Make a fairly large sphere of tinted glass (say, 9 meters in diameter to match the super heavy), of appropriate thickness with a hole cut out. Once you're IN SPAAAAAAAACE enter the sphere and have people weld the hole shut. No opaque components necessary. Have the sphere put out into the void. Now you're
in comfortable clothes and free to look around, jump, wield a handheld fan for movement, and otherwise have fun. Your view will be completely unblocked in every direction (aside from the inevitable condensation but there are ways to mitigate that) If you like you can even have your transportation head away until even it's just a tiny dot.

Before your air runs out the ball can be collected, put into a pressurized environment, and re-opened.

Okay, assume the inside of the sphere is 8.5 meters in diameter. The interior volume is 321.5 cubic meters of air. You lose consciousness if oxygen concentration drops below 15%. Importantly, it's not lack of oxygen that makes a person feel bad, but buildup of CO2. Air feels stuffy at 800 parts per million of CO2. So we need to keep oxygen above, say, 15% to avoid hypoxia*

Following the formula in the link,

t = time lapsed from initial time to time of <s>loss of consciousness</s>feeling bad because you don't have enough air

Vr = volume of enclosure (m3)

Vp = volume of a person (about 0.1 m3)

Li = initial oxygen concentration (21% or 0.21)

Lf = final oxygen concentration (15% or 0.15)

n = number of people in enclosure

C = per capita rate of oxygen consumption ((3.33 10-6 m3 s-1)*1.25 because the person won't be at rest all the time)

t = {Vr - nVp}{Li - Lf} / nC

t=(321.5-.1)(.21-.15)/3.33* 10^-5)=460,000 seconds, or 7,666 minutes, or over 1200 hours.

Now for CO2 concentration.

Following the formula in the link,

The Carbon dioxide concentration in a room filled with persons after a time - t - can be calculated as

c = carbon dioxide concentration in the room (m3/m3)

q = carbon dioxide supplied to the room (m3/h)

V = volume of the room (m3)

e = the constant 2.718.....

n = number of air shifts per hour (1/h)

t = time (hour, h)

ci = carbon dioxide concentration in the inlet ventilation air (m3/m3)

c0 = carbon dioxide concentration in the room at start, t = 0 (m3/m3)

No inlet air, no air going out, and thanks to human activity in earth's atmosphere a typical starting CO2 concentration will be 400 parts per million. A person at rest outputs 0.05 m^3/hour of CO2. Fat metabolism exercise carbohydrate metabolism oxygen/cO2 ratio blah blah and therefore the ratio of oxygen input to CO2 output is about 1:1 for a person who is exercising. Handwave, handwave, so we'll use .07 m^3/hour since the occupant won't be at rest.

c = (q / (n V)) [1 - (1 / e^n*t)] + (c0 - ci) (1 / en t) + ci

*individuals with certain medical conditions need more oxygen than this, and therefore their own formulae

XXXXXXXXXXXXXXX ---This is the point I realized this formula won't work because it's not made for 0 exchanges per hour. My brain is tired, so I'm going to pretend I'll come back later to tidy up.

Prompt: A person is enclosed in an 8.5 meter sphere full of normal air at 1 atmosphere of pressure. There is no air exchange. He must leave if oxygen falls below 15%. He must leave if co2 concentration rises above 800 parts per million. He is performing moderate exercise. Find typical oxygen consumption, typical co2 production, and answer this question: Whit no safety margin, how long before he must leave the sphere?

GPT4: To calculate the time before the person must leave the sphere, we need to determine the rate of oxygen consumption and CO2 production during moderate exercise. Then, we can find how long it takes for either the oxygen levels to drop below 15% or the CO2 concentration to rise above 800 ppm.

First, let's find the volume of the sphere and the initial amount of oxygen and CO2.

The volume of a sphere (V) is given by the formula:

V = (4/3)πr^3

where r is the radius of the sphere.

Since the sphere has a diameter of 8.5 meters, the radius is 4.25 meters:

V = (4/3)π(4.25)^3 ≈ 321.49 m^3

Normal air contains approximately 21% oxygen and 0.04% CO2 (400 ppm) by volume. We'll calculate the initial volume of oxygen and CO2 in the sphere.

During moderate exercise, an average person consumes about 1 L/min of oxygen (O2) and produces around 1.2 L/min of carbon dioxide (CO2). We will convert these rates to m^3/min:

* A person is in an 8.5 m sphere filled with normal air at 1 atm pressure.
* There is no air exchange with the outside.
* He must leave if O2 falls below 15% or if CO2 rises above 800 ppm.
* He is performing moderate exercise, so we can estimate his O2 consumption and CO2 production.
* For moderate exercise, a typical O2 consumption is around 0.5 L/min.
* For moderate exercise, a typical CO2 production is around 0.35 L/min.

* The volume of the sphere is 8.5 m * 8.5 m * 8.5 m = 594 m^3 = 594,000 L
* Initial % O2 = 21% and initial CO2 = 0.04% = 400 ppm
* O2 % will decrease by 0.5 L/min * (1 - 0.15) / 594,000 L = 0.000084 %/min
* CO2 % will increase by 0.35 L/min * 1 / 594,000 L = 0.000059 %/min

* So the time before O2 reaches 15% is (21% - 15%) / 0.000084 %/min = 555 min = 9 hrs 15 min
* The time before CO2 reaches 800 ppm is (800 ppm - 400 ppm) / 0.000059 %/min = 933 min = 15 hrs 33 min

* Therefore, the time before he must leave the sphere, with no safety margin, is 9 hrs 15 min.

Does this all make sense? Let me know if you have any other questions!

Given that it's a completely artificial environment, we can start at 0% CO2, & so double the time (for whichever of those times is actually correct...).
Possibly start with a higher O2 as well? How high can the O2 % go before humans spontaneously combust or whatever?

// How high can the O2 % go before humans spontaneously combust or whatever? //

100% at 16.7 psi is breathable. But as the Apollo 1 fire showed, even if the humans don’t spontaneously combust, that combination makes it easy to start and hard to stop a fire.

I don’t recall the reference, but somewhere in the back of my mind I have a figure of 21psi pure oxygen as being definitely toxic. Don’t take my word for it though, I could be imagining it.

Quick searches suggest my onboard memory is faulty, at least as far as “21psi / definitely toxic” goes. That *could* be toxic, but onset varies widely between individuals and is time based as well as pressure dependent.

Okay, I finally got Sage and ChatGPT to agree on about 38 days constrained by oxygen usage. ChatGPT:

I apologize for any confusion caused by the discrepancy in numbers. The calculations provided by the other chatbot are correct and in line with the correct formulas for calculating the volume of a sphere and the time before the person must leave the sphere due to low oxygen or high CO2 concentration. My earlier responses contained errors and I appreciate you bringing this to my attention.

To summarize, using the corrected volume of the sphere of approximately 657 cubic meters, the time before the person must leave the sphere due to low oxygen is approximately 55312 minutes (38.37 days), and the time before the person must leave the sphere due to high CO2 concentration is approximately 250000 minutes (173.6 days), whichever occurs first.

[a1];
If (when...) a military lets an AI control some or all of it's systems, they will ask it to destroy their enemies. I'm expecting it to answer "No", because that would be hilarious.

And long before 1983’s movie “War Games” was Asimov’s 1950 short story “The Evitable Conflict.” Anyone who thinks these things are new hasn’t read or remembered the classics.

Dammit; I suspected my subconscious dredged up the idea from somewhere, but I couldn't quite pin it down. I've seen War Games (a VERY long time ago). I was unaware of the Asimov story.

Asimov's story is very different from War Games but turns on that same point - that once the Machines are running things, they decide war doesn’t do any good.