1-Day Rotation = 1-Gravity?

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CAllenDoudna

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At what diameter does one rotation in 24 hours allow us to feel one gravity of centrifigal force standing on the walls of a sphere, cylinder, or wheel? I have seen the formula but can't make heads or tails of it. Extrapolating from smaller sizes I can guess that it must be close to 50 kilometers (30 miles) but quite frankly I don't know and I really hate to always be offering a guess.

This would seem to be the ideal size for an "O'Neil" type Space Colony:

http://settlement.arc.nasa.gov/70sArt/art.html

We could have one Valley and one Window and the sun could rise over the eastern edge of the Valley , travel across a normal day-time sky, set over the western edge of the Valley and as the Window rotated away from the sun and out to the stars we would have a normal nightime sky.

If anybody has the answer I'd sure appreciate it.
 
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Mee_n_Mac

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CAllenDoudna":1vvdb68y said:
At what diameter does one rotation in 24 hours allow us to feel one gravity of centrifigal force standing on the walls of a sphere, cylinder, or wheel? I have seen the formula but can't make heads or tails of it. Extrapolating from smaller sizes I can guess that it must be close to 50 kilometers (30 miles) but quite frankly I don't know and I really hate to always be offering a guess.

This would seem to be the ideal size for an "O'Neil" type Space Colony:

http://settlement.arc.nasa.gov/70sArt/art.html

We could have one Valley and one Window and the sun could rise over the eastern edge of the Valley , travel across a normal day-time sky, set over the western edge of the Valley and as the Window rotated away from the sun and out to the stars we would have a normal nightime sky.

If anybody has the answer I'd sure appreciate it.

The force you want equals 1G, or 32.2 ft/sec[super]2[/super]. For a rotating ring (or cylinder) this force = v[super]2[/super]/r : with v = tangential speed at the rings radius and r being it's radius. So with v = distance/time and distance = 2*pi*r ; time = (24hrs*60mins/hr*60secs/min) ... this gets you a diameter of about 2,306,320 miles. But someone check my math to be sure. Naturally for a sphere this only works at it's equator. You'd get less "gravity" as you go towards either pole.

EIT : OK, duh, you asked for a diameter so when I got the radius above I naturally divided (?!?) by 2. Should have multiplied by 2. So revised number is now above.
 
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Saiph

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I work it out below, but the resulting ring is ~20% larger than earths orbit about the sun. Yeah, surprised me too (thinking I may be missing a decimal point...but I'm not seeing it...). I was guessing something large, but not quite that large. Say you want to do it in 60seconds, you're looking at only 900m radius. 6 seconds is about 9 meters.

The key factor: The really slow rotation rate. Bring that down, and you don't need a ring nearly so large. The derived equation to tell you the radius of a ring needed, given the time of rotation is: R = A*T^2/4pi^2 (A in this case being 1g, or 9.8m/s^2)

on a side note, I just worked out the centripetal force of a ring at geosynchronous orbit (i.e. what's the acceleration of a ring at geostationary orbit, rotating 1/day..the sci-fi holy grail of space docks...is only 0.18m/s^2)

Working out the details if you're interested:

Centripetal acceleration is A = v^2/R, and we want that to equal 9.8m/s^2, and you're asking for the radius required for it to require only on rotation in a day.

So V = Distance/Time, time in this case is 24 hrs = . So whats Distance? It's the circumference of a circle, so 2*(pi)*R.

so, plugging things in we get A=(2piR)^2/(T^2*R) = 4*pi^2*R^2/(T^2*R) = 4*pi^2*R/T^2

Rearranging so that R is solved for (not A) we get: R = A*T^2/4pi^2

A = 9.8 m/s^2
T = 24hrs = 86400 seconds --> T^2 = 7464960000s^2
4pi^2 = 39.4

A quick check to make sure the units come out right...R is in m, AT in m/s^2 * S^2 giving us M...so that works out (don't want a distance in seconds or velocity after all).

Plugging the numbers in...
39.4*R/7464960000 = 9.8
R=90*7464960000/39.4 = 1856766700 meters - 1856766 km...round out to significant figures (make it neat and tidy) and we get ~1,850,000 km radius

To make that number more meaningful, earth's radius is only ~6,000 km...so we're looking at a ring ~310,000 times the earths radius...

Still to big to really get your head around. An astronomical unit (distance between the earth and the sun) is ~150,000,000 km .... So this ring is just a bit larger than earths orbit....spinning at 1 revolution a day.



Looks like I was right, double checking turned up this site that has an online calculator. Just turn 1/day into rotations per minute (1/(24hrs*60min)...or 0.00069 rpm and set the mass to 1kg. Mess with the radius until you get 9.8N of force (which is 1g acceleration on 1kg of mass)...which matches my huge radius.
 
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CAllenDoudna

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Well, facts are facts. I have a hard time accepting those figures since at something like 1 kilometer you need one minute and at 6 km you only need about 20 minutes and I think 100 meters requires 20 seconds.

Larger than Earth or Earth's orbit doesn't sound like a very practical size. There goes a beautiful dream.
 
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Saiph

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1g acceleration is quite a bit of acceleration, and I was a bit skeptical too, until I tried smaller sizes. It gives reasonable figures for small rotation periods, that fit with things like the spinning carnival rides and such.
 
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aphh

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For a spinning space application for artificial gravity the important consideration is that the environment has to be comfortable for the astronauts. Quite a bit of thought has been given to this already, because in space the artificial gravity could be obtained by tethering two spacecrafts together using a cable and making them spin.

Empirically it has been determined that a diameter of roughly one mile would be suitable for artificial gravity application. This might be doable, but the tether would need to withstand the force of two spacecrafts trying to break away at 1 G acceleration.

If an astronaut weighs 75 kg, he would stress the tether with the force of F = ma, 75 kg * 9.8 m/s^2 = 735 Newtons. Multiply with the mass of the spacecrafts and you'll see that the tether has to be very strong. A mile of strong tether would be considerable mass in itself.

For rotational period for a diameter of one mile at 1 G centripetal acceleration we will first calculate the speed of the outer rim; a = v^2/r, v = 89 m/s.

Circumference divided by the speed of the outer rim gives rotational period of 5024 m / 89 m/s = 56 seconds.

Note also that even a small amount of artificial gravity might help on a long journey, as it would make the living conditions resemble those on the earth. Looking out the window might make your head spin, though.
 
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CAllenDoudna

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It will be a thousand years before we have the engineering skill to build something with a diamater of a couple million kilometers--and by then it may not seem worth it. Here's my speculation, though:

We could have a double-hulled tube a 1.8 million kilometers in diameter with the inner hull being clear and 5 km "above" the outer hull. ("Above" being from the perspective of someone standing on the outer hull.) Between the two hulls we would have atmospere with weather, clouds, ect. and of course a blue sky. The diameter of the outer hull would be 5.8 million km. 2.5 million km would be a "Valley" and the remaining 3.3 million km would be a "Window". The sun would rise each morning at 0600 and set each evening at 2000 giving us 14 hours of daylight and 10 hours of night. The tube would be about 12 million km long and instead of the "Valley" running the entire length it would shift every million km so that each 8-hour shift would have four Valleys. If the North Pole of axis rotation were tilted to the North Star we would have a change of seasons just as on Earth and there would be a span of climates from Polar to Tropical.

The term "Valley" is merely a carry-over from 1970s designs. We would no doubt want a border of 100 km of ocean perhaps 10 meters deep (only serious divers go below 10 meters) around each section of "Valley" and this 100 km wide ocean would further wind throughout the inside of the "valley" to create sizable islands perhaps on the order of Britain with mountains and rivers.

Transportation would need to be a million km an hour to be practical. Maglev could do that.
 
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csmyth3025

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According to the Wikipedia article on artificial gravity found here:

http://en.wikipedia.org/wiki/Artificial_gravity

It is generally believed that at 2 rpm or less no adverse effects from the Coriolis forces will occur, at higher rates some people can become accustomed to it and some do not, but at rates above 7 rpm few if any can become accustomed.

For the sake of economy and practicality, most designs for producing artificial gravity in a "space ark" or a space station rely on this figure (2 rpm) as the limiting value for rotation. There is no real need for a 24 hour rotational period and, as the previous calculations illustrate, there are very good reasons why such a slow period of rotation is impractical if one is trying to achieve an artificial gravity of 1g.

Using 1g (9.8 m/sec^2) and 2 rpm as a baseline and the formula given previously:

R = A*T^2/4pi^2

A = 9.8 m/s^2
T = 30 sec (at 2 rpm) = 30 seconds --> T^2 = 900s^2
4pi^2 = 39.4

If my math is correct, the radius =~224 meters and the diameter of the (cylinder or torus)=~450 meters.

For comparison, the Integrated Truss Structure of the ISS (the spine that holds all the solar arrays, etc) is about 109 meters long. For a 1g environment at 2 rpm, a space station would have to have to have an enclosed diameter about 4 times this size.

Chris
 
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MeteorWayne

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CAllenDoudna said:
It will be a thousand years before we have the engineering skill to build something with a diamater of a couple million kilometers--and by then it may not seem worth it.

It'll be a hell of a lot longer before someone is willing to pay for it :)
 
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csmyth3025

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It's estimated that the ISS has cost ~$150 billion (US) to construct over a 30 year period. Going by the very naive assumption that a space station four times as large would cost four times as much, a minimal 1g rotating station will cost ~$600 billion.

For comparison, according to Wikipedia the cost of the (US) interstate highway system is estimated as follows:

Although construction on the Interstate Highway System continues, I-70 through Glenwood Canyon (completed in 1992) is often cited as the completion of the originally planned system.[14][15] The initial cost estimate for the system was $25 billion over 12 years; it ended up costing $114 billion (adjusted for inflation, $425 billion in 2006 dollars[16]) and taking 35 years to complete.

Of course if we propose to Congress a project estimated to cost $600 billion, we can, by the track record given by the Interstate Highway System, expect it to actually cost 17 times as much (~$10.2 trillion). That would be a pretty hard sell even if we could split the cost with other countries.

Lest you get too discouraged that these numbers are just too high to even think about, Wikipedia offers the following information concerning the recent Wall Street bailout:

The 2008 federal budget submitted by the president is $2.9 trillion, meaning a $700 billion bailout would constitute a 24% increase to $3.6 trillion, which would in fact far exceed the $3.1 trillion 2009 budget. The total government commitment and proposed commitments so far in its current and proposed bailouts is reportedly $1 trillion compared to the $14 trillion United States economy.

If you think that this is just a special "one time deal", consider that - according to Wikipedia - the 2010 US defense budget is about $685 billion (that's for one year) and:

Defense-related expenditures outside of the Department of Defense constitute between $216 billion and $361 billion in additional spending, bringing the total for defense spending to between $880 billion and $1.03 trillion in fiscal year 2010.

Chris
 
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MeteorWayne

Guest
Folks, please let's recall the purpose of the Science 101 forum:

"This forum is for those seeking an understanding of basic, generally-recognized scientific concepts, with a desire to discuss them intelligently without aggressively debating them. Teachers and students are welcome! "

Budget stuff belongs in Space Business and Technology


Thanx

Moderator Meteor Wayne
 
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CAllenDoudna

Guest
Okay, Wayne, but there was an engineering reason for wanting one-day rotation to get one gravity: In one of the more popular designs proposed by Dr. Gerard K. O'Neil (Island III) in the 1970s a 4-mile diameter cylinder with the walls divided into 3 windows and 3 valleys each 2-miles wide would have sunlight reflected through the windows to the valleys below by mirrors hinged at one end and held out by cables at the other, looking something like a partially pealed banana.

I'm just a flunky factory worker currently washing dishes, but in my experience centrifugal force has a spiral to it so those mirrors would not hang straight out as pictured but would bend. They would need to be a lot stiffer than planned and as I thought about how heavy the bracing would need to be considering the load it would have to resist would increase by the square of the distance. . .
whistle!

However, if diameter could be increased to the point where one gravity was obtained by one day of rotation it would not only look more natural but would be simpler and would avoid the stress on the mirrors which would no longer be needed.

A note on cost: It would seem the cheapest production technique would be to melt sand or rock from the Moon or Mercury using sunlight focussed by mirrors and then using a tank of gas (air) to blow that into a jar as a glass blower would do. That should be cheap and cover with plastic and it would be 5 times stronger than steel. Add dirt for a valley and home sweet home. This should be a cheap way to produce O'Neil's Island II spheres.
 
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rfoshaug

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If a 24hr rotation of an object less than or equal to the diameter of the Earth would give a 1G centrifugal force, people living at the equator would need something to hold on to in order not to be trown off our planet. :lol:
 
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CAllenDoudna

Guest
rfoshaug":1h72o8se said:
If a 24hr rotation of an object less than or equal to the diameter of the Earth would give a 1G centrifugal force, people living at the equator would need something to hold on to in order not to be trown off our planet. :lol:

Okay, that makes sense.
 
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csmyth3025

Guest
rfoshaug":2rh318qv said:
If a 24hr rotation of an object less than or equal to the diameter of the Earth would give a 1G centrifugal force, people living at the equator would need something to hold on to in order not to be trown off our planet. :lol:

Your point is well taken. As it turns out, the velocity of a person standing on the Equator is about 465 meters/sec. due to the rotation of the Earth. The escape velocity for that same person to be "thrown off our planet" and break free of Earth's gravity is about 11,186 meters per second.

Mother nature has made sure that we all keep our feet solidly planted on good old terra firma. I'm not sure she anticipated our interest in rocketry, though. :lol:

Chris
 
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eburacum45

Guest
The concept of a spinning structure which produces one gravity of centrifugal force has been explored by Iain M Banks, the Scottish science fiction writer. The structures he describes are called Orbitals, and are ring-shaped objects around three million kilometres across (as CAllenDoudna quite rightly calculated).
http://en.wikipedia.org/wiki/Orbital_(The_Culture)

But there is a problem; spinning something that large, that fast would tear the ring apart if it were made of any physical matter. The largest any spinning ring could get without tearing itself apart would be around three thousand kilometers in diameter for 1 gravity of centrifugal force; this would be using a ring of pure carbon nanotube with no payload.

So Mr Bank's orbitals could not be built in the real world, unless some exotic hyperstrong form of matter were discovered.
 
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