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If Ravi's theory with the plexiglass pans out, I don't think you need any software to re-create this scene.

Yeah, the video playing behind the plexiglass needs to be produced somehow; but could be done physically with a cardboard wing and saucepan lid.

Your use of quotation marks is misleading, I did not write this. I don't think you've understood the point I was trying to get across.

The ISW report had some interesting things to say again, after a few days of "business as usual". Russian sources reported that stores in occupied Melitopol and Volnovakha are beginning to transition to using the Russian ruble. British Defense Intelligence reported that the ruble will be used in...

My comment isn't about threats at all. Moscow was wary that changing alliances in Europe would impact its strategic security, and it needed reassurance that lifting the iron curtain and allowing East Germany to join the West would not come to unbalance Europe. But that did happen, and the...

You're asking for pre-2007 plexiglass and a pre-2007 desk lamp?

I was hoping to have @johne1618 figure that out. My point is that we're very small compared to the universe, and his dismissing that unknown chance as being too small is just argument from incredulity.

and that's not what my source claimed, so we are in agreement promises serve a function even when they're not legally binding (and you explained why NATO could not legally offer such a treaty without redesigning its foundation), and there may still be consequences when they are broken.

If they're all of different aliens, your analogy fails.

That's about the same, right? So how do you know how big P(A) is? There are 10^24 stars in the universe, so if there are 10,000 alien civilisations out there, that's exactly that 1:10^20 chance that one of them visits us. And what do you say to the three concerns I raised in that post (#20)?

Congratulations on getting out of the rabbit hole! Reading this, I'm horrified by your parents, who made your life much more difficult than it should have been. It's a huge achievement to gain independence from them! Obviously nobody is ever wholly independent of their past, but each day is now...

Source: https://mobile.twitter.com/mbk_center/status/1520172075352219649

The 2020 election fraud convictions I've seen concerned Republicans, though my sample is somewhat biased. I wonder how the filmmakers gained access to that cellphone data—or if they indeed did.

Your next bigger screenshot (120 frames) has the same pattern tiled with some alterations (marked green). Most surprisingly, some camera perspectives are mirrored (marked yellow and purple); I can't imagine any honest motive to use mirrored surveillance footage. When you pay attention to the...

It uses 12 different camera perspectives for 30 tiles, and only 4 of these are unique. Camera 3 is used 6 times, cameras 4 and 7 are used 4 times each, cameras 1 and 2 are used 3 times, cameras 5, 6 and 8 are used 2 times, and cameras 9 to 12 are unique (or panning?)

That's where it is at first: @Mechanik then pans over the trail, and after that, it's here: From the on, it stays in view, vertically centered at first, moving towards the right edge, then towards the left edge; and then, approximately from the "flash" comment, it moves towards the bottom center.

@johne1618 's argument boils down to "if a UFO believer sees 10 reports that he's 99% confident in, he's going to believe with 50% confidence that we actually have alien visitors, unless the probability of aliens visiting us is actually smaller than 1:100,000,000,000,000,000,000". One problem...

That's equivalent to P( { RCE_i } | A )×P(A) = P( { RCE_i } | Not[A] )×P(not[A]) You also wrote, "P( { RCE_i } | A ), to equal one" and I think you were arguing P( { RCE_i } | Not[A] ) should be 0.01^10 = 10^-20 ? Let's plug this in: P( { RCE_i } | A )×P(A) = P( { RCE_i } | Not[A] )×P(not[A])...

indeed

W=Witness saw an alien ^W=Witness did not see an alien (lie, hallucination, hoax, or no report) CE=Witness reports a close encounter We want to know the probability for a witness having seen an alien, given they reported one. P(W|CE)=P(CE|W)×P(W)/P(CE) P(CE|W) is the probability that someone...