Usually photographs are not allowed on-site, but nobody has ever stopped me from sketching, so here’s the refined result of a series of pen sketches I took of a preserved 737 skull. Please forgive any inaccuracies, I am not the greatest aerobiologist.
I got a good view of the enormous radar melons (those big concavities in the nose). Like an electric eel the plane can generate a large voltage, but it does so across modified nerves that act as wires, focusing the resulting EM waves using bones shaped like a parabolic dish. This is partly used for signaling but supposedly they can detect bad weather with it.
Those little holes on either side are the angle of attack sensors, basically just whiskers that detect local airflow.
All that bone looks really heavy, but it’s mostly hollow, really weird for a mammal! I put the grey around it to show where the outllne of the plane is with all its skin and blubber.
Someday I’d love to sketch an Airbus skull, but they’re more common in Europe and I’m in the states. Maybe someday I’ll get a chance.
Oh hey, I just sketched out this a380 airbus last week, when they were taking it apart after it had been decommissioned (by natural cause).
i think more haunted houses should have haunted clawfoot bathtubs that move and exist as separate, distinct demonically possessed entities
Fun fact! Haunted bathtubs are an example of coevolution with haunted houses! As houses became more elaborate, haunted water fixtures changed to match the opulence. As such, the haunted hut and its symbiotic counterpart, the haunted well, gradually changed into the more inviting haunted estate with the haunted tub now being an internal parasite instead of an external one. You are correct, more haunted houses SHOULD have haunted clawfoot bathtubs because they promote the health of the house and indicate a functional paranormal ecosystem.
this website is so utterly detached from reality i love it
Let S(n) be the statement “every herd of n sheep is monochromatic” - let’s work out what the one color is later. Without some careful thought, each statement S(n) may be true or may equally well be false - but we’ll convince ourselves here that S(n) is true regardless of what number n represents.
It’s easy to prove that the statement S(1) is true: all the sheep in a herd of size 1 are all the same color.
Next, let’s show that S(k) implies S(k+1) - i.e. we’ll temporarily decide to believe S(k) for some given integer k, and then notice, as a consequence, that we now inescapably believe S(k+1).
To this end, consider an arbitrary herd H which contains k+1 sheep. Assume, without loss of generality, that H contains two special sheep, named Archibald and Beatrice (renaming some sheep if necessary).
Let A be the sub-herd of H consisting of all sheep in H except Archibald.
Let B be the sub-herd of H consisting of all sheep in H except Beatrice.
Both A and B are herds of size k, and S(k) would imply that herd A is monochromatic. Likewise, S(k) says herd B is monochromatic. Lastly, every sheep aside from Archibald and Beatrice is in BOTH herds, so the color of herd A has to be the SAME as the color of herd B.
We have thus shown that all sheep in H are the same color, assuming only that we believe S(k). But herd H was to represent an arbitrary herd of k+1 sheep! That is, we have used S(k) to convince ourselves that every herd of k+1 sheep is monochromatic. Thus we have shown that S(k) implies S(k+1).
The consequence is that we now believe S(n) for any n. For instance, to prove that a herd of 10 sheep is all the same color, we convince ourselves of the truth of S(10). But we’ve shown that S(10) follows from S(9), which follows from S(8), and so on… and since we believe S(1), we’re good.
There are surely finitely many sheep in the world. Suppose there are N of them. The statement S(N), which we now believe, asserts that they’re all the same color, so we just have to figure out what that color is. Luckily, that’s easy, because I happen to have a cute little plush sheep which is pink. The argument did not distinguish between adorable little plush sheep and living meat-sheep. Thus every sheep is the same color as my plush sheep: pink.
There you go!
Obviously there is a flaw in the above argument somewhere, and the exercise is to find it. We give this exercise to undergraduate math students when they are learning a proof technique called “induction”. The mistake I have (deliberately) made in the above argument is slightly subtle and students tend to make it all the time.
This is why it took four hours.
goddamn it i’ve been reading this out loud for the last however many minutes and i’m mad about it all over again
you think you’re mad now, wait till you hear about how every positive integer is fascinating.
Let me make a guess, the mistake is the assumption of S(A) and S(B) being equal, there is a distinction for a reason
professor @birdthatlookslikeastick, what is your response to this bold conjecture?
i don’t think that’s quite correct - A and B are indeed distinct herds, as you say, but the size of each is k. The statement S(k) does apply to both - as well as any other herd of size k.
But you’re very close, and you’re right to be suspicious of this part of the argument. Let’s inspect the reasoning about herd A and herd B more carefully. It goes like this:
1) Herd A is monochromatic,
2) Herd B is also monochromatic, possibly a different color than herd A,
3) BUT there’s some sheep common to both herd A and herd B, so in fact all of herd H would have to be monochromatic.
Steps 1) and 2) are legit, but there is a problem with step 3…
Let S(n) be the statement “every herd of n sheep is monochromatic” - let’s work out what the one color is later. Without some careful thought, each statement S(n) may be true or may equally well be false - but we’ll convince ourselves here that S(n) is true regardless of what number n represents.
It’s easy to prove that the statement S(1) is true: all the sheep in a herd of size 1 are all the same color.
Next, let’s show that S(k) implies S(k+1) - i.e. we’ll temporarily decide to believe S(k) for some given integer k, and then notice, as a consequence, that we now inescapably believe S(k+1).
To this end, consider an arbitrary herd H which contains k+1 sheep. Assume, without loss of generality, that H contains two special sheep, named Archibald and Beatrice (renaming some sheep if necessary).
Let A be the sub-herd of H consisting of all sheep in H except Archibald.
Let B be the sub-herd of H consisting of all sheep in H except Beatrice.
Both A and B are herds of size k, and S(k) would imply that herd A is monochromatic. Likewise, S(k) says herd B is monochromatic. Lastly, every sheep aside from Archibald and Beatrice is in BOTH herds, so the color of herd A has to be the SAME as the color of herd B.
We have thus shown that all sheep in H are the same color, assuming only that we believe S(k). But herd H was to represent an arbitrary herd of k+1 sheep! That is, we have used S(k) to convince ourselves that every herd of k+1 sheep is monochromatic. Thus we have shown that S(k) implies S(k+1).
The consequence is that we now believe S(n) for any n. For instance, to prove that a herd of 10 sheep is all the same color, we convince ourselves of the truth of S(10). But we’ve shown that S(10) follows from S(9), which follows from S(8), and so on… and since we believe S(1), we’re good.
There are surely finitely many sheep in the world. Suppose there are N of them. The statement S(N), which we now believe, asserts that they’re all the same color, so we just have to figure out what that color is. Luckily, that’s easy, because I happen to have a cute little plush sheep which is pink. The argument did not distinguish between adorable little plush sheep and living meat-sheep. Thus every sheep is the same color as my plush sheep: pink.
There you go!
Obviously there is a flaw in the above argument somewhere, and the exercise is to find it. We give this exercise to undergraduate math students when they are learning a proof technique called “induction”. The mistake I have (deliberately) made in the above argument is slightly subtle and students tend to make it all the time.
This is why it took four hours.
goddamn it i’ve been reading this out loud for the last however many minutes and i’m mad about it all over again
you think you’re mad now, wait till you hear about how every positive integer is fascinating.
Let S(n) be the statement “every herd of n sheep is monochromatic” - let’s work out what the one color is later. Without some careful thought, each statement S(n) may be true or may equally well be false - but we’ll convince ourselves here that S(n) is true regardless of what number n represents.
It’s easy to prove that the statement S(1) is true: all the sheep in a herd of size 1 are all the same color.
Next, let’s show that S(k) implies S(k+1) - i.e. we’ll temporarily decide to believe S(k) for some given integer k, and then notice, as a consequence, that we now inescapably believe S(k+1).
To this end, consider an arbitrary herd H which contains k+1 sheep. Assume, without loss of generality, that H contains two special sheep, named Archibald and Beatrice (renaming some sheep if necessary).
Let A be the sub-herd of H consisting of all sheep in H except Archibald.
Let B be the sub-herd of H consisting of all sheep in H except Beatrice.
Both A and B are herds of size k, and S(k) would imply that herd A is monochromatic. Likewise, S(k) says herd B is monochromatic. Lastly, every sheep aside from Archibald and Beatrice is in BOTH herds, so the color of herd A has to be the SAME as the color of herd B.
We have thus shown that all sheep in H are the same color, assuming only that we believe S(k). But herd H was to represent an arbitrary herd of k+1 sheep! That is, we have used S(k) to convince ourselves that every herd of k+1 sheep is monochromatic. Thus we have shown that S(k) implies S(k+1).
The consequence is that we now believe S(n) for any n. For instance, to prove that a herd of 10 sheep is all the same color, we convince ourselves of the truth of S(10). But we’ve shown that S(10) follows from S(9), which follows from S(8), and so on… and since we believe S(1), we’re good.
There are surely finitely many sheep in the world. Suppose there are N of them. The statement S(N), which we now believe, asserts that they’re all the same color, so we just have to figure out what that color is. Luckily, that’s easy, because I happen to have a cute little plush sheep which is pink. The argument did not distinguish between adorable little plush sheep and living meat-sheep. Thus every sheep is the same color as my plush sheep: pink.
There you go!
Obviously there is a flaw in the above argument somewhere, and the exercise is to find it. We give this exercise to undergraduate math students when they are learning a proof technique called “induction”. The mistake I have (deliberately) made in the above argument is slightly subtle and students tend to make it all the time.
Yo of all songs from that era why is it that Istanbul not Constantinople is one of the ones that survived to be known for decades after? Why is a fucking song written for the 500th anniversary of the fall of Constantinople not completely unknown today? Who even listened to it when it was recorded like why
the reason is that its a fucking banger, patrick. theres no conspiracy, its just good
