Question |
| Asked by: |
Luis Gonzalez |
| Subject: |
Centrifugal Force in Precession’s Orbit ?? |
| Question: |
What additional phenomena can we expect from gyros IF precession’s orbit does indeed produce centrifugal force?
We can expect precession of a gyro released at an angle above the horizon will travel “Faster” than one released at an angle below the horizontal.
This difference would not be perceptible, under Earth’s gravity-induced torque, in a standard gyro.
However, in a planet 10 times more massive than Earth, its corresponding greater gravity would assumedly produce an exponentially greater centrifuge.
At some angles “Above” the horizontal, this greater centrifuge should produce yet greater torque (augmenting gravity’s torque), thus producing even faster precession.
There is no telling how fast precession would seek to accelerate, under the resulting torque condition when it exceeds some yet unknown gravity-threshold.
On the other hand (in the more massive planet), if we released the same gyro from an angles “Below” the horizontal position, the effect would be reversed.
The powerful centrifuge would work “Opposite” to the planet’s gravity, thus reducing the net torque upon the gyro.
This effect would reduce the velocity of precession, perhaps until some balance would be met.
The following questions come to mind:
Can centrifugal torque ever exceed a planet’s gravity torque? At what level?
If so, what effects would it have on gyros released at angles above, and below the horizontal (if the centrifugal-torque exceeds gravitational-torque)?
Regards,
Luis G |
| Date: |
6 January 2013
|
| report abuse
|
|
Answers (Ordered by Date)
|
| Answer: |
Blaze - 06/01/2013 22:03:05
|
| | Hi Luis.
Some good questions. The centrifugal force would be greatest for a gravity powered gyro when the gyro arm is horizontal for any given precession speed. When the arm is above horizontal the centrifugal force would be trying to do a "vectored push down" on the flywheel end of the arm and would indeed provide extra torque on the arm which would produce a faster "natural" precession rpm. When the arm is below horizontal the centrifugal force would be trying to "vectored lift" the flywheel end of the arm which would reduce the gravity couple and therefore would reduce the "natural" precession rpm.
The centrifugal force when the arm is at 45 degrees above or below horizontal would be about 70.7 percent of what it is at horizontal. Of course the centrifugal force would vary for every angle above or below horizontal for a fixed precession rpm. If you plotted the calculated centrifugal force for varying angles at a fixed precession rpm, the graph would be 1/2 a sinusoid.
The centrifugal force would be MOST effective at moving the gyro arm when the gyro arm is near vertical, up or down, however, this is where the centrifugal force is the least. The centrifugal force would be LEAST effective at moving the gyro arm when the gyro arm is near horizontal, however, this is where the centrifugal force is the greatest
Certainly, the magnitude of centrifugal acceleration can exceed the magnitude of gravitational acceleration. It depends largely on the system parameters but orbit velocity plays a big part. It wouldn't happen on typical table top models because the orbit velocities are just too low. Even my giant gyro system with a 24 inch arm is too small to get the velocities needed. You would need a fairly large system to get centrifugal acceleration to exceed gravitational acceleration. However, from the previous paragraphs you can see that is isn't straight forward on what the results would be due to the vectored input the centrifugal acceleration has. One would have to do some calculations to get the corrects answers. Of course on a system where the pivot hub is powered it would be easy to get the orbit velocities needed.
I am fairly certain that you know everything I just wrote, however, the whole idea of centrifugal force during "steady state" precession is something I have only recently started to look at. I haven't had time to look at all the effects on a precessing gyro yet.
best to all,
Blaze
|
| Report Abuse |
| Answer: |
Blaze - 08/01/2013 02:25:15
|
| | Luis: "what effects would it have on gyros released at angles above, and below the horizontal (if the centrifugal-torque exceeds gravitational-torque)?"
For arm angles above horizontal, any centrifugal torque (CT) would add to the gravitation torque which would increase the precession rpm. If the downward component of the CT were exactly equal to gravity the gyro should precess at twice the "normal" rate. However since it is now precessing faster the CT would be greater which would mean that the precession rpm would increase some more. This would be a declining positive feedback which would peak at some precession rpm. This should be a relatively smooth progression to its final precession rpm.
For arm angles below horizontal the opposite would "sort of" happen. In this case the CT would be countering the gravity torque and the gyro would rise and the precession rpm would slow. However, since the precession rpm is now slower the CT would also be less so the gyro would fall and the precession rpm would speed up again (sounds a lot like nutation). There is also the varying radius to contend with in this scenario. At some point the torques would come to a balance and the gyro would precess steadily.
Aren't gyros fun?
Blaze
|
| Report Abuse |
| Answer: |
Momentus - 10/01/2013 17:07:27
|
| |
Do the simple experiment first. Measure the radial force exerted here on earth.
Note the absence of centripetal accelleration. That is THE mystery.
Louis is aware of this. He starts the post with IF.
Momentus
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 10/01/2013 19:12:05
|
| | Hi Blaze,
We can be sure the feedback ends when the gyro passes the horizontal.
However, why would the “Above Horizontal” feedback decline? (Why wouldn’t it accelerate?)
After all it keeps piling centrifuge upon centrifuge.
Regards,
Luis G
|
| Report Abuse |
| Answer: |
Blaze - 11/01/2013 02:05:15
|
| | Hi Luis. When the arm is at an angle above horizontal only the downward component of the centrifugal force adds to "gravity" torque. The radial component doesn't help. So the feedback declines as the arm approaches horizontal because the downward component approaches zero. Actually if you started with the arm at nearly vertical and progressed to horizontal and graphed the downward component I would guess that you would get some sort of curve because the arm length, velocity and downward component are all changing as you approach horizontal.
cheers,
Blaze
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 11/01/2013 02:30:36
|
| | Hi Blaze,
Excellent analysis.
Dear All,
Anyone who has absorbed the gyro related information that has evolved in this forum during the last 5 years, should know that the unusual behaviors (phenomena) of gyros do NOT apply one hundred percent (100%).
One basic example of this percentage issue is that orbiting-precession occurs at 90 degrees to the tilting torque… but in fact, it also moves downward (in the direction of the torque) at a slower rate.
Some portion of the motion is at 90 degrees and some portion is in the same direction as the tilting torque, and these proportions vary depending on a number of variables.
We have conducted and observed a number of experiments, which in some cases appear to prove (and in others disprove) that the motions in precession do or do not manifest “Newtonian” attributes.
Here again we have a situation involving multiple variables coming into play.
Thanks to Blaze, we are now able to focus on what appears to be an overwhelming factor (centrifuge) whose effect varies “Exponentially”, as another factor (angular velocity of precession) changes in a “Linear” fashion.
Blaze’s observation magnifies our focus on the most significant conjecture toward the viability of producing linear-propulsion by manipulating angular motions (does precession have Newtonian qualities?).
Success in this endeavor will probably require figuring out ways to increase the desired effects (i.e. increasing the percentages for effects we desire in our devices).
Regards,
LuisG
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 13/01/2013 03:11:47
|
| | Did I miss the main point?
All I am saying is that Orbiting-precession’s centrifuge is not 100% there, or not there.
The next logical conclusion is that there are ways to adjust or mitigate the centrifugal effects, though only to a degree.
The answer is plain to see in the math.
Regards, LuisG
|
| Report Abuse |
| Answer: |
Harry K. - 13/01/2013 12:00:17
|
| | Hello Luis,
Centrifuge is always present and does not disappear at all even if some in this forum are convinced about. Only the effect of centrifuge will be varying between zero and maximum, as Blaze and you already correctly explained.
As we all know, gyro systems will not be affected by single forces but by torques. Thus the centrifugal force will create a torque with its vertical lever arm component to the gyro´s pivot. If the gyro orbits above the horizontal plane, centrifugal torque supports the tilting torque and if the gyro orbits below horizontal plane, centrifugal torque works against the tilting torque.
If the (overhung) gyro orbits above horizontal plane, centrifugal force will have its maximum at horizontal position and its minimum at upper vertical position. However, the centrifugal torque is zero in both positions:
- zero at horizontal plane because the lever arm is zero although centrifugal force is at maximum
- zero at vertical position because centrifugal force is zero although the lever arm length is at minimum
That means that the maxium centifugal force will be at a position somewhere between horizontal and vertical plane. I guess at 45 degrees but without knowing,
Anyway, it`s not trivial to calculate the exact precession speed of an orbiting gyro with the overhelmed centrifugal torque although it is easy to present formula to calculate the single torques by a given angle alpha about horizontal plane:
- Tilting torque of an overhung gyro = m(gyro) * g(gravitation) * cos(alpha) * R(radius piivot)
- Centrifugal force = m(gyro) * W(Omega precession)^2 * cos(alpha) * R(radius pivot)
- Tilting torque by centrifuge = centrifugal force * sin(alpha) * R(radius pivot)
I`m sure you are aware of these calculation formula.
Now the problem is, that centrifugal force is a function of the precession speed given by the origin gravitational tilting torque and which must be present at first to induce the overhelming centrifugal force effect. Thus it is not easy or not possible at all to find a formula to calculate the real precession speed with the influence of centrifugal force.
And beside overhelming effects caused by centrifuge there wiil be additional overhelming effects such as conservation of angular momentum (rotation speed will be increased if the rotation mass moves to the pivot and vice versa) and the fact that different sizes of rotated masses as well as different rotation speed of masses will be overhelmed in different manner caused by the stored energy in form of angular momentum. These effects likewise apply to the rotation speed of the gyro`s fly wheel as well as to the induced precession speed.
As mentioned before it is not trivial to consider all possible influences to calculate the real outcome how an overhung gyro system will behave how it behave. However, one have to be aware of these effects!
|
| Report Abuse |
| Answer: |
Harry K. - 13/01/2013 12:00:19
|
| | Hello Luis,
Centrifuge is always present and does not disappear at all even if some in this forum are convinced about. Only the effect of centrifuge will be varying between zero and maximum, as Blaze and you already correctly explained.
As we all know, gyro systems will not be affected by single forces but by torques. Thus the centrifugal force will create a torque with its vertical lever arm component to the gyro´s pivot. If the gyro orbits above the horizontal plane, centrifugal torque supports the tilting torque and if the gyro orbits below horizontal plane, centrifugal torque works against the tilting torque.
If the (overhung) gyro orbits above horizontal plane, centrifugal force will have its maximum at horizontal position and its minimum at upper vertical position. However, the centrifugal torque is zero in both positions:
- zero at horizontal plane because the lever arm is zero although centrifugal force is at maximum
- zero at vertical position because centrifugal force is zero although the lever arm length is at minimum
That means that the maxium centifugal force will be at a position somewhere between horizontal and vertical plane. I guess at 45 degrees but without knowing,
Anyway, it`s not trivial to calculate the exact precession speed of an orbiting gyro with the overhelmed centrifugal torque although it is easy to present formula to calculate the single torques by a given angle alpha about horizontal plane:
- Tilting torque of an overhung gyro = m(gyro) * g(gravitation) * cos(alpha) * R(radius piivot)
- Centrifugal force = m(gyro) * W(Omega precession)^2 * cos(alpha) * R(radius pivot)
- Tilting torque by centrifuge = centrifugal force * sin(alpha) * R(radius pivot)
I`m sure you are aware of these calculation formula.
Now the problem is, that centrifugal force is a function of the precession speed given by the origin gravitational tilting torque and which must be present at first to induce the overhelming centrifugal force effect. Thus it is not easy or not possible at all to find a formula to calculate the real precession speed with the influence of centrifugal force.
And beside overhelming effects caused by centrifuge there wiil be additional overhelming effects such as conservation of angular momentum (rotation speed will be increased if the rotation mass moves to the pivot and vice versa) and the fact that different sizes of rotated masses as well as different rotation speed of masses will be overhelmed in different manner caused by the stored energy in form of angular momentum. These effects likewise apply to the rotation speed of the gyro`s fly wheel as well as to the induced precession speed.
As mentioned before it is not trivial to consider all possible influences to calculate the real outcome how an overhung gyro system will behave how it behave. However, one have to be aware of these effects!
|
| Report Abuse |
| Answer: |
Harry K. - 13/01/2013 12:22:32
|
| | Sorry for sending twice!
And I have discoverde a type error:
"- zero at vertical position because centrifugal force is zero although the lever arm length is at minimum"
must be read as:
- zero at vertical position because centrifugal force is zero although the lever arm length is at MAXIMUM"
Best regards,
Harry
|
| Report Abuse |
| Answer: |
Blaze - 13/01/2013 19:04:44
|
| | Very good analysis Harry. As you indicated, this is not a simple calculation because of all the variables involved which are constantly varying between vertical and horizontal arm angles. You brought up a good point about the "figure skater" effect that I hadn't thought of which will also vary the precession speed and therefore, the amount of centrifugal force.
I agree that centrifuge should always be present in varying amounts depending on the arm angle, precession speed, etc., for any arm angles that are not vertical.
Blaze
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 14/01/2013 01:04:57
|
| | Thank you Blaze and Harry,
Good comments.
Well thought-out perspectives are always helpful toward accurate perception of the overall dynamics.
Regards,
LuisG
|
| Report Abuse |
| Answer: |
Glenn Hawkins - 14/01/2013 18:44:00
|
| | Can centrifugal torque ever exceed a planet’s gravity torque?
No. This is about force causing force. When gravity is acting to cause centrifuge to react, the condition must obey the third law of motion. For every action there is an equal and opposite reaction.
regards,
Glenn
|
| Report Abuse |
| Answer: |
Glenn Hawkins - 14/01/2013 19:48:14
|
| | Hello gentlemen,
This is all so wrong. First it should be stated that the angle the shaft points, as if it were an arrow, is not the direction of centrifuge. Centrifuge has a center of rotation, like the center of a flat rotating plate, but this solid shape is not necessary. The center can be hollow like the center of a rotating ring. With these gyroscope are tilted at angles, the condition is like a ‘Y’ shaped rotation. As the base of the ‘Y’ is rotated the center of rotation will be in the center between the two upward projections. Centrifuge will be generated directly outward from the center of that empty space. It will not matter if the upward shafts have spinning disk on them or not.
Also a rotating vertical wheel is so balance as a part of nature, that although the front of the rotation is a condition where matter is being hurled downward toward gravity, while the rear half is hurling upward defying gravity. But a spinning wheel is self contained. No matter at what angle it is set during rotation, no outside directional force is released relating to the angle.
Always precession is in balance. Think of a bull’s eye painted on the floor with graduated rings. Put a gyroscope in the center. If you raise, or lower the tilt, notice that the center of the gyroscope moves in, or out from the center of the bull’s eye. The effect is the same as shorting, or lengthen the shaft.
Angular momentum seeks to maintain its speed. When the shaft is long the radius of rotation is long and the path around the revolution the gyro travels is long. When the shaft is short the path of a revolution is short. In order for the speed of mass moving around the circumference to be maintained, it must revolve slowly when the shaft is long and fast when the shaft is slow. Same speed is maintained that way.
One other balance: the length of the shaft has like a pry bar effect. The longer it is the faster it falls and the faster it precesses, because there is greater augmented force downward.
The spinning ice scatter: The explanations are true to rotation, but precession has breaks. Both would be carried for the same reasons in a circle my angular momentum, but deflections occurring in the changing directions of the spinning wheels resist preceding in precession and must be continuously forced—unlike rotation.
Most of the downward force of gravity is transmitted to life, or hold the gyro upward through a series of torques . That amount of tilting force, very modestly remaining, is what powers precession. That maybe why precession collision is less than hoped for and why the gyro falls, if it’s precession is hindered only slightly.
If math does not follow these mechanics, somehow there must be errors. Pardon me for not cleaning this up.
Happy trails!
Regards Glenn,
|
| Report Abuse |
| Answer: |
Glenn Hawkins - 14/01/2013 19:58:37
|
| | I meant: In order for the speed of mass moving around the circumference to be maintained, it must revolve slowly when the shaft is long and fast when the shaft is SHORT. Same speed is maintained that way. HOWEVER AS THE SHAFT IS LONGER, MORE PRYING GRAVITY FORCE IS ADDED AND THIS IS THE REASON FOR HIGHER SPEED WITH A LONGER SHAFT.
I have a book, play and scrip to finish and no time. I just got caught up in following this and probably did not do a good enough job putting in my two cents worth. Sorry.
Buy all,
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 16/01/2013 02:11:52
|
| | Thanks all for your perspectives.
Regards,
LuisG
|
| Report Abuse |
| Answer: |
Blaze - 16/01/2013 22:10:11
|
| | Hi Glenn. As the flywheel is pulled outward by centrifuge, it pulls down on the arm (for arm angles above horizontal), thereby providing additional torque that adds to gravity torque.
Blaze
|
| Report Abuse |
| Answer: |
Glenn Hawkins - 17/01/2013 00:57:45
|
| | Hi Blaze,
You are very smart, but I like you anyway : )
These little ditties might be fun! I hope you like them.
1) In distant free space an ‘I’ beam has been placed in alignment from the earth toward the moon. A rope is tied to the vertical end of the bean (upward end) toward the moon. The rope is pulled directly outward at a right angle. How does the ‘I’ beam respond?
2) If the earth end of the beam (bottom end) had a hinge attached to a floating pedestal and the rope at top were pulled at a right angle, what would be the effect on the pedestal?
regards Glenn,
|
| Report Abuse |
| Answer: |
Blaze - 17/01/2013 02:28:29
|
| | Hi Glenn. The earth moon system you described is quit different from a gyroscope sitting on a table.
Here are the assumptions:
- gyro pivot firmly attached to an unmoveable object
- gyro precessing with arm at 45 degree inclined angle (45 degree above horizontal)
The centrifugal force is radially outward as you say however the only way that the gyro weight, the flywheel, can move outward is by moving down as it is moving outward because the arm is attached to the pivot with a hinge. Therefore the outward force will cause some downward torque on the arm because of the mechanical arrangement. If the arm was locked in position then the flywheel would not move outward.
The "hinged/arm at 45 degrees" arrangement on the gyro described is not an unusual mechanical arrangement as it is commonly in hydraulics actuation of heavy loads. The "outward" force is supplied by a hydraulic cylinder. The mechanism I am referring to is commonly used to lift a truck box, like on a farm truck.
So with the arm at an inclined 45 degree angle the centrifugal force will cause the flywheel to move outward which will cause the arm to try to move downward which will indeed cause additional torque on the arm. This additional torque will cause the gyro to precess faster and cause more torque on the arm until everything balances out. The extra precession speed will cause increased "second twist" which will "hold up" the gyro, as you indicated. But the point is that and inclined angle will put additional torque on the arm.
regards,
Blaze
|
| Report Abuse |
| Answer: |
Glenn Hawkins - 17/01/2013 10:29:53
|
| | Yes Blaze. Thank you. You are right. There are interesting exceptions. Without the pedestal of course, but also without the concave top of the pedestal to act as an anchor; or exactly to the point, without an anchoring hinge much more substantial than the cup used, my examples would be correct. Centrifuge if strong enough would become linear and pull the gyroscope away and if it accelerated fast enough the gyro would momentarily for a few degrees precess in the opposite direction and rise.
Now then. Let us accept it the way you explained it and see it for again your are correct.
Happy trails,
Glenn
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 19/01/2013 19:27:42
|
| | Hi Blaze,
The video in the link below does not appear to support that orbiting-precession has 100% centrifugal-force; how would you explain it? http://www.youtube.com/watch?v=nstIIZZadAM
Regards,
Luis G
|
| Report Abuse |
| Answer: |
Blaze - 19/01/2013 22:27:56
|
| | Hi Luis. Looking only at the centrifugal effects and based on some VERY CRUDE measurements I get a centrifugal force of about 1/12 newtons. Not very much and it wouldn't take much to overcome that little amount of force. That is the best I can do without actually having the apparatus available to me but I would think that I am easily within +/- 50% of the actual amount of centrifugal force being generated.
regards,
Blaze
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 20/01/2013 19:19:26
|
| | Hi Blaze,
To put some perspective on it, 1/12 Newton is about 0.0187341 Pounds of force or 0.6008 lb-ft/sec-sec.
As small as it may be, I would expect some counter-turn on the Styrofoam base.
(Assuming my calculations are correct)
Regards
LuisG
|
| Report Abuse |
| Answer: |
Blaze - 20/01/2013 20:51:47
|
| | It looks like there is some mass to the apparatus as it has batteries, framework, actuating linkage and either a solenoid or motor for moving the rotating mass back and forth. I would think this would add up to a pound or two. When I was doing my own experiments with ping pong balls I found some inconsistencies that I believe were due to ping pong ball "frictions" from the balls being too close together, although that only happened a couple of times for me because I didn't have a lot of ping pong balls to begin with. In the video I think that they consistently had the ping pong balls packed too tightly under the platform to allow proper freedom of movement and the balls rubbing together would/could be enough to prevent the small centrifugal force from producing any significant motion.
When doing the calculation for centrifugal force I measured the dimensions of the spinning mass (and calculated weight based on steel) and the radius and scaled them to the ping pong balls in the video. I measured the ping pong balls I have to get the proper scaling for "real life" sizing. From the video it seems that the 1/2 precession takes about 3 seconds. The approximate 1/12 newton force is very small and would be hard to measure or see any movement generated by it if there were any friction or resistance. I believe a much larger system would be required to see any real centrifugal force or a significantly higher precession rate would be required.
cheers,
Blaze
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 21/01/2013 18:00:33
|
| | Hi Blaze,
Inconsistencies caused by the ping pong balls would occur in random directions (the video results seem to require 100% same direction consistency, according to the consistent direction).
The video shows that the balls are not always packed tightly together, but the responses are consistent.
Also, most of the device framing is made from bent aluminum sheet that make it look thicker.
There are also some other very subtle motions in the video that are consistent.
Regards,
Luis G
|
| Report Abuse |
| Answer: |
Nitro - 21/01/2013 21:06:21
|
| | Dear Luis G
Re; the Austrian you tube video:- These Austrian guys really have got Laithwaite’s later design to work IMHO. I imagine that they must be university people as they have their faces blanked. Nutters in sheds like me don’t need to do this as we cannot be harmed by the “establishment” only by friends. Don’t worry youselves unduly that there is friction with the table – only those who have done nothing practical in this field will be unable to tell if there is “stick slip” involved. There isn’t.
Now you can see it is possible I can only add that there are more elegant and efficient ways to try for. Though they (the ones using a gyro’s ability to displace more than its own mass) all, in my experience, can only produce an impulse drive. That may be useful to shove a poorly aimed satellite back to its orbit but it ain’t goin’ to produce warp drive by any stretch of the shed dwellers imagination.
Kind regards
NM
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 21/01/2013 22:53:57
|
| | Thanks for the post Nitro.
My skeptic nature remains, as it did from the beginning.
On the other hand, I put no-limitations on the potential of a device that has yet to be successfully invented.
Not all spin-related research is public, and certainly that condition is reflected in this forum.
None of us know what the future will bring, or whether acceleration can be aggregated beyond one step at a time.
Thank you for your much appreciated experiments, contributions, and for Nitro’s Law.
Regards,
LuisG
|
| Report Abuse |
| Answer: |
Nitro - 23/01/2013 23:43:59
|
| | Hi Luis,
You are probably wise, based on the evidence of the past, to be a sceptic. Aggregating impulse into smooth thrust may yet prove possible. However, being an old hand at this you will notice that I restricted my comment to “my experience”.
I believe, though, you to be mistaken on both your comment that “a device .. has yet to be successfully invented” and that “None of us know what the future will bring, or whether acceleration can be aggregated”. Though my belief is probably just me being big headed as usual.
My Island has had half an inch of snow and looked very pretty but everything ground to a halt. Have you had a massive snow fall to like that to contend with?
Kind regards
NM
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 24/01/2013 02:33:51
|
| | Hi Nitro,
Some places have had a good bit of snow and others have had less than average.
We’ll see how the rest of the season plays out.
Regards,
LuisG
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 24/01/2013 21:57:16
|
| |
Hi Blaze and Harry,
As Harry said in this thread “gyro systems are Not affected by Forces, they are affected by Torques” (even though we all have a tendency to confuse these two words during our explanations).
I had some spare time and figured out the following:
Maximum Centrifugal Torque occurs at 30 degrees
Maximum Combined (Tilting + Centrifugal) Torque occurs at 15 degrees
Of course Maximum Centrifugal force occurs at zero degrees (horizontal)
Intuitively we think that these Torques Max-Out at 45 degrees, perhaps because the product of Sin (a) X Cos (a) maximizes at 45 degrees.
Thank for your input Harry and Blaze.
Regards,
LuisG
|
| Report Abuse |
| Answer: |
Blaze - 24/01/2013 22:20:08
|
| | Hi Luis. Thanks for calculating those angles. I was planning on getting to it but hadn't been able to just yet. I never would have guessed that the combined tilting and centrifugal torque would be a such a small angle of 15 degrees. Just goes to show that you have to do the math when working with apparatus which has all motions and movements curving.
So, would the same angles of 15 and 30 degrees apply for angles below horizontal? In other words, would those angles provide the maximum counter gravity couple?
cheers,
Blaze
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 25/01/2013 01:35:18
|
| | Hi Blaze,
I didn’t calculate angles below horizontal.
Regards,
Luis
|
| Report Abuse |
| Answer: |
Harry K. - 26/01/2013 13:20:48
|
| | Hi Luis,
"Maximum Centrifugal Torque occurs at 30 degrees
Maximum Combined (Tilting + Centrifugal) Torque occurs at 15 degrees
Of course Maximum Centrifugal force occurs at zero degrees (horizontal) "
I have also done some calculations which are not finished yet. I also find the "magical" angle at 30 degrees. Also the combined torque of tilting torque and centrifugal torque should be at 30 degrees. I'm curious how you find the 15 degrees angle?
Harry
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 27/01/2013 04:23:01
|
| | Hi Harry,
There is no reason that the Max combined sum of centrifugal and tilting torques should occur at 30 degrees.
Consider the following:
We know that Max Centrifugal torque occurs at 30 degrees.
We also know that Max tilting torque occurs very near zero degrees because that is where the value of Cos(a) is most near to equal “1” (look at the equations you presented).
Therefore common sense indicates that the sum of both equations will yield a Max angle that is less than 30 degrees.
Simply, (1) add the two equations, and (2) determine the angle where the new combined equation produces the highest value. It’s just math.
Regards,
LuisG
|
| Report Abuse |
| Answer: |
Harry K. - 27/01/2013 13:31:34
|
| | Hi Luis,
Thank you for your explanation. I`ve calculated the tilting torque, precession velocity Omega, centrifugal force and the sum of centrifugal- plus tilting torque for angles between 0 and 90 degrees in 5 degrees steps and inserted the results in a value table.
The analysis of the results showing the following:
1. The maximum centrifugal torque is at 30 degrees.
2. The sum of centrifugal- and tilting torque has it maximum at 30 degrees.
3. It seems that at 30 degrees the value of centrifugal torque is exactly the square root of "3".
4. At 60 degrees centrifugal torque and tilting torque are equal.
In my point of view your assumption that the maximum torque of centrifugal- and tilting torque must be in the range between 0 and 30 degrees is not correct.
The intersection point of both functions will indicate the balance point of centrifugal- and tilting torque at an certain angle. That means that both torques at this certain angle are equal, in the same way at it is at 60 degrees. Remember that we have this already discussed in my balancy point theorie thread.
At the moment I`m not able to find a formula to calculate the real total torque of this gravity overhunging gyro system, because there are too much variables which affect each other. For instance the calculated sum of centrifugal- and tilting torque will give the value for the new tilting torque, which in turn causes a higher precession velocity, which in turn causes a higher centrifugal force, which in turn causes a higher centrifugal force, and so on.
I'm afraid I`m to dumb to find the correct formula to calculate these things. Maybe you have an idea?
Anyway, at the "magical" angle of 30 degrees the overhunging gyro should be accelerate by itself, assumed the absence of any kind of friction or mass inertia.
Now I have to fly to Madrid for a business trip. It would be nice if you could come with me for translating issues... ;-)
Have a nice Sunday.
Harry
|
| Report Abuse |
| Answer: |
Harry K. - 27/01/2013 13:35:51
|
| | Correction:
"3. It seems that at 30 degrees the value of centrifugal torque is exactly the square root of "3"."
must read as:
3. It seems that at 30 degrees the value of centrifugal torque is exactly the square root of "3" * tilting torque (1,73 * tilting torque).
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 28/01/2013 00:12:25
|
| | Hi Harry,
Enjoy your trip to Madrid; it is a fun, classy city.
I am sure you will figure out why the max combined-torque Cannot coincide with the max centrifugal-torque (even at 30 degrees), considering that the max tilting-torque occurs at near zero degrees (as Cos(a) max is at zero degrees).
Is it not obvious that the values of both torques will skew each other toward their own value?
Also, my calculations tell me that the centrifugal-torque and tilting-torque do not become equal at 60 degrees.
My common sense also tells me that this torques cannot be said to be consistently equal at “any given angle” (this become obvious when you see that the value of the torques are for the most part at least one order of magnitude apart, except when they are both near zero around 90 degrees).
It’s too bad our calculations do not agree.
If you give me one month warning, I can meet you next time you are in Madrid; as long as your company arranges and pays the hotel bill (I usually stay at Hotel Wellington on Calle de Velasquez).
Have fun.
Regards,
LuisG
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 30/01/2013 01:58:52
|
| | Hi Blaze,
Regarding torques below the horizontal:
The Maximum (negative) Centrifugal-torque also occurs at -30 degrees (as expected).
However the Maximum combined Centrifugal Torque + Tilting Torque (below the horizontal) occurs at near zero degrees (not at 15 or 30 degrees).
Please let me know if your calculations show something different to any of my calculations.
Regards,
LuisG
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 03/02/2013 17:51:41
|
| | Hi Harry,
Calculating the angle locations where Centrifugal and where Total orbiting-precession maximizes is much easier than calculating the Actual orbiting-precession Velocities.
As we know, the actual orbit-velocity (for each angle) is affected by centrifugal-feedback, which makes it more difficult to calculate (the max is not as difficult to calculate).
Regards,
LuisG
|
| Report Abuse |
| Answer: |
Harry K. - 04/02/2013 11:50:12
|
| | Hi Luis,
You are correct. Important is the fact, that the centrifugal feedback causes a precession acceleration! Thus precession or orbit velocity increases with time, assumed there are no counter torques by friction and angular momentum.
I have found an equation to calculate the precession acceleration which keeps constant for a given angle. This fact is very interesting.
Anyway, at an angle of 30 degrees above horizontal plane provides the maximum feedback torque, because centrifugal TORQUE has it maximum at this angle and the acceleration occurs only due to the centrifugal torque.
Regards,
Harry
|
| Report Abuse |
| Answer: |
Blaze - 05/02/2013 02:43:02
|
| | Well, guys, I am not getting the same answer as the two of you (Luis & Harry). I get a maximum tilting torque from centrifuge at 36 degrees (I only did the calculation for every 3 degrees, not every degree).
I believe the reason that Luis and Harry get 30 degrees is because the formula that Harry provided was used by both of you, and I believe that it has an error. I did not use that formula but developed it independently on my own. The formula provided was as follows:
"- Centrifugal force = m(gyro) * W(Omega precession)^2 * cos(alpha) * R(radius pivot)
- Tilting torque by centrifuge = centrifugal force * sin(alpha) * R(radius pivot)"
If you use this you will be multiplying cos(alpha) and sin(alpha), which I believe would be incorrect. I believe the correct way to do it would be as follows:
Tilting torque by centrifuge = m(gyro) * W(Omega precession)^2 * sin(alpha) * R(radius pivot)*R(radius pivot)
When I multiply my results by cos(alpha) I also get the maximum tilting force due to centrifuge at 30 degrees, which, again, I believe may be incorrect. Of course, I could be the one that is incorrect.
What do you guys think?
The other thing I noticed in my calculations is that for various sized systems and especially for high speed systems (50 to 100 G's) the total tilting force changes, moving generally to a higher number of degrees. This would seem to make sense as the centrifugal force is a squared function so the high speed systems should be different from slow speed systems.
regards,
Blaze
|
| Report Abuse |
| Answer: |
Blaze - 05/02/2013 02:54:57
|
| | Woops! Typo. Make that:
Tilting torque by centrifuge = (m(gyro) * W(Omega precession)^2 * sin(alpha) / R(radius pivot))*R(radius pivot)
Blaze
|
| Report Abuse |
| Answer: |
Harry K. - 05/02/2013 11:30:44
|
| | Hello Blaze,
The by me provided formula can be find in any related physic book:
Centrifugal force Fz = m * R * w^2
Because the orbiting radius will change according the angle above horizontal plane, the correct equation of Fz in an orbiting system can be calculated by:
Fz = m * R * cos (alpha) * w^2 (1)
The tilting torque created by centrifugal force FZ can be calculated by:
Tilting torque Mz = Fz * R * sin (alpha) (2)
Equation (1) in equation (2):
Mz = m * R * cos (alpha) * w^2 * R * sin (alpha)
<-> Mz = m * R^2 * w^2 * cos (alpha) * sin (alpha)
Your provided formula:
"Tilting torque by centrifuge = (m(gyro) * W(Omega precession)^2 * sin(alpha) / R(radius pivot))*R(radius pivot)"
is wrong.
Thus your asumption:
"The other thing I noticed in my calculations is that for various sized systems and especially for high speed systems (50 to 100 G's) the total tilting force changes, moving generally to a higher number of degrees. This would seem to make sense as the centrifugal force is a squared function so the high speed systems should be different from slow speed systems."
is unfortunately also wrong. The angle of maximum torque by the influence of centrifugal torque remain constant, independently of size or other parameters of the gyro system.
Regards,
Harry
|
| Report Abuse |
| Answer: |
Blaze - 06/02/2013 01:19:51
|
| | Hi Harry. I see where I made a mistake and I also see where some of the confusion is coming from. I am going through my calculations again.
You are using angular velocity and I am using tangential velocity. When using angular velocity your equation 1 is as you say. When using tangential velocity your equation 1 would become F=m*(v^2)/r, where both v and r are modified by cos(alpha) for the angle being calculated. With either formula, this calculation provides the horizontal component of the centrifugal force. Your equation 2 is the calculation for the vertical component of the centrifugal force, the tilting torque. Of course, there is a vertical component only because the gyro arm is at an angle above (or below) horizontal.
However, in either case, wouldn't the tilting torque due to centrifuge use the tan function, not sine? When drawing a force triangle, the gyro arm is the hypotenuse, your equation 1 is the "adjacent side" and the "opposite side" is the vertical component (the tilting force). Since equation 1 is the adjacent side, not the hypotenuse, wouldn't the proper calculation to get the vertical component be using the tan function?
regards,
Blaze
|
| Report Abuse |
| Answer: |
Harry K. - 06/02/2013 18:25:51
|
| | Hello Blace,
One hint, use angular velocity instead of tangential velocity. This makes calculation work much easier!
The centrifugal force does not have a vertical component but only a horizontal component.
The distance of this centrifugal force in horizontal plane to the centre of hub rotation can be calculated by R (hypthenuse) * cos (alpha) which is identical with the adjacent side of angle alpha.
The resultant tilting torque is given by the horizontal centrifugal force component and the resultant lever arm given by angle alpha. Thus the lever arm length can be calculated either by R (hypothenuse) * sin (alpha) or by R * cos (alpha) (adjacent) * tan (alpha). I think the first choice is easier.
The resultant tilting torque, created by the horizontal centrifugal force component AND the vertical lever arm created by angle alpha, creates the vertical downward component which has to be addef to the tilting torque caused by gravity.
If angle alpha is below horizontal plane, the resultant tilting torque of centrifugal force has to be subtract from gravity tilting torque.
I hope this issue is now more clear. If not, please let me know.
Regards,
Harry
|
| Report Abuse |
| Answer: |
Blaze - 07/02/2013 03:16:27
|
| | Hi Harry. Yes, I understand what you said. After rereading what you said I see that we are both saying the same thing in different ways.
Harry & Luis:
However, I cannot figure how you get 30 degrees as the maximum tilting torque due to centrifuge. I have done it using your equations. I have done using my equations. I get the same numbers both ways and both ways I get the maximum tilting torque due to centrifuge is at 45 degrees, not 30 degrees. I will check my math yet again.
I also get an angle that changes for the total tilting torque depending on the input torque applied (gravity or a multiple of gravity), especially if I use 10+ G's for input torque (this speeds up the precession rate which increases the centrifuge by a squared amount for a given arm angle). To me this makes sense as there is a squared velocity term in the equation for centrifugal force. For a 10 G input torque, the centrifugal force is 100 times that of a 1 G input torque. With a 20 G input torque, the maximum tilting force due to centrifuge is larger than the actual input torque. Of course this is somewhat dependent on the other system parameters as well, but even for a relatively small system this is the case.
For example using an input torque of 200 m/s/s, a 2 pound flywheel (solid disk shape) with a 2 inch arm radius, a 2.5 inch flywheel radius spinning at 1400 rpm I get the total tilting torque at 15 degrees. However if I change the arm radius to 4 inches and keep everything else the same I get the total tilting torque at 39 degrees. Now increase the input torque to 2000 m/s/s with the 4 inch arm radius and the total tilting torque angle becomes 45 degrees. I have only done these calculations for every three degrees so the actual total tilting torque angles may be out by a degree or so.
Again, I have done this using your equations and mine and the numbers are the same but the angle for total tilting torque does not stay at a constant angle.
So, where am I going wrong?
regards,
Blaze
|
| Report Abuse |
| Answer: |
Harry K. - 08/02/2013 08:49:14
|
| | Hello Blaze,
I cannot verify your calculations because I don.t know what you mean with input torque = 200m/s/s?
Anyway, if you increase the tilting torque manually and shorten the orbiting radius, the influence of tilting torque by centrifuge decreases. As a result the angle of maximum tilting torque is getting smaller. However, the maximum angle of maximum tilting torque will not exceed 30 degrees.
Regards,
Harry
|
| Report Abuse |
| Answer: |
Blaze - 08/02/2013 18:50:25
|
| | Hi Harry.
Input torque was a poor choice of words. What I meant was instead of using gravity at 9.81 m/s/s I used 200m/s/s for calculations. Does that make more sense?
Using 200 m/s/s instead of 9.81 m/s/s means that the gyro will precess at about 20 times the rate normally it would for any given angle (196.2 m/s/s would be exactly 20 times). This would increase the centrifugal force by about 400 times due to the squared term in the equation. The tiltiling torque due to centrifugal force would still be zero at 0 and 90 degrees however, if you were to graph the torque due to centrifuge, the highest value on the graph would be much greater. That means that rate of climb of the torqe due to centrifuge from zero to its peak has changed, it it more rapid, but the rate of change of the tilting torque due to gravity (or 20 times gravity in this case) hasn't changed, so the total of the two tilting torques will change for a given angle. At least that is the way I see it right now.
I have gone over my spreadsheet calculations five times and still can't find my error. So, I am going to do the calculations from scratch by hand with my calculator (not with the spreadsheet) for both 30 and 45 degree angles as a way to check my math. It might help me discover my error. I usually don't have this much trouble discovering when or where I have made a mistake.
thanks for your help,
Blaze
|
| Report Abuse |
| Answer: |
Blaze - 09/02/2013 02:34:09
|
| | Well guys, I have now officially entered the "Twilight Zone".
I just did the calculations by hand for 30 degrees and 45 degrees and I get 45 degrees as being the larger tilting torque due to centrifugal force. I checked the calculation twice. Then I did this calculation for both angles in a separate spread sheet (not the one I was using before) and got the same thing.
I would suggest that you both do the calculation by hand for both 30 degrees and 45 degrees and see what you come up with.
regards,
Blaze
|
| Report Abuse |
| Answer: |
Blaze - 09/02/2013 02:35:52
|
| | P.S. I used the equations Harry provided (which are correct) to do the calculations by hand.
regards,
Blaze
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 09/02/2013 02:47:03
|
| | Hi Blaze,
The Twilight Zone would be easy to recover from…the gyro-zone, now there is something to be afraid of…
I will try to find some time this week-end to check my figures.
Cheers,
LuisG
|
| Report Abuse |
| Answer: |
Blaze - 09/02/2013 03:03:48
|
| | Hi Luis. Thanks in advance for checking it out.
From Harry's formula previously in this thread,
"Mz = m * R^2 * w^2 * cos (alpha) * sin (alpha)"
it would appear that everything before the cos(alpha) and sin (alpha) is a constant for a given angle. Therefore, if you multiply cos(30) by sin(30) to get an answer, and then separately multiply cos(45) by sin(45) to get an answer, and compare the two answers, that should tell you right away which one is bigger. When I do that, I get the answer for 45 degrees as being larger.
regards,
Blaze
|
| Report Abuse |
| Answer: |
Blaze - 09/02/2013 03:08:21
|
| | Hi Luis. Thanks in advance for checking it out.
From Harry's formula previously in this thread,
"Mz = m * R^2 * w^2 * cos (alpha) * sin (alpha)"
it would appear that everything before the cos(alpha) and sin (alpha) is a constant for a given angle. Therefore, if you multiply cos(30) by sin(30) to get an answer, and then separately multiply cos(45) by sin(45) to get an answer, and compare the two answers, that should tell you right away which one is bigger. When I do that, I get the answer for 45 degrees as being larger.
regards,
Blaze
|
| Report Abuse |
| Answer: |
Harry K. - 09/02/2013 09:26:02
|
| | Hi Blaze,
“ it would appear that everything before the cos(alpha) and sin (alpha) is a constant for a given angle. “
No, this is not the case, because precession velocity wp = Tilting torque Mt / angular momentum of spinning Gyro Lg is varying as well (Mt is varying according sinus function).
This is the error.
Regards,
Harry
|
| Report Abuse |
| Answer: |
Harry K. - 09/02/2013 09:30:24
|
| | Correction:
(Mt is varying according sinus function)
Must read as:
(Mt is varying according cosinus function)
|
| Report Abuse |
| Answer: |
Blaze - 09/02/2013 13:34:30
|
| | Well, it looks like I am one of the rare people who actually find there way out of the Twilight Zone and maybe even the "Gryo Zone". I found my mistake. Woooo Hoooo ! I woke up at 5:30 this morning realizing what I had done wrong.
As expected, it was something I was consistently overlooking. When calculating the precession speed (w), I didn't modify the arm length by cos(alpha). I did modify the arm length everywhere else in the equations, just not when I calculated the precession speed. I have corrected that error and found, just as Luis and Harry, that the maximum tilting torque due to centrifugal force is at 30 degrees. Doing the calculation by hand did help me find my error.
However, I still find that the total tilting force (centrifugal tilting force + tilting force due to gravity or a multiple of gravity) does vary from 0 degrees when using a small fraction of a G to the maximum of 30 degrees when using a large multiple of G.
thanks for your help and patience guys,
Blaze
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 09/02/2013 18:46:11
|
| | Hi Blaze,
Below is an equation that I used as an alternate way to check my figures, as the 30 degree MAX mark for Centrifugal Torque emerged from the BASIC equations presented by Harry, no more and no less.
My guess is that we can now agree on an exact single equation for Centrifugal Torque (i.e. the resulting equation once we have expanded “Omega” into Harry’s Centrifugal Force equation, and then combined that resulting equation into the one for Centrifugal Torque).
My single resulting equation for Centrifugal Torque (Tc) is as follows:
Tc = (M^3 * R^4 * G^2 * Cos^3 * Sin) / (m^2 * r^4 * w^2)
The terms involved should be obvious (“w” is the angular velocity of the flywheel). Please excuse the simplified “Sin” and “Cos”.
This equation brings the basic factors out in the open, making it relatively easy to determine what can happen to centrifugal-torque when we vary the magnitude of any-one, or any-combination of basic factors.
Note also that the factor (Cos^3 * Sin) is the only one that changes with the angle.
Best Regards,
LuisG
|
| Report Abuse |
| Answer: |
Harry K. - 09/02/2013 19:18:33
|
| | Hi Luis and Blaze,
I'm happy to read that we 3 guys are in agreement (did this ever happened before?)!
I will check the equation presented by Luis but I'm sure he is correct.
Now we have reached the first step, let's see if we can go ahead to reach further steps.
Best regards,
Harry
|
| Report Abuse |
| Answer: |
Harry K. - 09/02/2013 19:18:35
|
| | Hi Luis and Blaze,
I'm happy to read that we 3 guys are in agreement (did this ever happened before?)!
I will check the equation presented by Luis but I'm sure he is correct.
Now we have reached the first step, let's see if we can go ahead to reach further steps.
Best regards,
Harry
|
| Report Abuse |
| Answer: |
Blaze - 09/02/2013 23:49:12
|
| | Hi Luis. I agree with your formula. I get the same thing. Very nice, good work Luis. This formula helps to see small where changes to various parameters can make large changes to Tc due to the second, third and fourth power terms.
Interesting to note that you have mass in the top and bottom of the Tc equation. The value of this mass is usually ASSUMED to be the same mass (ie: the flywheel mass) especially when looking at the standard gyro precession equation. So one might assume that you could reduce the Tc equation to a single mass term on the top. That would be an incorrect assumption because the mass in the numerator and the denominator is in fact NEVER the same value. This is because there is always dead weight which is in the top mass term and not the bottom. When looking at the equation for gyro precession speed, if you have as much dead weight (axle, flywheel shielding, etc.) as spinning weight you will actually double the precession speed (mass in numerator is twice the mass in the denominator).
This extra dead weight mass also affects Tc, and that effect is NOT linear as you increase the dead weight.
cheers,
Blaze
|
| Report Abuse |
| Answer: |
Harry K. - 10/02/2013 09:44:40
|
| | Hello Luis,
A small correction of your presented equation, because mass inertia of a flat disk shaped flywheel JG = mG * rG^2 / 2:
Tc = (M^3 * R^4 * G^2 * Cos^3 * Sin) *4/ (m^2 * r^4 * w^2)
It is a good idea to separate dead mass md and spinning mass mg and thus this should be indicated in te equation more clearly:
Tc = ((md+mg)^3 * R^4 * G^2 * Cos^3 * Sin) *4/ (mg^2 * r^4 * w^2)
The sum of acting tilting torque of an overhunging gyro system can be calculated by:
Ts = Tc+Tg
Ts = ((md+mg)^3 * R^4 * G^2 * Cos^3 * Sin) *4/ (mg^2 * r^4 * w^2) + ((md+mg) * G * R * cos)
However, Tc and thus Ts increases during time. But this is the next step.
Regards,
Harry
|
| Report Abuse |
| Answer: |
Blaze - 10/02/2013 15:13:43
|
| | Hi Harry & Luis. The Tc equation presented by Luis is for the ideal flywheel shape. It is the "standard" Tc equation. Any shape other than the ideal flywheel shape (hockey puck, cone, rod, ball, etc.) would require a slight modification to the standard Tc equation as they all have different moments of inertia. In reality, even the standard Tc equation would have to be modified slightly because an ideal flywheel shape can't be made.
cheers,
Blaze
|
| Report Abuse |
| Answer: |
Harry K. - 10/02/2013 15:28:31
|
| | Hi Luis & Blaze,
The modified equation from Luis in my last post isn`t exactly correct, because I have assumed that the dead weight mass has the same distance as the overhunging flywheel but that is not the case in reality. Thus we have to take into account the centrifugal- and gravity component of the additional dead weight mass with the rotation distance to its centre of mass. This makes the equation unfortunately a bit more comprehensive.
I suggest to give names for each values:
a - angle alpha
G - earth gravity acceleration = 9,81 m/s^2
R - radius overhunging flywheel
Tg - tilting torque gyro (Tg = mg * G * R * cos(a))
Rg - radius spinning mass of the flywheel
mg - flywheel mass
Jg - inertial mass of flywheel (flat disk shape: Jg = mg * Rg^2 / 2)
wg - angular velocity of flywheel
Lg - angular momentum of flywheel (Lg = Jg * wg)
wp - precession velocity (wp =
md - dead weight mass
Rd - distance centre of mass of dead weight mass
Jd - inertial mass of dead weight (Jd = md * Rd^2)
Ld - angular momentum of dead wight mass (Ld = Jd / wp)
Tc - tilting torque by centrifugal force of flywheel
Tdg - tilting torque by dead weight mass - gravity component
Tdc - ilting torque by dead weight mass - centrifugal component
Td - tilting torque by dead weight mass
T - sum of all acting torques (T = Tg + Tc + Tdg + Tdc)
Calculation formula:
1. Tg = mg * G * R * cos(a)
2. Tc = mg * R^4 * G^2 * cos(a)^3 * Sin(a)) *4/ (Rg^4 * wg^2)
3. Tdg = md * G * Rd * cos(a)
4. Tdc = md * Rd^4 * G^2 * cos(a)^3 * Sin(a)) / (Rg^4 * wg^2)
5. T = Tg + Tc + Tdg + Tdc
->Inserted equations 1-4 and simplified:
T = (G^2 * cos(a)^3 * sin(a) * ((mg^2 * r^4) + (md^2 * Rd^4)) / (Rg^4 * wg^2)) + (cos(a) * G * ((mg * R) + (md * Rd)))
6. wp = T / Lg = T / (mg * Rg^2 / 2 * wg)
Hope i didn't overlooked something...
Best regards,
Harry
|
| Report Abuse |
| Answer: |
Harry K. - 10/02/2013 15:45:51
|
| | Blaze,
In principle you are right. However, we talking all the time from spinning "flywheels" and not from two opposite spinning mass points. Thus I think it is better to implement the disk shape factor into the formula. If one forgot it, the calaculation is at least wrong by factor 4.
However do it as you pleases but dn't forget it!
Regards,
Harry
|
| Report Abuse |
| Answer: |
Blaze - 10/02/2013 16:07:02
|
| | Hi Harry, Luis and all.
"Tc and thus Ts increases during time"
True but not quite complete. It is far more complicated than that. For a real world gyro, the dominant factor causing droop is friction. Although the asymptotic acceleration to steady state plays a part, it would quickly become insignificant compared to the friction that is increasing with increasing precession speed. This is friction is from friction at the pivot, friction in the flywheel bearings, windage, etc.
For inclination angles, as the fly wheel droops with time towards horizontal, the Tg increases. This is independent of precession speed.
However for inclination angles from 90 degrees to 30 degrees Tc increases and for inclination angles from 30 degrees to horizontal Tc decreases. Then there is the fact that the precession speed is increasing during droop so for inclination angles from 90 degrees to 30 degrees Tc will increase even more. For inclination angles from 30 degrees to horizontal the the addition precession speed adds to Tc and although Tc still decreases, it does so less quickly.
Although I have not yet done the math for declination angles, I believe it would be mostly the opposite that happens except that Tc is subtracted from Tg. Where it gets interesting is when you modify the system parameters to produce more Tc than Tg. This would cause the gyro to possibly rise and definitely cause the precession speed to slow or even stall, which would cause the gyro to drop and start to precess again. This bouncing motion (sounds something like nutation) should eventually damp itself out.
So the question that immediately comes to mind is can you get the same bouncing motion for inclination angles and if so how? I believe the answer is yes but for different reasons than for declination angles. Any one got an opinion on that?
best to all,
Blaze
|
| Report Abuse |
| Answer: |
Blaze - 10/02/2013 16:29:56
|
| | Hi Harry. Well, that equation is getting ugly now.
I would like to make a suggestion. Instead of:
Tg - tilting torque gyro (Tg = mg * G * R * cos(a))
Rg - radius spinning mass of the flywheel
mg - flywheel mass
Jg - inertial mass of flywheel (flat disk shape: Jg = mg * Rg^2 / 2)
wg - angular velocity of flywheel
Lg - angular momentum of flywheel (Lg = Jg * wg)
I would much prefer to use the subscript "f" for the flywheel parameters rather than "g" because you are are also using the "g" subscript elsewhere for gravity and it gets confusing.
I would suggest the following:
Tf - tilting torque gyro (Tg = mg * G * R * cos(a))
Rf - radius spinning mass of the flywheel
mf - flywheel mass
Jf - inertial mass of flywheel (flat disk shape: Jg = mg * Rg^2 / 2)
wf - angular velocity of flywheel
Lf - angular momentum of flywheel (Lg = Jg * wg)
I believe this would cause less potential confusion.
regards,
Blaze
|
| Report Abuse |
| Answer: |
Blaze - 10/02/2013 16:34:53
|
| | corrections:
leave Tg labeled as Tg, but Tg would be defined as Tg - tilting torque gyro (Tg = mf * G * R * cos(a))
Rf - radius spinning mass of the flywheel
mf - flywheel mass
Jf - inertial mass of flywheel (flat disk shape: Jf = mf * Rf^2 / 2)
wf - angular velocity of flywheel
Lf - angular momentum of flywheel (Lf = Jf * wf)
regards,
Blaze
|
| Report Abuse |
| Answer: |
Harry K. - 10/02/2013 23:27:43
|
| | Hi Blaze,
Your renaming of the variables are acceptable for me.
Regarding your statements about influence of friction, etc. I have stated im my posting from 04.02.2013:
“ Thus precession or orbit velocity increases with time, assumed there are no counter torques by friction and angular momentum“.
It's clear that we have to consider counter torques caused by friction and energy stored in form of angular momentum in precession velocity. Otherwise we had found a perpetuum mobile which accelerates itself without energy input.
The angle of 30 degrees does only indicate the gyro position of lowest resistance or maximum output reaction in regard of precession velocity, nothing more and nothing less.
Let's go ahead step by step.
Regards,
Harry
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 24/02/2013 19:02:57
|
| | Hi Guys,
How is the equations pursuit coming along?
I have been working on a somewhat different equation, which at first glance appears to be Unrelated to the ones in this thread, as the new equation involves Laithewaite’s gyro-on-a-string experiment (Christmas Lecture #9). (I will probably put it in a different thread to prevent premature mix-up).
My goal is to compare the forces discussed in this thread to the forces handled by the Simple equation I am currently working on.
(The challenging part is presenting the logic in a compelling and easy to understand manner).
This comparison should enable us to determine whether there is a discrepancy in one of the force symmetries.
Such a discrepancy would provide a trail as to where and how angular motions may spill into linear acceleration (that should be interesting).
Is this perhaps what is happening in Nitro’s interesting trolley experiment? Nitro has made a connection between his trolley and pendulum experiments (of course all the dots will need to be properly connected).
This whole gyro pursuit is fascinating (whether it is possible or not), and we sould remain skeptic until proven.
Regards,
Luis G
|
| Report Abuse |
| Answer: |
Blaze - 24/02/2013 19:52:10
|
| | Hi Luis. I have also been thinking about gyros precessing on strings. Although I have not had time to go into any equations on this subject just yet, I would think that the centrifugal force of the gyro mass is causing the string to "cone". The amount of potential energy gained (height the gyro climbs against gravity) due to the string coning should equal the amount of centrifugal force due to the precession rpm.
Just some of my initial thoughts,
cheers,
Blaze
|
| Report Abuse |
| Answer: |
Blaze - 24/02/2013 19:55:42
|
| | P.S.
perhaps "equal" is a poor choice of words, "related" may be a better word.
Blaze
|
| Report Abuse |
| Answer: |
Harry K. - 24/02/2013 20:40:17
|
| | Hi Luis,
I have continued last weekend the equation work and added the aceleration part. Thus I'm now able to calculate the hub velocity after a defined time period, however, without consideration of any energy losses (this will be the next step).
I agree with you to start a new thread to present your new equation concept.
Regards,
Harry
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 27/02/2013 03:04:36
|
| | Hi Harry,
It sounds like great work.
I am currently working a new contract that requires much of my time, and have also found an exciting new hobby.
Hi Blaze,
Sometimes it requires time to find things that then become obvious.
Regards,
Luis G
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 02/03/2013 17:56:25
|
| | Hi Guys,
The new contract is growing, requiring many interesting components (the new hobby, in electronics, is also evolving quickly).
Meanwhile, I manipulated my gyro-on-string Equation to predict the Base-Radii of Coning formed by hanging gyros; this was a very satisfying discovery.
Of course you need to weigh and measure the components involved, including the gyro-spin-rate, and length of the tether-string.
The rewarding part is in discovering (with me it’s all about the equations and the simple experiments that prove them true).
I had though sharing would have been fun too, but it has proven to be exactly the opposite.
So, I will stop posting my discoveries (which should be OK for everyone).
I know that with a little thinking Harry, Blaze, Ravi, Momentus et al will have no trouble deriving the right equations too (these equations show that all gyroscopic actions on a string have equal-and-opposite reactions).
Those who have no interest or mathematical ability should have no use for them.
Let’s continue doing what we enjoy.
Regards,
Luis G
|
| Report Abuse |
| Answer: |
Blaze - 02/03/2013 19:39:06
|
| | Luis, it will not be ok with me if you stop posting. I value your insight and knowledge, even if it doesn't always agree with my ideas. That is ok because more than once I have gained further understanding from your comments.
If you stop contributing to this forum, it will be a great loss.
Sincerely,
Blaze
|
| Report Abuse |
| Answer: |
Glenn Hawkins - 03/03/2013 00:06:00
|
| | Sometimes I get down.
Post your equations
People read here who never reply.
You have been heard.
This for you. http://www.youtube.com/watch?v=V4iieg4j-4Q
|
| Report Abuse |
| Add an Answer >> |
