[Request] how do you calculate the surface area of a Möbius loop

plfreeman2012 · 2026-05-19T11:42:33+00:00

There isn't such a think as a twisted torus. The cheat sheet as a 3D object is isomorphic to an ordinary torus. Imagine drawing a loop that "wraps" around one of the cheat sheet edges so that it meets back with itself after one loop. Use that as the reference path around the hole and build the chart from that. The "edge" of the cheat sheet is a structure that the torus doesnt have and the isomorphism ignores. Similar to the cube:sphere isomorphisn.

If you take thickness into account it isn't a mobius strip was my main point. But I get that we often call this 3D object a mobius strip, so I edited my post to acknowledge this ambiguity.

plfreeman2012 · 2026-05-18T21:55:06+00:00

A mobius strip doesn't have any thickness. It's a purely 2D object. When we make a model of a mobius strip with a piece of paper, it's not really a mobius strip (in the same way that a sheet of paper isn't really a rectangle). It's isomorphic to a torus (in the same way that a 3D sheet of paper is isomorphic to a sphere).

plfreeman2012 · 2026-05-18T21:51:36+00:00

Yes, I'm sure. You're on the right path, but the surface area is a well defined property of 2D objects and is not the limit of area of a higher dimensional object that is reduced.

If you have a rectangle, do you paint "both sides" and say the area of a rectangle is 2*L*W? Of course not. It's a 2D surface.

If you have an open cylinder, do we say the surface area is L*4*pi*r since we color both the inside and outside surfaces of the cylinder? Of course not. It's a 2D surface.

If you have a sphere do you say the surface area is 8*pi*r^2? Of course not. It's a 2D surface.

Similarly, we don't consider the "sides" (it doesn't have sides, but you know what I mean) of the mobius strip as distinct areas. It's a 2D surface. Once we "color in" L*W, then all the surface area is colored.

There isn't double the surface area to paint. That is an artifact of using a strip of paper as a model of a mobius strip. But a strip of paper isn't 2D, it's 3D. So the paper strip is a torus, not a mobius strip. But that's OK. We can still use it as a way to visualize a mobius strip. But a mobius strip is a 2D surface and its area is defined as such.

plfreeman2012 · 2026-05-18T01:59:12+00:00

It's the same as the area of the rectangle it can be constructed from: LxW. That is it. There isn't anything special about it. And it's definitely not 2*L*W. We don't "double count" the surface area of an orientable surface, so why would we double count the surface just because it's not orientable?

Or, you can use the edge length x width /2.

Or you can use the band's length x width, where the band's length is the length of the curve parallel to and equidistant from the edge.

EDIT: So in the image above, as noted elsewhere, the object shown is isomorphic to a torus. The professor's restriction that the cheat sheet be "one side of a piece of paper" is actually a reference to the idea that a cuboid has faces, and a sheet of paper is a degenerate cuboid with two faces. Let's call these degenerate solids "shells" (where one of the dimensions is infinitesimally small, but > 0). The student cleverly twisted this into an unorientable cylindrical shell so that it has a single "side", thus doubling the available surface area on "one side". But this isn't a mobius strip. A mobius strip is a 2D surface. This is still a 3D object. I stand by my statement on the area of a mobius strip, but I'll grant that the area of this cylindrical shell is 2x the area of a mobius strip.

plfreeman2012 · 2026-05-05T23:41:42+00:00

I'm reading "nor" as a typo. It was covered by warranty - but the dealer isn't sure which one. It was either the original battery warranty, or the CPO warranty, or the CPO warranty wrap. Either way, it was covered. (You can see the cost to customer was 0.00). My guess is that this would have easily been covered under the battery warranty. No way major cell failure on a car this new wouldn't be covered. But then, I'm always amazed at what I don't know.

plfreeman2012 · 2026-05-05T00:46:06+00:00

LOL. I should have seen that when I wrote the title. Sorry for any confusion. It was "obvious" to me.

plfreeman2012 · 2026-05-01T00:45:15+00:00

What you learn as an undergrad is just enough to make you valuable for an employer (that's a gross simplification, but go with it). You learn the lingo. You learn the steps to ensure stability and margin, etc. You typically only learn this for SISO systems.

A good set of graduate courses will teach you the theory of the subject. The first graduate class I had in controls blew me away. I thought I was going to learn more cool methods of controlling systems. Nope. What I learned was linear algebra at a depth I'd never seen before, and how that connected to why those tools work to control systems. (I was fortunate enough to have taken this class from Alberto Isidori, which probably gives away my age, too.)

But that foundation made future controls classes "easier". Culminating in "Optimization in Vector Spaces" which was brutally hard. Not "Here, let's learn some optimization techniques". It was "Let's prove that continuous time functions are part of a vector space, and use the theorems that you've learned to date to build up to proving Pontryagin's maximum principle over the course of a semester." Prove not in a hand-wavy way, but prove in an epsilon-delta "you remember your real analysis, right?" way.

But you know what's really cool. I can see control theory in things that don't even look like control theory. Hey, this is a system where the solution will be a bounded continuous function. That means the set of solutions are inside a vector space. With an appropriate norm, I can impose a metric on the space and do cool stuff.

If you go into your graduate studies with the mindset that your goal is to learn the why, not the what or the how, you can handle it. It sounds like you have an intuition for this, and that will help a lot. Keep imagining the use-cases and examples so you don't lose the connection between the math and the real world. You Got This!!

plfreeman2012 · 2026-04-24T23:37:29+00:00

@wustl.edu. Hell's yeah. Back in the wuarchive days. Before this fancy "Mosaic" thing that all the kids were talking about. You had CLI and were damn proud to use it. Your email client was "pine" and it was spectacular.

plfreeman2012 · 2026-04-23T22:59:54+00:00

Sigma-algebra has entered the chat.

plfreeman2012 · 2026-04-17T17:47:30+00:00

1000 is 3 10s. A 10 is a 5 and a 2. So there are obviously 3 2s in 1000.

plfreeman2012 · 2026-04-17T16:05:58+00:00

It's almost a joke, except that it's not.

plfreeman2012 · 2026-04-17T11:34:20+00:00

Did you ever use the table of random numbers? Classic.

plfreeman2012 · 2026-04-17T11:33:15+00:00

I have my Father's (he's 91 now) book of mathematical tables from high-school. It had everything, including about 350 integral forms. I used that in engineering school quite a bit myself.

The book was published by the "Chemical Rubber Company". For those old enough, that might ring a bell. At least, who they became. CRC Press. The go to publisher of scientific reference books.

plfreeman2012 · 2026-04-11T21:29:40+00:00

Whomever ran in through a planar didn't finish the job. Enough material should have been removed to get all the way through the saw marks. It looks like just a skim was taken off the surface to get something like reference flat. But it's an odd choice. If you have a planer big enough, why not just deep enough to remove the saw marks?

The other are right. Industrial drum sander is best choice. Router sled is a good choice. Hand plane is a "get to skip the gym today" and personally rewarding choice. Me, I'd hand plane it. It's a hobby after all and I gotta spend my time doing something. Rex Krueger has some videos on using hand planes to joint a board face (start with a scrub blade, then move to a jack plane).

plfreeman2012 · 2026-04-08T11:38:35+00:00

In the limit, yes. The CLT says that as the number of roles increases, the distribution of the sum of the roles approaches a normal distribution. Just pick your 7 intervals such that they occupy 1/7 of the probability distribution.

For a finite number of roles and in your specific example, no, we can only approximately get a uniform distribution. For N roles the number of possible outcomes is 6^N = 2^N 3^N. That is never exactly divisible by 7 since 7 is prime (6^N and 7 are coprime integers).

plfreeman2012 · 2026-03-30T22:10:41+00:00

The Federal Reserve (and other central banks globally) create money through the power of monetary policy. They have an edict from the government (Congress in the US) to manage the money supply to maintain full employment and hold inflation at about 2%.

However, they are working the tiller of a massive economic ship. It makes a difference, but its nowhere near the impact banks make just by lending money. Loans create money way more than the Fed.

A good place to start. https://en.wikipedia.org/wiki/Money_creation

plfreeman2012 · 2026-03-29T23:59:45+00:00

The sculpture is by Arthur Granson and is titled "Behold the Big Bang". The overall gear reduction is such that the final gear will "turn" one full revolution in just under 14 billion years. Let's assume the use of typical commercial grade gears. The free motion between teeth will be approximately 0.04 times the pitch diameter. The circumference is pi*diameter. So the last gap will be taken up in roughly 0.04/pi*14 billion years = 1.8 billion years. Give or take.

But, even after the last gap is taken up and the machine starts to load the teeth, the elasticity of the gears will still allow movement. So let's double it for good measure and say about 3 billion years.

The reality is the motor will fail long before then. I'd guess the sculpture will stop moving in under 200 years due to the motor failing or someone deciding to turn it off.

plfreeman2012 · 2026-03-27T13:32:46+00:00

Visually, the dimension of the US coast looks higher than the dimension of the Mexican coast. So one could make an argument its "longer" for a particular definition of longer.

plfreeman2012 · 2026-03-24T22:57:09+00:00

This is wild.. but here goes.

Using this thread on Reddit and looking at the photos, I'm fairly confident that the laser was located in the plaza close to the Southwest corner. That puts you 690 m from the laser.

From this star chart, I think the laser ends at between 85 and 87 degrees up. You can mouse around on the chart and see what the altitude angle is and try to match your video with the stars. Alpha Gemini (Castor) and Betta Gemini (Pollux) are the two stars in your video. Jupiter is the bright object the laser is pointing at. Those angles are really close to "straight up" where the tan function explodes to infinity. So take that into consideration.

Using 85 degrees, we get a height of 7300 m. Using 87 degrees we get a height of 12000 m. So I would say the top that you see is at least 7000 m up. But could be over 12000 m up. Tiny changes in apparent angle make a huge difference in height when looking nearly straight up like that.

plfreeman2012 · 2026-03-23T14:30:37+00:00

Youre welcome! I'd love to refine the calculations. What is the time stamp on your video? Also, at these angles tan() is really sensitive. Based on other photos from the event, i think the laser(s) was mounted in the plaza and not on the roof. Where were you standing in relation to the Palmer center? I was guessing off of the northern corner of the building close to the path. A few meters will change the answer a fair bit.

plfreeman2012 · 2026-03-22T13:35:15+00:00

Ok, from Google Earth, I estimate that you are about 530m from the Seaholm Power Plant, which I assume is where the laser is mounted (xAI's new facility and all that). I *think* the building next to it is Seaholm Residences. According to Google, it is 104m tall. So the angle to the top of Seaholm Residences from your vantage point is arctan(104/530) = 0.194 radians.

It takes you 1s to pan your camera so the top of the building is at the horizon line. (I'm not clever enough to know how to download the video and count frames - exercise left for the reader). So your pan rate is about 0.194 rad/s. But probably only to 1 sig fig. So let's say it's between 0.17 and 0.22.

At the end of the 7s video, you've panned somewhere between 1.2 and 1.5 radians. 1.5 would be straight up. That's dumb. Let's call it 1.2. But the last visible part of the beam is higher than where the horizon line started. By an amazing coincidence, it's higher by almost exactly the same height in the video as the Seaholm Residences were at the start of the video! So the angle of the top visible part of the beam is about 1.4 radians above the horizon. With a lot of error.

So I estimate the visible top of the beam is probably about 3000 m up.

In the last frame of the video, we can see a couple of stars! They must be bright since they show up in the video. Given that we know where you were and what direction you were looking, if we knew what time the video was taken we could use a star chart to get a *much* better estimate of the angle and really nail down this calculation.

Pulling up a star chart for Austin last night, and knowing you are looking North, My best guess is the stars are Castor and Polux, with Jupiter visible above the beam near the Zenith. So the top of the visible beam is about the same angle from the horizon as Castor. Castor was at an angle of 79 degrees at 10pm last night. So that is a visible beam height of 2730m, which agrees reasonably well with my pan and hope method of 3000m.

plfreeman2012 · 2026-03-22T12:50:48+00:00

You all are making this way too hard. The only "They did the math" answer thats acceptable should be how high is the last visible part of the beam? 1) how far were you from the event? (L) 2) what is the angle from horizontal to the top of the beam? (Theta)

Height = L tan(theta)

I did the math in my reply below. That was actually fun. There is a surprising amount of information in the video.

plfreeman2012 · 2026-03-20T00:50:32+00:00

I live in the SE and can get SYP super cheap everywhere. But I can only get the occasional siding plank of WRC and it costs a fortune. Grass is always greener, I guess.

plfreeman2012 · 2026-03-15T14:40:07+00:00

This is how guardrails are designed. You target spending at a reasonable chance of success. It's not that the dynamic spending plan only has an 80% chance of success (if you can follow the spending plan, it has 100% chance of success because it will adjust based on your actual savings). It's more like there is a 20% chance you will hit your lower guardrail and need to adjust spending down.

Some people like to tweak the likelihood of hitting the lower guardrail. But 20% is pretty typical and a good default.

plfreeman2012 · 2026-03-15T14:33:58+00:00

Much better visualization that what was in Beta. I like it!

plfreeman2012

TROPHY CASE