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Comment: Re:That's a ton of bandwidth (Score 1) 328

by PlaneShaper (#38655684) Attached to: Almost 1 In 3 US Warplanes Is a Drone

It could easily be 500 megabits/s. What most people here seem to fail to realize is that the Global Hawk is not some camcorder with an MPEG-2 stream. The RQ-4 collects high-resolution *still-frame* imagery.

FMV streams are some video typically ranging from 1 Mbps to 10 Mbps -- i.e. they record no more and no different than a digital camcorder. The RQ-4 records National Image Transmission Format (NITF) images rather than a video stream, which might range from hundreds of MB to a few GB per image, depending on resolution, and it can collect those images at some frequency. These images are encoded with geospatial data and would probably need to be lossless. It has a high-powered, high-resolution digital camera, rather than a camcorder, not significantly different in concept than the imaging sensors employed on the U-2 or SR-71. The sensor(s) on the RQ-4, if used as robustly as possible, could very likely over-saturate even the 500 Mbps link.

This article from the actual Air Force: http://www.af.mil/news/story.asp?id=123185754 has the RQ-4 collecting 400-700 images in 14 hours. This article: http://air-attack.com/page/54/RQ-4-Global-Hawk.html claims the RQ-4 can collect 1,900 images in 24 hours. Just doing the math on that first estimate from the Air Force, 550 images at a low end of 500MB per image is still almost 50 Mbps. That's an order of magnitude lower than the 500 Mbps claim in the PDF file (which, you'll note the claim comes from the Navy, NOT from Wired). However, the numbers I used are the low estimate. At the high estimate, 1,900 images/day at 3 GB/image, you're looking at a 528 Mbps requirement. Pretty spot on.

Comment: Re:It needs what??? (Score 5, Informative) 328

by PlaneShaper (#38650966) Attached to: Almost 1 In 3 US Warplanes Is a Drone

On page 17 of the actual report (page 22 of the PDF file), it says "a single Global Hawk...'requires 500Mbps bandwidth...'" So yes, somewhere between there and the Wired story, someone miscapitalized the B. That statistic is cited within the report as being from the Department of the Navy.

Comment: Re:Source on Gamification (Score 0) 111

by PlaneShaper (#36634292) Attached to: Current Social Games Aren't Fun, Says MUD Co-Creator

if people didn't enjoy the games, wouldn't they move on and find other ways to amuse themselves?

If people didn't enjoy abusive relationships, wouldn't they move on and find other ways to amuse themselves? What's being discussed by "gamification" is really the exploitation of human social behavior to accept a situation as an obligation instead of a choice.

When *you* have to press buttons so that *your friend* can obtain a reward or satisfaction from a game, then not doing so is letting your friend down. It's quite possible that the experience can be fun and rewarding for both of you. But TFA is essentially arguing that current social games are placing a much greater design emphasis on setting up the "obligation trap" than actually making the experience fun so that you both would choose to stay. And they are doing this with their bets hedged on this being a more profitable system...which is currently panning out in their favor.

Comment: Re:all that wave particle jazz (Score 0) 370

by PlaneShaper (#36252460) Attached to: 10-Year Study Reveals Electron Shape

Mostly, but the electron doesn't have to have a slightly + charge to have a dipole moment, just a "dent in the sphere." A pole is a point location where field lines emanate or coalesce. You can imagine that a dented sphere, even if negatively charged, if it were spinning, would result in a small electric dipole moment thanks to distortions in the field lines due to the dent. Those distortions can be mathematically expressed as multi-pole moments. (dipole, quadrupole, etc).

The research here is essentially saying, "hey, we performed some measurements, and it doesn't look like there is an electric dipole moment large enough to validate certain predictions." The article describes this by calling the electron very round, which is the analogy. They basically set an upper bound on what an electron's electric dipole moment could possibly be (if it even has one).

Comment: Innocent Until Suspected of Guilt, eh? (Score 0) 220

by PlaneShaper (#36048356) Attached to: A Court's Weak Argument For Blocking IP Subpoenas

I admit, I got about halfway through this willing to concede some aspects of earlier points could have been made stronger by the judge. Until I got here:

"I certainly don't want to take the position that anyone who doesn't deny their guilt is guilty — but we shouldn't assume that they're innocent, either. Come on — it's not that hard for a copyright holder to join a p2p network or find a Torrent tracker, and get a list of the IP address of a few users who are sharing or downloading their content illegally. Is it that hard to believe that most of the users on that list are probably doing what VPR said they were doing?"

Assumption of innocence isn't just not a thing we shouldn't be doing, it is the very thing required of us to, in fact, do. Especially when the situation in question makes it easy to presume guilt without evidence, then it is even more important we not assume guilt just because of our ability to believe the accusation.

Identities should most certainly NOT go to the corporation asking for them. Instead, the list of IP address should be passed to the appropriate authorities who can then conduct legitimate investigations as needed to obtain further evidence of any possible crime. The corporation should not receive any identity information about suspects until investigations conclude a crime has taken place.

Comment: This is Relevant to Me (Score 2, Interesting) 140

by PlaneShaper (#34800352) Attached to: It's Surprisingly Hard To Notice When Moving Objects Change

I won't necessarily claim to be one here, since anyone can claim anything on the internet. But I wonder how well experienced MTI analysts (http://en.wikipedia.org/wiki/Moving_target_indication) would perform against these videos. Quite literally, it is the job of an MTI analyst to distinguish coherent from non-coherent changes across a vast number of moving dots displaying all sorts of crazy behaviors.

Additionally, I would prefer to see this experiment run under different conditions, such as having the videos begin with dots in apparent motion for many seconds, then having them stop moving. I do think the results of the experiment are damaged by having the motion segment (3 secs) be significantly shorter than the non-motion segment (5 secs) and always happening after the eyes and brain have adjusted to the lack of apparent motion.

I think another big problem with the videos provided is that the motion segment alters direction 9 times (counting first and last) within a short window -- this may not result in the human mind having difficulty seeing change in moving objects, but a difficulty in adjusting the perspective of the total scene to something observing rapid fluxuation in velocity. (I.E. The circle constantly rotates back and forth, preventing the brain from "getting used to" the scene, whereas when the dots are stationary and only changing in specific property, they remain in this configuration for a much longer period of time).

It's a interesting topic, but research could have focused more on overall configuration of how viewers were presented with the experiment. There isn't enough information present to draw an accurate conclusion from these observed results.

Comment: If a Vote for Him is a Vote for Apple... (Score -1, Flamebait) 97

by PlaneShaper (#34194554) Attached to: Australian State Govt. To Fund iPads For Doctors

I'm not Australian, but I'm pretty sure that if a candidate promised me a free iPad just for my vote, that's the first person I would not be voting for. Not because I disagree with the tactic, but because I don't want an iPad...or anything related to Apple for that matter.

Might as well give me more underwear and socks for Christmas, at least I can patch those myself if they get a tear.

Comment: Re:Well? (Score 1) 981

by PlaneShaper (#32732486) Attached to: The Tuesday Birthday Problem

Exactly what I'm talking about!

Because out of the 14 BG/GB combinations, 50% of them would be telling you about their *girl*

That pulls 7 out of the total set, and leaves 7/14 being the probability for 2 boys.

I cheated though, because I've already read this blog:
http://blog.tanyakhovanova.com/?p=221

It's from a couple months ago and summarizes several solutions also seen in this thread. But it's easier to get people to pick up the concept that it's just a trick question designed to get you to determine your dataset.

Comment: Re:Well? (Score 1) 981

by PlaneShaper (#32731362) Attached to: The Tuesday Birthday Problem

I know I'm replying to my own post, but I wanted to point out that the result would be 13/27 if you *asked* the parent if they had a son born on Tuesday. In this case, the question is posed by the parent, meaning they made the conscious choice (assumed 50%) to pick Tuesday over the day of the week of their other son in cases with 2 boys born on different days.

The real riddle isn't the idea that you need to interpret a set of data, that's (relatively) easy. It's actually identifying which set of data you have presented to you.

Comment: Re:Well? (Score 0) 981

by PlaneShaper (#32731102) Attached to: The Tuesday Birthday Problem

Let's only look at the families with (at least one) tuesday boy:

    B1B2
B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7
    B3B2
    B4B2
    B5B2
    B6B2
    B7B2

B2G1 B2G2 B2G3 B2G4 B2G5 B2G6 B2G7

    G1B2
    G2B2
    G3B2
    G4B2
    G5B2
    G6B2
    G7B2

Of these 27 families, 13 have another boy. So P = 13/27.

Only half of the parents with two boys born on a different day would give you information about the boy born on Tuesday.

Therefore, in these 12 cases: B2B1, B1B2, B2B3, B3B2, B2B4, B4B2, B2B5, B5B2, B2B6, B6B2, B2B7, B7B2...

Only 50% of the parents would tell you they had a son born on Tuesday, the other 50% would give you a different day of the week. That reduces 6 cases from both the set of 13 BB sets and the set of 27 parents telling you about their kids.

This leaves you with 7/21, or 1/3.

Comment: Re:Let's try it without reading TFA (Score 1) 981

by PlaneShaper (#32730850) Attached to: The Tuesday Birthday Problem

"I have two children, one of whom is a boy born on a Tuesday. What's the probability that my other child is a boy?"

I have a boy, there is 1/3 probability that the other is also a boy.

I question your logic.

In a group of 1,000 sibling pairs, the following is the "perfect" breakout:
250 - Older Boy, Younger Boy
250 - Older Boy, Younger Girl
250 - Older Girl, Younger Boy
250 - Older Girl, Younger Girl

Once the statement is made, "I have a boy," it eliminates the two girl set, leaving only 750 sets of siblings evenly divided into thirds.

Comment: Re:Let's try it without reading TFA (Score 1) 981

by PlaneShaper (#32730056) Attached to: The Tuesday Birthday Problem

13/27 doesn't scale. In the set of all parents with two children, where at least one of which is a boy, it is assured that a boy is born on a particular day of the week.

If those parents each came up to you individually and provided you the statement, "I have two children, one of whom is a boy born on [day of week]," then every single parent would be assigned a probability of 13/27 for the other child to be a boy. (Assume that with a large enough sample size, the cluster of parents with two boys born on separate days of the week would evenly split their decision of which day of the week to inform you of.)

The analysis given argues that each case would have a 13/27ths probability of the child you were not informed of being a boy. Yet the group as a whole is determined to only consist of 33.33% parents with two boys.

On an individual basis, if the group was a perfect sample containing 189 cases, split evenly among the 7 days of the week, then 91 cases would be classified as having 2 boys using a 13/27 probability, when in actuality only 63 cases would have two boys.

Since the individual probability that has been assigned doesn't scale to the known probability of the entire group, then it must be incorrect. The probability the other child is a boy is 1/3, the day of the week is meaningless information.

You can bring any calculator you like to the midterm, as long as it doesn't dim the lights when you turn it on. -- Hepler, Systems Design 182

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