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Boosted Orbital Tether and Orbital Runway upgrades


MatterBeam

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Hi @Snark

Genuine question about converting angular to linear momentum. I'm probably missing something but I can't quite see what.

Thought experiment. I have a flywheel with a paddle attached to the rim. I place a ball bearing at rest in a position where it will be hit by the paddle. Flywheel rotates, paddle comes around and hits the ball bearing, imparting linear momentum to it. 

Clearly there's been a net gain in linear momentum and intuition tells me that the impulse exerted by the ball on the paddle (Newton 3) slows it down slightly, thus reducing it's angular momentum. Thus angular momentum has been converted to linear momentum.

On a related thought, what about a water wheel. Linear momentum of water flowing through the mill race is converted to angular momentum of the spinning wheel - which is kind of the point of a water wheel?

What am I missing here? Or am I taking your comment out of context?

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26 minutes ago, KSK said:

Hi @Snark

Genuine question about converting angular to linear momentum. I'm probably missing something but I can't quite see what.

Thought experiment. I have a flywheel with a paddle attached to the rim. I place a ball bearing at rest in a position where it will be hit by the paddle. Flywheel rotates, paddle comes around and hits the ball bearing, imparting linear momentum to it. 

Clearly there's been a net gain in linear momentum and intuition tells me that the impulse exerted by the ball on the paddle (Newton 3) slows it down slightly, thus reducing it's angular momentum. Thus angular momentum has been converted to linear momentum.

On a related thought, what about a water wheel. Linear momentum of water flowing through the mill race is converted to angular momentum of the spinning wheel - which is kind of the point of a water wheel?

What am I missing here? Or am I taking your comment out of context?

Conversion between the two is impossible.

In each case above the linear and angular momentum of the system is conserved. For instance, in your flywheel paddle example, angular momentum of the system as a whole is conserved as the loss of angular momentum in the flywheel is balanced by the angular momentum of the ball bearing (because particles moving in a straight line still have angular momentum with respect to a fixed point). Linear momentum is conserved because in the collision, the impulse that slows the rotation in the flywheel also causes it to move in the opposite direction to the ball (in the case that the flywheel is just spinning in free space with no mounting system, otherwise the momentum is transferred through the mount to whatever the wheel is mounted to).

Edited by Steel
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17 minutes ago, Steel said:

because particles moving in a straight line still have angular momentum with respect to a fixed point

OK, that's what I was missing. Not sure I really grok it (as in fit it into my mental picture of what's going on) but it makes everything that @Snark was saying make sense. Thanks.

 

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3 hours ago, KSK said:

Hi @Snark

Genuine question about converting angular to linear momentum. I'm probably missing something but I can't quite see what.

Thought experiment. I have a flywheel with a paddle attached to the rim. I place a ball bearing at rest in a position where it will be hit by the paddle. Flywheel rotates, paddle comes around and hits the ball bearing, imparting linear momentum to it. 

Clearly there's been a net gain in linear momentum and intuition tells me that the impulse exerted by the ball on the paddle (Newton 3) slows it down slightly, thus reducing it's angular momentum. Thus angular momentum has been converted to linear momentum.

On a related thought, what about a water wheel. Linear momentum of water flowing through the mill race is converted to angular momentum of the spinning wheel - which is kind of the point of a water wheel?

What am I missing here? Or am I taking your comment out of context?

Its an difference between momentum and energy, the waterwheel takes energy from the water, the waterwheel angular momentum is constant if it run by constant speed. 
On earth you can easy gain or get rid of angular momentum by interacting with earth. This don't work in space, you can gain or loose angular momentum with rocket engines, flywheels only store it. 

Now you could store energy by spinning up an flywheels with an tether connected to it. slower ship grab tether, and the tether wind out and flywheel spins up storing lots of the energy from the different speeds. 
You can then bleed of the reaction wheel over time, it would be very smart to use two counter rotating ones. 
 

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4 hours ago, KSK said:

Thought experiment. I have a flywheel with a paddle attached to the rim. I place a ball bearing at rest in a position where it will be hit by the paddle. Flywheel rotates, paddle comes around and hits the ball bearing, imparting linear momentum to it. 

Clearly there's been a net gain in linear momentum and intuition tells me that the impulse exerted by the ball on the paddle (Newton 3) slows it down slightly, thus reducing it's angular momentum. Thus angular momentum has been converted to linear momentum.

Excellent explanation by @Steel above.  :)  No gain in linear momentum, because the wheel goes one way and the ball goes the other:  when the wheel whacks the ball, it moves the wheel, too.  No gain in angular momentum, either, because in the frame of reference of wheel + ball, the wheel has transferred some of its angular momentum to the ball.

Here's a similar example that may also help to illustrate.  You have a spinning wheel that has a dozen heavy weights attached to the rim, spaced 30 degrees apart.  At some point it "lets go" of one of the weights, which goes whizzing away.  Where did the linear momentum come from?  Answer:  From the wheel.  When it lets go, it causes the wheel to go in the other direction.

Be very, very careful any time you find yourself thinking of "converting" between linear and angular momentum; they really are different, they actually have different units on them.  "Converting" between the two would be the same as converting meters to kilograms.  It's a meaningless statement.

It's not the same as when you talk about, for example, converting between kinetic and potential energy, because energy is energy and it's the same stuff, with the same units, just in different forms.  Momentum's not like that-- it's not the same stuff in two different forms, it's two different things altogether.

5 hours ago, KSK said:

On a related thought, what about a water wheel. Linear momentum of water flowing through the mill race is converted to angular momentum of the spinning wheel - which is kind of the point of a water wheel?

The missing ingredient there is that the water wheel is rigidly mounted on an axle which is rigidly attached to the Earth.  So the net linear momentum (of water + wheel + Earth) stays constant, and the net angular momentum (of water + wheel + Earth) stays constant, too.

 

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19 hours ago, Steel said:

Conversion between the two is impossible.

I'd assume you would have a second (possibly more) flywheel to counterbalance the angular momentum.  And possibly have your momentum-killing thrusters on a long pole to provide more torque.

My real issue is the idea of adding thousands of m/s of delta-v via a tether (I can't see it making sense otherwise).  For each second it takes to reel in your spacecraft, you need to supply 1km of tether per 1000m/s difference between your spacecraft and the "orbital runway".  It looks like you've built a space elevator with just a fraction of the performance, with probably higher requirements on the tether than the elevator.

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On 05/02/2017 at 9:09 AM, natsirt721 said:

Man, I thought the death of the Dean Drive in the 50's really nailed this point down but boy was I wrong.

I was under the impression that a flywheel can apply torque through friction which converts its rotation into linear motion of the tether.

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4 hours ago, MatterBeam said:

I was under the impression that a flywheel can apply torque through friction which converts its rotation into linear motion of the tether.

It does, but it does so while conserving both linear and angular momentum. You can do whatever you like between rotational energy and linear kinetic energy as long as the two momenta are conserved.

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Linear Momentum or 'momentum' to layfolk is the product of mass (kg) and velocity (m/s).  Units are kg*m/s.  By nature, linear momentum of a system is conserved.

Angular Momentum is the product of a rotating body's moment of inertia (kg*m2) and its angular velocity (rad/s).  Units are kg*m2/s.  Angular momentum of a system is also conserved.  It is important to note that angular momentum is measured with respect to an axis, and for simplicity we take that axis to be that of the flywheel.

From a strictly dimensional analysis standpoint, we can see that these two terms do not describe comparable quantities, but that glaring error aside, let us consider the simplest system possible, that is a 100,000 kg flywheel-tether apparatus and and 10,000 kg spaceship braking on the tether.  This scenario also assumes that at no point propellant is expended to change the velocity of either the flywheel or the spacecraft.  The flywheel is moving east at 8,000 m/s and the ship east at 7,000 m/s.  

The linear momentum of the ship is 10,000 kg * 7,000 m/s = 70,000,000 kg*m/s.  Likewise, the flywheel has linear momentum of 100,000 kg * 8,000 m/s = 800,000,000 kg*m/s.  The total momentum of the system is 870,000,000 kg*m/s.

The angular momentum of the flywheel requires the moment of inertia, which in this case is m*r2/2 for a cylinder rotating about its axis.  For a 5 m cylinder at 100 rad/s the angular momentum is 125,000,000 kg*m2/s.  The angular momentum of the ship is a little different, but it boils down to the linear momentum of the ship times the tangent distance to the axis of rotation.  This distance is going to be the radius of the tether spool, which for simplicity we will assume to be the same as the flywheel.  For a ship moving (with respect to the flywheel, our reference axis) at 1,000 m/s west, the angular momentum is -1,000 m/s * 10,000 kg * 5 m = -50,000,000 kg*m2/s.  The total angular momentum of the system is 125,000,000 - 50,000,000 = 75,000,000 kg*m2/s.  

Both of these quantities must remain constant, because the laws of physics say so.  No matter what you do with changes in geometry, friction, or any other internal forces, both linear and angular momenta will stay the same for the system.  Now, lets examine another state involving the same system.  We want to accelerate the ship by 1,000 m/s to match the velocity of the flywheel , so we trade angular momentum of the flywheel for angular momentum of the ship.  The final angular momentum of the ship = m * v * r = 0, because then the ship and the flywheel will have the same velocity and therefore velocity relative to the axis is 0.  So the ship gains 50,000,000 kg*m2/s in angular momentum (which is transferred to the flywheels via the tether), and the flywheel loses 50,000,000 kg*m2/s, bringing the new totals to 0 kg*m2/s for the ship and  75,000,000 kg*m2/s for the flywheel.  Total is still 75,000,000 kg*m2/s - check; however this doesn't work out.

If the ship and flywheel are now moving at 8,000 m/s, then the linear momentum of the flywheel is still 800,000,000 kg*m/s, but the linear momentum of the ship is now 10,000 kg * 8,000 m/s = 80,000,000 kg*m/s.  The total linear momentum of the system is 880,000,000 kg*m*s, or 10,000,000 higher than it was.  You've created linear momentum seeming out of nowhere, even though for a closed system linear momentum is constant.  This is impossible.  What actually occurs is, during the angular momentum exchange the flywheel will decelerate by 100 m/s in order to keep linear momentum constant.

If we try this again but include the linear momentum exchange, we see that the final relative velocity of the flywheel and ship is now +100 m/s, because the ship is going 8,000 m/s and the flywheel 7,900 m/s.  The final angular momentum of the ship is 5,000,000 kg*m2/s, a net gain of 55,000,000 kg*m2/s.  As before, the flywheel takes this up, bringing the new totals to 5,000,000 kg*m2/s for the ship and 70,000,000 for the flywheel, a total of 75,000,000 kg*m2/s - check.

The linear momentum of the ship is 80,000,000 kg*m/s as calculated before, and the linear momentum of the flywheel is 100,000 kg *  7,900 m/s = 790,000,000 kg*m/s.  Total linear momentum is 870,000,000 kg*m/s - double-check. Both linear and angular momenta are conserved.

As you can see, even using rotational energy from a flywheel to increase the kinetic energy (and therefore linear momentum) of the spacecraft still results in a deceleration of the flywheel itself, because the total linear momentum of the system is independent of of the angular momentum of the flywheel, but dependent on the linear momentum of the spacecraft.  If the spacecraft accelerates, the station has to decelerate, and there is no way around that.

 

 

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I actually did my Masters Dissertation on Momentum Exchange Tethers. More specifically, the design of a capture system for them.

The tether system itself can be either boosted by interacting with the earth's magnetic field, without use of propellant (this is slow, but very doable with solar or nuclear power). The other way the tether system can be boosted is by using it not only to catch payloads from earth, but also return payloads from the moon. The proposed use for my system was to supply a lunar colony. The lunar colony would use a railgun or coilgun to fire payloads on an earth intercept. Slowing them down would speed up the tether system, which would then drop them into a descent trajectory once it had captured enough momentum. It would then have enough momentum to catch a suborbital payload from earth of similar mass without itself reentering. A combination of capturing returning payloads from the moon and electrodynamic tether propulsion would allow the station to theoretically operate without additional reaction mass.

The system we came up with for capture was essentially a giant net. It would be held "open" by centrifugal force, by spinning the net attachment point about the long axis of the tether. We could get about 3km/s of delta-V this way without subjecting the payload to excessive G forces, and using existing materials (Kevlar for most things)

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12 hours ago, peadar1987 said:

I actually did my Masters Dissertation on Momentum Exchange Tethers. More specifically, the design of a capture system for them.

The tether system itself can be either boosted by interacting with the earth's magnetic field, without use of propellant (this is slow, but very doable with solar or nuclear power). The other way the tether system can be boosted is by using it not only to catch payloads from earth, but also return payloads from the moon. The proposed use for my system was to supply a lunar colony. The lunar colony would use a railgun or coilgun to fire payloads on an earth intercept. Slowing them down would speed up the tether system, which would then drop them into a descent trajectory once it had captured enough momentum. It would then have enough momentum to catch a suborbital payload from earth of similar mass without itself reentering. A combination of capturing returning payloads from the moon and electrodynamic tether propulsion would allow the station to theoretically operate without additional reaction mass.

The system we came up with for capture was essentially a giant net. It would be held "open" by centrifugal force, by spinning the net attachment point about the long axis of the tether. We could get about 3km/s of delta-V this way without subjecting the payload to excessive G forces, and using existing materials (Kevlar for most things)

3km/s sounds close to the specific velocity of a tapered tether. It is about 2.25km/s in an untapered section of Zylon.

Could I ask a question related to the concept of braking a payload into orbit? More specifically, the part where the payload connects to the tether to start breaking.

Here is a long discussion of the various designs I came up with to solve the problem of docking a 7.34km/s object to a 0m/s object: http://bbs.stardestroyer.net/viewtopic.php?f=5&t=166030. I settled on the concept of a hook being braked by a series of airbags, slowing down enough to be bolted through and form a connection between the main tether and a second tether from the payload. The continuous connection between the tethers allows the regular braking process to begin.

Some images:

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The railgun was deemed as un-necessary, as the heatshield is lightweight enough to be pre-attached to the hook. 

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Sorry, that's a typo, should read 2km/s of Delta-V.

The simplest answer is that you spin the tether up before you catch the payload. That way the relative velocity between the hub and the payload might be 2km/s, but the relative velocity between the tip and the payload is close to zero at the intercept point. Once you've captured at low relative velocity (we allowed for up to 100m/s due to navigational inaccuracy), your acceleration relative to the hub is done over an entire quarter revolution of the tether, reducing forces.

I can't think offhand of how you'd be able to capture something moving at 8km/s relative velocity without destroying either it or the station. Airbags are one possibility, but that requires mass to be brought up to the facility every time you capture something, and also will pose the problem of creating debris. I haven't run the numbers, but I'd say you would need a lot of airbags as well, just judging by the distance a reentering capsule travels through the atmosphere before it is slowed down. Again without numbers, I would say that you would need a mass of air much greater than the mass of the payload, as all you are doing is transferring momentum from the payload to the air in the bag, and you are not going to transfer anything close to 100% as the air escapes.

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1 hour ago, peadar1987 said:

Sorry, that's a typo, should read 2km/s of Delta-V.

The simplest answer is that you spin the tether up before you catch the payload. That way the relative velocity between the hub and the payload might be 2km/s, but the relative velocity between the tip and the payload is close to zero at the intercept point. Once you've captured at low relative velocity (we allowed for up to 100m/s due to navigational inaccuracy), your acceleration relative to the hub is done over an entire quarter revolution of the tether, reducing forces.

I can't think offhand of how you'd be able to capture something moving at 8km/s relative velocity without destroying either it or the station. Airbags are one possibility, but that requires mass to be brought up to the facility every time you capture something, and also will pose the problem of creating debris. I haven't run the numbers, but I'd say you would need a lot of airbags as well, just judging by the distance a reentering capsule travels through the atmosphere before it is slowed down. Again without numbers, I would say that you would need a mass of air much greater than the mass of the payload, as all you are doing is transferring momentum from the payload to the air in the bag, and you are not going to transfer anything close to 100% as the air escapes.

In the thread I linked, I did propose a spinning capture system, but the window between 0 and 100m/s relative velocity as the tether rotates away from tangent is as small as 6 milliseconds. A 100km radius tether improves this to about a tenth of a second, but either way, it is a very impractical solution.

Airbags for the whole payload might be a solution. Instead of sitting on a station, there's a better use for them. Bring up 1kg of gas into orbit, then attach it to very efficient electric rocket to put it into an elliptic retrograde orbit, with the apoapsis at Lunar altitude and the periapsis at 200km. Encounter velocity at the periapsis can be as high as 12km/s.

Have the gas slam into a heatshield to push a payload from a parabolic trajectory to orbital velocity. The 1kg of gas is worth 3.5kg of propellant in this scenario. If we use an inverted funnel instead of a heatshield, the gas will converge, compress and rebound with most of its velocity, becoming an effective 2450Isp rocket engine.

A 10 ton payload starting at 0m/s and ending at 7.8km/s has a virtual rocket engine of 2450 to 856Isp. It would need 6.2 tons of gas to reach orbit, instead of 28 tons of rocket fuel. 

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5 hours ago, MatterBeam said:

In the thread I linked, I did propose a spinning capture system, but the window between 0 and 100m/s relative velocity as the tether rotates away from tangent is as small as 6 milliseconds. A 100km radius tether improves this to about a tenth of a second, but either way, it is a very impractical solution.

Airbags for the whole payload might be a solution. Instead of sitting on a station, there's a better use for them. Bring up 1kg of gas into orbit, then attach it to very efficient electric rocket to put it into an elliptic retrograde orbit, with the apoapsis at Lunar altitude and the periapsis at 200km. Encounter velocity at the periapsis can be as high as 12km/s.

Have the gas slam into a heatshield to push a payload from a parabolic trajectory to orbital velocity. The 1kg of gas is worth 3.5kg of propellant in this scenario. If we use an inverted funnel instead of a heatshield, the gas will converge, compress and rebound with most of its velocity, becoming an effective 2450Isp rocket engine.

A 10 ton payload starting at 0m/s and ending at 7.8km/s has a virtual rocket engine of 2450 to 856Isp. It would need 6.2 tons of gas to reach orbit, instead of 28 tons of rocket fuel. 

Are you sure? My data said that with a 2km/s relative velocity and a 100km tether you would have a window of about 3.5 seconds when the relative velocity of the payload relative to the tip was below 100m/s. Not ideal, but not insurmountable.

I'm having trouble visualising what you're proposing with your retrograde airbags solution. Surely if you're picking up a payload from a suborbital trajectory you want it to be hit by something moving prograde? Are you suggesting essentially using an orbital version of the Pelton wheel, turning the flow of gas back in the direction it came from?

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17 hours ago, peadar1987 said:

Are you sure? My data said that with a 2km/s relative velocity and a 100km tether you would have a window of about 3.5 seconds when the relative velocity of the payload relative to the tip was below 100m/s. Not ideal, but not insurmountable.

I'm having trouble visualising what you're proposing with your retrograde airbags solution. Surely if you're picking up a payload from a suborbital trajectory you want it to be hit by something moving prograde? Are you suggesting essentially using an orbital version of the Pelton wheel, turning the flow of gas back in the direction it came from?

The design I calculated tried to go for zero relative velocity at the lower tip and 14km/s at the upper tip, leading to an extremely short window for rendezvous....
...although I can see I design where a series of 2km/s wheels throw the payload to each other for a final velocity of 8km/s.

Sorry, the retrograde is a mistake. Lunar return velocity is 12km/s in the prograde direction.

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23 hours ago, FleshJeb said:

Who originated that idea? I'm only familiar with it from David Brin, who probably pulled from Poul Anderson.

Tank Farm Dynamo was published in 1983. The earliest reference I found in a very brief search of scientific literature came out in 1978, and was published by one P.R. Williamson.

On 10/02/2017 at 1:31 PM, MatterBeam said:

The design I calculated tried to go for zero relative velocity at the lower tip and 14km/s at the upper tip, leading to an extremely short window for rendezvous....
...although I can see I design where a series of 2km/s wheels throw the payload to each other for a final velocity of 8km/s.

Sorry, the retrograde is a mistake. Lunar return velocity is 12km/s in the prograde direction.

Well you don't absolutely need zero relative velocity or a cascade of tethers, 2km/s is still not to be sniffed at in terms of saving fuel.

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