Getting Away from it all

I was thinking, the other day, about a getaway cabin, a bit of solitude far from the madding crowd. OK, sitting here on my 120 acres in the middle of Alaska that madding crowd isn’t too close but none the less….

So! Where? I thought how ‘bout up there? A space habitat at L4 or L5 is a bit beyond mine and most folks reach but how about the say 18 miles up in stratosphere?

Almost outer space but no need for a rocket to get there, tether to a hydrogen filled balloon and just float up. 18 miles, your above the weather, all the thunder lightning, clouds, politicians, snow and rain is below, nothing but smooth sailing. Each and every day is a sunny day. Air’s there quite enough to float a simplified (Got gravity, no need to centrifuge.), modified (Cool but not black space cold, insulate but not all that much more than my Alaska house.) O’Neill Habitat (Gerard K. O'Neill, The High Frontier: Human Colonies in Space, 1976.). Yep air’s there but a bit thin, so will need to pump more within.

OK, a few minor details to work out like, food, water, power, tub or shower, etc., -surprisingly it just might be doable.

So, so, again so! I started doing proof of concept calculations; an arbitrary payload, say four tons, what would it take to get it 18 miles up and floating there?

I started my calcs, after half an hour, much if it spent walking back and forth to my book shelves looking up formulas and constants I haven’t considered for well more than twenty years, I wondered how fast an AI might deal with it.

So, so I decided to ask that so and so, ChatGPT, the available to any and all AI, ‘Hey bro, whatsup with puttin’ it up’?

The following is our conversation. Note it took me three queries (As with any computer system if you put garbage in you get garbage out.) to get my answer. The conversation took maybe as much as five minutes, most of that me thinking or typing. I’d say ChatGBT went through all his calcs and replies in well less than half a minute. It would have taken me hours checking numerous reference books for constants and formulas, doing the calculations and checking them thrice, ChatGBT less than a minute. Needless to say, though he likes metric and minor trouble with English units, close enough for government work and I’m quite impressed!

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Putting an O’Neil modified O’Neill Habitat in the stratosphere

using chatbot;

ChatGPT

Ask ChatGPT any question. For more accurate and detailed answers: provide context, crucial details, and tell the bot exactly what you need.

You

How many cubic feet of hydrogen is needed to lift 4 tons to an elevation of 18 miles?

ChatGPT

To calculate how many cubic feet of hydrogen are needed to lift 4 tons to an elevation of 18 miles, we first need to consider how buoyancy works and the lifting capacity of hydrogen.

Step 1: Calculate the Lift Provided by Hydrogen

Hydrogen has a lifting capacity (buoyancy) because it is less dense than air. The density of hydrogen at standard conditions is about 0.08988 grams per liter, while the average density of air is approximately 1.225 kg/m³ (or about 0.001225 g/L).

Buoyant Force

The buoyant force can be determined by Archimedes' principle, which states:

[ F_b = V \cdot (ρ_{air} - ρ_{hydrogen}) \cdot g ]

Where:

• (F_b) = Buoyant force in newtons

• (V) = Volume of hydrogen in cubic meters (to be converted to cubic feet later)

• (ρ_{air}) and (ρ_{hydrogen}) = Densities of air and hydrogen respectively in kg/m³

• (g) = Acceleration due to gravity (approximately (9.81 , m/s²))

Given:

• (ρ_{air} \approx 1.225 , kg/m³)

• (ρ_{hydrogen} \approx 0.08988 , grams/L = 0.08988 , kg/m³)

The difference in density is:

[ ρ_{air} - ρ_{hydrogen} = 1.225 , kg/m³ - 0.08988 , kg/m³ \approx 1.13512 , kg/m³ ]

Step 2: Calculate the Volume of Hydrogen Needed to Lift 4 Tons

1 ton = 1000 kg, so 4 tons = 4000 kg.

The weight of 4000 kg in newtons is:

[ Weight = mass \cdot g = 4000 , kg \cdot 9.81 , m/s² \approx 39240 , N ]

To find the volume of hydrogen required, we set the buoyant force equal to the weight:

[ 39240 , N = V \cdot (1.13512 , kg/m³) \cdot 9.81 , m/s² ]

Rearranging to find (V):

[ V = \frac{39240 , N}{(1.13512 , kg/m³) \cdot 9.81 , m/s²} \approx \frac{39240}{11.1160} \approx 3521.04 , m³ ]

Step 3: Convert Cubic Meters to Cubic Feet

1 cubic meter is approximately equal to 35.3147 cubic feet.

[ V_{cuft} = 3521.04 , m³ \cdot 35.3147 , \frac{ft³}{m³} \approx 124835.43 , ft³ ]

Conclusion

Approximately 124,835 cubic feet of hydrogen is needed to lift 4 tons to an elevation of 18 miles.

You

what is the diameter of a 124,835 cubic foot balloon ?

ChatGPT

To find the diameter of a balloon that has a volume of 124,835 cubic feet, we can use the formula for the volume of a sphere:

[ V = \frac{4}{3} \pi r^3 ]

where ( V ) is the volume and ( r ) is the radius. First, we'll solve for the radius and then find the diameter.

1. Rearranging the formula to solve for ( r ):

[ r^3 = \frac{3V}{4\pi} ]

2. Plugging in the volume ( V = 124,835 ) cubic feet:

[ r^3 = \frac{3 \times 124,835}{4\pi} ]

3. Calculating it:

[ r^3 \approx \frac{374,505}{12.5664} \approx 29,850.24 ]

4. Taking the cube root to find ( r ):

[ r \approx \sqrt[3]{29,850.24} \approx 30.05 \text{ feet} ]

5. The diameter ( d ) is twice the radius:

[ d = 2r \approx 2 \times 30.05 \approx 60.10 \text{ feet} ]

So, the diameter of the balloon is approximately 60.1 feet.

You

How much weight does a 60 foot diameter hydrogen filled balloon displace at an elevation of 18 miles?

ChatGPT

To calculate the weight of the air displaced by a hydrogen-filled balloon, we can use Archimedes' principle, which states that the weight of the fluid displaced is equal to the weight of the object.

Step 1: Calculate the volume of the balloon

The volume ( V ) of a sphere is calculated using the formula:

[ V = \frac{4}{3} \pi r^3 ]

Given that the diameter of the balloon is 60 feet, the radius ( r ) is:

[ r = \frac{60 \text{ feet}}{2} = 30 \text{ feet} ]

Now, substituting ( r ) into the volume formula:

[ V = \frac{4}{3} \pi (30 \text{ feet})^3 = \frac{4}{3} \pi (27000 \text{ ft}^3) ] [ V \approx \frac{4}{3} \pi (27000) \approx 113097 \text{ cubic feet} ]

Step 2: Determine the density of air at 18 miles altitude

The density of air decreases with altitude. At sea level, the average density of air is about ( 0.075 \text{ lb/ft}^3 ), but at 18 miles (about 29 km or 18,000 meters) in the stratosphere, the density is significantly lower.

At 18,000 meters, the density of air is approximately ( 0.0009 \text{ lb/ft}^3 ).

Step 3: Calculate the weight of the displaced air

To find the weight of the air displaced by the balloon, we multiply the volume of the balloon by the density of the air:

[ \text{Weight of displaced air} = V \times \text{Density of air} ]

Substituting in our values:

[ \text{Weight of displaced air} = 113097 \text{ ft}^3 \times 0.0009 \text{ lb/ft}^3 ] [ \text{Weight of displaced air} \approx 101.7873 \text{ lb} ]

Conclusion

Therefore, the weight of the air displaced by the 60-foot diameter hydrogen-filled balloon at an elevation of 18 miles is approximately 101.79 pounds.

You

What is the diameter of a hydrogen filled balloon displacing 4 tons at an elevation of 18 miles?

ChatGPT

To determine the diameter of a hydrogen-filled balloon that displaces 4 tons (or 8000 pounds) at an elevation of 18 miles, we need to use some basic principles of buoyancy and the properties of hydrogen compared to the surrounding air.

Step 1: Calculate the Volume of Air Displaced

The buoyant force on the balloon is equal to the weight of the air it displaces. According to Archimedes' principle:

[ \text{Buoyant Force} = \text{Weight of Displaced Air} = \text{Weight of the Balloon} + \text{Weight of the Payload} ]

In this case, the weight to displace is 4 tons (8000 pounds).

Step 2: Determine Air Density at 18 Miles

The air density decreases with altitude. At an elevation of about 18 miles (approximately 29 km), the average air density is significantly lower than at sea level. The air density at sea level is approximately (1.225 , \text{kg/m}^3). At 18 miles up, the air density is around (0.1 , \text{kg/m}^3) - this can vary widely based on atmospheric conditions.

Step 3: Calculate Volume of Air Displaced

To calculate the volume of air displaced, we can use the weight of air and its density:

[ \text{Weight of displaced air} = \text{Density of air} \times \text{Volume} ]

Rearranging gives:

[ \text{Volume} = \frac{\text{Weight of displaced air}}{\text{Density of air}} ]

Converting 8000 pounds to kilograms (1 pound ≈ 0.453592 kg), we have:

[ 8000 , \text{lbs} \approx 8000 \times 0.453592 \approx 3628.74 , \text{kg} ]

Using the air density at 18 miles:

[ \text{Volume} = \frac{3628.74 , \text{kg}}{0.1 , \text{kg/m}^3} = 36287.4 , \text{m}^3 ]

Step 4: Calculate the Diameter of the Balloon

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

[ V = \frac{4}{3} \pi r^3 ]

Rearranging to solve for the radius (r):

[ r = \left(\frac{3V}{4\pi}\right)^{1/3} ]

Substituting in our volume:

[ r = \left(\frac{3 \times 36287.4}{4\pi}\right)^{1/3} ]

Calculating this:

[ r \approx \left(\frac{108862.2}{12.5664}\right)^{1/3} \approx (8674.10)^{1/3} \approx 20.5 , \text{m} ]

Step 5: Find the Diameter

The diameter (d) is twice the radius:

[ d = 2r \approx 2 \times 20.5 \approx 41 , \text{m} ]

Conclusion

The diameter of a hydrogen-filled balloon displacing 4 tons at an elevation of 18 miles is approximately 41 meters.

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Fun thought experiment!

Of course my 4 ton arbitrary load is the total load including the weight of the balloon, tethers, etc.

However scaling up from a 41 meter balloon diameter (36287.4 cubic meters or 1,281,477 cubic feet), is no problem,the tech's there, the Graf Zeppelin's gas capacity was over 3 million cubic feet.

It seems, assuming me and ChatGPT are correct, if someone wanted to, say for example put a close to space science module/habitat or a closer to vacuum chip factory up, this is actually a could do. The materials, equipment and systems needed are pretty much off the shelf, nothing exotic.

Hum, I’m wondering…

Hey ChatGPT, How thick do stainless steel walls need be if I build a 15 psi inside cabin at the bottom of the Mariana Trench where the outside pressure is around 16,000 psi?