I get about a 3.5 degree beam width (at -3 dB). Comparatively, Mars is tiny, even at its maximum which is about 22.5 arc seconds this year.<br /><br />Still, it would seem that Stilley's antenna would have to be motor driven if he needs to observe Mars for any length of time. Objects in the sky move at half the rate of the hour hand on a clock, going 180 degrees (from 3:00 to 9:00) in 12 hours instead of six, but that 3-1/2 degrees is about 15 minutes of observation.<br />SiGraybeardhttps://www.blogger.com/profile/00280583031339062059noreply@blogger.comtag:blogger.com,1999:blog-1592992209402300549.post-47708422581403789482020-10-10T23:00:05.008-04:002020-10-10T23:00:05.008-04:00I'm curious as to why the frequency is there i...I'm curious as to why the frequency is there in your path-loss equation?<br /><br />The 1/r^2 term ( 20 log(d) ) makes sense, but why would there be a dependence on frequency in a vacuum where nothing is absorbing the beam? It isn't some sort of sender-gain thing because you don't have any sender details.<br /><br />MadRocketSciAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-1592992209402300549.post-51851398969276193842020-10-10T22:22:01.421-04:002020-10-10T22:22:01.421-04:00At that dish size and frequency, what would be the...At that dish size and frequency, what would be the angular coverage?Ritchiehttps://www.blogger.com/profile/08075903551422300106noreply@blogger.com