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7 May 2024 04:32
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Welcome to the gyroscope forum. If you have a question about gyroscopes in general,
want to know how they work, or what they can be used for then you can leave your question here for others to answer.
You may also be able to help others by answering some of the questions on the site.
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Question |
| Asked by: |
Dave Brown |
| Subject: |
I'll stop polluting your threads but... |
| Question: |
Please read here from time to time.
You can move any posts of mine you don't like into this thread.
Lets forget gravity for now, and let it help us use friction for forward movement.
Have you tried you projects on their side on a wheeled buggy?
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| Date: |
15 August 2005
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Answers (Ordered by Date)
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| Answer: |
dave brown - 15/08/2005 20:58:55
| | | scratch that.
do the solid bar test with 2 or 4 gyros.
spin it's support slowly on a center post so you can put a cap and keep it from going too high.
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dave brown - 15/08/2005 21:03:50
| | | The longer the bar, the more the leverage, the less force needed to hold it up, the less energy is needed to get it up.
put the equipment needed to be lifted in the spinning rings of the gyros, that is where you must put the weight. The lifting force at the support is lessened by the length of the bars. Smaller thrust. (negative thrust? lets find out)
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dave brown - 15/08/2005 21:19:36
| | | You may not see it weigh less, but you will see it go up without it weighing more.
It is not antigravity, it is anti-acceleration.
instead of accelerating to the ground, it is precessing.
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dave brown - 15/08/2005 22:02:31
| | | I stated that wrong, and am working on the next post.
Actually, it is anti change.
flip it on it's side and put it into action on a moving object and it will keep it moving.
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Dave Brown - 15/08/2005 22:47:14
| | | Maybe not, no tests done.
Now, they say a gyro has no torque.
It doesn't show it maybe, but don't stick your fingers in the spokes!
That's where torque is tranduced in to precession.
ft/lbs of torque... The lbs are transferred to the support,
the ft are transferred into precission.
Do the bicycle wheel in the hands thing.
You have to turn to keep the axis level, but now
turn faster, and push down on the un-supported end as hard as you need to keep the axis straight. Gets pretty hairy.
- i kept the wheel like so | with my body and did it.
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dave brown - 15/08/2005 23:00:44
| | | oh, the slower the gyro spins,
the faster you have to turn to keep up with the down force.
spin it fast enough, and maybe you won't have to turn?
- must be related to wheel diameter too.
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Dave Brown - 16/08/2005 11:35:07
| | | Thanks for you post/reply Sandy.
I wish I still had age youth, but my youthfullness in gyros is reality. :D
What I feel there is to work on:
1) Forward movement
We need gravity and it's effects to push against, a gyro is just too darn self centered, pun intended.
This requires a go-cart, with a peddle-spun gyro in the same orientation as the front wheel of a bicycle to a cyclist.
- The drive sprocket would have to be on a round coupling to allow pivot in relation to the axis. A tensioning gear would also be needed.
- Each end of the axis would be levered to their own go-cart wheel, offset from center, so that when it forces, it is in the right direction for forward movement, and so that the up-down action is not relayed back to the gyro.
- To achieve gyro propulsion, a downward force would have to be applied downward, first on one side, then the other.
-- Here is where you would prefer longer shafts and larger downward force as the gyro's mass and spin increase.
*** Notes: Hills would cause no problem, but turning would be hectic.
2) This thing stores energy.
See lect. #4 @ 4:55. He helps precission and it stores his energy by lifting the weight.
Now imagine not a weight, but a spring, or many springs one after the other.
- Now do that with multi gimbles, needing only to aid the outer most ring.
-- The strength of the spring combos are again gyro weight and rotation speed dependant. There would be a necessity for locking in and/or out of springs at different intervals, I'm guessing.
Once the system is sprung and locked, the gyro could be allowed to spin down.
- When the output is required, just spin-up the gyro first.
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Dave Brown - 16/08/2005 11:42:58
| | | addition to my last post.
2) The gimbles may need locks too.
I am picturing this at a water fall, although I know electric energy is efficient, there is some waste.
The water could be ported to spin the gyro through the supporting arms, as well as having centrifugal port closing; as well as the driving force for storing it's energy.
Then figure a way to connect it to a lightning bolt.
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dave brown - 16/08/2005 12:12:19
| | | 3) torque to linear motion
Put a gyro in a barrel shaped housing with it's axis through the center of the top and bottom of the barrel, and locked to prevent lateral movement within the barrel. Put an aerodynamic point at one end of the barrel.
Slide the barrel into a tube, that may or may not be able to pivot vertically, and rotate it briskly 180 degrees; may need a shorter barrel if it's still in the tube.
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dave brown - 16/08/2005 12:21:20
| | | addition
3) If a precessing gyro is used to spin the tube, and the original gyro does fly out the end, what we have done is redirect gravity's force.
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dave brown - 16/08/2005 12:25:05
| | | addition
3) if we can then design a path to use gravity to return the flung gyro to it's original position in the barrel, to be fired in the same direction, we get or propulsion.
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dave brown - 16/08/2005 12:30:46
| | | addition 3
3) scale this down to aligned-spin particles in a magnetic flinging tube, and we don't need to return anything.
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dave brown - 16/08/2005 12:52:57
| | | Something I don't think has been worded as such:
If we try to give a gyro centrifugal force, it will convert that too.
Sounding like the ulitmate force tranducer:
- Now we just have to get a path from here to there, like that game of pipes, we have to redirect and redirect.
- My only concern with changing or maintaining the gyro's rotational speed, is if the force required to do so changes. Other than bearing friction, unless it becomes really large.
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dave brown - 16/08/2005 13:56:48
| | | 4) Low torque, ok
I need links for static generators.
- I need to know the force required to turn one
-- it's voltage produced
-- it's amperage produced
The voltage will be dropped to run a motor which will spin the gyro, precess the gyro, and hopefully supply a next-larger gyro. (Requires a dc-ac converter. Transformers are also efficient.)
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dave brown - 16/08/2005 17:34:56
| | | Thanks again for post. :)
I'm getting rods welded to setup my first little project. :)
Sandy, feel free to email me directly, I would like that... you may not! question question question lol
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dave brown - 16/08/2005 21:42:13
| | | Finally, I'm on to something.
Tests will be done, first.
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dave brown - 18/08/2005 00:59:11
| | | Brain hurts! lol, i quit for a bit.
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dave brown - 07/09/2005 08:27:03
| | | Yes, I was a-miss on some points and in time, realized more.
The 1g of force constantly applied to one end would accelerate the gyro through space causing precission,
- It would also cause it's axis of ratation to change swiftly enough, as seen in lecture #8, near the end.
(Why are you doing it? I think the same effect as on earth is trying to be set up, but in space, without all the frictions: Air, support to table... Cool.)
- Although, and not that it matters i guess, I still feel like something is missing; Should the statement have been: There is a force equivalent to 1g for the whole gyroscope applied to one end....?
-- Just that the concept of 1g is eluding me.
Due to lecture #8 and some thinking, I realized that the force is not only down on the gyro but, up on the supported end.
- It is the up force that causes the change in the axis' orientation. The normal force.
- The whole gyro wants to fall, well, go straight through space/time, but the the support is torquing one end of it's axis.
-- Without the support, it would accelerate downward. a=mF ( F=ma :)
-- With the the support, it does not have one end falling, no, it effectively has one end being forced up, and it's axis of rotation is changing in relation to it's center of mass, but at the same radius as before from the pivot point.
Now, if the Force of the mass is being detected by a scale under the support, where did the acceleration go? The gyro is still hanging in 'mid-air', so now I/we need a formula for acceleration equalling precession times the gyro's properties.
--- I think I now know why it is called centripital acceleration.
Does the gyro's speed change other than due to fricition and air losses?
I also retried the bicycle wheel gyro and forced it in the direction of precession. the gyro rpm slowed.
I then forced against precesion and there was no easily noticeable change in rpm.
hmmm :)
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dave brown - 07/09/2005 08:28:27
| | | Nah, no bonus :(
There is still the torque of the weight to be converted.
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dave brown - 07/09/2005 08:57:52
| | | I realize I still haven't said why it precesses, and the direction it does.
If the vertical gyro you are looking at flat-on from the free end is rotating clockwise,
and knowing that it is the upwards torque of the support that causes precession,
then it is only the upper and lower points of the gyro that deflect the most,
- that feel the torque the most.
as you get closer to the left and right sides,
- they only feel a twisting action, but nothing to deviate them from their happy little plane of spin.
As the upper deflection is towards you to the right, and the lower away from you to the left;
- the gyro is torquing, free end left.
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dave brown - 07/09/2005 17:29:23
| | | Interesting read that captures my ill-feel with force.
http://www.physicstoday.org/vol-57/iss-10/p11.html
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dave brown - 07/09/2005 18:01:02
| | | About 1g... well, the g part.
What I'm understanding about gravity from the general theory of relativity is that:
they are nolonger saying that 2 bodies in spacetime are attracted towards each other but, that
the earth, for example, changes the space time around it and
the moon, for example, is changed from it's straight path through spacetime by this change,
and also the earth is also moved by the moon's effect on spacetime.
(The earth-moon pair must have a wobble, to a distant observer.)
They are still using the word gravity, but they must be redefining it; changing it's meaning.
- I wish they'd say stuff like that in english!!!
What does that have to do with gyro.s? Well if I don't know what I'm trying to gyro out of, how the heck do I work in the right way?
- (I'd much have wings at 10 000 feet than swimming trunks, ya know? ;)
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dave brown - 07/09/2005 18:54:44
| | | hmmm, that said, then:
Imagine a gyroscope travelling through spacetime 'over' the Earth towards the Sun, and it's speed and proximity are such that it will hit the Earth on the Sunward side, 90 degrees of orbit after it's initial tangential 'fly-over' begins.
It's axis of rotation is toward the center of the Sun.
Now,
if gravity is 2 bodies pulling at each other, the same side of the axis of rotation should still be pointing Sunward but,
if gravity is the warping of spacetime as stated in the last post, that same side of the axis should, to us, turn 'downwards', if not completely around towards the Earth, because it's straight path has been warped by the Earth.
- ya see? to the gyro it is going in a straight line, but the Earth warps the line.
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dave brown - 08/09/2005 22:09:03
| | | I asked someone the above question, and they replied that in both instances, the axis would be horizontal to the surface of the earth when landing, but!!!
That, I figure, was made under the assumption that gravity, as we know it, is 2 bodies directly attracting each other.
I didn't realize that it may have answered my question, that gravity is not acting as we believe it is. The gyro, today, may land with a horizontal-to-earth's-surface axis, but that falls under the warping of spacetime, and not direct attraction.
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dave brown - 08/09/2005 22:11:30
| | | Yes, there is always the possibility that both are happening and we have to figure out the split.
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dave brown - 08/09/2005 22:18:37
| | | Again, my belief:
Gravity is 100%:
2 bodies attracting?
- gyro lands with same axis side still towards the sun.
Spacetime warp?
- gyro lands with same axis side towards the earth.
I have in mind the non-rotating moon going in a straight line, but always the same point on the moon leading.
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dave brown - 09/09/2005 05:28:16
| | | Bicycle wheel, slowing down:
It is not the friction causing the slowing, it is the 'forced precession',
cause when I stop forcing, the speed returns to about normal.
That said, a flywheel must have a percentage of it's spin stored elsewhere even under normal precession, I believe you refer to that as a static system, as it is now accepted here that there is no such thing as 'forced precession'.
- I shall assume, for now, that it is stored in the form of 'flywheel oscillation'.
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dave brown - 09/09/2005 05:33:58
| | | Bicycle wheel, slowing down:
Forgot to add,
I forced the axis of spin to remain horizontal and
ofcourse kept the radius the same as I turned.
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dave brown - 11/09/2005 06:38:08
| | | A formula a came up with.
Please please tell me where I can find it, if it is not the first time you have seen it.
- I can post full explanation, proof, and whatever you want to know about it.
-- Forget it, I'll post the full thing now.
simplify calculations for Circular: Vector(1D), Area(2D), Volume(3D)...
In my search to find the the flip side of Pi, a never ending number,
I found the following:
Formula for circles in any Dimension:
diameter ^ (dimension being worked in: 1D, 2D, 3D...) *
( π )
( _______________________________________ )
( 2 * (dimension being worked in: 1D, 2D, 3D...) )
simplified form: d^dimension * ( π / 2 * dimension )
The birth of the second part of the above equation is explained in
the following example.
2D, Area:
A circle with a diameter of 4 has an area of... π * r^2 = 12.5664
I like 1, so
I found how many circles of diameter 1 were needed to equal the area
of that 4d circle.
The answer is 16: 12.5664 / 0.7854
a) 16 is also the square of the diameter.
b) 0.7854 = area of circle with a diameter of 1.
c) 0.7854 happens to be π / 4 or
π / (2 * (dimension in question, in this case 2D)) =
π / 2 * 2 = π / 4
3D, Volume:
I did the same check with 3D: (4/3) * π * r^3 and
I found that it takes d^3 circles with a diameter of 1.
so again
4^3 = 64,
Volume = 64 * 0.5256(Volume for dia. of 1) = @33.5103
π / 2 * 3 = 0.5256
1D, Vector:
Please note this is supposed to be simple, it is 1D.
- and it relates to circles.
If you travel straight for 4 feet, how many circular feet have you traveled?
4^1 * (π / 2 * 1) = 4 * (π / 2) = 2π
What does this mean?
well we know that the circumference is π * d,
and we don't need to go in a complete circle,
we only need half the circle, hence C = 4π, 4π / 2 = 2π.
The 1D example is used to prove the validity of the formula.
so
1D gives us the circular or arc vector in feet
2D gives us the area in square feet
3D gives us the volume in cubic feet
4D gives us the ?(warp of)? spacetime in ?(quad feet)?
- is it the radius of, or area of, or volume of... who knows.
- but 4D may give radius of,
- then 5D area of,
- and 6d the volume of, warping. (still thinking about what the 4D, 5D... is really telling me.)
If the 4D is tested on a circle with a d of 4, against a circle with a d of 1,
it would take 256 for the circle with a d of 1 to fill or do the same as
the circle with a d of 4.
- assuming the 4th dimension is spacetime,
as it is accepted that space and time are together the 4th dimension,
then what is going on that it takes 256 of the circle with a d of 1
to do the same as the circle with a d of 4?
π = Pi (to be clear :)
If comparing a circle with a d of 1 with that of one with a d of 4 then:
in 1D we need 4 to go the same distance: π/2 vs. 2π ... (4^1=4)
in 2D we need 16 to cover the same area: π/4 vs. 4π ... (4^2=16)
in 3D we need 64 to fill the same volume: π/6 vs. 64π/6 ... (4^3=64)
in 4D we need 256 to xxxx? the same xxxx?: π/8 vs. 256π/8 ... (4^4=256)
Teachers will not like this but:
2D, area:
Take any diameter, square it, multiply by π/4 = area
3D, volume:
Take any diameter, cube it, multiply by π/6 = volume
MINE, COPIED FROM POST ON ANOTHER BOARD:
There is a link from another thread that explains the beach sand going into the sand box.
It is under the title of topology. http://en.wikipedia.org/wiki/Topology
I'm refering to the line that reads:
However, it is not possible to deform a sphere into a circle by a bicontinuous one-to-one transformation.
If you take a square foot and stand in the middle of one side, looking toward the opposite side, there is an infinite number of steps. But,
if you set a step-size, to keep things 'real', set by the smallest known particle/wave/energy-level, there is now a finite number of steps.
It would then be possible to transform a 2D representation into a bunch of lines.
- good for getting a reference frame from which to 'see' what is going on. (something I do alot of.)
Now back to the sand. To start:
1 layer of sand(2D + 3D estimate based on 2D diameter) 2 ft^2 into a 3D box volume of 2ft^3.
- I won't use 1 for dimension lengths. Based on my formula for circles, a diameter of 1 is the only one that gets smaller as we prgress into the next dimension.
-- re: 1D gives circular vector π/2 ft, 2D gives area π/4 ft^2, 3D gives volume π/6 ft^3...
(makes me think of water, being the only one to expand below 4C.)
CHANGE: We know there is a guesstimated volume fill limit based on the dimensions of the 2D sand, so:
2 ft^1 * 2 ft^2 = 2(ft^1 * ft^2) = volume of 2 ft^3
ADDITION: volume / volume of sand grains = how many grains.
Now in the 3D to 4D example,
there must be a limit, but what?
Mathematically it is, if dealing with circles and trusting my formula:
2 ft^1, so:
2 ft^1 * 2 ft^3 = ?spacetime? of 2 ft^4 of sand
But that number will only hold true if I can get a relation between all dimensions.
CHANGE: - hmm found something here to help get the 4th dimension wording. It's in the wording.
This is here to help in understanding why I no longer deal with the size of the grains of sand in 4D:
0D: Can not divide by zero. Not yet part of the 'real' world. See quantum physics?
We first had to calculate the physical dimensions of the mass-x grains of sand we would work with.
- quantum physics I presume.(based on strong, weak, whatever forces.)
-- I will think of that as quantum dimensions for now.(Amplitude, Spin...)
1D: Used to calculate the boundary lines/vectors of pysical space to fit siz-x grains of sand for diameter y ft.
CHANGE: ---- not yet a tangiable dimension. remember sand has Second dimension.
2D: Used to calculate the boundary area of pysical space to fit size-x grains of sand for diameter y ft^2.
---- Not yet a tangiable dimension. remember sand has Third dimension.
3D: Used to calculate the boundary volume of pysical space to fit size-x grains of sand for diameter y ft^3.
---- Now a tangiable dimension. remember sand occupies 4D spacetime, or
------ is spacetime the start of something larger? One that covers 2, 3, or more dimensions?
a reach here?
4D: Used to calculate the boundary lines/vectors of spacetime that is affected by mass-x sand * grains in diameter y ft^3.
5D: Used to calculate the boundary area of spacetime that is affected by mass-x sand * grains in diameter y ft^3.
6D: Used to calculate the boundary volume of spacetime that is affected by mass-x sand * grains in diameter y ft^3.
The following is babbling, but I left it in.
after writing all that,
I'm not sure now if the 4th dimension is looking at a larger or smaller scale of the sand.
- it may be looking at the quantum properties of circle with diameter y.
-- so much to think about.
is spacetime the 4th dimension? it can't be the 1st? or just another?
OR
is the 4th dimension, like I say, just on a smaller physical/energy scale but felt farther out, like gravity?
Can't see it, but it's effects are surely seen in real time on large pysical items.
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dave brown - 11/09/2005 07:01:10
| | | Gyro spin-down
Starts vertical
precesses to horizontal
from that point on, the free end scribes a spiral.
Reminds me of those negative's, the pics of some particle, forget which, from science tests.
Please inform me of which? :)
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dave brown - 11/09/2005 07:15:34
| | | Force that causes precession,
on a toy gyro on stand.
After reading this thread:
http://www.gyroscopes.org/forum/questions.asp?id=459
I feel it's time to protest which force is causing the precession.
I looked at this video:
http://phys23p.sl.psu.edu/phys_anim/mech/gyro_s1_nu_avi.html
and again saw now arrow for the Normal force, up through the pedestal.
Yes, gravity acting on the mass of the gyro system will gives the magnitude of the force, but only that.
Put a pedestal under each end of the gyro; same force, no precession.
The torquing force is caused through / by the pedestal, the Normal force.
Does it matter?
- Not really, you still have to push down on one end or up on the other.
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dave brown - 11/09/2005 07:22:23
| | | Forces on the pedestal.
Have a video or experience with a large based pedestal,
on an equal sized platform,
balanced dead center on a short pointed shaft?
I'm guessing the platform, being on a short shaft, would
touch the table it's on at 90 degrees lag of precessing end.
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dave brown - 13/09/2005 01:25:18
| | | Gyro Nutshell
-0--.
1) the Mass of the Flywheel will let us calculate the weight force.
2) the Shaft it spins about, and the Pedestal, will allow interaction with the Normal force.
--- a non-spinning Flywheel will torque the Shaft down at the point connected to the Pedestal.
---- 5lbs @ 2 ft gives 10 ft/lbs of ccw torque.
3) a spinning Flywheel will change the torque from ccw of the Shaft about Pedestal, to cw of the Shaft about dead-Center of Mass of the Flywheel, to keep it's axis of rotation unchanged.
--- the weight of the system is felt straight down on the Pedestal point, as it is no longer the center of torque, but tangential to the torque. the Normal force is clearly seen doing it's thing, opposing gravity. (the Normal force may not equal the system's dead weight force, as there is a downward vector of system movement.)
--- the inertia of the Shaft vs. the Inertia of the Flywheel dictates how much the axis of rotation is deflect ccw in direction of the applied force.
--- the Normal force is 5lbs at 2 feet = 10 lbs. the Flywheel is experiencing @10 ft/lbs of torque.
a spinning Flywheel wants to fall straight down. the Shaft on the Pedestal is what allows us to experience this spectacular view of the laws of forces in action.
due to the action-reaction at the top and bottom of the Flywheel-Shaft points, the orbit of the matter in the Flywheel is changed on the first half spin.
--- that puts a torque on the Shaft about the horizontal, precession.
Not to confuse that issue but, forcing in the same direction as precession, to me, creates 2 precession reactions with one vector. just like our g force and hand force are 2 forces with one vector. My point is that both forces are needed for the one direction of precission.
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dave brown - 13/09/2005 01:52:32
| | | Gyro Nutshell
forgot to add:
Precession around the horizontal? Possible?
it is at that moment being force straight down, evenly, across the top bearings.
- there may be something there we can use, or should avoid.
-- I notice that gyroball is close to but not at the horizontal. re: supported, i seem to see, top one side, bottom on the other.
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dave brown - 13/09/2005 02:26:33
| | | Gyro Nutshell
Loss of Flywheel spin due to precession.
As the shortest/fastest path between 2 points is a straight line,
the change of orbit is a longer/slower path.
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dave brown - 14/09/2005 01:18:18
| | | bugging out for a while.
gyro's seem closer to the 4th dimension than propulsion.
off to tackle quantum physics.
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dave brown - 07/10/2005 06:17:22
| | | Particle reaction explained?
There is a physics experiment I recall, where a particle goes into a liquid medium and spirals in on itself.
If you look at the pick on this page:
http://www.geocities.com/xulfrepus/index.html
- which is not working at the moment :)
it shows flux flatening out as 2 of the others transmit a force.
If you imagine them parabolling to one end instead of flattening out, it would mimic the effects we see of the 'offset center of mass' flywheel, when it hits the liquid.
Just a thought. :)
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dave brown - 07/10/2005 06:18:30
| | | More specifically, on this page:
http://www.geocities.com/xulfrepus/docs/fluxi.html
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