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# Bolo "AntiGravity" Device

Extra-Atmospheric Lift ; Why Flying Saucers Spin
 (+1, -4) [vote for, against]

At any given orbital height, there is a certain horizontal speed required to avoid hitting the ground. If we tether 2 heavy globes together (for the time being assuming a long, magically strong tether) and spin them up far past orbital velocity then, at any given point in time, delta-V > orbital-velocity so they should "float" or even rise.

Works for donuts too of course.

 — FlyingToaster, Nov 07 2009

I wouldn't bet against him having anit-gravity... [Jinbish, Nov 08 2009]

I'm not convinced: the center of mass of the bolo would still fall to Earth. If the tethers were long enough the bolos might still be in space at this point.
 — sninctown, Nov 07 2009

Couldn't the floating endpoints pull up the center with them?
 — jutta, Nov 08 2009

I don't understand.
Are the bolos spinning vertically or horizontally?
What is dv >? Delta V?
 — 2 fries shy of a happy meal, Nov 08 2009

If you put the lid on a container fast enough you can trap light inside it.
 — normzone, Nov 08 2009

 — csea, Nov 08 2009

This is either a joke or a desperate plea for education.
 — MikeD, Nov 08 2009

I couldn't work my brain around it for some reason, and decided to share.
 — FlyingToaster, Nov 08 2009

 [marked-for-bad-science] seconded. It's wrong, but not //widely known to be wrong//. Seems to happen a lot around here.

 As I understand this the bolo is spinning in a horizontal plane. So, the endpoints/weights/globes are pulling horizontally in opposite directions, keeping tension on the tether.

To make it clearer, let's get rid of the spinning:
Imagine two rocket engines attached to opposite ends of a long pole, aimed at each other, so the smoke and flames go toward the middle of the pole, and the rocket engines are trying to stretch the pole. The thrust from the rocket engines is like centripetal force from spinning the bolo.
The rocket stick won't go anywhere-- the rocket thrusts cancel out. If the stick is <i/ really> long, the rockets engines and ends of the stick will be above the atmosphere, but the middle of the stick will still be resting on the ground.
 — sninctown, Nov 08 2009

yes but there is no point on your "rocket stick" contraption at escape velocity.
 — FlyingToaster, Nov 08 2009

 I seem to remember being taught, a long time ago, that the difference between speed and velocity was that velocity was in a particular direction. For example, if I'm driving north at 90kph, then I have a northward velocity of 90kph, and an eastward velocity of 0kph. If I then hang a rather dangerous left, keeping 90kph on the speedo, my northward velocity becomes 0kph and my eastward velocity becomes -90kph.

 Now, my point is this; if an object is moving in a circle, then its velocity in any given direction averages zero. Zero is less than escape velocity.

Does that clear things up at all?
 — pertinax, Nov 08 2009

 velocity at any given point in time is the direction, which is the tangent to the circle being scribed, and speed at which it is travelling... for some reason I keep expecting it to rise (if the speed > orbital velocity) while losing rotational energy to compensate, until equilibrium.

//object moving in a circle// you mean like an... orbit ?
 — FlyingToaster, Nov 08 2009

Well for me it does.
Other than the conservation of angular momentum thing. If the bolas were spinning fast enough, (much faster than escape velocity), then they would not remain in a horizontal plane with Earth because they would be unable to tilt as they orbited, no?
 — 2 fries shy of a happy meal, Nov 08 2009

Well, [21Quest], after seeing his dismissal of those inconsequential and incompetent henchmen, you're a braver man than I...
{You're not wrong though. I've not seen Bloodsport in *ages*}
 — Jinbish, Nov 08 2009

//there is no point on your "rocket stick" contraption at escape velocity.// Orbital velocity and escape velocity are different. Also, orbiting works when an object is traveling fast enough in a straight line that the Earth curves away underneath it as fast as the object can fall: if gravity turned off, an orbiting object would fly away from the planet. With the bolo, if gravity turned off, it would just hang there spinning, so gravity must cause it to fall. Are you saying that if you have a bolo out in deep space, spinning away, it will produce thrust? It would have to for this idea to work. Does this make sense to you?
 — sninctown, Nov 08 2009

 Woo! This is a fun idea to think about.

 So, suppose orbital velocity is 8km/s. If you fire a weight eastwards at 10km/s, it will rise relative to the surface of the earth (ie, it moves to a higher orbit, at which orbital velocity is 10km/s). Likewise, if you fire a weight westwards at 10km/s, it will rise. Equally true for north, south or any other direction.

 So, if you have two masses orbiting eachother, boloesquely, such that the tangential velocity of each is 10km/s, they too should rise.

 But no. Can't be. Why not?

 Because orbit involves not just the motion of the mass, but the curvature of the earth.

 Imagine you're standing on the north pole. You throw a mass eastwards (you know what I mean, don't quibble) at 8km/s, from a height of (say) 5m above the ground. It falls steadily, but by the time it has fallen 5m (one second later), it has travelled far enough eastwards that the ground has curved away from it by 5m, so it's no closer to the ground. This keeps happenning, and so it's in orbit. (If you threw it at 10km/s, then by the time it had fallen 5m, the ground would have curved away by more than 5m, and the object would have climbed relative to the earth's surface).

 Now, instead, imagine that the mass travels in a circle, boomeranguesquely. It starts travelling 10km/s eastwards, but within one second it has curved around and come back to where it started (ie, to the north pole). During this one second, the mass has fallen 5m, and the north pole is at the same altitude as it was before, so the mass simply hits the ground. No result.

 Likewise, if you throw the mass westwards and it again circles around, you get the same result: it comes back to the north pole by the time it has fallen the 5m, and it is in contact with the ground.

So, if you tie the two masses together with a string to make a bolo, you again get the same result. They just land on the ground. <<\tol><\antol,btotk>
 — MaxwellBuchanan, Nov 08 2009

 //you mean like an... orbit ?//

Oh. Excuse me while I wipe this egg off my face. ;-)
 — pertinax, Nov 08 2009

 //assuming you can use gravity to vary the spinning axis slightly.//

Do wha? Wouldn't it be simpler to assume that we can just turn off gravity? What do you mean?
 — MaxwellBuchanan, Nov 08 2009

Oh god. You're not Eric Laithwaite are you?
 — MaxwellBuchanan, Nov 08 2009

 I wouldn't put it past him.

 Basically, you're trying to use a gyroscope to achieve a reactionless drive. (It's not "perpetual motion", as long as you're prepared to lose kinetic energy stored in the flywheels.)

 People (including me) get very confused about gyroscopes because their motion seems counterintuitive but, as even Laithwaite accepted in the end, they obey Newton's laws of motion.

 So, you won't find it easy to convince people (like me) on paper, because the maths gets complex (literally) and because many more competent mathemeticians claim to have proven that it won't work. (And I believe many people, including Laithwaite, tried to demonstrate in practice and failed.)

 If you really think you've found something that nobody else could, then the best option is to build a demonstrator. It needn't be huge and needn't even lift itself - it just needs to demonstrate a measurable amount of lift.

Good luck!
 — MaxwellBuchanan, Nov 08 2009

 or you could accept the generally verified model for gravity:

 Spinning objects are no different from objects at rest re. gravity.

 All velocity is relative.

Inertial energy is simply a differential of relative rotational rate. It does not matter if the bola is spinning or holding still (relative to what absolute reference!) the influence of gravity is relative to distance.
 — WcW, Nov 08 2009

 //or you could accept the generally verified model //

 Exactly.

On the other hand....gyroscopes are fascinating. Although one can always bypass everything (correctly!) by cutting to the chase and saying "conservation of energy" or "Newton's laws" etc, it *is* fun to think about these things and try to actually understand what is going on. Hence, I am giving a resounding [ ] to this idea.
 — MaxwellBuchanan, Nov 08 2009

 I'm not certain, but pretty sure this won't work:

 In order for an object to achieve orbit, it has to have an orbital velocity tangential to the center of mass. As a result, any orbit will pass above a point above the center of mass and above a point on a line on the opposite side of the mass (distance can be variable, as in an elliptical orbit, and this ignores the N-body problem, but is basically true)

 These bolos have their primary instantaneous velocity in a line that is not tangential to the mass unless the tether passes through the center of the mass.

 As such, rather than the mass falling inwards and it's velocity keeping it from hitting the mass, it is falling downwards, and what keeps it from hitting the earth is the upwards component of the acceleration imparted by the tether (centripetal). While it is possible to get it fairly close to horizontal, it is not possible to get it truly horizontal (for the same reason it is impossible to get a rope perfectly straight by pulling on the ends). At 45 degrees, the bolo experiences 1g in the downward direction and 1 g inward, but a person in the bolo would experience sqrt(2) g at 45 degrees (sqrt(1^2+1^2)). As you get closer to horizontal, the inward force increases geometrically. Since the forces are never in the same plane, they never cancel out. Thus you never get microgavity.

(This assumes the tether was anchored at the center. If it isn't, then the entire collection has no upward velocity, and just hits ground fairly quickly)
 — MechE, Nov 08 2009

 Can anyone who actually knows physics please answer for me why a spinning body in the shape of a sphere, containing other spheres which each spin in a different plane, would not resist being tilted in any direction to the point where gravity is no longer strong enough to keep this sphere from trying to remain still as the rest of the universe zips on by? It wouldn't be anti-gravity, it would be...pro-inertia.

I know it must be a stupid question but I honestly don't understand why it wouldn't work and my subconscious keeps working at it like a loose tooth, so if anyone is able to explain it to me in simple enough terms then maybe I can convince it to just bloody let it go already.
 — 2 fries shy of a happy meal, Nov 08 2009

 [2fries] you could move it, you just couldn't tilt it. I imagine your contraption inside a sphere that doesnt spin.

For fun you could note that if you put it on a flat table it will slo-o-o-wly roll off as it stays in the same orientation while the Earth spins (the basis for my "Egocentric Clock")
 — FlyingToaster, Nov 08 2009

 //note that if you put it on a flat table it will slo-o-o-wly roll off //

Ah, but see that's the point I can't seem to let go of though.
If this one device spins in multiple directions at once quickly enough it will resist any tilting. Not just as the Earth turns, but as the milky way turns, and whatever the milky way revolves around turns.
 — 2 fries shy of a happy meal, Nov 08 2009

yahbut angular velocity is *very* little. The Earth takes a year to revolve around the Sun... imagine how long it takes the Sun to revolve around the centre of the Milky Way.
 — FlyingToaster, Nov 08 2009

 [2fries] has it right.

 A gyroscope (ie, anything that spins) doesn't resist *translation* (ie, moving left, right, up, down, back, forth). It only resists *rotation* (twisting) in some directions.

 Why does it resist twisting? Well, here is a half of an answer.

 Imagine a bicycle wheel, with its axis vertical (ie, the wheel is on its side). Ignore the mass of the spokes. Imagine that the rim (the only mass involved) is made up of a great many small segments, each with some mass (say, 10grams).

 Suppose the wheel is spinning very fast indeed - say 100m/s (ie, that's the speed of any given part of the rim). Now just focus your attention on one segment of the wheel (paint it red) as it goes around. Look at the wheel from the side.

 Suppose that you look at the wheel just as the red segment is passing closest to you, travelling from left to right at 100m/s. Over a short instant of time, it is moving in roughly a straight line (ie, we can ignore the whole curve business).

 Now suppose that, at that instant, you try to tilt the axle. Let's imagine that the bottom of the axle is resting on the ground (held in some little indentation), and you try to push the top of the axle leftward, just at the moment the "red" segment is passing in front of you.

 Before you tilt the axle, the red segment is moving horizontally at 100m/s. If you could tilt the axle by 20 degrees, the wheel would tilt and the red segment would now be moving diagonally upwards instead, still at 100m/s.

 So, to tilt the axle, you have to provide enough force to change the direction of a 10gram mass going at 100m/s, from horizontal to a 20 degree upward angle. Once you've tilted it, that 10gram mass is going to be moving upwards at about 34m/s (and horizontally at about 94m/s).

 In other words, by tilting the axle 20 degrees, you are trying to give that 10 gram mass an upwards velocity of 34m/s - and that takes a considerable force.

 The same is true for other segments of the wheel, of course. All those toward the "front" of the wheel have to be deflected from moving left-to-right to moving right- and-up (just like the red segment); whilst all those at the "back" of the wheel have to be deflected from moving right-to-left to moving left-and- down (requiring, again, a considerable force to give them 34m/s of downwardness).

 In other words, when you tilt the axle of the wheel, you are trying to make lots of wheel-rim-segments, each of which is moving very fast, suddenly change their direction of movement. This needs a very large force - roughly (not quite) as if a mass equal to the whole wheel (say, a block of lead of the same mass) were hurtling toward you at 100m/s and you tried to deflect it sideways by 20 degrees.

 OK, so why doesn't the wheel resist *translation* (movement without rotation)?

 Suppose now that the wheel is still spinning horizontally, and you try to lift it up (by grabbing the top end of the verticle axle and lifting) at, say, 1m/s.

 Now imagine your red segment as it goes past horizontally. If you want to lift the wheel, you need to apply enough force to give that segment (and all the others, of course) an upward velocity of 1m/s, but the fact that they are moving horizontally as well makes no difference. You might as well be lifting a stationary wheel - no "extra" force is needed simply because the wheel is spinning.

 Final ultra brief summary: in *tilting* a flywheel, you're trying to instaneously change the *direction* of a large number of very rapidly moving masses. In *moving* a flywheel, you're nudging all the masses one way without changing their direction of spinny movement.

Hope that makes sense, and apologies to true gyroscopists for leaving out all the complexity and pis. Eric Laithwaite would be spinning in his grave (yet, he would not rise).
 — MaxwellBuchanan, Nov 08 2009

 Ferzackerly so.

 However, I was trying (by way of a minor digression) to explain why a flywheel is easy to move, difficult to twist. (I mentioned the bullet in this context because, at the velocities of real bullets, the earth is effectlvely flat; I hadn't meant to get into orbital mechanics there, though you're absolutely right about the earth-falling-away- from-a-fast- enough- bullet bit.

[EDIT: I changed the original long annotation to remove a confusing and out-of-context analogy]
 — MaxwellBuchanan, Nov 08 2009

)
 — MaxwellBuchanan, Nov 08 2009

[EDIT: I removed my annotation where I asked about the confusing and out-of-context analogy]
 — jutta, Nov 08 2009

the spinning object smply resists the change in velocity in the same way that an object with linear motion tends to resist changes in the direction of that linear motion. In this case the "resistance" to changes in direction comes in the form of a large amount of force being required to change the direction and a resultant loss of speed.
 — WcW, Nov 08 2009

If the magic cable in between these spinning masses is sufficiently long and they are spinning sufficiently fast, then the near-planar masses would defy gravity at an arbitrarily high altitude, given the curvature of the earth.
 — RayfordSteele, Nov 10 2009

 /spinning vertically/[2 fries shy of a happy meal]

In pitching ball sports it is possible to generate a riser (magnus effect). Would a vertical bolas rotation in a forward motion of travel have a climbing assist? The bolas rotational circumference would have a periodically changing front surface area.
 — wjt, Nov 10 2009

 // the near-planar masses would defy gravity at an arbitrarily high altitude, given the curvature of the earth.//

 Actually, no, they wouldn't.

 The only way to make this work is if the two masses are on opposite sides of the earth, which will make it inconvenient to tie them together with a rope.

 And, even if we cut the earth in half and put some sort of yoyo-axle through the middle to make this possible, you would find that the tangential velocity needed to keep the masses aloft would be exactly the same as their normal orbital velocity.

This idea is just good old fashioned wrong.
 — MaxwellBuchanan, Nov 10 2009

Ok so what the poster needs to do is re-post this idea adding a complicated but pointless twist (CVT, air bearings, Deep sea RO, LTA craft, etc.) and a new title. Be creative "Orbital Bolas II: Gravity Vengeance" or "Son of Bola Vs Newton"
 — WcW, Nov 10 2009

 Picture this: you have a tallish tower at the North Pole. At the top of this tallish tower is a magic-mega-motor which is spinning two equally-weighted balls at the opposite ends of some magically-strong length of rope that is, say, half the diameter of the earth.

 If the balls spin fast enough, then their inertia largely overcomes the force of gravity, approximating level at infinitely fast, so that if you were to measure their distance from the earth at any instantaneous point, it would be taller than the tower, due to the simple hypotenuse.

That is what I meant.
 — RayfordSteele, Mar 30 2010

At this point in time I'd like to point out that this thought-provoking post ;D is much better than the recent spate of "Who needs a runway? . . ." cheap knockoff echos.
 — FlyingToaster, Mar 08 2013

Who needs originality? Every idea can be derivative crap.
 — ytk, Mar 08 2013

 [annotate]

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