Space Station

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steviep187

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Anybody know where the space station is now? Also when is the best time to view it? <div class="Discussion_UserSignature"> <p><font size="6" color="#0000ff"><strong><em><br /><img id="7841257c-c435-495d-9b40-a4a2ae809e40" src="http://sitelife.space.com/ver1.0/Content/images/store/8/11/7841257c-c435-495d-9b40-a4a2ae809e40.Large.jpg" alt="blog post photo" /><br /></em></strong></font></p> </div>
 
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prossa_15

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I recently viewed Space Station narrated by Tom Cruise, and was wondering why in all the shots you see of the Station the background is a black void? Why don't you see ANY stars?
 
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qso1

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Basic photographic optical physics.<br /><br />If you have ever tried to view the stars at night in a city with a lot of lights, you'll notice you see very little in the way of stars. To see this, look at a patch of sky from a fairly well lit area of town and look at that patch for a minute or two to see how your eye adjusts until more stars become visible.<br /><br />Now look at the same patch of sky from a poorly lit area, look for a minute and see how many more stars you see. Your eye has to keep the iris open enough to allow feeble starlight to be seen.<br /><br />You can easily test this by taking a picture at night with a 35mm or digital camera. Take it on a lit street look at the image once taken and see how many stars show up in the image.<br /><br />The ISS shots are the same as any other visible light daytime image so to speak. The station reflects a lot of light and the Earth when in the shot reflects even more. To set the camera to where the stars would be visible would actually cause the footage to wash out due to oversaturation by the much brighter reflected light from station and Earth.<br /><br />From the shuttle and ISS, much footage has been taken where stars can be seen but its when either vehicle is on the night side of Earth. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>
 
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harmonicaman

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<b>qso1 - </b><br /><br />That was an excellent reply!<br /><br />I would just like to add that this exact same problem was encountered by the Apollo astronauts on the Moon. The stars rarely appeared in any of their photographs because the film they used was too fast and they had to deal with the very bright direct Sunlight.<br /><br />The "Missing Stars" in the Moon photos has been offered as "Proof" that the Moon landings were an elaborate hoax; but the fact is, the stars are just too dim to be picked up on the film due to the overwhelming brightness of the Moon's Sunlit surface.
 
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toothferry

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I really don't have anything to add.. except that I own a copy of that documentary, Space Station, on DVD <img src="/images/icons/cool.gif" />
 
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prossa_15

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do you know of any links to see pictures of the stars from the astronuats/cosmonuats point of view {ie.; pics from the shuttles or the ISS} Thanks for the help
 
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qso1

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I'd start by googling NASA, KSC, and other space related sites but I'm not sure how much still imagery they have that was shot on the night side of Earth. My footage came from old recordings I did of NASA select stuff which also included some footage of lightning storm strikes on Earth as viewed from shuttle. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>
 
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thegustav

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Ok... So I'm no fancy smart astrologist, physics, or space expert... But I have a little project I'm doing....<br /><br />Lets say I have a space station... There's 3 connected rings on top of each other, all around (and connected to) a large center sphere....<br /><br />Is there anyway to create some form of gravity within the rings?<br /><br />Maybe using centrifugal force I hear from people... Then maybe putting my floors on the side of the ring....? If this is an option, about how fast would these rings have to be spinning around the center object? The whole station is about 150m in diameter.... Yes, it's big...<br /><br />Any other way to create gravity? Or shall I just plan on installing magnets in the floor, and attaching magnets to all the furniture and have people wear... Magnet shoes....<br /><br />:)<br /><br />Thanks for any help,
 
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willpittenger

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Spinning the craft would simulate gravity. However, the larger the diameter, the better. If you make it too small, you get problems with the feet encountering .9g while the head has only .5g. This can cause serious orientation and health problems. <div class="Discussion_UserSignature"> <hr style="margin-top:0.5em;margin-bottom:0.5em" />Will Pittenger<hr style="margin-top:0.5em;margin-bottom:0.5em" />Add this user box to your Wikipedia User Page to show your support for the SDC forums: <div style="margin-left:1em">{{User:Will Pittenger/User Boxes/Space.com Account}}</div> </div>
 
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qso1

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Welcome to SDC and if I understood your question correctly, the answer is yes. Recall the wheeled station in "2001 A Space Odyssey". This station rotated which generates centrifical force that causes objects to be pressed against the walls so to speak.<br /><br />In the illustration, you see two stick people who are standing on the wall or floor of a ring or wheel shaped station. The yellow arrows indicate direction the ring is rotating and the axis of rotation is at the center. This configuration can consist of as many rings as required so long as they are docked together for spinning. The rings cannot be at angles to each other and get the same spin generated gravity which is why the have to be docked at the central hub in a stacked fashion. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>
 
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qso1

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It can also work for partial ringed stations of the configuration shown here. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>
 
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willpittenger

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As noted though, you should be careful of the minimum diameter for the G forces you will be simulating.<br /><br />Qso1, how large are we talking to simulate 1g without problems? <div class="Discussion_UserSignature"> <hr style="margin-top:0.5em;margin-bottom:0.5em" />Will Pittenger<hr style="margin-top:0.5em;margin-bottom:0.5em" />Add this user box to your Wikipedia User Page to show your support for the SDC forums: <div style="margin-left:1em">{{User:Will Pittenger/User Boxes/Space.com Account}}</div> </div>
 
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qso1

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Last I calculated, it was 1,200 feet with a 1 RPM rotation. I'll have to recheck my references but this produced about 1 G and maintained a slow enough rotation to keep vestibular problems from occuring. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>
 
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thegustav

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So a minimum diameter of 1,200 ft and rotating the rings would provide decent gravity? At 1 RPM.... Would the thing be spinning around...<br /><br />What, 40 or so mph?<br /><br />Velcro idea is interesting. <img src="/images/icons/smile.gif" /> Thanks for the input.... Anybody feels like adding ideas feel free.
 
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qso1

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Thats the diameter I recall. I haven't calculated the actual 1 RPM speed but its probably somewhat less than 40 mph. Of course, you could cut the diameter in half and have a station with .5 to .6 Gs. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>
 
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qso1

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If the rotational axis is zero "G"...the 1,200 foot distance is 1 "G"...how would it be non linear?<br /><br />Anyway, here is the formula from page 208 of "Introduction To Space, The Science Of Spaceflight" by Thomas d. Damon. Published in 1995.<br /><br />Fc = mv2/r<br /><br />Where m is the mass of the object, v is velocity and r is the radius of the circle.<br /><br />f = ma<br /><br />Relates to the accelleration of the object to the applied force. Compare the above equations, noting acceleration due to centripetal force must be...<br /><br />ac = v2/r<br /><br />Solving for v we get...<br /><br />It gets dicey here because it appears to be a long division symbol which I can't type on this KB. It then specifies a rotating colony 5,280 ft diameter which yields the radius of 2,640 ft in the formula below. If 1 "G" is desired, ac must equal 32 ft/sec squared or gravity acceleration on earth. The equation below works out the velocity of the object at the rim of the wheel where I positioned my stick figures.<br /><br />v = division symbol over 2,640 X 32 = 290 ft/sec = 198 mph.<br /><br />Period of rotation is next. Locate the circumference of the wheel, which will give the distance travelled in one rotation. In this case, the circumference is 5,280 feet.<br /><br />5,280 ft X pie = 16,600 ft<br /><br />Divide distance by speed.<br /><br />16,600 divided by 290 ft/sec = 57 seconds.<br /><br />A 1 mile diameter whell shaped station will provide 1 "G" at 1 rpm at the rim.<br /><br />I came up with 1,200 feet by increasing the rotation to get maybe .8 Gs. This because it seemed oddly enough to be more practical to build a 1,200 ft diameter ring than a 1 mile diameter one. I have to rework this again to see what the exact values were. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>
 
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qso1

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IIRC, velcro was a culprit in the Apollo fire because of its flammability in a pure oxygen environment. It is apparently much less flammable in a mixed gas atmosphere (78 N 21 O2) at sea level pressure. It's used in huge amounts aboard the shuttle and ISS. <div class="Discussion_UserSignature"> <p><strong>My borrowed quote for the time being:</strong></p><p><em>There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened.</em></p> </div>
 
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spacester

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Sigh. Once more into the breech.<br /><br />http://uplink.space.com/showthreaded.php?Cat=&Board=sciastro&Number=604361&page=&view=&sb=&o=<br /><br />Studies indicate that the fastest rotation without causing excessive Coreolis force problems is 3 rpm (revolutions per minute).<br /><br />If we use that number, we find that the calculation becomes very very simple, you just divide the radius in meters by 100 to get the decimal fraction of Earth's Gravity<br /><br />100 meter radius yields 1.00 "gee" (Earth)<br />38 meter radius yields 0.38 "gee" (Mars)<br />16 meter radius yields 0.16 "gee" (Moon)<br /><br />Your 150 m diameter station has a radius of 75 m so a 3 rpm rate of rotation would simulate 0.75 or 75% or 3/4 of Earth's normal gravity.<br /><br />Here's the formula:<br />G = [R * [(pi*rpm) / 30]^2] / 9.81<br />OR<br />R = (9.81 * G) / [(pi*rpm) / 30]^2<br />Where:<br />G = Decimal fraction of Earth gravity<br />R = Radius from center of rotation in meters<br />pi = 3.14159<br />rpm = revolutions per minute <div class="Discussion_UserSignature"> </div>
 
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willpittenger

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<blockquote><font class="small">In reply to:</font><hr /><p>how would it be non linear?<p><hr /></p></p></blockquote><br />You may have misunderstood me. I was stating that just because radius X and spin rate Y result in A g's, you shouldn't assume that radius X/2 and spin rate Y/2 results in A/2 g's. <div class="Discussion_UserSignature"> <hr style="margin-top:0.5em;margin-bottom:0.5em" />Will Pittenger<hr style="margin-top:0.5em;margin-bottom:0.5em" />Add this user box to your Wikipedia User Page to show your support for the SDC forums: <div style="margin-left:1em">{{User:Will Pittenger/User Boxes/Space.com Account}}</div> </div>
 
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