text:
# 94

### Scenario

You're not certain that aliens are real. Even after receiving that email from your friend who works at NASA, with the subject line "What do you make of this?" and a single binary file attached, you're still not certain. It's rather unlikely that aliens would be near enough to Earth to make contact with humanity, and unlikelier still that of all the humans on Earth, you would end up being the one tasked with deciphering their message. You are a professional cryptanalyst, but there are many of those in the world. On the other hand, it's already been months since April Fools, and your friend isn't really the prankster type.

### Rules

No need to use spoiler text in the comments: This is a collaborative contest, where working together is encouraged. (The contest is between all of you and the difficulty of the problem.)

Success criteria: The people of LessWrong will be victorious if they can fully describe the process that generated the message, ideally by presenting a short program that generates the message, but a sufficiently precise verbal description is fine too.

The timeframe of the contest is about 2 weeks. The code that generated the message will be revealed on Tuesday July 12.

### Why is this interesting?

This contest is, of course, based on That Alien Message. Recently there was some discussion under I No Longer Believe Intelligence to be "Magical" about how realistic that scenario actually was. Not the part where the entire universe was a simulation by aliens, but the part where humanity was able to figure out the physics of the universe 1 layer up by just looking at a few frames of video. Sure it may be child's play for Solomonoff Induction, but do bounded agents really stand a chance? This contest should provide some experimental evidence.

# 94

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Solution:

#!/usr/bin/env python3

import struct

a = [1.3, 2.1, -0.5] # initialize data
L = 2095 # total length
i = 0

while len(a)<L:
if abs(a[i]) > 2.0 or abs(a[i+1]) > 2.0:
a.append(a[i]/2)
i += 1
else:
a.append(a[i] * a[i+1] + 1.0)
a.append(a[i] - a[i+1])
a.append(a[i+1] - a[i])
i += 2

f = open("out_reproduced.bin","wb") # save in binary
f.write(struct.pack('d'*L,*a)) # use IEEE 754 double format
f.close()

Then one can check that the produced file is identical:

\$ sha1sum *bin
70d32ee20ffa21e39acf04490f562552e11c6ab7  out.bin
70d32ee20ffa21e39acf04490f562552e11c6ab7  out_reproduced.bin

edit: How I found the solution: I found some of the other comments helpful, especially from gjm (although I did not read all). In particular, interpreting the data as a sequence of 64-bit floating point numbers saved me a lot of time. Also gjm's mention of the pattern a, -a, b, c, -c, d was an inspiration. If you look at the first couple of numbers, you can see that they are sometimes half of an earlier number. Playing around further with the numbers I eventually found the patterns a[i] * a[i+1] + 1.0 and a[i] - a[i+1]. It remaine...

Replicated!

I have mixed thoughts on this.

I was delighted to see someone else put forth an challenge, and impressed with the amount of people who took it up.

I'm disappointed though that the file used a trivial encoding. When I first saw the comments suggesting it was just all doubles, I was really hoping that it wouldn't turn out to be that.

I think maybe where the disconnect is occurring is that in the original That Alien Message post, the story starts with aliens deliberately sending a message to humanity to decode, as this thread did here. It is explicitly described as such:

From the first 96 bits, then, it becomes clear that this pattern is not an optimal, compressed encoding of anything.  The obvious thought is that the sequence is meant to convey instructions for decoding a compressed message to follow...

But when I argued against the capability of decoding binary files in the I No Longer Believe Intelligence To Be Magical thread, that argument was on a tangent -- is it possible to decode an arbitrary binary files? I specifically ruled out trivial encodings in my reasoning. I listed the features that make a file difficult to decode. A huge issue is ambiguity because in almost all...

1blf5mo
Your interlocutor in the other thread seemed to suggest that they were busy until mid-July or so. Perhaps you could take this into account when posting. I agree that IEEE754 doubles was quite an unrealistic choice, and too easy. However, the other extreme of having a binary blob with no structure at all being manifest seems like it would not make for an interesting challenge. Ideally, there should be several layers of structure to be understood, like in the example of a "picture of an apple", where understanding the file encoding is not the only thing one can do.
1anonymousaisafety4mo
I have posted my file here https://www.lesswrong.com/posts/BMDfYGWcsjAKzNXGz/eavesdropping-on-aliens-a-data-decoding-challenge. [https://www.lesswrong.com/posts/BMDfYGWcsjAKzNXGz/eavesdropping-on-aliens-a-data-decoding-challenge.]
9Rafael Harth5mo
Interesting, I expected the solution to be simpler, and I'm surprised you found it this quickly given that it's this complex.
5Bucky5mo
This suggests a different question. For non-participants who are given the program which creates the data, what probability/timeframe to assign to success. On this one I think that I would have put a high probability to be solved but would have anticipated a longer timeframe.
3anonymousaisafety5mo
https://en.wikipedia.org/wiki/Kolmogorov_complexity [https://en.wikipedia.org/wiki/Kolmogorov_complexity] The fact that the program is so short indicates that the solution is simple. A complex solution would require a much longer program to specify it.
5Rafael Harth5mo
Kolmogorov Complexity isn't the right measure. I'm pretty sure this program is way simpler (according to you or me) than most uniformly sampled programs of equal KC that produce a string of equal length. And i wouldn't consider it short given how hard it would be to find a "typical" program with this KC.
1anonymousaisafety5mo
Why do you say that Kolmogorov complexity isn't the right measure? I am worried that you might have this backwards? Kolmogorov complexity describes the output, not the program. The output file has low Kolmogorov complexity because there exists a short computer program to describe it.
2Rafael Harth5mo
Let me rephrase then: most strings of same length and same KC will have much more intuitively complex programs-of-minimum-length that generated them. Why is the program intuitively simple? Well for example, suppose you have the following code in the else block: a.append(min(max(int(i+a[i]),2),i) i += 1 I think the resulting program has lower length (so whatever string it generates has lower KC), but it would be way harder to find. And I bet that a uniformly sampled string with the same KC will have a program that looks much worse than that. The kinds or programs that look normal to you or me are a heavily skewed sample, and this program does look reasonably normal.
5Bucky5mo
I don’t think this follows - your code is shorter in python but it includes 3 new built in functions which is hidden complexity. I do agree with the general point that KC isn’t a great measure of difficulty for humans - we are not exactly arbitrary encoders.
4gjm5mo
Why do you think that "would be way harder to find"? (My intuition goes exactly the other way.)
5Rafael Harth5mo
Eh. I agree with both replies that the example was bad. I wanted to do something unintuitive with pointer arithmetic, but the particular code probably doesn't capture it well, so it may in fact be easier to find. (Still totally stand by the broader point though.)
7gjm5mo
Lovely!
2Alex Vermillion5mo
Might be an obvious thought, but why did the aliens send us this gibberish. Is there an interpretation of this that makes sense?

Even more obviously, why do aliens adhere to the IEEE 754 standard? My interpretation is that the cryptanalyst from the post has indeed been pranked by their friend at NASA.

Interpreting each 8-byte block as an IEEE double-precision float (endianity whatever's default on x86-64, which IIRC is little-endian) yields all values in the range -4 .. 4.2. FFT of the resulting 2095 values gives (ignoring the nonzero DC term arising from the fact that the average isn't quite zero) is distinctly not-flat, indicating that although the numbers look pretty random they are definitely not -- they have more high-frequency and less low-frequency content than one would get from truly random numbers.

Here's what you get when you interpret them as (y,x) positions of points.

7Nnotm5mo
For what it's worth, colored by how soon in the sequence they appear (blue is early, red is late) (Also note I interpreted it as 2094 points, with each number first used in the x-dimension and then in the y-dimension): Note that one line near the top appears to be drawn twice, confirming if nothing else that it's not a rule that it's not a succession rule that only depends on the previous value, since the paths diverge afterwards. Still, comparing those two sections could be interesting.
1Nnotm5mo
The double line I was talking about is actually a triple line, at indices 366, 677, and 1244. The lines before come from fairly different places, and they diverge pretty quickly afterwards: However, just above it, there's another duplicate line, at indices 1038 and 1901: These start out closer together and also take a little bit longer to diverge. This might be indicative of a larger pattern that points that are close together and have similar histories tend to have their next steps close to each other as well.
6gjm5mo
Every number in the sequence which equals 1-(y/2)^2 for "nice" y to "good" accuracy is in fact on the quadratic; i.e., when a number is "approximately" 1-(y/2)^2 for a y that "could" come next, that y does in fact come next. BUT to make this so I am having to define "nice" and "good" fairly tightly; this in fact occurs for only 14 values early in the sequence, perhaps before roundoff error starts being a nuisance. (I am conjecturing that the way this thing is constructed might be that we start with a few initial values and then iterate some process like: "if (condition) then next number is - previous number; else if (condition) then next number is 2sqrt(1-previous); else if ...". If so, then we might start with "nice" numbers -- in fact the first two are 1.3 and 2.1 to machine precision -- but that "niceness" might be eroded as we calculate. This is all very handwavy and there is at least one kinda-obvious objection to it.)
5gjm5mo
A few reasons for suspecting that this sort of iterative process is in play: * It would fit with the relationship to "That Alien Message" where the idea is that each frame (here, each number) is the "next" state of affairs as something evolves according to simple rules. * There are many cases in which it sure looks as if one number is derived from its predecessor. (Or maybe from its successor.) * It's an obvious thing to do, if you want to set a challenge like this. Note that I am not necessarily conjecturing that each number is a function of its predecessor. There might be other internal state that we don't get to see.
1blf5mo
Do you see how such an iteration can produce the long-distance correlations I mention in a message below, between floats at positions that differ by a factor of2/e? It seems that this would require some explicit dependence on the index.
2Donald Hobson5mo
Its not exactly 2/e. Here is a plot of the "error" of those points. The x axis is the larger point. The y axis is the smaller point minus 2/e times the larger. So its within about 1% of 2/e, suggesting it might be a real thing, or might just be a coincidence.
3Rafael Harth5mo
Have you tried writing programs that create sequences like the above to see how close they get? (Asking to avoid redoing work if the answer is yes.)
2gjm5mo
Making explicit something I said in passing elsewhere: it seems not-impossible that the sequence might be best looked at in reverse order, since mapping x to 1-(x/2)^2 or (1-x^2)/2 is somewhat more natural than going the other way. However, it does look rather as if there might be some accumulation of roundoff error in the forward direction, which if true would argue for reading it forwards rather than backwards.
5ObserverSuns5mo
More structure emerges! Here's a plot of consecutive pairs of values (data[i], data[i+1]) such that data[i+1] = -data[i+2]. ![Consecutive values before a negation](https://i.imgur.com/2FRBhAz.png)
4dkirmani5mo
Has anyone tried a five-dimensional representation instead of a two-dimensional one? 2095 isn't divisible by 2 or by 3, but it is divisible by 5. Maybe our "aliens" have four spatial dimensions and one temporal.
5Rafael Harth5mo
I've tried highlighting every k-th point out of n with the same color for a bunch of n, but it all looks random. Right now, I've also tried using only 2 of 5 float values. It looks like a dead end, even though the idea is good. I think the data is 1-dimensional, the interesting part is how each number is transformed into the next, and the 2d representation just happens to catch that (e.g., if x is transformed into −x, it lands on the diagonal).
5Nnotm5mo
Second try: When looking at scatterplots of any 3 out of 5 of those dimensions and interpreting each 5-tuple of numbers as one point, you can see the same structures that are visible in the 2d plot, the parabola and a line - though the line becomes a plane if viewed from a different angle, and the parabola disappears if viewed from a different angle.
1Nnotm5mo
Looking at scatterplots of any 3 out of 5 of those dimensions, it looks pretty random, much less structure than in the 2d plot. Edit: Oh, wait, I've been using chunks of 419 numbers as the dimensions but should be interleaving them
4gjm5mo
(This has y increasing downwards.) The straight line comes from the (x,-x) pairs I remarked on, which make up ~2/3 of the dataset. The parabolic thing comes from values ...,a,b,... where a = 1-(b/2)^2 to something like full precision.
2moridinamael5mo
Ah, it's the Elden Ring
2Rafael Harth5mo
There are 1047 points. The points on the diagonal are distributed all across the list -- here are the first indices [3,6,9,10,13,16,18,21,25,27,28,31,34,38,41,43,46,49,52,55]... They're usually 3 apart, but sometimes 1, 2, 4, 5, or 6. The last one has index 1045.

By the way, it's definitely not aliens. Aliens would not be sending messages encoded using IEEE754 double-precision floats.

2blf5mo
Here is a rather clear sign that it is IEEE754 64 bit floats indeed. (Up to correctly setting the endianness of 8-byte chunks,) if we remove the first n bits from each chunk and count how many distinct values that takes, we find a clear phase transition at n=12, which corresponds to removing the sign bit and the 11 exponent bits. These first 12 bits take 22 different values, which (in binary) clearly cluster around 1024 and 3072, suggesting that the first bit is special. So without knowing about IEEE754 we could have in principle figured out the splitting into 1+11+52 bits. The few quadratic patterns we found have enough examples with each exponent to help understand the transitions between exponents and completely fix the format (including the implicit 1 in the significand?).
2dkirmani5mo
Alternative hypothesis: The first several bits (of each 64-bit chunk) are less chaotic than the middle bits due to repetitive border behavior of a 1-D cellular automaton. This hypothesis also accounts for the observation that the final seven bits of each chunk are always either 1000000 or 0111111. If you were instead removing the last n bits from each chunk, you'd find another clear phase transition at n=7, as the last seven bits only have two observed configurations.
1blf5mo
If you calculate the entropyp0log2(p0)+p1log(p1)of each of the 64 bit positions (wherep0andp1are the proportion of bits 0 and 1 among 2095 at that position), then you'll see that the entropy depends much more smoothly on position if we convert from little endian to big endian, namely if we sort the bits as 57,58,...,64, then 49,50,...,56, then 41,42,...,48 and so on until 1,...,8. That doesn't sound like a very natural boundary behaviour of an automaton, unless it is then encoded as little endian for some reason.
2dkirmani5mo
Now that I know that, I've updated towards the "float64" area of hypothesis space. But in defense of the "cellular automaton" hypotheses, just look at the bitmap! Ordered initial conditions evolving into (spatially-clumped) chaos, with at least one lateral border exhibiting repetitive behavior:

The first few numbers are mostly "nice". 1.3, 2.1, -0.5, 0.65, 1.05, 0.675, -1.15, 1.15, 1.70875, 0.375, -0.375, -0.3225, -2.3, 2.3. Next after that is 1.64078125, which is not so nice but still pretty nice.

These are 13/10, 21/10, -1/2, 13/20, 21/20, 27/40, -23/20, 23/20, 1367/800, 3/8, -3/8, -129/400, -23/10, 23/10, 10501/6400, ...

A few obvious remarks about these: denominators are all powers of 2 times powers of 5; 13 and 21 are Fibonacci numbers; shortly after 13/10 and 21/10 we have 13/20 and 21/20; we have three cases of x followed by -x.

5gjm5mo
Slightly less than 1/3 of all numbers are followed by their negatives. Usually it goes a, -a, b, c, -c, d, etc., but (1) the first x,-x pair isn't until the 7th/8th numbers, and (2) sometimes there is a longer gap and (3) there are a few places where there's a shorter gap and we go x, -x, y, -y.
3Measure5mo
Most of the breaks in the (a, -a, b, c, -c, d) pattern look like either the +/- pair was skipped entirely or the unpaired value was skipped. My guess is the complete message consists of alternating "packets" of either a single value or a +/- pair, and each packet has some chance to be omitted (or they are deterministically omitted according to some pattern).
4gjm5mo
The number 23/20 appears three times. Each time it appears it is followed by a different number, and that different number is generally "unusually nasty compared with others so far". First time, at (0-based) index 7: followed by 1367/800, first with a denominator > 40. Second time, at (0-based) index 21: followed by 3.1883919921875 ~= 1632.4567/2^9, first with a denominator > 6400. Third time, at (0-based) index 48: followed by 2.4439140274654036, dunno what this "really" is, first with no obvious powers-of-2-and-5 denominator. [EDITED to fix an error.]
2Measure5mo
2.4439140274654036 might be (3³x19×3671×10631)/(2¹⁹x5⁶) with some incorrect rounding (2.4439140274658203125).
2Measure5mo
Value[71] is exactly half of value[49]. (and this again follows a 23)
2gjm5mo
The ".4567" seems just slightly suspicious. That third number is quite close to 5005/2^11. If we multiply it by 2^11/1001, the number we actually get is 5.00013579245669; that decimal tail also seems just slightly suspicious. 1, 3, 5, 7, 9, 2, 4, 6, approximately-8. This could all be mere pareidolia, obviously.
2gjm5mo
Early in the sequence (i.e., before roundoff error has much effect, if there's something iterative going on) it seems like a surprising number of our numbers have denominators that are (power of 2) x 10000. As if there's something happening to 4 decimal places, alongside something happening that involves exact powers of 2. (Cf. Measure's conjecture that something is keeping track of numerators and denominators separately.) This is all super-handwavy and I don't have a really concrete hypothesis to offer. [EDITED to add:] The apparent accumulating roundoff is itself evidence against this, because it means that after a while our numbers are not powers of 2 over 10000. So I'll be quite surprised if this turns out to be anything other than delusion. I'm leaving it here just in case it gives anyone useful ideas.
2gjm5mo
These aren't all quite correctly rounded. E.g., the 12th number is about -0.3225 but it isn't the nearest IEEE754 doublefloat to -0.3225. I suspect these are "honest" rounding errors (i.e., the values were produced by some sort of computation with roundoff error) rather than there being extra hidden information lurking in the low bits.
4Donald Hobson5mo
x followed by -x is common. cases where 1−xn=(xn+1/2)2are also common.
4Measure5mo
Hypothesis: the generator separately tracks the numerator and denominator and uses the xₙ₊₁ = 2*sqrt(1 - xₙ) rule exactly when this will result in both the numerator and denominator remaining integers.
5Donald Hobson5mo
Here is a plot of denominator ( of the closest fraction with denominator < 1000,000,000) This looks exactly what you would expect if you started with a number that happened to be a fraction, and applied a process like squaring or adding that made the denominator bigger and bigger. This also indicates the sequence was computed from start to end, not the other way around.
4Donald Hobson5mo
Wellxn+1=−2√1−xnappears around as often. If this were true, there must be something aiming towards simplicity. (A huge numerator + denominator are unlikely to be squares)
3Measure5mo
The second relation never occurs when xₙ is the negation of the previous xₙ₋₁. Furthermore, the second relation is always followed by xₙ₊₁ = -xₙ (i.e. there is never a "skipped pair" pattern break immediately following). This means that the skips are unlikely to be random.
1blf5mo
Whenever1−xn=(xn+1/2)2, this quantity is at most 4. I'm finding also around 50 instances of1−2xn=(xn+1)2∈[1,4](namely1≤|xn+1|≤2), with againxn+2=−xn+1.
2gjm5mo
It doesn't look as if there are a lot of other such relationships to be found. There are a few probably-coincidental simple linear relationships between consecutive numbers very near the start. There are lots ofy=−x, quite a lot of4x =4−y2, one maybe-coincidental14−x=2−y2, one maybe-coincidentalx=−4(1−y)2, some2x =1−y2, and I'm not seeing anything else among the first 400 pairs. [EDITED to add:] But I have only looked at relationships where all the key numbers are powers of 2; maybe I should be considering things with 5s in as well. (Or maybe it's 2s and 10s rather than 2s and 5s.)
2gjm5mo
There is one place where−xn=4(1−xn+1)2. I wonder whether there are many other relationships of broadly similar shape.
2gjm5mo
The distribution of the numbers is distinctly not-uniform. Might be normal. Mean is about 0.157, standard deviation about 1.615. (I doubt these are really pi/20 and the golden ratio, but those would be consistent with the data :-).) Again, clearly not independent given how the FFT looks.
2gjm5mo
Pretty sure it's not normal. The middle isn't bulgy enough. It's hard to be very confident with relatively few numbers, but I eyeballed the histogram against several batches of normals with the same mean and variance, and it's well outside what looks to me like the space of plausible histograms. I haven't bothered attempting any actual proper statistical tests. (Also, of course, if the numbers are generated by the sort of iterative process I imagine, it seems like it would be quite difficult to arrange for them to be much like normally distributed.)

If you plot the correlation of consecutive pairs of bytes, it looks like this. The bright dot is 102, which repeats itself a fair bit at the start. The horizontal and vertical lines are 63,64,191,192 as every 8th number is one of those four values. Most of the space is 0. when a pattern appears once, it usually appears more than once. 2,4,6,8,10,12 all appear frequently.

Just wanted to say I think this is a cool experiment and look forward to seeing how it plays out.

(This is somewhat betting on the contest turns out to be a well crafted experience/goal, so I guess also specifically looking to see the debrief on that)

Without looking at the data, I’m giving a (very rough) ~5% probability that this is a version of the Arecibo Message.

7dkirmani5mo
I turned the message into an Arecibo-style image [https://i.ibb.co/2qkQW7b/poss-pictogram.png]. There are patterns there, but I can't ascribe meaning to them yet. Good luck!
5dkirmani5mo
Update: I took the row-wise xor, and eliminated the redundancy on the right margin: The full image is available here [https://i.ibb.co/CtxpF0G/poss-pictogram-v-d.png].
2Alex Vermillion5mo
Can you explain more how you got this? I'm trying to figure out why the left hand side of the full picture has a binary 01010101 going almost the whole way (after the header) Nevermind, I got it! Break the bits into groups of 64
2dkirmani5mo
Yeah, I originally uploaded this [https://i.ibb.co/WvPN41z/poss-pictogram-v-sd.png] version by accident, which is the same as the above image, but the lines that go [0,0,0, .... ,0,1,0] are so common that I removed them and represented them as a single bit on the left.
2Alex Vermillion5mo
The bar on the right hand side, if you take it as high and low, has an interesting property. Pairing each high-low, it goes 1 1 10 2 8 1 3 2 4 2 15 1 8 1 3 2 15 2 ... 12 1 5 1 8 3 9 3 26 2 1 2 4 2 3 1 3 2 9 The second group (0s) is almost always pretty short, while the first group isn't as bounded.

1. We know the file was generated by code
2. We know that code was written by a human 2a. That human was trying to simulate an extraterrestrial signal
3. We know that this is a puzzle, though not a "fair" puzzle like those found in escape rooms or that MIT contest. 3a. An exhaustive brainstorming of different domains of solution is warranted.

Well, looking at the binary data reveals some patterns; the first 6 bytes of each block of 8 start out highly repetitive, though this trend lessens later in the file. This is what it looks like as an image.

8dkirmani5mo
Edit: The y-axis is inverted. The graph shows differences when it should show similarities. * Each bit of the message is 55% likely to be the the same as its predecessor; the bits of the message are autocorrelated. * If we split the message into 64-bit chunks, each bit in a given chunk has a 68% chance of being the same as the corresponding bit in the previous chunk.
4dkirmani5mo
If you split the message into 64-bit chunks, the last 6 bits of each chunk are always identical. That is, they're either "000000" or "111111". I don't think there's a third spatial dimension to the data, as chunking by n*64 doesn't yield any substantial change in autocorrelation for n > 1.
2dkirmani5mo
Oh, and the 7th-last bit (index [57]) is always the opposite of the final six bits. I interpret this as either being due to some cellular automaton rule, or that the "aliens" have given us six redundant bits in order to help us figure out the 64-bit reading frame.
1dkirmani5mo
There's something up with the eighth bit (index [7]) of every 64-bit chunk. It has a remarkably low turnover rate (23%) when compared to its next-door neighbors. Bits [48:51] also have low turnover rates (22%-25%), but the eighth bit's low turnover rate uniquely persists when extended to context windows with lengths up to 20 chunks.
2dkirmani5mo
The last seven bits of every 64-bit chunk actually carry only one bit of information. The bit right before these seven (index [56]) has an abnormally high turnover rate w.r.t. next-door neighbors(66%). Part of me wants to attribute this to some cellular automaton rule. But isn't it interesting that, in a chunk, the eighth bit is unusually stable, while the eighth-last bit is unusually volatile? Some weird kind of symmetry at play here.
4dkirmani5mo
Looking at the image, two conspicuous vertical bars emerge.

Thanks for doing this. I was considering offering my own version of this challenge, but it would have just been the last 50% of a file in a standard media format. That may stump one unmotivated person but probably would be easy for a large community to crack. (But what an update it would be if LWers failed that!) Anyway, I didn't have the time to do the challenge justice, which is why I didn't bother. So again, thanks!

Adding a .txt extension and translating to English produced some fun messages. :-)

2dkirmani5mo
I failed to replicate this finding.
4Mitchell_Porter5mo
Rename to out.bin.txt, view in Chrome, translate from Chinese to English, and you too may "recognize the spicy 5, recognize evil".
6Rafael Harth5mo
Swallowing Hot.?Swallowing Hot.@ 嗫Swallowing Hot.?Swallowing Hot.?Locker Cabinet??ffffff詩ffffff.?wup= W ? ? = I wish the key ffffff 纅ffffff @q= I wish @ ?wu p= W ? p= W繩徛 (\銺? { 铽qi { 铽進?ffffff驩ffffff ?V0┯ @dffffτ?dffffτ緿韜zhen 43333 43333 @(~尮k罱? { 罱巷 { 罱巷?xu= Zhukey ffffff and fffff @V0┯侚?| 0 ?徛 (\ ? (\劭ffffff ffffff ? 1 " @0L ?0L 遵頠 镢?dffff?dffffshrinking 龾澑采?潽 q琙 ?掽 q琙锟囪tBf.@山管@ ?山閇@陈 / -2兯啩Wsheng @醯sheng . Swimming cool? (For convenience :-))

If you let w be the first of the negation pairs, and q come before it. So the sequence goes ..., q, w, -w, ...

Then a plot of q (x axis) and w(y axis) looks like.

All points are within the orange line (which represents absolute(w)<-0.5*q+2.5 )

Note also that this graph appears approximately symmetric about the x axis. You can't tell between ...q,w,-w...  and ...q, -w, w, ... The extra information about which item of the pair is positive resides elsewhere.

For comparison, the exact same graph, but using an unfiltered list of all points, looks like...

Here is the image you get if you interpret the data as floats, consider the ratio of all possible pairs, and color those with a simple ratio blue.

The pattern of horrizontal and vertical lines in the top left corner is straightforward, those numbers are individually simple fractions, so their ratio is too. The main diagonal is just numbers having a simple 1:1 ratio with themselves. The shallower diagonal lines tell us something interesting.

The non_diagonal line that is closest to diagonal means some process must be refering back to values around 73% a...

3Donald Hobson5mo
What are the values along that line. Shows you multiply by 1/2, -1/2, 2 or -2.
1blf5mo
These simple ratios are "always"±2n, see my comment https://www.lesswrong.com/posts/dFFdAdwnoKmHGGksW/contest-an-alien-message?commentId=Nz2XKbjbzGysDdS4Z [https://www.lesswrong.com/posts/dFFdAdwnoKmHGGksW/contest-an-alien-message?commentId=Nz2XKbjbzGysDdS4Z] for a proposal that 0.73 is close to2/e(which I am not completely convinced by).

Misc. notes:

• As we've all discovered, the data is most productively viewed as a sequence of 2095 8-byte blocks.
• The eightth byte in each block takes the values 64, 63, 192, and 191.  64 and 192 are much less common than 63 and 191.
• The seventh byte takes a value between 0 and 16 for 64/192 rows, weighted to be more common at the 0 end of the scale. For 63/191 rows, it takes a value between ??? and 256, strongly weighted to be more common at the 256 end of the scale (the lowest is 97 but there's nothing special about that number so the generator probably
...

Some thoughts for people looking at this:

• It's common for binary schemas to distinguish between headers and data. There could be a single header at the start of the file, or there could be multiple headers throughout the file with data following each header.
• There's often checksums on the header, and sometimes on the data too. It's common for the checksums to follow the respective thing being checksummed, i.e. the last bytes of the header are a checksum, or the last bytes after the data are a checksum. 16-bit and 32-bit CRCs are common.
• If the data represents
...

I'm treating the message as a list of 2095 chunks of 64 bits.  Let d(i,j) be the Hamming distance between the i-th and j-th chunk.  The pairs (i,j) that have low Hamming distance (namely differ by few bits) cluster around straight lines with ratios j/i very close to integer powers of 2/e (I see features at least from (2/e)^-8 to (2/e)^8).

1blf5mo
This observation is clearer when treating the 64-bit chunks simply as double-precision IEEE754 floating points. Then the set of pairs(i,j)for whichxi/ xjis±2nfor somenclearly draws lines with slopes close to powers of2/e. But they don't seem quite straight, so the slope is not so clear. In any case there is some pretty big long-distance correlation betweenxiandxjwith rather different indices. (Note that if we explain the first linej≃(2/e)ithen the other powers are clearly consequences.)

Hex dump of the first chunk of the file:

00000000: cdcc cccc cccc f43f cdcc cccc cccc 0040  .......?.......@
00000010: 0000 0000 0000 e0bf cdcc cccc cccc e43f  ...............?
00000020: cdcc cccc cccc f03f 9a99 9999 9999 e53f  .......?.......?
00000030: 6666 6666 6666 f2bf 6666 6666 6666 f23f  ffffff..ffffff.?
00000040: d8a3 703d 0a57 fb3f 0000 0000 0000 d83f  ..p=.W.?.......?
00000050: 0000 0000 0000 d8bf a070 3d0a d7a3 d4bf  .........p=.....
00000060: 6666 6666 6666 02c0 6666 6666 6666 0240  ffffff..ffffff.@
00000070: 713d 0ad7 a340 fa3f d8a3 703d 0a57
...
5gjm5mo
I believe all this bit-level structure is a consequence of the values being IEEE754 double-precision values, many of them for fairly "simple" numbers, often with simple arithmetical relationships between consecutive numbers.
7jaspax5mo
The nature of binary representations of floating-point is that nice bit-patterns make for round numbers and vice-versa, so I'm not sure that we can conclude a lot from that. The fact that the floating-point interpretation of the data results in numbers that cluster around certain values is telling, but could still be a red-herring. Part of my reluctance to endorse that theory is narrative: we were told that this is a simulated alien message, and what are the odds that aliens have independently invented double-precision floating point? In any case, I'm reading those threads attentively, but in the meantime I'm going to pursue some hunches of my own.
2jaspax5mo
Actually, the opener is quite a bit more structured than that, even: it's three 4-byte sequences where the bytes are all identical or differ in only one bit, followed by a different 4-byte sequence. There is probably something really obvious going on here, but I need to stare at it a bit before it jumps out at me. ETA: Switching to binary since there's no reason to assume that the hexadecimal representation is particularly useful here.
6jaspax5mo
Okay, here's something interesting. Showing binary representation, in blocks of 8 bytes: 00000000: 11001101 11001100 11001100 11001100 11001100 11001100 11110100 00111111 .......? 00000008: 11001101 11001100 11001100 11001100 11001100 11001100 00000000 01000000 .......@ 00000010: 00000000 00000000 00000000 00000000 00000000 00000000 11100000 10111111 ........ 00000018: 11001101 11001100 11001100 11001100 11001100 11001100 11100100 00111111 .......? 00000020: 11001101 11001100 11001100 11001100 11001100 11001100 11110000 00111111 .......? 00000028: 10011010 10011001 10011001 10011001 10011001 10011001 11100101 00111111 .......? 00000030: 01100110 01100110 01100110 01100110 01100110 01100110 11110010 10111111 ffffff.. 00000038: 01100110 01100110 01100110 01100110 01100110 01100110 11110010 00111111 ffffff.? 00000040: 11011000 10100011 01110000 00111101 00001010 01010111 11111011 00111111 ..p=.W.? 00000048: 00000000 00000000 00000000 00000000 00000000 00000000 11011000 00111111 .......? 00000050: 00000000 00000000 00000000 00000000 00000000 00000000 11011000 10111111 ........ 00000058: 10100000 01110000 00111101 00001010 11010111 10100011 11010100 10111111 .p=..... 00000060: 01100110 01100110 01100110 01100110 01100110 01100110 00000010 11000000 ffffff.. 00000068: 01100110 01100110 01100110 01100110 01100110 01100110 00000010 01000000 ffffff.@ The obvious pattern now is that every 8 bytes there is a repeated sequence of 6 bits which are all the same. Despite my initial protestations that the latter part of the file is less regular, this pattern holds throughout the entire file. The majority of the time the pattern is 111111, but there are a decent number of ones which are 000000 as well.

1. A deliberate message in an alien language
2. Sensor data from a pulsar, comet, or other astronomical event
3. An encoded image in an alien encoding scheme
4. A mathematical "test pattern" - possibly to serve as a key for future messages
5dkirmani5mo
The output of a 1-dimensional cellular automaton, with a context window of more than one row. Or non-deterministic behavior.

The zipped version of the message is 59.47% the size of the orginal. DEFLATE (the zipping algorithm) works by using shorter encodings for commonly-used bytes, and by eliminating duplicated strings. Possible next steps include investigating the frequencies of various bytes (histograms, Huffman trees).

1dkirmani5mo
Bytewise tallies: array([156, 117, 71, 90, 114, 95, 65, 59, 97, 49, 80, 60, 58, 52, 37, 57, 80, 28, 79, 53, 58, 53, 76, 31, 77, 30, 47, 41, 50, 33, 30, 28, 43, 31, 43, 36, 59, 38, 33, 37, 52, 41, 50, 37, 33, 55, 75, 52, 59, 50, 37, 44, 34, 23, 27, 51, 41, 15, 21, 46, 55, 47, 24, 903, 335, 59, 76, 38, 47, 37, 102, 47, 81, 50, 68, 33, 45, 28, 65, 38, 76, 55, 30, 42, 61, 43, 70, 74, 54, 39, 52, 48, 37, 37, 19, 49, 97, 65, 31, 55, 70, 48, 131, 60, 62, 42, 60, 64, 42, 42, 39, 39, 71, 37, 55, 49, 77, 38, 66, 23, 66, 45, 52, 37, 47, 35, 80, 43, 78, 46, 45, 36, 21, 46, 80, 38, 43, 46, 84, 47, 49, 39, 46, 54, 51, 19, 43, 32, 66, 28, 48, 56, 59, 40, 62, 41, 48, 37, 47, 20, 57, 29, 61, 26, 45, 34, 76, 74, 21, 67, 66, 30, 38, 43, 49, 56, 82, 41, 42, 54, 63, 43, 53, 74, 35, 49, 53, 47, 71, 56, 46, 798, 335, 45, 52, 65, 50, 53, 32, 52, 55, 28, 66, 35, 60, 65, 73, 48, 58, 85, 37, 58, 65, 72, 64, 58, 51, 42, 62, 59, 60, 35, 82, 55, 60, 80, 68, 77, 101, 57, 95, 73, 62, 45, 73, 45, 109, 36, 88, 84, 133, 157, 85, 117, 142, 123, 100, 95, 112, 80, 96, 82, 135, 107, 82, 84])

If (like me) you're having a hard time reading the .bin format, here's a plaintext version of it in hexadecimal.

Interpreting the data as unsigned 8-bit integers and plotting it as an image with width 8 results in this (only the first few rows shown):

The rest of the image looks pretty similar. There is a almost continuous high-intensity column (yellow, the second-to-last column), and the values in the first 6 columns repeat exactly in the next row pretty often, but not always.

[+][comment deleted]5mo 10


text:
Difference between revisions of "2001 AMC 12 Problems/Problem 3"

The following problem is from both the 2001 AMC 12 #3 and 2001 AMC 10 #9, so both problems redirect to this page.

Problem

The state income tax where Kristin lives is levied at the rate of $p\%$ of the first $\textdollar 28000$ of annual income plus $(p + 2)\%$ of any amount above $\textdollar 28000$. Kristin noticed that the state income tax she paid amounted to $(p + 0.25)\%$ of her annual income. What was her annual income?

$\text{(A)}\,\textdollar 28000 \qquad \text{(B)}\,\textdollar 32000 \qquad \text{(C)}\,\textdollar 35000 \qquad \text{(D)}\,\textdollar 42000 \qquad \text{(E)}\,\textdollar 56000$

Solution

Solution 1

Let the income amount be denoted by $A$.

We know that $\frac{A(p+.25)}{100}=\frac{28000p}{100}+\frac{(p+2)(A-28000)}{100}$.

We can now try to solve for $A$:

$(p+.25)A=28000p+Ap+2A-28000p-56000$

$.25A=2A-56000$

$A=32000$

So the answer is $\boxed{B}$

Solution 2

Let $A$ be Kristin's annual income. Notice that

$A = p\%\cdot28000 + (p + 2)%\cdot(A - 28000) = [p%\cdot28000 + p%\cdot(A - 28000)] + 2%\cdot(A - 28000) =$ (Error compiling LaTeX. ! LaTeX Error: \begin{equation*} on input line 20 ended by \end{document}.)

format: plain


text:
2007ApJ...671.1944M

 other querymodes : Identifierquery Coordinatequery Criteriaquery Referencequery Basicquery Scriptsubmission TAP Outputoptions Help

 Query : 2007ApJ...671.1944M

2007ApJ...671.1944M - Astrophys. J., 671, 1944-1958 (2007/December-3)

Spectropolarimetry of the type IIb supernova 2001ig.

MAUND J.R., WHEELER J.C., PATAT F., WANG L., BAADE D. and HOFLICH P.A.

Abstract (from CDS):

We present spectropolarimetric observations of the Type IIb SN 2001ig in NGC 7424; conducted with the ESO VLT FORS1 on 2001 December 16, 2002 January 3, and 2002 August 16, or 13, 31, and 256 days postexplosion. These observations are at three different stages of the SN evolution: (1) the hydrogen-rich photospheric phase, (2) the Type II to Type Ib transitional phase, and (3) the nebular phase. At each of these stages, the observations show remarkably different polarization properties as a function of wavelength. We show that the degree of interstellar polarization is 0.17%. The low intrinsic polarization (∼0.2%) at the first epoch is consistent with an almost spherical (<10% deviation from spherical symmetry) hydrogen-dominated ejecta. Similar to SN 1987A and to Type IIP SNe, a sharp increase in the degree of the polarization (∼1%) is observed when the outer hydrogen layer becomes optically thin by day 31; only at this epoch is the polarization well described by a dominant axis.'' The polarization angle of the data shows a rotation through ∼40° between the first and second epochs, indicating that the asymmetries of the first epoch were not directly coupled with those observed at the second epoch. For the most polarized lines, we observe wavelength-dependent loop structures in addition to the dominant axis on the Q-U plane. We show that the polarization properties of Type IIb SNe are roughly similar to one another, but with significant differences arising due to line blending effects especially with the high velocities observed for SN 2001ig. This suggests that the geometry of SN 2001ig is related to SN 1993J and that these events may have arisen from a similar binary progenitor system.

Journal keyword(s): Stars: Supernovae: General - supernovae: individual (SN 2001ig) - Techniques: Polarimetric

Full paper

 Number of rows : 19
N Identifier Otype ICRS (J2000)
RA
ICRS (J2000)
DEC
Mag U Mag B Mag V Mag R Mag I Sp type #ref
1850 - 2022
#notes
1 SN 2002ap SN* 01 36 23.85 +15 45 13.2   13.11 14.54     SNIc-BL 536 1
2 SN 2004dt SN* 02 02 12.3 -00 06 02           SNIa 134 1
3 SN 2001el SN* 03 44 30.60 -44 38 23.4   12.81       SNIa 254 1
4 GD 50 WD* 03 48 50.1859724280 -00 58 32.290187664 12.596 13.787 14.063 14.210 14.388 DA1.2 250 1
5 SN 2003bg SN* 04 10 59.43 -31 24 50.4     15.0     SNIc 140 1
6 SN 1999em SN* 04 41 27.04 -02 51 45.2   13.79 13.7     SNIIP 635 1
7 SN 1987A SN* 05 35 28.020 -69 16 11.07           SNIIpec 4738 2
8 HD 49798 HXB 06 48 04.7001531360 -44 18 58.436036028 6.758 8.017 8.287 8.391 8.543 sdO6 307 0
9 2MASS J07371703+6535579 HXB 07 37 17.0432126424 +65 35 57.826001220           SNIIP 253 1
10 SN 1993J SN* 09 55 24.77476 +69 01 13.7026   10.8 12.0     SNIIb 1344 1
11 SN 1996cb SN* 11 03 41.98 +28 54 13.7   14.22       SNIIb 104 1
12 NAME Virgo Cluster ClG 12 26 32.1 +12 43 24           ~ 6316 0
13 SN 1987K SN* 12 43 41.17 +16 23 44.9     14.2     SNIIb: 143 1
14 SN 1997X SN* 12 48 14.28 -03 19 58.5   14.75 13.5     SNIb 65 1
15 CD-38 10980 WD* 16 23 33.8380498896 -39 13 46.161043572 9.916 10.885 11.029 10.936 10.841 DA2 466 0
16 NGC 7424 GiG 22 57 18.3924511080 -41 04 14.124670032   10.88 10.38 10.02 10.9 ~ 207 0
17 SN 2001ig SN* 22 57 30.69 -41 02 25.9     14.5     SNIIb 159 1
18 EGGR 264 WD* 23 19 35.3881861344 -17 05 28.465573380 13.279 14.132 14.111 14.07 14.049 DB3 114 0
19 NAME Cas A SNR 23 23 24.000 +58 48 54.00     14.30     ~ 2629 1

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2022.09.26-15:12:24


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# Manifolds / Lie Groups - confusing notation

1. Mar 16, 2008

### GSpeight

I'm reading over my Lie groups notes and in them, in the introductory section on manifolds, I've written that $F_{\star}$ is a commonly used notation for $d_{x}F$ and so the chain rule $d_{x}{G \circ F}=d_{F(x)}G \circ d_{x}F$ can be written $(G\circ F)_{\star}=G_{\star}\circ F_{\star}$

Is what I've written correct? To me this seems horribly confusing since it neglects to mention where you are taking the differential. Should it instead be that $F_{\star}$ is the map from M to $d_{x}F$. On second thoughts this doesn't make total sense either...

He's gone on to make definitions like:

A vector field X on a Lie group G is called left-invariant if, for all g,h in G, $(L_{g})_{\star}X_{h}=X_{gh}=X_{L_{g}(h)}$ where $L_{g}$ is the left multiplication map by g ,which I'm finding difficult to understand with my current definition of $F_{\star}$.

Thanks for any replies.

Last edited: Mar 17, 2008


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Symbolic non-commutative algebra

Consider 3 non-commutative elements $$A,B,C$$ and we have the relation $$CB=\alpha BC$$.

I want to compute things like : $$(A\otimes BC ) ( B \otimes B) = AB \otimes \alpha B^2C$$

Is there any package doing this ?

• Check the NCAlgebra package (I am not very familiar with it so I can't give an answer, but it's the biggest/oldest package dealing with such problems). – Szabolcs Mar 21 at 8:25
• try www.github.com/NCAlgebra/NC – Slepecky Mamut Mar 21 at 8:39
• This package deals with non commutative algebra but it doesn't seem to treat tensor products – Nico045 Mar 21 at 10:16
• You don't need a package. Just give Mathematica the rules for working with non-commutative multiplication and tensor product. – Somos Mar 21 at 14:46

Using Dot for your noncommutative multiplication, we can encode your relation with:
B /: C . B := α B . C

Using KroneckerProduct for your "⊗" operation, the output you want to get can then be obtained with:
KroneckerProduct[A, B.C].KroneckerProduct[B, B] //TensorExpand


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# 1 & -1 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{cc} \frac{1}{10} & 0 \\ \frac{1}{2} & 1 \end{array}\right] P \text { fupp } \] zeros in the second row of the left hi fore. does not exist. transformations, find the inverse of 17. 2.

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 Question Text 1 & -1 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{cc} \frac{1}{10} & 0 \\ \frac{1}{2} & 1 \end{array}\right] P \text { fupp } \] zeros in the second row of the left hi fore. does not exist. transformations, find the inverse of 17. 2. Updated On Dec 5, 2022 Topic Algebra Subject Mathematics Class Class 12 Answer Type Video solution: 1 Upvotes 91 Avg. Video Duration 5 min


text:
# Rear Bear Vs. The Russian Bear

### Help Support HomeBuiltAirplanes.com:

#### JimCovington

##### Well-Known Member
Question? Did the 140lbs include instruments and gear?
Yes, if you call it that. It's a foot-launched glider. The carbon dragon lands on a wheel, the magic dragon (I think - it's been a while) lands on a skid. The magic dragon is lighter. You'll have to add more weight for a wheel (maybe) and more weight for engine instruments.

Think about it for a second - you're talking about removing HALF the weight from one of the lightest rigid airframes in the world. It's a VERY light airframe. This plane was planned by people who weighed each part to the TENTH of an ounce - and you're going to remove 70 lbs - ONE THOUSAND+ ounces.

The weight's just not there to remove.

You say you'll get the weight from the wing. The wing is designed to weigh 68.6 lbs. That means you can THROW AWAY THE WING from a plane that is SEVENTEEN INCHES WIDE at the seat and you STILL are over your limit. Of course, with no wing, your aerodynamic profile will suffer somewhat. :gig:

BTW, you can't just ad an engine without an engine mount. That energy has to be transmitted to the airframe somehow. Ooops, more unplanned weight!

Are you beginning to grasp just how impossible your 100lb high-AR high-performance plane is?

Put the BD-5 on a really serious diet.
Again - WHAT are you removing? It's already one of the lightest airplanes in the world! You keep saying you'll "remove weight" but haven't offered anything concrete - just your opinion that skilled designers & builders in front of you have been ignoring one of the MOST important factors when building for years. I don't buy it.

#### Starman

Tony

##### Moderator
Autoreply has said and I believe him that about ½ of the weight factor in a glider is for flutter. At 1/3 the speed the airplane should need a lot less mass to combat flutter.
It's a bit more complicated than that, but the counter-torsion structure is much heavier than the whole spar is.
No PRSU. Probably 16lbs installed. 15Hp on a self launch hang glider needs a re-drive. At the slow speeds that the hang glider flies at it needs the extra mass flow for good thrust. At a faster speed you don't need as much prop diameter to do the trick.
I've been reading about airplanes my hole life, and I've got reasonably good recall, I know it will work. Showing you why is a different story.
Let's look at those remarks for a while.
If you're making a claim you have to prove it, not me to disprove it. I and many others are polite enough to tell you friendly that your ideas (not all of them, but most and most combinations of ideas) are not realistic.
I tell you, it's impossible to have a 16 lbs, 16 HP installed aircraft engine including prop.
Glider pilots are willing to trade or compromise everything, including steep maintenance, reliability and their wives to loose weight. A turbo (30HP-ish foldable engine) is in many environments almost a requirement. Again, lower weight to whatever price and that price is very steep.
So, we have a two-stroke engine plus prop, driven by a belt for a total weight of 37 kg (80lbs). That's the minimum weight, no electric starter (windmilling), no engine mounts (that's not included in the weight), no battery, just an engine, prop (fixed pitch, LIGHT) and a reduction drive. You won't get rid of it since that'll ruin your prop properties, but just assume we can ditch that weight too, that might get us to 35 kg (75 lbs).
How in heavens sake are you going to scale that down to 16 lbs?

I've been discussing with Autists/Aspergers a couple of times and had the pleasure (without sarcasm) to meet (some of) them in real life afterwards as well. Some are capable of more or less understanding the excellent points Topaz wrote, most don't. That majority usually ends up pretty sad "because no-one understands them" and that's a self-reinforcing attitude you don't get away with in real life.

There's a good reason I've told you several times before: "go out and prove us wrong" since reality is the ultimate judge, not the way you communicate, perceive or think, or the way you type some words over the internet

#### Starman

##### Well-Known Member
Very good! I was thinking of a shape like this for my low aspect ratio ultralight because it's the lightest possible shape for the wing area. With this you could meet FAR103 with a 100 pound plane, not counting engine and fuel and maybe landing gear. Are we counting those things yet?

The problem with that design is that there's no view of the scenery from the cockpit, the cockpit should be at the leading edge for good visibility, with the engine at the back for balance.

Here's one with better visibility:

#### JimCovington

##### Well-Known Member
Here's one with better visibility:
There are many who think that's a highly underrated design that had the misfortune to appear at the beginning of the jet age when just about everything with a prop was considered old-fashioned. Since it required big props to neutralize the large wingtip vortices and could't be converted to a pure jet, it got left behind.

Charles Zimmerman and his "Skimmer" | Flight Journal | Find Articles at BNET

It was designed for a 425mph top speed and a 20!mph landing speed. That 20:1 ratio beats the snot out of the 4:1 that's the goal of many designers today.

#### pie_row

##### Well-Known Member
Nothing taken out of context. .
You changed what you wrote and what you quoted of me after the fact. And so this statement of yours is a lie. And in doing this you abused you moderator privileges.

Last edited:

#### JimCovington

##### Well-Known Member
I think that I've found a way to get what I see in my head about airplanes out past the blender. Autocad. I watched someone use a 3D graphics program and I can use that to get the stuff in my head out.
Excellent! I'm really looking forward to it. There are also some basic aeroengineering programs that allow you to draw planes *and* test them.

http://www.homebuiltairplanes.com/forums/design-structures-cutting-edge-technology/5025-what-software-do-you-use.html

#### Mac790

##### Well-Known Member
8 lbs engine... well, prop, PSRU and respective fairing are all made of Helium?
Theoretically it could be possible, check out this little turboprop

Power 5.62KW (7.5hp)
Diameter 86mm (core engine)
Length 326mm (12.8in)
Engine Weight 1750g (3lbs 13oz)
Ancillaries Weight 200g (7oz)
Propeller RPM range 1200 - 9000 RPM
Wren Turbines Ltd | 44 Turboprop

Seb

#### Attachments

• 39.2 KB Views: 1,021

#### JimCovington

##### Well-Known Member
Theoretically it could be possible, check out this little turboprop

Power 5.62KW (7.5hp)
Diameter 86mm (core engine)
Length 326mm (12.8in)
Engine Weight 1750g (3lbs 13oz)
Ancillaries Weight 200g (7oz)
Propeller RPM range 1200 - 9000 RPM
Wren Turbines Ltd | 44 Turboprop

Seb
So cute! You measure its weight in grams, its fuel flow in ounces, and its lifespan in minutes.

#### pie_row

##### Well-Known Member
As for engines, most European engines are about 1/3rd the size of an American, use about 1/3rd of the fuel and have more torque and power at much lower weight and size. A V8 is nice in a dragster, but otherwise completely the wrong way to go for a normal car. It's unwise at least to have a 3, 4 or 5 liter V8 when a 1.5 liter L4 with turbo does the same with all the aforementioned benefits..
I love it when someone argues and actual believes both sides of the same argument. I really do.

What a lot of people really want is a 454 chev V-8 from before emission conrtols went in. (7.5L)

#### Mac790

##### Well-Known Member
... and its lifespan in minutes.
Rather in hours, it seems that the maintenance is required after 50 hours or 500 million revolutions. Model Jet Club - Turbines.
The Wren Company has more than 20 years of experience with those engines, so I think that they can build an engine which last more than few minutes, plus I really doubt that someone would pay 3000-6000$for so unreliable engine. Also I have more belief in those engines than in 50lbs titanium structure. This idea isn't new. [video=youtube;sEMFhEGgCUM]http://www.youtube.com/watch?v=sEMFhEGgCUM&feature=fvw[/video] Seb Last edited by a moderator: #### JimCovington ##### Well-Known Member Rather in hours, it seems that the maintenance is required after 50 hours or 500 million revolutions. Model Jet Club - Turbines. The Wren Company has more than 20 years of experience with those engines, so I think that they can build an engine which last more than few minutes, plus I really doubt that someone would pay 3000-6000$ for so unreliable engine.
Seb
1) Sure people would pay that for an unreliable engine - it's going on a model airplane, after all. Crashes are a part of the hobby.

2) Their website says they've been doing it for 10 years, not 20. How many engines have they made?

3) The Wren factory site (not the distributor) says that "maintenance" may be required after 25 hours, not 50. Hmmm...your numbers aren't looking so good.

4) It's not maintenance, it's a bearing change - which is an overhaul. Changing oil is maintenance. Changing bearings is an overhaul.

So let's sum up - We have a engine that offers less time between OVERHAULS than most engines do between OIL CHANGES. It's got a TBO of 25 hours, which is 1500 minutes - less than the number of HOURS that most aircraft engine manufacturers allow.

So yes, its lifespan is measured in minutes.

And I know it's not a new idea - at least TWO!!! human-carrying aircraft have been powered by these engines. But that doesn't mean it's a practical - or even good - idea.

I am sure more airplanes will be built with these engines. But I'm also sure they'll be just novelty* acts for the foreseeable future.

*The two flying (that I know of) are novelties.

1) The jet wing requires launching from a mothership and cannot climb from sea level due to fuel/payload constraints.

2) The jet cri-cri uses almost 5lb of fuel PER MINUTE at full throttle. It's fuel load is barely FAA-legal.

2


text:
# Chessboard and Circle calibration provide radically different results

I have a 35mm nominal focal length camera that I'm trying to calibrate using a PyQt front end I'm building onto OpenCV's python library. It finds and displays matched points. Everything works except for the returns on cv2.calibrateCamera() when using a 4 x 11 asymmetric circle grid and cv2.findCirclesGrid().

Using a 6 x 9 chessboard pattern, I obtain a focal length (converted to mm) of 35.8 mm. I'd accept this for now with the 35mm lens I'm using. When I use the same camera/lens and a circle grid pattern I obtain a focal length of 505.9 mm. This is clearly wrong. The k and p coefficients are also enormous. What am I missing? See code below.

Edited down to show only relevant bits. Some variables are defined elsewhere.

# termination criteria
criteria = (cv2.TERM_CRITERIA_EPS + cv2.TERM_CRITERIA_MAX_ITER, 30, 0.001)

# prepare object points, like (0,0,0), (1,0,0), (2,0,0) ....,(6,5,0)
params = []
def objParams (grid):
if grid == "Checkerboard":
param1 = np.zeros((6 * 9, 3), np.float32)
param2 = np.mgrid[0:9,0:6].T.reshape(-1,2)
elif grid == "Circle Grid":
param1 = np.zeros((4 * 11, 3), np.float32)
param2 = np.mgrid[0:4,0:11].T.reshape(-1,2)
params.append(param1)
params.append(param2)
objParams(pattern)
objp = params[0]
objp[:, :2] = params[1]

# Arrays to store object points and image points from all the images.
objpoints = []  # 3d point in real world space
imgpoints = []  # 2d points in image plane.

images = glob.glob(folder + '/*.jpg')
i = 0
for fname in images:
smimg = cv2.resize(img, (0, 0), fx=scaleTo, fy=scaleTo)
gray = cv2.cvtColor(smimg, cv2.COLOR_BGR2GRAY)
# step counter
i += 1
# Find the chess board corners
if pattern == 'Checkerboard':
print("Now finding corners on image " + str(i) + ".")
ret, corners = cv2.findChessboardCorners(gray, (9, 6), None)
# If found, add object points, image points
if ret == True:
print("Corners found on image " + str(i) + ".")
objpoints.append(objp)
cv2.cornerSubPix(gray,corners,(11,11),(-1,-1),criteria)
imgpoints.append(corners)
print("Drawing corners on image " + str(i) + ".")
# Draw and display the corners
cv2.drawChessboardCorners(smimg, (9, 6), corners, ret)
if saveMarked == True:
cv2.imwrite(os.path.join(outfolder, os.path.basename(fname)), smimg)
cv2.imshow(os.path.basename(fname), smimg)
cv2.waitKey(500)
cv2.destroyAllWindows()
elif pattern == 'Circle Grid':
print("Now finding circles on image " + str(i) + ".")
ret, circles = cv2.findCirclesGrid(gray, (4,11), flags = cv2.CALIB_CB_ASYMMETRIC_GRID)
# If found, add object points, image points
if ret == True:
print("Circles found on image " + str(i) + ".")
objpoints.append(objp)
imgpoints.append(circles)
# Draw and display the circles
cv2.drawChessboardCorners(smimg, (4, 11), circles, ret)
if saveMarked == True:
cv2.imwrite(os.path.join(outfolder, os.path.basename(fname)), smimg)
cv2.imshow(os.path.basename(fname), smimg)
cv2.waitKey(500)
cv2.destroyAllWindows()

ret, mtx, dist, rvecs, tvecs = cv2.calibrateCamera(objpoints, imgpoints, gray.shape[::-1], paramMtx, None, None)

EDIT: Here are the outputs as well as my pretty print statements:

Chessboard ...

edit retag close merge delete

How many images did you use for the calibration process ?

I know only the basics with Python, so maybe I missed the information, how did you construct your object point ? For asymetric grid, the formula is different as you can see it here: https://github.com/Itseez/opencv/blob/3.1.0/samples/cpp/calibration.cpp#L125.

Also, I don't see in your code if you have supply the real square size ? Otherwise, you will be dependent of a scale factor.

You can look at this tutorial for C++: Camera Calibration with OpenCV.

( 2016-02-24 10:44:52 -0500 )edit

I used 35 images for each pattern. I think you are correct that the object point vectors aren't being constructed correctly. I only know the basics of C++ so I'm having trouble translating that. The arrays are set up in the objParams() function near the top as 0's. Values are appended to these arrays following the check on whether or not points were found, if ret == True: I'm pretty much building on this tutorial (which didn't work initially, for the record), but it's a little light on details. I'm not quite sure how I should construct an object point array for asymmetric circles.

As for the square size, I'm just scaling it for now

( 2016-02-24 12:38:30 -0500 )edit

Sort by » oldest newest most voted

I'm posting this as a late answer, for folks coming from Google.

The problem is almost certainly that your object points are incorrect. What you've done is symmetric grid (like a chessboard) which assumes the same number of grid points per row. The asymmetric pattern doesn't. This code is adapted from the C++ source. It's worth plotting your object points to double check that it actually looks like your pattern:

objectPoints= []
grid_size = 0.03 # 3cm, or whatever
rows, cols = 4, 11

for i in range(cols):
for j in range(rows):
objectPoints.append( (i*grid_size, (2*j + i%2)*grid_size, 0) )

objectPoints= np.array(objectPoints).astype('float32')

scatter(objectPoints[:,0], objectPoints[:,1])

Note that the rows are staggered in the pattern, that's where the asymmetry comes from. The grid_size is the x-spacing between the circles. The loop plots a 4x11 grid, but shifts alternate columns down by grid_size. Here's a plot of the pattern as generated:

more

Official site

GitHub

Wiki

Documentation


text:
This is an opinion, so feel free to disagree.

Whenever you have an enumeration (whenever you list stuff out) that’s written inline with commas, break it off into an explicit list, either unordered (bullet points) or ordered (with 1., 2., etc.).

This makes the enumeration obvious and easier to read.

## Examples of Comma Separated Enumerations

### Contrived example

A homeomorphism is a bijective, continuous function, whose inverse is also continuous.

### Actual Example from a Friend’s Blog

His blog no longer has this, but the original form was:

Pure functions are often defined as those without side effects. Side effects tend to be handwaved as “things that affect hidden state”, “things that are confusing”, “things that are bad that you can do in C++ because C++ is bad”, and other such bogeymen.

## Rewritten forms

### Homeomorphisms

A homeomorphism $f$ is a function that is:

• bijective
• continuous
• $f^{-1}$ is also continuous.

### Side Effects

Pure functions are often defined as those without side effects.

Side effects tend to be handwaved as:

• “things that affect hidden state”
• “things that are confusing”
• “things that are bad that you can do in C++ because C++ is bad”
• other such bogeymen.

## Caveats

It’s important to put the redundant info in the header for each list, or else it sounds weird when read.

### Example

A homeomorphism is:

• a bijective function
• a continuous function
• $f^{-1}$ is also continuous.

Repeating “a function” sounds weird.

## Takeaway

Go forth and write lists that people can read at a glance. Don’t make your readers work as much as possible, because God knows they’re lazy.

## Errors

If you find any errors, or if I’m just plain wrong, or you want to tell me something, feel free to tell me.


text:
# Cauchy density function for Brownian motion

Let $$\{W(t):t\geq0\}$$ be a Brownian motion, and let $$\{\mathcal{F}_{t},t\geq0\}$$ be its natural filtration. Let $$\{W_{2}(t):t\geq0\}$$ be a Brownian motion, independent of $$\{W(t):t\geq0\}$$. Denote, for $$a>0$$, $$\tau_{a}=\inf\{t\geq0:W(t)=a\}.$$ Using that the probability density of the first hitting time of $$a>0$$ for Brownian motion is given by $$f(t)= \left\{ \begin{array}{ll} \displaystyle\frac{ae^{-a^2/2t}}{\sqrt{2\pi t^{3}}}&\text{if }t>0\\ 0&\text{otherwise}. \end{array} \right.$$ Show that the probability denisty of $$W_{2}(\tau_{a})$$ is given by the Cauchy density function $$f(y)=\frac{a}{\pi(a^{2}+y^{2})},\quad y\in\mathbb{R}.$$

First, I do not fully understand the question. Is the Cauchy density function the probability density function of the second Brownian motion at time $$\tau_{a}$$, when the first Brownian motion is stopped? The distribution of a Brownian motion should have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case.

"Is the Cauchy density function the probability function of the second Brownian motion at time $$\tau_a$$, when the first Brownian motion is stopped?" Yes.
"The probability of the a Brownian motion should only have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case." As $$a$$ increases, so will $$\tau_a$$ overall; and although a Cauchy distribution doesn't have finite variance, the distribution does get more dispersed as $$a$$ increases. So there doesn't seem to be a conflict with your understanding that the variance of Brownian motion is increasing in time.
And the solution to the problem: \begin{align} f_{W_2(\tau_a)}(y)dy &=P[W_2(\tau_a)\in(y,y+dy)]\\ &=\int_{t=0}^\infty P[W_2(\tau_a)\in(y,y+dy)|\tau_a=t]f_{\tau_a}(t)dt\\ &=\int_{t=0}^\infty \frac{1}{\sqrt{2\pi t}}e^{-\frac{y^2}{2t}}dy\frac{1}{\sqrt{2\pi t^3}} ae^{-\frac{a^2}{2t}}dt\\ &= \left(\int_{0}^\infty \frac{a}{2\pi t^2}e^{-\frac{y^2+a^2}{2t}}dt\right)dy,\quad\text{let }s=\frac{1}{t},\\ &= \left(\int_\infty^0\frac{as^2}{2\pi}e^{-\frac{y^2+a^2}{2}s}\left(-\frac{1}{s^2}\right)ds \right)dy\\ &=\left( \int_0^\infty\frac{a}{2\pi}e^{-\frac{y^2+a^2}{2}s}ds \right)dy\\ & = \frac{a}{\pi(a^2+y^2)}dy. \end{align}


text:
# Algebra/Order of Operations

 Algebra ← Arithmetic Order of Operations Real_Numbers →

The Order of Operations is used when doing expressions with more than one operation (e.g., ×, +, -). These are rules so you only get one answer all the time.

Example: When faced with ${\displaystyle 4+2\times 3}$, how do you proceed?

There are two ways:

${\displaystyle 4+2\times 3=(4+2)\times 3}$

${\displaystyle 4+2\times 3=6\times 3}$

${\displaystyle 4+2\times 3=18}$

or

${\displaystyle 4+2\times 3=4+(2\times 3)}$

${\displaystyle 4+2\times 3=4+6}$

${\displaystyle 4+2\times 3=10}$

This is confusing, so which is correct? (Parentheses, "(" and ")" are used to show what to do first)

In order to communicate using mathematical expressions we must agree on an order of operations so that each expression has only one value.

For the above example all mathematicians agree the correct answer is 10.

You're probably wondering what this order is.

## The Standard Order of Operations

Evaluate expressions in this order.

• Parentheses or Brackets (evaluate what's inside them)
• Exponents
• Multiplication and/or division from left to right
• Addition and/or subtraction from left to right

### An Easy Way of Remembering

Use this memory tool to help remember the order! Please Excuse My Dear Annoying Sister It is also commonly called by its acronym, PEMDAS.

An alternative form of this is; Brackets Indices Division or Multiplication Addition or Subtraction (BIDMAS).

Yet another way of remembering this is Brackets Orders Division Multiplication Addition Subtraction (BODMAS)

or Bring Our Dear Mother Along Saturday

## Examples

Order of Operations - Examples
Expression Evaluation Operation
4 × 2 + 1 = 4 × 2 + 1 Multiplication
= 9
12 - 9 ÷ 3 = 12 - 9 ÷ 3 Division
= 12 - 3 Subtraction
= 9
2 × 9 ÷ 3 = 2 × 9 ÷ 3 Left to Right
= 18 ÷ 3 division
= 6
9 ÷ 3 × 3 = 9 ÷ 3 × 3 Left to Right
= 3 ×3 multiplication
= 9
3 + 12 ÷ (5 - 2) = 3 + 12 ÷ (5 - 2) Parentheses
= 3 + 12 ÷ 3 Division
= 7
7 × 10 - (2 × 4)2 = 7 × 10 - (2 × 4)2 Parentheses
= 7 × 10 - 82 Exponents
= 7 × 10 - 64 Multiplication
= 70 - 64 Subtraction
= 6

## Practice Problems

1

Evaluate the numerical expression

 ${\displaystyle 2+4*3=}$

2

Evaluate the numerical expression

 ${\displaystyle 2*4+3=}$

3

Evaluate the numerical expression

 ${\displaystyle (2+4)*3=}$

4

Evaluate the numerical expression

 ${\displaystyle 9^{2}+1-7*(8+4)/2=}$


text:
# How prove this “Trigonometric interpolation”?

let $f(x)$ show that $$f(x)\approx\sum_{i=0}^{2n-1}f(t_{j})L_{j}(t)$$

where $$t_{j}=\dfrac{\pi}{n}j$$ $$L_{j}(t)=\dfrac{1}{2n}\left[1+2\displaystyle\sum_{m=1}^{n-1}\cos{m(t-t_{j})}+\cos{n(t-t_{j})}\right], j=0,1,2,\cdots,2n-1$$

It is say that “Trigonometric interpolation”,But I find sometimes, I can't find this and can't anywhere have solution,But this is from china book,I think this have solution,Have you someone can help me,Thank you very much.

is from:

This is based on the Discrete Fourier Transform for a function on the sample points $t_j$ for $j\in\{0, \ldots, 2n-1\}$. Let $e_m(t_j) = \exp(imt_j)$ for $m\in \mathbb{Z}$. Note that $e_{m+2n}=e_m$ on the sample points $t_j$. For a function $f$ the DFT $\hat{f}$ is then expressed by
$$\hat{f}(m) = \sum_{j=0}^{2n-1} f(t_j) e_{-m}(t_j)$$
$$\begin{eqnarray} 2n f(t_k) &=& \sum_{m=0}^{2n-1}\hat{f}(m)e_m(t_k)\\ &=& \sum_{j=0}^{2n-1}f(t_j)\sum_{m=0}^{2n-1}e_{-m}(t_j)e_m(t_k)\\ &=& \sum_{j=0}^{2n-1}f(t_j)\sum_{m=0}^{2n-1}e_m(t_k-t_j)\\ &=& \sum_{j=0}^{2n-1}f(t_j)\left(1+\cos n(t_k-t_j) + \sum_{m=1}^{n-1}e_m(t_k-t_j) + \sum_{m=1}^{n-1}e_{2n - m}(t_k-t_j)\right)\\ &=& \sum_{j=0}^{2n-1}f(t_j)\left(1+\cos n(t_k-t_j) + 2\sum_{m=1}^{n-1}\cos m(t_k-t_j)\right) \end{eqnarray}$$
The interpolation simply replaces $t_k$ by an arbitrary $x$ to get a periodic real function with period $2\pi$. This interpolation uses the low frequency information from the sampled signal.


text:
-1

# How would I make a spawn system, as seen in obbies? (Checkpoints)

I'm having trouble making a obby spawn which means anytime you go to the next level, I want the player to hit the spawn then move on so when they die they dont go back to level 1, Example: Say i was on level 104, then I finally died then go back to level 1. If you don't understand, Here's more information Say, When I die on level 9, I don't(or players) don't wanna go back to level 1 with there hard work.

1
Unclear 1776
7 years ago

As long as you have SpawnLocations with different TeamColors, Neutral disabled, and AllowTeamChangeOnTouch enabled, you can have your checkpoints. This will give you SpawnLocations that allows players sharing the same TeamColor as the SpawnLocation to spawn there.

Also make sure that each player's Neutral property is set to false to enable proper team changing!

If you're interested in players saving their level forever, you should look into Data Persistence and Data Store.

0
How'd I use that? Roboy5857 20 — 7y
0
DataStore Service Lightning_Game27 217 — 10mo
0

It is a free model. Basically, when your player is created, the leaderstat is made and when you respawn, your CFrame of your Torso is moved to 2 units above the part that has the name the same as the leaderstat. There is a script in the parts (not SpawnLocations) that see if you completed that stage - 1 and if you did, your leaderstat is changed to that parts name. No script editing, but feel free to make your own.

0

A man walks into a bar

0

ROBLOX created a model for a checkpoint here's the script because you may get into some viruses in the toolbox

local Players = game:GetService("Players")
local ServerStorage = game:GetService("ServerStorage")

local checkpoint = script.Parent

function onTouched(hit)
if hit and hit.Parent and hit.Parent:FindFirstChildOfClass("Humanoid") then
local player = Players:GetPlayerFromCharacter(hit.Parent)
local checkpointData = ServerStorage:FindFirstChild("CheckpointData")
if not checkpointData then
checkpointData = Instance.new("Folder")
checkpointData.Name = "CheckpointData"
checkpointData.Parent = ServerStorage
end

local userIdString = tostring(player.UserId)
local checkpointValue = checkpointData:FindFirstChild(userIdString)
if not checkpointValue then
checkpointValue = Instance.new("ObjectValue")
checkpointValue.Name = userIdString
checkpointValue.Parent = checkpointData

wait()
local storedCheckpoint = ServerStorage.CheckpointData[userIdString].Value
character:MoveTo(storedCheckpoint.Position + Vector3.new(math.random(-4, 4), 4, math.random(-4, 4)))
end)
end

checkpointValue.Value = checkpoint
end
end

checkpoint.Touched:Connect(onTouched)


text:
# Calculate the coordinates of the third vertex of triangle given the other two and the length of edges

Using the solution from the_candyman

the_candyman (https://math.stackexchange.com/users/51370/the-candyman), Calculate the coordinates of the third vertex of triangle given the other two and the length of edges in the cheapest computational way, URL (version: 2017-02-22): https://math.stackexchange.com/q/2156910

Let $A = (x_A, y_A)$ and $B = (x_B,y_B)$ the known vertices of your triangle. Let's call $d_{AB}$, $d_{BC}$ and $d_{CA}$ the lengths of each side.

1. Translate your points subtracting $x_A$ and $y_A$ so that $A$ corresponds with the origin. That is:

$$A' = (0, 0), B' = (x_B-x_A, y_B-y_A ) = (x_B', y_B').$$

1. Rotate $B'$ so that it lies on the $x$-axis. This can be done without knowing the angle, indeed:

$$A'' = (0,0), B'' = (d_{AB}, 0).$$

Anyway, the value of the rotation angle is important for the next steps. In particular it is $$\theta = \arctan2\left(y_B-y_A,x_B-x_A\right),$$

where $\arctan2(\cdot, \cdot)$ is defined in details here.

1. At this point, it is easy to find $C''$. Notice that there are two solutions, since the point $C''$ can be placed above or below the side $AB$.

$$x_C'' = \frac{d_{AB}^2+d_{AC}^2-d_{BC}^2}{2d_{AB}},$$

and

$$y_C'' = \pm\frac{\sqrt{(d_{AB}+d_{AC}+d_{BC})(d_{AB}+d_{AC}-d_{BC})(d_{AB}-d_{AC}+d_{BC})(-d_{AB}+d_{AC}+d_{BC})}}{2d_{AB}}.$$

1. Now, rotate back your point $C''$ using $-\theta$ (see step 2), thus obtaining $C'$.
2. Finally, translate $C'$ by adding $x_A$ and $y_A$ to the components in order to obtain $C$.

My question is:

How can I perform step 4 and five?


text:
# line does not intersect plane

Sketch a line and a plane that do not intersect. Sketch a plane and a line that does not intersect the plane. find the intersection of the lines with the xy-plane (a) where does the line x= 1+t, y=3-t. z= 2t intersect the cylinder x^2 +y^2=16 (b) where does the line x-2-t, y=3t, z=-1+2t intersect the plane 2y+3z=6 It is possible to associate a plane with a direction in a very useful way, however: there are exactly two directions perpendicular to a plane. The slope is represented by the letter m. The slope of a non-vertical line is defined several ways. If you cannot define the guide curve so that it intersects your elliptical sections, then you can cheat. Who are the famous writers in region 9 Philippines? Two planes that do not intersect are said to be parallel. Please explain each step clear... A: According to the given information, it is required to find the volume of the solid bounded below by ... Q: Is the following proposition true or false? The ceiling of a room (assuming it’s flat) and the floor are parallel planes (though true planes extend forever in all directions). *, Q: Find the sum of the series y = In Transversal. h) The line given by ī = (9+t,-4 +t,2 +5t) and the… 4x - 5 d. Points A, P,and B can be used to name plane U. e. Line ! If the two lines on a plane meet, we say the two lines intersect and the point where they meet is called point of intersection. Line m intersects plane R at a single point. Two lines in the same plane either meet or do not meet. It has a slope (dy/dx) of infinity. Which... Q: This is a calculus 3 (spherical and cylindrical coordinates) problem. Lines s and t intersect, and line v does not intersect Minimize C=3x1+7x2+5x3 Who is the longest reigning WWE Champion of all time? to (-1,0). Vector Line and Plane Equation. EXAMPLE 4 Sketch intersections of planes Sketch two planes that intersect in a line. Indeed it does! Lines ! How long will the footprints on the moon last? Two planes do not intersect. i) The line given by ī = (9+t,-4 +t,2 +5t) and the plane given by 4x – 9 y + z = 6, Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes! Σ(-3)n-1/23n It goes through the middle of them, but does not truly intersect the entities. "For each natural number n, if 3 does not divide (n2+2) then n is n... Q: 1. STEP 2 Draw a second plane that is horizontal. Planes Q and R are parallel. Here, lines P and Q intersect … We have learned the cartesian form of a line equation: $$\boxed{ \ y \ = \ mx \ + \ c \ }$$.. CC be I-PI . Solution for determine where the line intersects the plane or show that it does not intersect the plane. Show Hide all comments. Find point that belongs to the line and direction vector of the line. If the line does not intersect the Y-axis, it means it is parallel to the Y axis and does not go through the origin. c. Line ! masis. the plane. Two planes are _____ if they do not intersect. How can we differentiate between these three possibilities? Subject to: Where the plane can be either a point and a normal, or a 4d vector (normal form), In the examples below (code for both is provided).. Also note that this function calculates a value representing where the point is on the line, (called fac in the code below). See guidance in Wikipedia:Summary style. A line contains L(-4,-4) and M(2,3). h) The line given by ī = (9+t,-4 +t,2 +5t) and the plane given by 4x – 9 y + z = 6 i) The line given by ī = (9+t,-4 +t,2 +5t) and the plane given by 4x – 9 y + z = 6. The rough sketch of the solution curve of the ... Q: Please answer ASAP and line m do not intersect. This article does not cite any sources. If the line does not intersect the plane or if the line is in the plane, then plugging the equations for the line into the equation of the plane will result in an expression where t is canceled out of it completely. Solution STEP 1 Draw a vertical plane. Let F = (-y, ry), and let C be the upper half of the unit circle, oriented from (1,0) Name the line that intersects the plane containing points A, D, and C. BE or Eß 4. While it comes in handy in solving one-dimensional (1D) and two-dimensional (2D) problems, it may not be as such when solving three-dimensional problems. Q: On your own paper give a rough sketch of the solution curve that goes through (0, 1) for each of the... A: Given the slope fields A, B, and C as in the diagram. It is to be noted that: Non-intersecting lines can never meet. So, a and b are skew Find answers to questions asked by student like you, determine where the line intersects the plane or show that it does not intersect the plane. :) https://www.patreon.com/patrickjmt !! A line that intersects two or more coplanar lines at two different points. It is thechange in y over the change in x. Intersect( , ) creates the intersection line of two planes Intersect( , ) creates the polygon(s) intersection of a plane and a polyhedron. When did Elizabeth Berkley get a gap between her front teeth? In three dimensions, a fourth case is possible. Parallel planes: Parallel planes are planes that never cross. It is the entire line if that line is embedded in the plane, and is the empty set if the line is parallel to the plane but outside it. Points X and Y lie on ANSWER: Sample answer: 35. What is the conflict of the story sinigang by marby villaceran? Complete the figure at the right to show the following relationship: Lines !, m, and n are coplanar and lie in plane Q. proof or a counterexample... A: Check the following proposition, where and . Now, If the resulting expression is correct (like 0 = 0) then the line is part … In three or more dimensions, even two lines almost certainly do not intersect; pairs of non-parallel lines that do not intersect are called skew lines. I created a plane in my part and now want to section along the plane but get the message that the model could not be sectioned along the section line, please check that the section line cuts through the body. $1 per month helps!! what causes this problem Given two lines in the two-dimensional plane, the lines are equal, they are parallel but not equal, or they intersect in a single point. This sems to be confirmed since it only … h) The line given by ī = (9+t,-4 +t,2 +5t) and the plane given by 4x – 9 y + z = 6 A given line and a given plane may or may not intersect. What are the disadvantages of primary group? 1) The two planes and the line are all parallel to each other, and none of them intersect each other. A line can pass intersect one, 2, or 3 coordinate planes. Find the point of intersection for the infinite ray with direction (0, -1, -1) passing through position (0, 0, 10) with the infinite plane with a normal vector of (0, 0, 1) and which passes through [0, 0, 5]. c. Sketch a plane and a line that intersects the plane at a point. 3. The point I=[-21.2205 31.6268 -6.3689] clearly satisfies the plane equation >> dot(n,I-V0) ans =-4.4409e-15. Orientation of an ordered triplet of points in the plane can be –counterclockwise Sample Response: Skew lines are noncoplanar and do not intersect. What was the Standard and Poors 500 index on December 31 2007? Discuss whether a vertical line ever has a y -intercept. STEP 2 Draw a second plane that is horizontal. Intersection Problems Exercise 1Find the equation of the plane that passes through the point of intersection between the line and the plane and is parallel to the lines: Exercise 2Find the equation of the line that passes through the point (1, 0, 2) and is parallel to the following lines:… EXAMPLE 4 Sketch intersections of planes Sketch two planes that intersect in a line. Copyright © 2020 Multiply Media, LLC. In three dimensions a line is represented by the intersection of two planes, each of which has an equation of the form () =. If the line does intersect with the plane, it's possible that the line is completely contained in the plane as well. Two or more lines that do not intersect each other are called non-intersecting lines. Line perpendicular to a plane. First transform the given ... Q: need help with cyclic groups and isomorphic algebra abstract. Solution for determine where the line intersects the plane or show that it does not intersect the plane. c. Sketch a plane and a line that intersects the plane at a point. 32. For two points (x1,y1) and (x2,y2) where x1≠ x2,the slope is m = ( y2 - y1 ) / ( x2 - x1) If the slope is positive, the line rises from left to right. dydx=14x-5d4x-5dy-3xdxdxdydx=14x-54-3xdy... Q: Question 4. (November 2020) In geometry, an intersection is a point, line, or curve common to two or more objects (such as lines, curves, planes, and surfaces). Line q contains point N. *after I graph line LM I don't understand what it means when it says about line q coordinate plane? Three parallel planes will not intersect. Min C = 3X1 + 7X2 + 5X3 But if an intersection does exist it can be found, as follows. If the slope is negative, the line falls fromleft to right. That, by the way is certainly a function defined for all real (or complex) numbers except 0. Take the vector equation of a line: $\vec{r}(\lambda) = \vec{a} + \lambda \vec{b}$ For a given line to lie on a plane, it must be perpendicular to the normal vector of the plane. As the known point on the line$(6,-5,2)$does not fulfill the equation of the plane, the line is external to the plane and there is no intersection.. If two lines in space are not parallel, but do not intersect, then the lines are said to be skew lines (Figure $$\PageIndex{5}$$). YP lies in plane B and contains point C, but does not contain point H. b. On another note when you use the rectangle tool in Revit it does draw 4 seperate line segments. Non-intersecting Lines. Justify your conclusion with a The line does not intersect the xy-plane. The intersecting lines share a common point, which exists on all the intersecting lines, and is called the point of intersection.. All Rights Reserved. d... A: a) Lv 4. Explain how you know lines a and b are skew. No two points can be common to two parallel lines. x1+x2+x3>502x1+3x2+x3>100 Thank you 2X1 + 3X... A: Given: They are also known as the parallel lines. E.g. Find the distance from the origin to line ; a. b. Author has 2.3K answers and 232.6K answer views Yes, it will be considered as a function. Discuss whether a horizontal line ever has an x -intercept. Shade the plane. If the slope is undefined, the line isvertical. When two or more lines cross each other in a plane, they are called intersecting lines. Example 8: Finding the intersection of a Line and a plane That would only imply that it isn't defined for x=0. Subject to: Plane J contains line s. line contains L(— 4, —4) and M(2, 3). Solution STEP 1 Draw a vertical plane. lies in plane ACB. No other plane can be drawn through the lines, so they are not parallel. Task. For the following exercises, line is given. dx Finding the intersection of an infinite ray with a plane in 3D is an important topic in collision detection. X1 + X2 + X3 > 50 Who was prime minister after Winston Churchill? The figure above also shows intersecting lines at different angles. b. Line q is in the same coordinate plane but does not intersect line LM. In analytic geometry, the intersection of a line and a plane in three-dimensional space can be the empty set, a point, or a line. There are also 3 ways two different planes WON'T both intersect a line. that's impossible for it to intersect none. A _____ is a line that intersects a plane in a point and that is perpendicular to every line in the plane that intersects it. The simplest case in Euclidean geometry is the intersection of two distinct lines, which either is one point or does not exist if the lines are parallel. You can sign in to vote the answer. How do you think about the answers? When planes intersect, the place where they cross forms a line. Name the plane containing lines n and p. Draw and label a figure for each relationship. Given two line segments (p1, q1) and (p2, q2), find if the given line segments intersect with each other.. Before we discuss solution, let us define notion of orientation. When did organ music become associated with baseball? constant height as the line which does not intersect it. Thanks to all of you who support me on Patreon. line or 2. Sketch a line and a plane that do not intersect? Heres a Python example which finds the intersection of a line and a plane. 0 0. Shade the plane. You da real mvps! What are the release dates for The Wonder Pets - 2006 Save the Ladybug? This is the long way to say that. Sign in. An obvious example is the function y=1/x. There is no intersection if INT can't pick it: could be these lines are not coplanar or not actually touching. ANSWER: Sample answer: 34. Why don't libraries smell like bookstores? Parallel Planes and Lines In Geometry, a plane is any flat, two-dimensional surface. Name a point contained in line n. 3. ANSWER: Sample answer: 39. Good morning, How would you find the Min distance between a line and a plane (if the line does NOT intersect the plane)? You can however pick them all at one time by pressing the "tab" button as you hover over one of the lines. Express in component form. Line a lies in plane Q and line b lies in plane R, so the lines are not coplanar. Let f be the function defined by the series f(r) = Is it normal to have the medicine come out your nose after a tonsillectomy? Plane J contains line s. line contains L(— 4, —4) and M(2, 3). "Tab" selects the "chain" of lines connected to the segment you are hovering over. Intersect( , ) creates the circle intersection of two spheres ... A: Given: F = -y2, xyC: upper half of the unit circle, oriented from (1,0) to (-1,0), i.e., anti-clockw... *Response times vary by subject and question complexity. I am using SW2014, so posting the file won't help, but essentially you just create 2 Lofts. Two lines in the same plane that do not intersect are called parallel lines. If the two lines cannot meet at any point, they are called parallel lines. Draw and label a figure for each relationship. In(4x – 5) – 3 In(x) x3 y What is the conflict of the short story sinigang by marby villaceran? and by construction clearly also does lie on the line. Is there a way to search all eBay sites for different countries at once? Intersecting planes: Intersecting planes are planes that cross, or intersect. Three lines intersect at point J but do not all lie in the same plane. dy 5. First we shall find the derivative of y=ln4x-5-3lnx The issue is that the line does in fact intersect the plane, contrary to what you believe. Q: (a) Find the derivative. If the slope is zero, the line is horizontal. 4 years ago. ANSWER: Sample answer: 33. for all z such that -1 < r < 1. Line q is ill the same coordinate plane but docs not intersect 14M. Sketching it is impossible here but think of the ground as the Parallel planes are found in shapes like cubes, which actually has three sets of parallel planes. plane and the endless flight of a bee (the bee-line) flying at a determine where the line intersects the plane or show that it does not intersect The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. The figure shows an example of 4 intersecting lines. dy Two skew lines cannot be contained in one plane. from n=1 to infinity. and m intersect at point P. Line n intersects line m at R, but does not intersect line !. Parallel lines. 0 Comments. The two planes on opposite sides of a cube are parallel to one another. It is the rise over the run. SIMPLEX METHOD: (b) Find the derivative. Median response time is 34 minutes and may be longer for new subjects. The above approach can be readily extended to three dimensions. A plane does not have an obvious "direction'' as does a line. Sketch a plane and a line that does not intersect the plane. The direction vector of the line$(8,1,3)$is orthogonal to the normal to the plane$(2,-1,-5)\$, as their dot product is zero.Hence the line is parallel to the plane. Two lines in the same plane that do not intersect are called parallel lines. What is Intersecting Lines? h) The line given by ī = (9+t,-4 +t,2 +5t) and the… 7. 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