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peto , in Would a mechanical counter pressure suit make you poop?

Probably not? I think the suit would have to hold all of you together to work otherwise you would risk parts of you following the waste.

I’m also pretty sure the vacuum doesn’t even top the list of things you need a space suit to protect you from. Eva suits do a lot of work.

CanadaPlus ,

In terms of engineering, it does. Micro meteorite protection and heat management can both be provided by normal garments. UV protection is obviously easy enough too. Breathing gas is a bit less convenient, but still not as tricky as making a suit that’s both rigid enough to reliably hold several PSI in and flexible enough to comfortably work in. That’s why the elastic suits are being researched like they are.

CanadaPlus , (edited ) in Would a mechanical counter pressure suit make you poop?

… That’s actually a good point. I’m guessing since the digestive tract is flexible and isn’t held open to the outside all the time, it wouldn’t cause problems with things deep inside. I also think it’s inevitable that if you did shit yourself in it, suction would kick in at some point and make it all a bit more dramatic. And then it would boil-freeze off into space, and be icy cold. That might still be better than pooping a sealed space suit, though.

smuuthbrane , in Would a mechanical counter pressure suit make you poop?
@smuuthbrane@sh.itjust.works avatar

Your body IS being constantly pressurized by the atmosphere, and your various sphincters are used to that. Presuming the suit doesn’t pressurize your body beyond what it’s used to (at which point breathing would be difficult), there should be no unexpected anal excretions due to the suit.

Omacitin ,

But the pressure from the atmosphere applies to both sides of the sphincter, resulting in zero net pressure. Unless the suit actually does press against the outside of the sphincter like it does the rest of the body, I think OP’s concern about the suit squeezing you like a tube of toothpaste is valid.

Maybe the suit only applies a few PSI instead of the full 14.7, which it seems like one’s sphincters would be able to withstand.

PlasticPigeon ,

Toothpaste poop tube space man.

Definitely would watch.

Giving it 4 out of 7 stars.

CanadaPlus ,

I think it’s like a third of an atmosphere or something. Enough to comfortably achieve the same partial pressure of oxygen as normal Earth air, by providing it pure.

Hexagon , in What are the problems with Copenhagen interpretation?

Not an expert, but I’d say the wave function collapse. How does it happen, and why?

Jeredin , (edited )

As I’m also a non-professional, I’d like to use your your comment to add my experience with studying quantum mechanics:

From all my studies of both math and lab experiments, intuitively and likely in reality, matter at the quantum level is made of vibrations, oscillations and standing waves of “SpaceTime.” The amplitude, frequency, position, magnetic moment (spin/charge), temperature, pressure and other properties are what we measure and thus describe particles and emergent phenomena like phonons and other quasi-particles.

So this all seems simple enough, we have mathematical descriptions and tools to measure with, what’s this whole issue with “observation” and how how far do we need to take it?

My simple answer is: whenever you see “observer”, translate it to interaction. This can be anything, so long as it interacts with the quantum system being “observed.” But what does this really mean, why does it matter so much? Go back to our wave properties and understand that anything quantum that interacts with anything else quantum is actually introducing their own wave properties and thus, allowing quantum interactions. That is, it’s likely impossible to use something with quantum wave properties (which everything has) to precisely measure something else with quantum wave properties and not have some level of wave disruptions - in other words, we cannot have precise measurements because the closer your quantum measuring tool tries to measure another object’s quantum property, the more the interactions influence the results. The Copenhagen perspective, as I’ve come to orient my understanding, is a question of: does the math reflecting these wave interactions/measurements of them, only mathematically describe it, or do we take the math literally and call it reality?

There are those in both camps and especially as a non-professional, I join the camp that says it only mathematically describes reality. Keep in mind, relativity of all objects makes it so even the very conditions of the experiment can skew results; the quantum level is extremely sensitive to its wave environment and even in a vacuum, the zero-point energy field exists. Also, keep in mind that just because you can’t precisely measure a given property doesn’t mean that you can’t have very good measurements of most/all properties; it’s only a matter of how badly you need to precisely know any given property.

There’s obviously more nuance, but I think the key thing that I want to impart is not to take quantum mathematics to literally, but it’s the best description and predicting tool that we have for that level of physics.

regdog , in Help me understand this effect

What effect are you talking about? Is there a frog clinging to the underside of that stone, and you try to coax him out with a flash light?

Brokkr , in Help me understand this effect

I’m guessing it’s an aluminum oxide abrasive? The abrasive is flourescing due to the little bit of uv coming out of the LEDs.

You might find this interesting, if you are grinding iron or steel then the grinding surface may not flouresce due to the iron bonding with the aluminum oxide.

9point6 ,

I saw “iron”, “aluminium” and “oxide” and I briefly assumed you were trolling until I looked again to check which one was the oxide.

troyunrau ,
@troyunrau@lemmy.ca avatar

This seems like a perfectly reasonable answer. OP! You could probably test this by changing the type of light you’re using. Try a red laser pointer as a control, and a black light wand (the sort they use to detect counterfiet bills), and see what happens.

Wogi OP ,

Sadly I have neither of these things available on hand to test that theory but I can at least confirm that the abrasive wheel is a ceramic alumina.

Actually NVM, I found a laser pointer and it has no effect, though it is admittedly quite dim.

troyunrau ,
@troyunrau@lemmy.ca avatar

Complete tangent, but alumina, aka aluminum oxide, is usually considered the second hardest naturally occurring material. When it is found in nature, it is given the mineral name corundum and is clear. But if there are some impurities in it, you can get colours. Red corundum is called Ruby, and blue is called Sapphire. In the beauty industry, the same material (mixed with magnetite) is called emery, and lends its name to emery board, and is used in nail files. In the tech industry, it’s used to make the extremely scratch resistant coating on most modern phone screens (basically nothing but diamond will scratch it).

You have subscribed to alumina facts. I’m sorry, the cat facts guy was busy.

Wogi OP ,

We also use emery paper to smooth out rough surface finishes on machined parts. None of my tools but some of the tools the other guys have in the shop use little Ruby beads as reference surfaces. Our wire EDM also uses Ruby for some critical parts.

You’ve been subscribed to machine facts, strap in it’s a bumpy ride

troyunrau ,
@troyunrau@lemmy.ca avatar

You. I like you.

Wogi OP ,

Fun fact, machinists do not have friends. Only mortal and natural enemies.

troyunrau ,
@troyunrau@lemmy.ca avatar

So… night shift?

Wogi OP ,

Well. SOMEONE’S gotta pick up day shifts mess!

troyunrau ,
@troyunrau@lemmy.ca avatar

Is there a machining community active somewhere on lemmy (yet)? I only dabble, but I like to sneak peaks at real folks fucking up, err, showing off their projects.

Wogi OP ,

sh.itjust.works/c/machinist

Not super active but we got shit to do

Ok that’s a lie but like we gotta LOOK like we got shit to do

CommunityLinkFixer Bot ,

Hi there! Looks like you linked to a Lemmy community using a URL instead of its name, which doesn’t work well for people on different instances. Try fixing it like this: !machinist

thantik ,

Works fine for me on a different instance. Maybe other instances should get their shit together instead?

Mango ,

Yo this is fascinating!

april , in What are the problems with Copenhagen interpretation?

It is not a well-defined theory. It does not say what measurement or an observer is.

Kepabar , in What are the problems with Copenhagen interpretation?

Primarily defining what extractly is an observer or the point of wave form collapse I suppose.

BalabakGuy , (edited ) in What is the average temperature of earth?
@BalabakGuy@lemmy.ml avatar

It’s not a stupidly easy question. It is extremely hard. Calculating the average temperature of all molecules on Earth is extremely hard due to the vast range of temperatures across different environments, from the Earth’s core to the atmosphere. Calculation isn’t the problem but collecting data is. You need to collect all the data that distributed widely on earth. Factors like altitude, latitude, and specific conditions in different regions making collecting an accurate data even harder.

AmalgamatedIllusions , (edited ) in Attempting to freeze matter under extreme pressure?

Most experimental research in matter under extreme pressures is concerned with recreating conditions within the interiors of planets and stars (the latter falls under the field of high energy density physics). The temperatures involved therefore tend to be very high. However, there’s no inherent conflict between high pressures and low temperatures, it’s just that temperature tends to increase when you compress something. Compress an ideal gas, for example, and it will heat up. Let it sit in its compressed state for a while though, and it will cool back down despite remaining under high pressure.

This is true for solids and liquids too (putting any phase transitions aside), though they are much less compressible. The core of the Earth will eventually cool too, though it’s currently kept at high temperature by the radioactive decay of heavy elements. Diamond anvil cells, however, can reach pressures exceeding those at the center of the earth in a laboratory setting, and some DACs can even be cooled to cryogenic temperatures. This figure on Wikipedia suggests cryo-DACs can be used to reach pressures up to 350 GPa at cryogenic temperatures. As an example, a quick search turns up a paper (arxiv version) that makes use of a DAC to study media at liquid nitrogen temperatures and pressures up to 10 GPa (~3% the pressure at the center of the Earth). Search around and I’m sure you can find others.

Aremel , in Attempting to freeze matter under extreme pressure?

You can get certain types of ice when you apply a certain temperature and pressure to water. en.m.wikipedia.org/wiki/Ice_IX

CanadaPlus , in ADP keeps running ads about solar flairs that force earth into a 25 hour day. Could a solar flair even do that, and what distasters would happen if Earth's orbit changed?

… What? No that’s crazy.

Cunningham’s law has already taken care of orbit vs. rotation, I see.

ShittyBeatlesFCPres , in What is the average temperature of earth?

One thing that might help is knowing that the crust is very thin, 1% of the Earth’s mass. The core is 15% and the Mantle is 84%. But we don’t have to be too exact — the earth’s mantle isn’t one temperature

The Mantle ranges from 1000°C near the crust and 3700°C near the mantle. The core is starts there and gets up to 6000°C.

So, just a simple formula weighted by mass might be:

.84*((3700°C+1000°C)/2) + .15*((3700°C+6000°C)/2)

Which equals 2701.5°C. That answer might be fine for a Lemmy but an academic paper would have major error bars.

I also asked ChatGPT and Bard, out of curiosity. ChatGPT was more conservative and told me it’s hard to get anything besides a rough estimate. Bard warned about all that too but then said probably between 3000°C and 4000°C.

So, I’d say about 3000°C. My quick and dirty little formula could easily be +/- 1000°C and fall in Bard’s range.

givesomefucks , in ADP keeps running ads about solar flairs that force earth into a 25 hour day. Could a solar flair even do that, and what distasters would happen if Earth's orbit changed?

Like, it would be daylight for 25 hours?

No, day/night is because the earth rotates. Solar flares have nothing to do with that.

Statistically tho we’re way over due for a massive solar flare. And it would pretty much blow every electrical transformer on the day side of the planet when it hits.

To replace all of them would be a massive undertaking and could potentially take years to restore electrical grids everywhere.

So they’re a concern, just no idea what you’re talking about with a 25 hour day.

Limonene , in Why are the graphs for the distribution of light from the Sun as a function of (a) frequency and (b) wavelength not exactly reversed?

There are 2 reasons:

  1. Those two graphs have different scales on the y-axis. One is Irradiance per nanometer of wavelength, and one is Irradiance per terahertz of frequency. Both graph’s y-axis are called “spectral irradiance”, despite being different things. This causes most of the distortion between the two graphs, and can even change the location of the absolute maximum.
  2. The graphs’ x-axis have different units. This causes some distortion too, but wouldn’t change the absolute maximum. It would help if they used a log scale in both cases, because wavelength and frequency are inversely related, so then the graphs could just be horizontally flipped.

So, look at the top graph (by wavelength), and see how much power is in that 1000-2000nm area. It’s still a lot, just spread out over a large area. It’s the same amount of power in the lower graph (by frequency) shoved into the much smaller area from 150THz to 300THz. Since it’s in a smaller area on the lower graph, it has more power-per-unit-of-x-axis.

BackOnMyBS OP ,
@BackOnMyBS@lemmy.world avatar

Thank you. I understand most of your comment, and it makes sense. However, I still don’t understand how the change of units in the y-axis would cause a different maximum. It seems to me that the y-axis for both use the same formula with their respective x-axes: W/m^2/x.

I’m not in STEM by the way.

Limonene ,

It’s because the wavelength and frequency are inversely related. When the wavelength is low and the frequency is high, the wavelength is also moving very slowly, compared to the frequency which is moving very quickly. Since the frequency is changing so quickly, the power-per-unit-frequency is lower at higher frequencies, and higher at lower frequencies (at least relative to the power-per-unit-wavelength).

Let me try and use a car analogy:

You’re driving home through Wisconsin, and you live on the border between Wisconsin and Minnesota. The mile markers on the road decrease as you go, reaching 0 at the state border, where you happen to live.

The cows along the highway are evenly distributed, so if you count the cows as you drive, but restart your count every mile when you see the mile marker, you will reach the same number of cows every mile.

Now, the frequency is inversely related to the mile number. The frequency in this case refers to your children in the back seat asking, “Are we there yet?” They know damn well how far it is to home, because they can just look at the mile markers. Regardless, their rate of asking increases as the mile markers go down. When you’re at mile marker 100, they ask once every 10 minutes. When you’re at mile marker 1, they ask 10 times per minute.

If you instead look at the number of cows between “Are we there yet?” asks, then you will find that the cows-per-ask is much different from the cows-per-mile. At high distances (low frequencies), the cows-per-ask is very high, while at low distances (high frequencies), the cows-per-ask is very low.

Now, the article is looking at power-per-unit-frequency, so you’d actually have to measure the rate in change of how often the kids ask “Are we there yet?” And that would give you a little different result. You might need calculus to correctly calculate the derivative of the number of asks. But hopefully this illustrates that you can get different results, by using a different per-thing to measure your value.

Kethal ,

This covers it all well, but I think a simple explanation is that although “W/m^2/x” looks the same on the axes, it’s not the same. f=1/w, so one axis is W/m^2/f and one is W/m^2*f. The article makes a big deal out of the differences as if the x axis were the only difference, but they’re just very different things being plotted.

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