@XanderHenderson true, in some its actually surprisingly easy like the F-14 Tomcat. In others, like an F-16 its very hard.

@Ted my location is more accurately given in my profile than it was in your map ;-)

@robjohn It will also depend on weather. Cold weather isn't as friendly as hot weather.

But in an F-22 raptor you can put it in a controlled spin

@ペガサスSeiya Yeah, I don't fly those.

GA, not military.

@TedShifrin we only go in the warmer times of the year.

I wasn't bothering to be that precise, @robjohn. I removed the link because when I clicked on it, the EV stuff was no longer there.

The favorite thing that I've flown is a Pitts S2B.

Lovely, lovely plane.

@TedShifrin Ah, it was easily found at the top of the page. It was not there in the mobile version of Google Maps.

@XanderHenderson yes, it was something stupid in the air. I shut off the engines, pulled in a steep climb, stalled the jet and put in full rudder. Got the F-14 to spin

I had fun coming back from Idyllwild to SD in a friend's Tesla. When I came down the mountain I had noticeably more charge than when I'd started, and the whole return trip to SD took something like 20% battery.

@ペガサスSeiya You are talking about a simulator.

I am not.

@XanderHenderson Try it with your aircraft and please film it. Title it "Mathematician tries to recover a flatspin (Gone Wrong!)"

@ペガサスSeiya No, because (1) I haven't flown in a decade, and (2) I am not an idiot.

@XanderHenderson (2) You aren't very curious*

@ペガサスSeiya About what it would be like to die in a flat spin? No.

I have no curiosity about that at all.

@XanderHenderson Hey, you can recover, that's the curiosity part. And, you can eject too

Dying isn't the only option

@ペガサスSeiya Depends on altitude.

Yeah well if you do it at 3000 feet good luck

My ex-wife got into a flat spin once in a Super Decathalon. Recovery happened when her flight instructor jumped over the seat and sat in her lap to change the CoG.

They lost a lot of altitude in that time.

This is not something I would want to experience first hand.

@XanderHenderson That instructor seems, rather suspicious

I have flown as a passenger in a friend's little Cessna, but I declined the opportunity of taking the rudder. :)

@ペガサスSeiya No, he's great.

@TedShifrin come on Ted, you should've done it

@XanderHenderson Planes with negative stability seem to be better at recovering from spins

And harder to spin too

@TedShifrin Really hard to mess up in a Cessna. They are designed to be super-duper stable.

E.g. it is crazy hard to get a Cessna-172 to stall and to stay in that stalled state. It really wants to recover.

Put me in a glider with a bunch of rocks to throw and I can probably shoot down an entire USAF squadron

@ペガサスSeiya It must be nice to be young and believe in your own infallibility and invulnerability...

@XanderHenderson it is. I want to enjoy it before I turn 40 and become just another grumpy old man

Outlook is confused...

Outlook moment

@Ted do you know how to explicitly describe a map $Gr(k,n)\rightarrow Gr(k(n-k),\infty)$ (Grassmannians, to be clear) classifying $TGr(k,n)$?

Nope. I never think about classifying spaces :)

You like unclassy spaces?

You know I’m not the least classy.

@TedShifrin Civic owner confirmed

@robjohn this kinda reminds me. Not too long ago I was ignorant enough to think a racist meant someone who likes to race

Hi

@TedShifrin I have a direct product of M with R^+, and M has a metric g. I was looking at a metric $r^2g + dr\otimes dr$ and another $g + r^{-2}\,dr\otimes dr$. Is there any easy way to show these are isometric or not? Ricci curvatures are the same, so I am at a loss.

So, a fried chicken is called fried chicken, because its done frying right? Then, why's a building called building when its done building? It should be called built

@ペガサスSeiya I think you made some typos in your sentences

@XanderHenderson I had no idea you live in HOL brook

high order logic people must live there

@geocalc33 hey :)

A student in my precalc class had a couple of good questions today! Yay!

His questions ultimately came down to "Is the intersection of two intervals always an interval?"

Which, it turns out, is slightly subtle, and depends on the precise definition one gives of an "interval".

And I am so happy that this student caught that.

Was the student you?

LOL

I was tutoring myself the other day, and a student of mine, yes it was me, had the greatest idea lol

No one gets the joke :|

dang...

@DLeftAdjointtoU No.

@anak I vote NOT. Note the second is isometric to $M\times \Bbb R^+$, no warped product at all.

@XanderHenderson cool. that issue comes up here more often than i would have expected.

@DLeftAdjointtoU I did? Where?

If someone makes fun of Xander, would it be called Xander Slander?

Hello, I've been stuck for a couple of day with a step in the following answer here and I would really appreciate if someone gave me a hint. In math.stackexchange.com/questions/2203755/… I struggle to understand, where he tensor the exact sequence with \otimes A_g.

It is clear to me that the resulting A-modules are those shown but i don't get how he claim that the morphism A_g \to \mathcal{O}_Z(Z)_h is indeed f^#(d(g)) but a priori is just f^# \otimes id_{A_g}.

I kinda hate asking it flat out here on the chat but at this point but this is my last resort

to recap the relevant content of the answer: starting from a closed immersion (f,f^#) : Z \to Spec A = X; one has an exact sequence 0 -> J -> A -> O_Z(Z) (second map is the inclusion of \ker f^#(X) into A, third map is f^#). Why is it the case that the tensor product with A_g yield a sequence whose third map A_g -> O_Z(Z)_h (where h = f^#(g)) is exactly f^#(d(g)) ?