The International Space Station

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Profile William Rothamel
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Message 1142723 - Posted: 19 Aug 2011, 20:00:18 UTC - in response to Message 1141545.  

It could have solar panel cladding and/or a small atomic reactor to drive a turbine. Cooling should be no problem in space. The wheel could generate artificial gravity by spinning. Vertigo may be a problem--might calculate size and speed for say .5 G for comfort and safety.
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Message 1142816 - Posted: 19 Aug 2011, 22:43:36 UTC

How the heck do you dock to a spinning space station? How do you have many docking ports for a spinning space station?

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Message 1142920 - Posted: 20 Aug 2011, 2:45:43 UTC - in response to Message 1142816.  
Last modified: 20 Aug 2011, 2:46:19 UTC

You spin the docking craft at the same rate as the space station. Since all motion is relative the two would not appear to be spinning as regards each other. In the center you would dock,there would be little or no G force in the middle of the spinning hub. Gravity would increase as you walked out to the rims of the station. You would be walking upside down on the "tire" side of the spinning wheel. you would want to be spinning fast enough to generate a comfortable level of gravity.

I do think that the inner ear would detect the motion and you might have to learn to overcome vertigo, I suspect. I presume that the wheels might have a diameter of 100 yards or so. So who can tell me what the rotational speed would have to be to generate one half g.
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Message 1142925 - Posted: 20 Aug 2011, 2:56:36 UTC - in response to Message 1142920.  

I come up with about 20 RPM for a half G. Need to check my work
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Message 1143005 - Posted: 20 Aug 2011, 8:45:32 UTC - in response to Message 1142925.  
Last modified: 20 Aug 2011, 9:00:32 UTC

I come up with about 20 RPM for a half G. Need to check my work

Where radius (r) is 50 m and g (1/2 standard gravity) = 4.9 m/s/s and v = rotational velocity:

and:

If I remember my uniform circular motion lessons from high-school correctly, g = V^2/r

then:

V = sqt(g*r)

v = sqt(4.9m/s/s * 50m) = 15.6m/s

Circumference = 2*pi*R = 2 * 3.14 * 50m = 314m

Period of rotation = Circumference / V = 314m / 15.6 m/s = 20s

So

60s / 20s = 3 RPM

. . . I think.
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Message 1143029 - Posted: 20 Aug 2011, 11:11:31 UTC - in response to Message 1143005.  
Last modified: 20 Aug 2011, 11:14:26 UTC

i think that you have the frequency of rotation (omega) and not the period. So you have .3 per sec or so which is 20 revolutions per minute ==roughly. This is what i get when i use the non-communist system of feet and G at 32 feet/sec/sec
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Message 1143098 - Posted: 20 Aug 2011, 16:33:48 UTC - in response to Message 1142920.  

You spin the docking craft at the same rate as the space station. Since all motion is relative the two would not appear to be spinning as regards each other.

Nice theory. Try in practice to get the center of gravity of both craft at the center of the docking port so there is no wobble. You likely can do it for the station with a bunch of water tanks and pumps, but the small docking craft isn't as easy to balance and it is the one that small amounts make for large wobbles. And you have at most two docking ports. A large station is going to need more than two docking ports to keep is supplied. Have you also considered how you are going to conduct space walks to do repairs outside the spinning station? Now those multi ton equipment packages not only have mass but also weight. Try pushing one into a capture lock so you can put a wrench on a bolt to put it in place, or releasing it and not having it fly away. There is a lot of physics that happens when you spin the station. Then there is the question of why spin it? Most everything we are doing today on the ISS is done there because it is zero g.

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Message 1143170 - Posted: 20 Aug 2011, 20:06:37 UTC - in response to Message 1143148.  

Most SCI FI films/books have stations rotating at about 3 rpm, so I think Kenzies maths is about right.

3 rpm is still rather fast to be stood looking out the window!

As already mentioned, maintaining an even balance on the wheel would be a difficult problem to efficiently solve, unless the rotating wheel is very significantly more massive than the mass of people moving around.


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Message 1143191 - Posted: 20 Aug 2011, 21:31:11 UTC - in response to Message 1143029.  
Last modified: 20 Aug 2011, 22:06:32 UTC

i think that you have the frequency of rotation (omega) and not the period. So you have .3 per sec or so which is 20 revolutions per minute ==roughly. This is what i get when i use the non-communist system of feet and G at 32 feet/sec/sec


Well, I'm not a physicist or engineer or anything so I've no option but to defer to your wisdom but by my (very amateurish) calculating:

Using your ancient measuring system and 20 revolution per minute calculation:

100 yard diameter = 300 feet.
50 yard radius = 150 feet

Circumference of the circle would be 300 ft * pi (3.14) = 942 ft

It is rotating 20 times per minute, which means that the outer edge of the circle is moving 314 feet per second.

G = v^2/r

G = 314ft/s^2/150ft

G = 657ft/s/s

Normal Earth gravity is 32 ft/s/s, so

657 / 32 = 20.5 G

As I see it, your space station spinning at 20 RPM is going to generate 20.5 times earth gravity on the rim.

So, obviously my calculations are in error somewhere, I just can't figure out where I goofed. :?
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Message 1143216 - Posted: 20 Aug 2011, 22:26:46 UTC - in response to Message 1143191.  

... Using your ancient measuring system and ...

Wasn't that the cause of one of the NASA Mars probes shooting off into oblivion?...


It is far easier for all calculations to contrive a system of measure where all your units of conversion become exactly the number one.

But then... It took a world war to get the USA to agree a consistent length of the inch against the rest of the world!

(Gee... Is that where the term "to give an inch" came from?)


;-)

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Message 1143661 - Posted: 22 Aug 2011, 0:45:11 UTC - in response to Message 1143029.  
Last modified: 22 Aug 2011, 0:45:40 UTC

i think that you have the frequency of rotation (omega) and not the period. So you have .3 per sec or so which is 20 revolutions per minute ==roughly. This is what i get when i use the non-communist system of feet and G at 32 feet/sec/sec


Still waiting and wondering how being on a space-station spinning at 20 RPM (once every 3 seconds) produces 1/2 G of acceleration. Newtonian physics indicates more along the lines of 20.5 G.

No answer?
Pure mathematics is, in its way, the poetry of logical ideas.

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Message 1143725 - Posted: 22 Aug 2011, 4:14:57 UTC - in response to Message 1143661.  

sorry,

I will sit down and figure the math more carefully and report back.

Try to do it tomorrow--too busy now and time for the rack.

What do we know about the likely hood of vertigo under these conditions.

regards,

Daddio
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Message 1143729 - Posted: 22 Aug 2011, 4:37:06 UTC - in response to Message 1143725.  
Last modified: 22 Aug 2011, 4:42:24 UTC

Yes I had a second wind. Rpm at 100 meter diameter comes out to right about 3 rpm for 1/2 G. Even though I have taught physics I still make mistakes when I try to do a quick calculation in my head. I came out with .3 per sec instead of 3 per minute before. Maybe it's those damn metric seconds that get me.

I see that it is thought that this should be below 2 Rpm for reasons of vertigo. Most can adapt to this , but not above 7 rpm. There seems to be little data on whether 1/2 G is enough to combat bone and muscle loss. (either with or without exercise in lower gravity.)

So: We would have to be a little bigger than 100 meters in diameter. What about cosmic ray protection ?
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Message 1143733 - Posted: 22 Aug 2011, 4:52:45 UTC - in response to Message 1143216.  

Listen youse guys: I am in favor of bringing back the imperial quart and the "Proper Pint" Could also use the imperial gallon right about now too. It seems to me that we in the Colonies should have a right to the old British system.
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Message 1143750 - Posted: 22 Aug 2011, 7:31:15 UTC - in response to Message 1143733.  

All you have to do is to board any airplane and you'll find yourself in the good old Imperial system.
Tullio
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Message 1143772 - Posted: 22 Aug 2011, 10:17:38 UTC
Last modified: 22 Aug 2011, 10:20:10 UTC

Also the nautical mile, used to calculate ships and aircraft speeds in knots, is different from the statute mile, which is different from the British mile. I wonder how American and Russian astronauts can cooperate on the ISS, the former using the Imperial system, the latter and their European and Japanese colleagues the metric system. A real Tower of Babel.
Tullio
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Message 1143777 - Posted: 22 Aug 2011, 10:46:30 UTC - in response to Message 1143772.  

To stay off-topic a little longer:

I conclude that the Aliens who set up our simulation had 12 fingers and toes and three feet and bought their whisky in 40 ounce bottles, That's right
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Message 1143867 - Posted: 22 Aug 2011, 15:58:34 UTC - in response to Message 1143854.  

I conclude that the Aliens who set up our simulation had 12 fingers and toes and three feet and bought their whisky in 40 ounce bottles, That's right

The yard, or 36 inches originated from the double Cubit. A cubit was the length of a forearm. The metre was originally intended to be one ten-millionth of the distance from the Earth's equator to the North Pole, at sea level.

In the UK up until 1980 standard size bottles were 26 2/3 fl ozs, they then changed to 750ml in 1980. in 1992 the EU the standard size from 750ml to 700ml.

40 Fl ozs is two pints, beer is sold in pints, whiskey is twice as strong, makes sense.

These alien programmers are clever 'aint they!

Therein lay many anachronisms including using weight to measure a volume!

Hence, call to a petrol (USA: automobile gas) station first thing in the morning to fill up whilst the tanks are coldest... (For practicality, fuel is dispensed by volume when actually for this example it should be dispensed by weight or even better, by energy content...)


Great care is needed to be meaningful about your units! (... As NASA has found out the very expensive way on various occasions...)

Keep searchin',
Martin


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Message 1144579 - Posted: 24 Aug 2011, 15:13:47 UTC

Retire the Shuttle and
http://news.yahoo.com/russian-space-ship-fails-reach-orbit-142022603.html
MOSCOW (AP) — An unmanned Russian supply ship bound for the International Space Station failed to reach its planned orbit Wednesday, and pieces of it fell in Siberia amid a thunderous explosion, officials said.
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Message 1144779 - Posted: 25 Aug 2011, 1:49:05 UTC

Russians have ever more financial backing than anyone else and their ship fails, interesting.


Mandtugai!
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