# How to quantify gyroscopic precession torque?

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• rcgldr
rcgldr
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In the Wikipedia article on counter-steering, it mentions a roll moment (torque) induced by precession:

https://en.wikipedia.org/wiki/Countersteering#Gyroscopic_effects

It states "The magnitude of this moment is proportional to the moment of inertia of the front wheel, its spin rate (forward motion), the rate that the rider turns the front wheel by applying a torque to the handlebars, and the cosine of the angle between the steering axis and the vertical." The Wikipedia article includes an example that calculates the precession related roll torque being about 12% of the total roll torque, most of which is due to lateral forces between pavement and contact patch. I don't understand how this is quantified, since it would seem that angular inertia about the roll axis would also be a factor, such as a 200 lb motorcycle versus a 400 lb motorcycle.

A classic example of gyroscopic precession is a gyro supported at one end of its axis, with gravity exerting a downwards roll torque, the gyro precessing about a vertical yaw axis, with an upwards counter-torque opposing gravity's downward torque so that the gyro remains horizontal (or nearly so) while precessing. The only torque exerted onto the gyro is the roll torque.

However, in addition to the downwards roll torque from gravity, say there is a yaw torque also applied that prevents precession. How much yaw torque would it take to stop the precession, and what would the reaction be? It would seem that if precession is prevented, then the gyro would simply drop due to the downwards roll torque from gravity, as if the gyro was not spinning, since the gyro's counter-torque to keep it horizontal is due to it's precession, and if the precession is prevented, there is no counter-torque.

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rcgldr said:
How much yaw torque would it take to stop the precession, and what would the reaction be?
You cannot stop the precession in the horizontal plane (spin axis yaw) with a pure vertical (yaw) torque. You need to apply a torque that cancels the horizontal torque from gravity, such that the total external torque (including gravity) has no horizontal component.

It might be simpler to consider the torques around the center of mass.

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rcgldr said:
In the Wikipedia article on counter-steering, it mentions a roll moment (torque) induced by precession:
You can’t get more work out than the one you put into the direct-steering mechanism.
Therefore, torque times rotational velocity in and out should be very similar, and a way to do your calculation.

Please, see:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/11-3-precession-of-a-gyroscope/

rcgldr said:
I don't understand how this is quantified, since it would seem that angular inertia about the roll axis would also be a factor, such as a 200 lb motorcycle versus a 400 lb motorcycle.
The resistive torque increases only as rotating velocity (forward speed of the bike) and/or moment of inertia of the front wheel increase.
The steering mechanism is not “aware” of the weight of the rest of the bike, which is pivoting about and following the trajectory of the steering column.

It is common to bend handlebars of relatively light racing motorcycles, due only to the necessary input torque to deviate a fast and light front wheel from its normal plane of rotation.
That input torque is much smaller in the case of a heavy cruiser moving at a slower velocity.

Consider also that when counter-steering stops, the opposite effect happens, which tend to stop the leaning or rolling movement of the chassis.

A.T. said:
You cannot stop the precession in the horizontal plane (spin axis yaw) with a pure vertical (yaw) torque.
So how does the screwdriver exert anything but a yaw torque in this video, from the 0:58 (link is set to 0:58) to 1:25 into the video, where the third time is the clearest example? Again, I don't know how to quantify the torque needed to stop the precession.

Later in the video, a precessing gyro is dropped and stops rotating about the yaw axis once in free fall, and the comments suggest this was unexpected, but the reason for the gyro not continuing to rotate about the yaw axis, is due to the angular momentum of spinning gyro.

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Lnewqban said:
The steering mechanism is not “aware” of the weight of the rest of the bike, which is pivoting about and following the trajectory of the steering column.
Motorcycles articles note that the amount of yaw torque exerted onto the handlebars needed to cause a motorcycle to change lean angle at some rate is a function of speed and roll axis angular inertia of the motorcycle. However, this could be due to the out-tracking component related to contact patch forces that need to be larger (requiring more counter-steering torque applied to handlebars ) for a heavier motorcycle, and unrelated to any gryo reactions.

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rcgldr said:
So how does the screwdriver exert anything but a yaw torque in this video, from the 0:58 where the link is set to to 1:25 into the video?
1) It's not clear how close to vertical the screwdriver is held, which combined with the shift of the base makes it unclear how close to horizontal the force by the screwdriver is.

2) The screwdriver is not the only thing exerting a torque around the center of mass here. The support does too. The base shift when the screwdriver hits the stop, indicates that the support force changes and so does it's torque around the CoM.

rcgldr said:
Again, I don't know how to quantify the torque needed to stop the precession.
The total applied torque around the center of mass (from screwdriver & support) equals the rate of change of angular momentum (which is approximately along the spin axis). To stop the precession of the axis in the horizontal plane that total applied torque must not have any horizontal component.

A.T. said:
1) It's not clear how close to vertical the screwdriver is held, which combined with the shift of the base makes it unclear how close to horizontal the force by the screwdriver is.
While the second try, is not quite vertical, the third try, seems pretty close vertical,

Back to the motorcycle case. The roll moment (torque) is related to the rate of change about the yaw axis, which would seem to require a second resistive torque to roll axis precession. The front tire is not free to precess due to lateral forces from the pavement and the angular inertial of the bike (it's not clear what effect the angular momentum of the rear tire has on this), and it would seem the combined effect of the two torques is what allows the front tire to be steered, rotated about the yaw axis. The effort to force the that rate of change about the yaw axis may be affected by the angular inertia of the bike, so perhaps that can be ignored if only considering the yaw rate of the front wheel, as opposed to the applied yaw torque.

rcgldr said:
While the second try, is not quite vertical, the third try, seems pretty close vertical,
Even if the screwdriver was vertical, the force by it can have a vertical component (friction). Even if the screw driver force was horizontal (no friction), the support force doesn't have to be vertical (see the base shifting).

For the spin axis to drop down, the total applied torque from screwdriver + support (change of angular momentum) must point down.

A.T. said:
For the spin axis to drop down, the total applied torque from screwdriver + support (change of angular momentum) must point down.
Gravity is already applying a downwards torque onto the gyro. In order for the gyro to fall, the gyro just has to stop exerting an upwards counter-torque, which is what happens if the gyro is prevented from precessing. I mention this in my original question. What I'm wondering is how much torque is required to stop precession, or something equivalent to an impulse, the torque x time.

rcgldr said:
Motorcycles articles note that the amount of yaw torque exerted onto the handlebars needed to cause a motorcycle to change lean angle at some rate is a function of speed and roll axis angular inertia of the motorcycle. However, this could be due to the out-tracking component related to contact patch forces that need to be larger (so a larger steering angle) for a heavier motorcycle, and unrelated to any gryo reactions.
In my practical experience, handlebar input yaw and chassis rolling are not directly linked regarding moments.
The former only deviates the linear trajectory, while the latter is driven by the subsequent centrifugal effect of the brief curvilinear trajectory on the center of mass (during the counter-steering).

For the same handlebar input (regarding angle range and rotational velocity) the rate of the roll only depends on the height of the CoM and the mass (moment of inertia).

For a quicker chassis roll to be induced, the angle range and/or rotational velocity of the steering input most increase (causing the centrifugal effect on the CoM to increase as well).
In that case, the reaction or opposition torque of the spinning front wheel (gyroscope) will be greater.

The roll of the chassis also causes precession on the rear wheel and crankshaft (6K to 12K rpm), which reaction is a lateral force on the steering column (yaw moment perpendicular to the road).

Please, see:
https://motochassis.com/Articles/Balance/BALANCE.htm

https://books.google.com/books?id=84hF-qoR5I8C&printsec=frontcover#v=onepage&q&f=true

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rcgldr said:
Gravity is already applying a downwards torque onto the gyro.
I was considering the torques around the center of mass, where the torque from gravity is zero. But if you use the support as reference point, the torque from gravity is a horizontal vector. It's not pointing downwards.

Make sure you understand torque and angular momentum as vector quantities:
http://hyperphysics.phy-astr.gsu.edu/hbase/rotv2.html

rcgldr said:
In order for the gyro to fall, the gyro just has to stop exerting an upwards counter-torque, which is what happens if the gyro is prevented from precessing.
The gyro isn't "prevented from precessing". The precession just changes direction, when the total external torque changes. For example if you apply an additonal force with the screwdriver.

rcgldr said:
What I'm wondering is how much torque is required to stop precession,
Again, think in terms of changing precession, not stoping it.

rcgldr said:
or something equivalent to an impulse, the torque x time.
Look at the definition of net torque. It is the rate of change of total angular momentum. If you know how you want the total angular momentum to change, then you know what net torque you have to apply. All this has to be done vectorially.

Lnewqban said:
For the same handlebar input (regarding angle range and rotational velocity) the rate of the roll only depends on the height of the CoM and the mass (moment of inertia).
The rate of roll depends on how much the front tire is counter-steered outwards and speed.
As mentioned in the motochassis web site, trail causes the front tire to steer into the direction the bike is leaned (roll axis). Not mentioned is that angular momentum of the front tire opposes the self-correcting steering response of trail, acting as a damper to prevent over-correction at normal speeds, but at high speed, 100+ mph, the angular momentum is so great that it dominates any trail response. For an imaginary bike with infinitely thin tires, this is described as capsize mode, where a bike falls inwards at a very slow rate. The wiki article on bike dynamics shows an Eigen value plot of the various stability modes (capsize mode on the right):

https://en.wikipedia.org/wiki/Bicycle_and_motorcycle_dynamics#Eigenvalues

For a real motorcycle with normal width tires, what happens at 100+ mph is the bike will tend to hold it's current lean angle (any rate of change in lean angle is so small that it's imperceptible). and the rider has to counter-steer inwards to straighten the bike back to vertical.

The other issue at high speeds is the relatively large amount of torque the rider has to apply to the handlebars to get the bile to change lean angle. This is noticeable at highway speeds, the amount of counter-steering torque needed to lean at 35 mph is much less than at 60 mph or faster. For the motorcycle racers taking turns at 200+ mph at Isle of Mann, they have to squeeze the tank with their legs to get enough leverage to apply enough torque on the handlebars to get the bikes to lean.

In the case of electric unicycles, at moderate speeds, the rider can just apply differential pressure to the pedals to get the EUC to tilt, which will then steer due to camber effect. At 30 to 50 mph, the EUC is barely tilted, but even that small amount of tilt requires the rider exerts an inwards force onto the outer upper pad and an outwards force on the outer spiked pedal in order to generate enough torque to force the EUC to tilt inwards:

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rcgldr said:
Gravity is already applying a downwards torque onto the gyro. In order for the gyro to fall, the gyro just has to stop exerting an upwards counter-torque, which is what happens if the gyro is prevented from precessing.

A.T. said:
I was considering the torques around the center of mass, where the torque from gravity is zero. But if you use the support as reference point, the torque from gravity is a horizontal vector. It's not pointing downwards.

The gyro isn't "prevented from precessing". The precession just changes direction, when the total external torque changes. For example if you apply an additional force with the screwdriver.
By downwards torque, I mean a roll torque that corresponds to the downwards force of gravity at the center of mass and the upwards force at the supporting post.

When gravity alone is acting on the gyro, the rate of precession is relatively slow. When the screwdriver stops the yaw rotation, the gyro drops straight down at the same rate (or nearly so) as it would if the gyro were not spinning. If it was precessing downwards, it would be doing so at a much slower rate of angular velocity.

rcgldr said:
By downwards torque, I mean a roll torque that corresponds to the downwards force of gravity at the center of mass and the upwards force at the supporting post.
Yes, but keep in mind that this torque vector is horizontal.
rcgldr said:
If it was precessing downwards, it would be doing so at a much slower rate of angular velocity.
Without knowing the exact forces and torques being applied by screwdriver & support you have no basis to say how fast the spin axis should change direction downwards. Per definition, how fast the angular momentum vector changes depends on how much net torque is being applied.

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rcgldr said:
The rate of roll depends on how much the front tire is counter-steered outwards and speed.
Many decades of riding and racing tell me otherwise.
It seems that our experiences do and will differ.

Going back to the estimation of the magnitude of the gyroscopic moments, we also have the case of tail-dragger airplanes turning to the left while the tail is being raised from the ground and the nose propeller is pitching down during take off.

Judging by the magnitude of the deviation of the tail from the movement direction in both phases, it seems to me that both moments, in and out, are of comparable magnitude and acting on the vertical and horizontal planes (at 90 degrees from each other).

Reference, if interested:
https://www.boldmethod.com/learn-to...takeoff-to-stay-on-centerline-during-takeoff/

rcgldr said:
The rate of roll depends on how much the front tire is counter-steered outwards and speed.

Lnewqban said:
Many decades of riding and racing tell me otherwise.
It seems that our experiences do and will differ.

The speed only affects the precession related roll torque (wiki article calls it roll moment), which in the wiki example, accounts for about 12% of the total roll torque that leans the bike. The primary 88% of that roll torque is due to the out-tracking of the tires and the lateral forces at the contact patches.

The more noticeable effect of speed is the amount of applied torque required to achieve a roll rate increases with speed. The other effect of high speed is precession being too slow and resulting in understeer due to opposing the torque from trail, but doesn't quantify it, which is what I'm looking for. As I posted previously, at moderate speeds, this precession acts as a damper, which doesn't prevent self-correction, but does reduce tendency to over-correct (wobble).

" ... At high forward speeds, the precession is usually too slow, contributing to an uncontrolled bike's tendency to understeer and eventually fall over without ever having reached the upright position. This instability is very slow ... "

https://en.wikipedia.org/wiki/Bicycle_and_motorcycle_dynamics#Gyroscopic_effects

This is also noted later in the Eigen values section, where this is described as capsize mode with infinitely thin tires. As I posted before, with real tires, instead of a slow fall inwards, the tendency at high speeds (100+ mph) is to hold the current lean angle (if there is a rate of change, it's so small that it's imperceptible).

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rcgldr said:
When gravity alone is acting on the gyro, the rate of precession is relatively slow. When the screwdriver stops the yaw rotation, the gyro drops straight down at the same rate (or nearly so) as it would if the gyro were not spinning. If it was precessing downwards, it would be doing so at a much slower rate of angular velocity.

A.T. said:
Without knowing the exact forces and torques being applied by screwdriver & support you have no basis to say how fast the spin axis should change direction downwards. Per definition, how fast the angular momentum vector changes depends on how much net torque is being applied.

The torque related to the screwdriver is significantly less than the torque related to gravity.

In this video a plastic soda straw size tube is used to change the precession rate to raise, lower, and drop the gryo. At 4:14 into the video, he stops the precession that just after spin up, and the gyro quickly drops (does not precess). At 4:45 into the video, you can see that the straw is not capable of lifting the gyro, it can't generate anywhere near as much force as gravity does, and the torque induced by that straw is much less than the torque induced by gravity.

The scale reading will only change when the gyro is accelerating upwards or downward during transitions, not when the gyro is steadily (or nearly so) moving upwards or downwards, which is why the title mentions "appears" to violate Newton's third law (from a video he references in his video). He should have clarified this point. There is a comment from ColinWatters (7 months ago) that specifically mentions this, comparing it to the sensation in a elevator were a person only feels the effects of acceleration or deceleration, not while moving at constant speed.

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rcgldr said:
The torque related to the screwdriver is significantly less than the torque related to gravity.
What matters is the total torque from screwdriver, support and gravity.

rcgldr said:
the gyro quickly drops (does not precess).
How does "precessing down" differ from "dropping" here.

rcgldr said:
The scale reading ...
The scale measures only one force component. It doesn't measure the other two force components, and none of the three torque components. You would need a 6-axis force/torque sensor for that.

A.T. said:
How does "precessing down" differ from "dropping" here.
The video shows the rate of precession due the torque from gravity, that the torque from the "straw" is much less than the torque from gravity, but the rate of "drop" about roll axis when the straw stops precession is greater than the rate of precession when the only torque is from gravity. The rate of precession is proportional to the applied torque. Torque about the roll axis due to gravity corresponds to precession about the yaw axis, while the smaller amount torque about the yaw axis due to straw corresponds to precession about the roll axis. The torque from the straw is smaller than the torque from gravity, yet the gyro "drops" at a faster rate than it precesses due to gravity alone.

With the gyro pointed to the right, using right hand rule, the angular momentum vector of the gyro points to the right, the gravity torque vector points away from the viewer, and the precession is in the direction of the gravity torque vector. The straw torque vector points down, and absent gravity, precession would be in the direction of the straw torque vector (down). Adding the two torque vectors results in a vector that points away from the viewer and slightly down, but the precession ends up being straight down and at a faster rate, not away and slightly down. This is part of what I'm asking about, the amount of torque needed to prevent precession is less than the torque that would otherwise induce precession, but I don't know how to quantify it (what the math would be).

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rcgldr said:
The torque from the straw is smaller than the torque from gravity,
We don't know this, given the vastly differnt lever arms. And you forgot the torque from the support again.

rcgldr said:
the torque from the "straw" is much less than the torque from gravity

A.T. said:
We don't know this, given the vastly different lever arms. And you forgot the torque from the support again.
Assume an idealized support free to rotate about any axis, so that it does not generate any torque.

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rcgldr said:
Ignoring friction and inertia, an ideal support allows the gyro to freely spin, precess, and|or move vertically, so an ideal support would not exert any torque.
This is only true for a ball joint (3 rotational degrees of freedom). The joint in the video looks like 2 DOF joint, and when the bar is constrained by hand it becomes a 1 DOF joint (hinge). Such a joint, even when frictions-less, can transmit any torque that is not parallel to its axis.

rcgldr said:
Again, I don't know the formula to calculate the amount of torque required to prevent precession
Again, you don't prevent precession, you divert precession. How much net torque is needed for this depends on how quickly you divert the precession. How much of that net torque is provided by a specific element (like the bar) depends on the geometry of your support joint.

A.T. said:
Again, you don't prevent precession
Missed an edit so posting here. There does seem to be a formula, but it is based on rate of rotation (about roll axis in my posts), not a torque (about roll axis).

After reading the Wikipedia article again, specifically "roll moment (torque) is proportional to the rate that the rider turns the front wheel", as opposed the amount of torque applied to the front wheel would seem to imply something similar for the case I'm wondering about, that the torque required to stop precession about the yaw axis is proportional to the rate that the gyro rotates about its roll axis and not the roll axis torque. This would explain why released gyro has to drop a bit to result in the torque in the direction of precession that initiates the precession. Once the rotation in the direction of precession starts, this starts the torque that opposes the torque related to gravity.

In the video where the precession was gradually slowed and stopped by a straw, the rate of rotation about the gyro's roll axis was small, so not much torque was required. In the video where the screwdriver is used to quickly stop precession, the gyro's rate of rotation quickly increased just before it hit the surface, and would have required more torque to stop the precession.

A secondary question is what happens if the precession stops? At physics stack exchange the given answer is that it just falls. In this video from Cal Tech, the gryo is counter-balanced so that there is a relatively small amount of roll torque, and when precession is stopped, it "falls" slowly due to the counter-balancing:

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rcgldr said:
A secondary question is what happens if the precession stops? At physics stack exchange the given answer is that it just falls. In this video from Cal Tech, the gryo is counter-balanced so that there is a relatively small amount of roll torque, and when precession is stopped, it "falls" slowly due to the counter-balancing:

In physics terms, "precession is stopped" actually means: additional torque is applied, which changes the total torque and thus the precession direction.

A.T. said:
In physics terms, "precession is stopped" actually means: additional torque is applied, which changes the total torque and thus the precession direction.
Quoting from that physics stack exchange post about preventing precession: "... obstacles prevent onset of precessing motion. Without that precessing motion there is no opposition to the exerted torque. As a consequence the wheel will just keep yielding to that torque."

https://physics.stackexchange.com/a/816506

The key point: the wheel (gyro) yields to the torque as opposed to precessing in a different direction,

What I was originally missing was that the torque that induces precession about a yaw axis is a reaction to rotation about the roll axis, not to the torque about the roll axis. Likewise the torque about the roll axis that opposes gravity is a reaction to the rotation about the yaw axis. If the rotation about the yaw axis is prevented, there is no torque to oppose the torque related to gravity, and the gyro just falls about the support pivot point.

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rcgldr said:
https://physics.stackexchange.com/a/816506

The key point: the wheel (gyro) yields to the torque as opposed to precessing in a different direction,
To change the angular momentum vector, you have to apply a net torque vector in the direction of that change. Inventing different names for that process cannot change that fact.

rcgldr said:
I don't know how to quantify the torque needed to stop the precession.
Maybe this is the right way to think about it:

When the gyro is doing the slow yaw-precession, the angular momentum vector is not exactly parallel to the spin axis of the gyro, but has a small additional vertical component due to the slow yaw-precession motion. To stop the yaw-precession you have to remove that vertical angular momentum component. I think this vertical component is the same as the angular momentum of the non-spinning gyro that performs only the slow yaw-precession motion, so that is the angular momentum your stopper has to transfer.

What happens next: Gravity still pumps horizontal angular momentum into the gyro, but since now this cannot go into the yaw-precession of the spin axis, it results in a different motion that is consistent with its angular momentum: When the spin axis is dropping vertically fast, the angular momentum vector has a quickly increasing horizontal offset from the spin axis, corresponding to the accumulating horizontal angular momentum from gravity. This means that you do not have to cancel the horizontal torque from gravity in order for the spin axis to drop vertically.

A.T. said:
cancel the horizontal torque from gravity in order for the spin axis to drop vertically.
Canceling (opposing) the horizontal torque from gravity is what occurs during precession. Gravity is exerting a horizontal torque, but a precessing gyro is not yielding to that torque and there is no horizontal component of rotation, only a vertical component.

Having a gyro pivoting about the end of one axis combined with the spinning of the gyro creates a relative velocity effect during precession. From an inertial frame of reference, when precessing, particles flowing through the bottom edge of the gyro have a greater magnitude of velocity than the particles at the top. The particles flowing through the bottom experience a greater amount of centripetal acceleration about the support pivot, and a coexisting greater amount of centrifugal reaction force than the particles at the top, resulting in a torque that opposes the horizontal torque from gravity. If the precession is prevented or stopped, then the magnitude of velocity of the particles flowing through the bottom and top edges is the same, eliminating the torque that opposes the horizontal torque of gravity, so the gyro yields to the horizontal torque of gravity.

This is a different situation than a gimbal mounted gyro where an inner gimbal is locked to hold a gyro's axis horizontal and prevent precession. I'm wondering how this affects the amount of vertical torque it takes to force the gyro to rotate at some rate about a vertical axis, and how much horizontal torque is applied to the inner gimbal that prevents precession.

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rcgldr said:
Canceling (opposing) the horizontal torque from gravity is what occurs during precession.
This is not what "canceling" usually means. If the gyro is precessing, then the net torque on it is not zero, so the torque from gravity is not cancelled. Precession is not a static equilibrium where all torques cancel.

This ambiguity of natural language makes those informal descriptions rather useless, when your goal is to quantify something. You should instead use vector math, and account for the offset between the spin-axis of the gyro, and the total angular momentum vector of the gyro, when the spin-axis itself changes direction.

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A.T. said:
This is not what "canceling" usually means.
In a steady precession, an internal roll torque reaction to yaw "cancels" an external roll torque related to gravity, and the gyro precesses with no vertical component of movement (assuming no losses). This doesn't change the fact that there is only one external torque that results in precession.

It's easier to get an idea of what is going on when there is nutation. When a gyro supported at one end of its axis is released, nutation is started. The gyro initially "drops a bit", and only after the drop has started does the gyro begin accelerating in the direction of precession. This is the beginning of nutation where the gyro lowers, yaw rate increases, gyro rises, yaw rate decreases, gyro lowers, ... . My assumption is that energy is conserved (assuming no losses), gravitational potential energy decreasing coexistent with gyro angular energy increasing and vice versa. The nutation dampens out, resulting in a steady yaw rate at some height with no vertical component of movement.

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rcgldr said:
...an internal roll torque reaction to yaw "cancels" an external roll torque...
"Internal torque canceling an external torque" is exactly the kind of confused talk that leads nowhere.

rcgldr said:
This doesn't change the fact that there is only one external torque that results in precession.
External torques and forces are all we care about.

rcgldr said:
The gyro initially "drops a bit", and only after the drop has started does the gyro begin accelerating in the direction of precession.
Because of angular momentum and energy conservation the gyro cannot just start precessing, while maintaining exactly the same spin rate and spin axis inclination. The precession motion itself has its own small angular momentum component around the vertical axis, which must be reflected by an equal but opposite change in the vertical angular momentum from the spin.

And it is that small angular momentum component around the vertical axis that you need to remove to stop the yaw-precession.

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rcgldr said:
The torque from the straw is smaller than the torque from gravity, yet the gyro "drops" at a faster rate than it precesses due to gravity alone.
Here a linear momentum analogy that might help to understand why it drops so fast:

Linear momentum:
The force of gravity acting on a body in circular orbit continuously changes direction as the body moves along the orbit, so the linear momentum transferred by gravity doesn't accumulate, it just goes into continuously changing the direction of the bodies linear momentum. When you prevent it from moving tangentially the force of gravity has a fixed direction, so the linear momentum transferred by gravity quickly accumulates, and the body drops to the center fast.

Angular momentum:
The torque from gravity acting on a horizontally precessing gyro continuously changes direction as the gyro precesses, so the angular momentum transferred by gravity doesn't accumulate, it just goes into continuously changing the direction of the gyros angular momentum. When you prevent it from precessing horizontally the torque from gravity has a fixed direction, so the angular momentum transferred by gravity quickly accumulates, and the spin axis drops down fast.

Lnewqban
A.T. said:
angular momentum ... gravity
Doing the math from the Wikipedia article, if there was no gravity, and instead the axis support allowed yaw rotation, but not roll rotation, then if a yaw torque was applied to the gyro until the same rate of yaw as gravity induced precession, then the induced roll torque exerted onto the support would have the same magnitude but opposite direction from the torque that gravity would exert if present. Once that same rate of yaw was reached, the yaw torque would go to zero, and there would be a Newton third law like pair of roll torques: the gyro exerting an "upwards' roll torque onto the support that is preventing roll rotation, coexistent with that support exerting a "downwards" roll torque onto the gyro.

The support would need a reactive external torque to prevent the roll rotation, such as being attached to some massive object. Once the yaw torque was removed, then the gyro's change in angular momentum would be balanced by an opposing change in angular momentum of the massive object, and the angular momentum of gyro and object would be conserved. In this closed system, there are no external torques, since the total angular momentum is not changing. This what I meant by the "internal torque" that opposes gravity in the original case.

rcgldr said:
"downwards" roll torque onto the gyro.
Calling the roll torque "downwards" is a bad idea, even if you put it in quotes. That torque vector is horizontal, and continuously changes direction in the horizontal plane during yaw-precression. This continuous change of torque direction is key to why angular momentum doesn't accumulate during yaw-precression, but does accumulate when yaw-precression is prevented (see post #32). Calling that torque "downwards" doesn't allow to distinguish these two cases, which makes it a particularly bad naming choice here.

rcgldr said:
... This what I meant by the "internal torque" that opposes gravity in the original case.
Whatever complicated rationale, based on comparing different scenarios, you had for this terminology, it just sounds dead wrong, to say that an internal torque balances an external one for a specific scenario. Just like with your "downward torque" above, your terminology choices make it really hard to follow you.

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A.T. said:
Calling the roll torque "downwards" is a bad idea
OK, but the key point was the torques involved with a gyro precessing at some specific rate about a vertical axis is the same for the gravity case or the support that only allows rotation about a vertical axis case, and in the support case, the yaw rate which is the same as the precession rate in the gravity case coexists with a roll torque that opposes the roll torque from the support, and the roll torque from the support is the same as the roll torque from gravity in the gravity case.

Last edited:

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