Coriolis Effect on Future Ships

[mrmorris]

I was thinking about this thread a couple of days ago when I thought of a problem with spin-based gravity that had never occurred to me before. Today I was checking out "Who's Online' and saw someone reading this thread so figured I'd post my thoughts....<br /><br />The 'gravity' imposed by a rotating system is essentially created by the momentum of the rotating surface (for the moment -- let's assume a bicycle-wheel station) providing acceleration to someone standing on the 'floor'. By contrast, gravity imposed by the planet Earth is created by the attraction between your body and the mass of the Earth. Gravity on Earth exists whether your feet are in contact with the ground or not. 'Gravity' created by rotation will only exist while you are in contact with the surface of the station. This leads to some interesting questions.<br /><br />1. If you 'jump' with sufficient force that your feet no longer contact the station floor -- essentially you have negated the accelleration provided by the roating station. Since you are no longer in contact with the station floor -- you should effectively be in zero-G... until you contact the station in some way and allow it to provide your mass with an acceleration vector again. Jogging -- as per 2001 seems not to be a possibility.<br /><br />2. Walking in the direction of spin will increase the acceleration of your body -- essentially making you heavier. Walking anti-spinward would decrease the acceleration vector of your body -- making you lighter. From spacester's calcs:<br /><br />Using 3 rpm, based on studies, we get <br />R = 38 meters for Mars gravity <br /><br />A 38M radius station circumference then is ~240 meters. Spinning at 3 RPM -- you are essentially being accelerated at 716 meters per minute or 42 km/hour. A brisk walk is about 8km/hour, so walking antispinward would reduce your weight by ~20% in Mars gravity or ~30% in lunar gravity. Mind you, you'd have to walk carefully enough to avoid bouncing -- or you'd en

 

[tap_sa]

<font color="yellow">" 'Gravity' created by rotation will only exist while you are in contact with the surface of the station."</font><br /><br />Umm ... no. You <i>feel</i> the gravity by rotation only when your feet push against the rotating surface, but if you jump a little or just lift your legs up quickly, the phenomenon causing you to feel gravity against the surface doesn't vanish and you will 'fall down'. That phenomenon is your body's desire to continue movement on a straight path (Newton 101) but since you are enclosed in a rotating space station you will swiftly meet it's floor again (ouch!).

 

[north_star_rising]

Depends on the size of the wheel. First, you must understand that the acceleration, or change in speed (distance per time, per time, so "meters/second/second") due to the Earth is the same no matter what object one is talking about. Galileo demonstrated this with the thought experiment that two bricks of the same stuff fall at the same rate, so what if they were tied together with a piece of string? Would they fall any faster because they were now one, doubly heavy object? No. They fall at the same speed. The only reason a feather floats slowly while a hammer falls is due to wind resistance and air drag. The Apollo 15 Commander David Scott demonstrated this on the moon, where there is virtually no atmosphere, using a hammer and feather. Same with being rotated, all objects experience different forces due to different weights, but undergo the same acceleration. <br /><br />Just in case you're REALLY curious, we can prove this mathematically: <br /><br />FG = gravitational force, due to Earth's gravity <br />g = acceleration due to gravitational force (always constant at same distance, see below), 9.8 meters/second/second <br />m1 = mass of any object, say, a person, roughly 30 kilograms <br />m2 = mass of any other object, like a pencil, roughly 0.01 kilograms <br />mE = mass of the Earth, roughly 5,970,000,000,000,000,000,000,000 kilograms <br />RE = radius of the Earth, 6,380,000 meters <br />G = gravitational constant, just so the numbers come out right, 0.000 000 000 0667 <br /><br />Newton's Second Law of Classical Physics: <br />F = m * a, <br />where a is the acceleration of a mass m due to a force F. <br /><br />Newton's Universal Law of Gravitation: <br />FG = G * m1 * m2 / r^2, <br />where the gravitational force FG between two masses, m1 and m2, is divided by the distance r between them squared, times a constant, G. <br /><br />We can modify Newton's Second Law for Earth's gravity like so: <br />FG = m1 * g, <br />with m1 representing our 30 kilogram person OR our 0.01 kilogram

 

[mrmorris]

<font color="yellow">"phenomenon is your body's desire to continue movement on a straight path (Newton 101) but since you are enclosed in a rotating space station you will swiftly meet it's floor again (ouch!). "</font><br /><br />One of us isn't thinking it all the way through. I still think I am. The 'gravity' is indeed caused by inertia. You have a potential energy which, if the station floor were to disappear, would send you in a straight line away from it. However -- having a velocity vector isn't enough to generate 'gravity' -- you also have to have acceleration. That accelleration comes from the fact that the station is spinning, and constantly accelerating you in a different direction. If you were to eliminate your acceleration vector, but *not* your velocity vector, the results would be as you describe. <br /><br />For example -- say I reach up to the 'ceiling' of the station, grab a convenient handhold, and pull myself up off the floor of the station to hang suspended from the 'ceiling'. If I then let go -- I will lose my <b>acceleration</b> vector -- but I still maintain the <b>velocity</b> vector that the station has already imposed on me. I will fall in a 'curving' path from the station's perspective as it rotates out from underneath me (it would be straight in a spatial sense) and hit the floor. (as you say... ouch)<br /><br />However, if you <b>jump</b> to to point where your feet leave the floor, you have effectively countered the velocity vector the station has imposed on your body (if the velocity vector weren't countered -- your feet wouldn't have left the floor). Simultaneously, you have eliminated your contact with the station and therefore the acceleration vector. Since both are gone -- you are in zero-G.

 

[tap_sa]

<font color="yellow">"However, if you jump to to point where your feet leave the floor, you have effectively countered the velocity vector the station has imposed on your body (if the velocity vector weren't countered -- your feet wouldn't have left the floor)."</font><br /><br />Which way you are jumping? If it's a normal jump 'up' towards the centerpoint of rotating station ie. opposite direction of the force pushing you down against the floor, then this is the point where you must brush up the big picture. When you spin along the station your velocity vector is not towards the floor but it's <i>tangent</i>, therefore no matter how hard you jump it won't cancel it, only increase your total velocity. You will feel momentary zero-G after you jump but really soon you will meet the floor again. <br /><br />Using your example, visualize you standing still having tangential velocity nearly twelve meters per second. Your body would like to continue forwards but the floor keeps changing directions and forces you too. Your acceleration vector is perpendicular to the velocity vector and pointing 'up'. You feel a sudden taste for zero-G and jump, all those hours of bicycle training have done wonders to your leg muscles and you achieve stellar 12 meters per second velocity 'up', same speed as your tangential. The moment your feet leave the floor your total velocity becomes sqrt(12^2 + 12^2) which is almost 17 meters per second. When your two velocities are added the direction of your total velocity is <i>in the big picture</i> 45 degrees up from the floor level. You won't meet the floor at 17m/s because in the big picture you will meet the floor at 45 degrees angle.<br /><br />While inside the rotating station the only way to achieve zero-G indefinitely is to jump 'sideways', directly against the rotation velocity. You correctly implied this in the second part you wrote, walking spinward and antispinward. But better be careful, now the station would be spinning around y

 

[mrmorris]

<font color="yellow">"When you spin along the station your velocity vector is not towards the floor but it's tangent..."</font><br /><br />Um... no. That's the velocity vector if the <b>space station</b>. If <b>your</b> velocity vector were tangental with the floor -- you'd be sliding along it -- not trying to go through it. <br /><br />Because the station is rotating and you are 'connected' to it, the direction of your velocity vector is constantly changing. The station's velocity in a tangental direction continues to rotate the angle of your velocity vector around in a circle -- like the hand of a clock.<br /><br />The difference in our understandings seems to revolve around which is the acceleration vector and which is the velocity vector. Part of the problem is that there's not *truly* an acceleration -- when taken to mean a change in velocity. The station is rotating at a constant velocity and your body is moving at a constant velocity. The 'accelleration' as such is simply the way you perceive the station 'pushing back' against the potential energy your body has due to angular momentum. <br /><br />Let me argue my case this way... if the floor of the space station vanished suddenly -- which direction would the poor spaceman fly? If his velocity vector is tangental to the station (i.e. parallel to the 'floor') -- he will fly off at a tangent to the space station. If his velocity vector is outward (i.e. at right angles to the floor) -- he will fly off directly away from the hub of the station. I'm pretty sure you would agree that the second is the case. If not -- we won't be agreeing on this thread. I'll take it as a given that you agree, and move on to state that this indicates definitively that the velocity vector of a person in the station is pointing <b>through</b> the floor of the station. This then would be what you have to counter in order to achieve zero-G.<br /><br />Because that rotation is contantly changing your velocity vector, you 'fe

 

[tap_sa]

<font color="yellow">"Let me argue my case this way... if the floor of the space station vanished suddenly -- which direction would the poor spaceman fly? If his velocity vector is tangental to the station (i.e. parallel to the 'floor') -- he will fly off at a tangent to the space station. If his velocity vector is outward (i.e. at right angles to the floor) -- he will fly off directly away from the hub of the station. I'm pretty sure you would agree that the second is the case."</font><br /><br />Please, test it. Your own empirical data is better than any arguing <img src="/images/icons/smile.gif" /> I just tested it to make sure laws of physics haven't changed, and it seems all is as it should be. Tie a small weight to a couple feet of string, rotate and release, inspect the direction it goes. You do agree that the system of weight and string is comparable to the astronaut and station floor, don't you?

 

[mrmorris]

<font color="yellow">" You do agree that the system of weight and string is comparable to the astronaut and station floor, don't you? "</font><br /><br />There are similarities, but the vectors forces are not being applied in the same fashion. I don't <b>know</b> that the two can be compared in that way. I know what you're referring to -- I'm aware of how a sling works.

 

[tap_sa]

<font color="yellow">I'm aware of how a sling works.</font><br /><br />A real sling ie. a small pouch with strings would the perfect comparison, there you don't have the doubt that the string going away along with the weight would somehow taint the experiment (which is similar to the poor astronaut flying away the station with broken off piece of the floor).<br /><br />OK, but let's try this. You said:<br /><br /><font color="yellow">"That's the velocity vector if the space station. If your velocity vector were tangental with the floor -- you'd be sliding along it -- not trying to go through it."</font><br /><br />You agree that a point in the station floor has a tangential velocity vector, pointing 'prograde' on the orbit that point is making, right? You are standing over that point, just being still, no jumping, you feel the acceleration pushing you towards the floor. You are not feeling sideway forces, you are not sliding anywhere even if you were wearing teflon slippers. You agree that this is how it would be if you were just standing, correct? Now comes $1000 question; remember the orbiting point on the floor surface, with tangentian velocity. If you are standing bare feet on that point, how come the cell in your feet hugging that point is not going anywhere? It keeps pushing against the rotating point in the floor. If two points, side by side so that they could be considered one point, in space follow the same path the same time, aren't their velocity vectors equal all the time?<br /><br />edit: Here's some more food for thought: Centripedal Acceleration.

 

[tap_sa]

From Wikipedia's article on acceleration:<br /><br /><i>"...Transverse acceleration (perpendicular to velocity) causes change in direction. If it is constant in magnitude and changing in direction with the velocity, we get a circular motion. ..."</i><br /><br />I did an illustration to show how the vectors of velocity and acceleration are.

 

[mrmorris]

<font color="yellow">"...Transverse acceleration (perpendicular to velocity)..."</font><br /><br />Ayup -- I recalled this on the drive home last night. I do lots of my best thinking in the car while I strive to avoid the maniacs driving the cars around me. It's been too long since my last physics class.<br /><br />On the positive side -- I was 100% correct in another area. I said that one of use wasn't thinking it all the way through. Eureka! I hit the nalil on the head on <b>that</b> one. Also -- while I hadn't worked out who it was -- I did narrow the non-thinking party down to two people. Given four billion-ish people in the world... that's pretty good. <img src="/images/icons/smile.gif" /> <br /><br />In addition -- we have established one of the cardinal rules of a rotating space station: 'Roller-blading anti-spinward is prohibited!'.<br /><br />I was wondering how increasing the radius of the space station would affect this issue (i.e. movement antispinward reduces the 'gravity'). As the radius gets larger, the 'period' of the revolution gets longer, but I didn't know if the edge velocity increased, decreased, or stayed constant. Working it out -- the larger the radius, the higher the edge velocity (despite the increase in rotational period), and the less pronounced the gravitation change of velocity against (or with) the spin of the station would be. Three stations which 'produce' ~.38 of Earth's gravity:<br /><br />38 meter radius<br />20 second period<br />12 m/s edge velocity<br /><br />100 meter radius<br />32.5 second period<br />19.33 m/s edge velocity<br /><br />200<br />46 second period<br />27.2 m/s edge velocity<br /><br />So -- one more reason for making the radius as large as possible... (albeit the day when *any* such rotating station is possible, whatever the size, will be a glorious one).

 

[tap_sa]

<font color="yellow">"albeit the day when *any* such rotating station is possible, whatever the size, will be a glorious one"</font><br /><br />Indeed ... I'd like to see real life Babylon 1 built before kicking the bucket but with current pace (or lack thereof) looks like gotta live to be Methuselah to achieve that.<br /><br />IMO it's probable that first rotating station will be mass-countermass instead of whole ring, despite the problem you summarized well. Maybe the system would be easier to manage if elastic tether is replaced by truss. A movable centerpiece for docking, it could be moved to the exact center of gravity.<br /><br />As you have pointed out, all is well as long as nobody moves, at least not from one end of the station to the other, or move cargo from and to the docked spacecraft. These activities would shift the cg and that would be dangerous if there's a massive craft docked. I wonder if the consumable liquids could be used as stabilazing ballast? For instances LOX tanks in both end sections with enough ullage to trim the cg. When person would move from one end to the other, same mass of LOX would be pumped vice versa. Too easy, what am I missing?

 

[mrmorris]

<font color="yellow">"When person would move from one end to the other, same mass of LOX would be pumped vice versa. Too easy, what am I missing"</font><br /><br />Well it's a time-based equation. The masses will have to shift in real-time. If you use water as ballast and (whole lot easier to pump water from place to place than LOX), and you do something like that, but the water shift is always a few minutes *behind* the movements of the shift of mass 'X' that it's responding to... it's liable to do more harm than good.<br /><br />However -- I could see a system (indeed I have thought of such for a bicycle-wheel station). Especially if you use a turbopump equivalent to what is used on the shuttle (i.e. able to shift swimming-pools worth of water in seconds). Have the entire BWS wired with accelerometers and a series of water tanks located around the rim of the BWS, plus a biggie in the center. A computer program detects acceleration changes due to mass shifts and pumps water in the apporpriate direction to counter. If the computer, software, and pumps are fast enough -- this can be essentially real-time.