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Electric unicycle and bicycle dynamics - gyro effects on steering


rcgldr

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1 hour ago, mrelwood said:

gyroscopic precession only exhibits itself while the roll/tilt of the wheel changes.

Or yaw changes, which would tend to tilt a wheel upwards, but the rider holds a wheel at a fixed tilt angle during a turn, preventing precession, and precession response is normally low because at slow speed the angular momentum is small and at high speed the rate of yaw is small. 

As I commented before, a rider has to adjust tilt to balance for lean angle, and if there is some precession related torque, it wouldn't feel any different than having to adjust for camber effect. I'm also thinking that the torque used by the rider to hold a fixed tilt angle reduces any yaw | precession effect. 

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11 hours ago, mrelwood said:

What the test confirms is what @rcgldr was saying from the beginning, that gyroscopic precession only exhibits itself while the roll/tilt of the wheel changes. No matter how hard I'd have to push the wheel to retain a roll angle and fight the camber effect or gravity, precession only takes place as the roll angle changes. And the magnitude of the precession is notable only when the roll/tilt angle changes rapidly.

By my rough estimate, I rarely tilt/roll the wheel during riding as fast as I did in this test. So I'd say that precession rarely participates in creating my turns. 

Thank you for making the video. 

What you said can be another way of saying torque.

Precession responds to torque. The higher the input torque, the higher the output torque. But since the angular momentum of the wheel is not high enough, in your first try (low input torque), you were not able to readily observed output torque. But, in your second try, you applied a large enough torque, and as a result you were able to observed the reaction.

Analogous to linear motion, the higher the applied torque, the higher the angular acceleration. And as a result, an increase in angular velocity, and angular displacement. 

But because it is torque and not angular velocity, it is possible to observe very small input angular displacement, and be able to observe a precessional effect. 

A good example would be counter steering on a motorcycle at very high speeds. You just have to push on the inner handlebar, the fork doesn't really rotate much (very small angular displacement), but that can make the bike lean significantly.

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2 hours ago, techyiam said:

Precession responds to torque.

A good example would be counter steering on a motorcycle at very high speeds.

Precession is a displacement response to torque. Using a helicopter rotor as an example a rotor blades maximum angle of attack occurs about 90 degrees ahead maximum displacement. On an EUC in a turn, part of the riders tilt input torque prevents the yaw torque to roll displacement, and probably reduces the effective yaw torque since part of that inwards roll torque causes the wheel to precess in the direction of yaw. 

When counter-steering on a motorcycle at high speed, precession accounts for about 1/8th of the roll response, as most of the roll response is due to out-tracking, the outwards lateral force from the pavement onto both tires due to the front wheel being steered outwards. The steering angle at high speed is so small that a rider may not be aware of it, but it's there.

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1 hour ago, rcgldr said:

Precession is a displacement response to torque.

How would you have an angular displacement response without corresponding torque since there is moment of inertia?

1 hour ago, rcgldr said:

On an EUC in a turn, part of the riders tilt input torque prevents the yaw torque to roll displacement, and probably reduces the effective yaw torque since part of that inwards roll torque causes the wheel to precess in the direction of yaw. 

In order to generate the yaw torque, the euc would have to be accelerating as it negotiates the turn (apply input torque). Riders who have powerful bikes and want to accelerate hard out of a turn will experience the outcome of this effect. Going back to euc's, the rider just have to lean more to compensate the outward roll torque. However, most euc wheels don't generate that much angular momentum to start with, and most riders don't accelerate hard while cornering. As to inwards roll torque exerted by the rider, the precessional yaw effect is not noticeable to the rider, as I have observed.

 

1 hour ago, rcgldr said:

When counter-steering on a motorcycle at high speed, precession accounts for about 1/8th of the roll response, as most of the roll response is due to out-tracking, the outwards lateral force from the pavement onto both tires due to the front wheel being steered outwards. The steering angle at high speed is so small that a rider may not be aware of it, but it's there.

At high speeds negotiating a tight turn, I would disagree.

It is possible to counter-steer and have the wheel pointing away from the turn, and which would cause the bike to lean into the turn which in turn steer the front wheel towards the turn. But it doesn't have to be that way. It also possible for the front tire not steer outwards and yet the bike leans inwards.

There are bikes out there that would go wide the instant you let up on the bar input.

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4 hours ago, techyiam said:

Yaw toque - How would you have an angular displacement response without corresponding torque since there is moment of inertia?

There are bikes out there that would go wide the instant you let up on the bar input.

Yaw torque is due to camber effect (which normally occurs at constant speed in a typical EUC turn), as explained in this earlier comment.

https://forum.electricunicycle.org/topic/24483-electric-unicycle-and-bicycle-dynamics-gyro-effects-on-steering/?do=findComment&comment=384205

At moderate speed, a bike tends to straighten up and go wide if you let up on counter-steering inputs to hold a lean angle. At high speeds (100+ mph), this tendency goes away, and the bike tends to hold the current line and lean angle if you let up on the handle bars. At high speeds, If the tires were infinitely thin, the bike would be falling inwards at a very slow rate called "capsize" mode or "capsize speed", but with real tires, the tendency is to hold the current lean angle, or the rate in lean angle change is so small that it's imperceptible. The reason for this is angular momentum of the front tire resists any change in steering angle, interfering with the self-correcting response of trail to lean angle to steer inwards enough to return to vertical. At moderate speeds, this resistance acts as a damper, preventing over-correction. At high speeds, the angular momentum is so great that it overwhelms trail response, and the bike holds the current line and lean angle with no steering torque applied to handlebars. I've experienced this myself. It takes as much counter-steering effort to straighten up as it does to lean.

Wiki article mentions the ratio of precession induced torque versus out-tracking, on some bike at around 40 mph, with precession accounting for about 12% of the leaning (roll axis) torque. 

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

A 1 g turn at 100 mph requires only a 0.15 degree steering angle, so it may feel like the front tire isn't steering, since it is steering by such a small amount. 

 

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16 hours ago, rcgldr said:

Yaw torque is due to camber effect (which normally occurs at constant speed in a typical EUC turn), as explained in this earlier comment.

It is a lateral force, not a torque. As the leaned wheel rolls forward, due to the round profile of the tire, the laterally tapered tread lays a continuous curved tire track on the ground, which at the contact patch generates a lateral force which changes the direction of motion. A yaw torque would cause a twist.

16 hours ago, rcgldr said:

At moderate speed, a bike tends to straighten up and go wide if you let up on counter-steering inputs to hold a lean angle. At high speeds (100+ mph), this tendency goes away, and the bike tends to hold the current line and lean angle if you let up on the handle bars.

I did qualify with "a tight turn". So for an average rider, he would not be going over 100 mph in a tight turn. Maybe Rossi on a race track, but that is beyond the scope.

22 hours ago, techyiam said:

At high speeds negotiating a tight turn, I would disagree.

In any case, I think the context is going in the wrong direction. I was replying to the following.

23 hours ago, rcgldr said:

When counter-steering on a motorcycle at high speed, precession accounts for about 1/8th of the roll response, as most of the roll response is due to out-tracking, the outwards lateral force from the pavement onto both tires due to the front wheel being steered outwards.

What I wanted to say was that I have observed that it is possible to counter steer without steering the front wheel outwards (I am assuming you mean pointing in the wrong direction). You can actually see that front wheel is pointed in the right direction. 

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3 hours ago, techyiam said:

It is a lateral force, not a torque. As the leaned wheel rolls forward, due to the round profile of the tire, the laterally tapered tread lays a continuous curved tire track on the ground, which at the contact patch generates a lateral force which changes the direction of motion. A yaw torque would cause a twist.

I did qualify with "a tight turn". So for an average rider, he would not be going over 100 mph in a tight turn. Maybe Rossi on a race track, but that is beyond the scope.

In any case, I think the context is going in the wrong direction. I was replying to the following.

What I wanted to say was that I have observed that it is possible to counter steer without steering the front wheel outwards (I am assuming you mean pointing in the wrong direction). You can actually see that front wheel is pointed in the right direction. 

"laterally tapered tread lays a continuous curved tire track on the ground" - since the track is curved, there is a rotation about the yaw axis. For the rotating wheel, there's a component of angular momentum perpendicular to the yaw axis, which would require a yaw torque in order to yaw the wheel.

"high speed - tight turn" - I missed that, but it seems to be a conflict. A 1 g turn at any speed is about as tight as that turn is going to be. At 40 mph, the radius of the turn would be about 107 feet. 

Motorcycle counter-steer:

 

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14 minutes ago, rcgldr said:

which would require a yaw torque in order to yaw the wheel.

On the contact patch, the reaction from the ground didn't twist the wheel. The reaction applied a force to enable the wheel to change direction. 

19 minutes ago, rcgldr said:

"high speed - tight turn" - I missed that, but it seems to be a conflict.

It depends on one's definition of high speed, and we are talking about on public roads on street legal motorcycles.

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14 minutes ago, techyiam said:

On the contact patch, the reaction from the ground didn't twist the wheel. The reaction applied a force to enable the wheel to change direction. 

It depends on one's definition of high speed, and we are talking about on public roads on street legal motorcycles.

Contact patch - motorcycle or unicycle? On a motorcycle, any camber effect from the front tire is negated by the rear tire (it doesn't skid sideways) and vice versa, and tracking (the front tire angled different than rear tire) causes a bike to lean and turn. There is some yaw torque when turning, but only to overcome the component of angular momentum perpendicular to the yaw (vertical) axis  (both tires and the rotating engine parts). For a motorcycle, the yaw torque is due to a small relative difference in the lateral forces at the front and rear tire.

On a unicycle, camber effect causes a wheel to turn in response to a tilt. The turning (yawing) requires a torque to change the component of angular momentum perpendicular to yaw. This can only happen if there are Newton third law pair of yaw torques, the tire exerting an outwards yaw torque on the ground, the ground exerting an inwards yaw torque on the tire.

"high speed - tight turn" - I used an example of a 40 mph turn at 1 g (a racer like 45 degree lean angle), with a 107 foot radius. With a more reasonable 1/2 g turn (30 degree lean angle), the radius would be 214 feet. (radius = speed^2 / (lateral acceleration)).

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31 minutes ago, rcgldr said:

On a unicycle, camber effect causes a wheel to turn in response to a tilt. The turning (yawing) requires a torque to change the component of angular momentum perpendicular to yaw.

If we are talking about precession than yes, tilt torque would produce a yaw torque. 

But let say a wheel is ridden slowly so that precessional effects are negligible. The wheel is tilted to turn, will there be yaw torque using your definition of camber effects?

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18 minutes ago, techyiam said:

If we are talking about precession than yes, tilt torque would produce a yaw torque. 

Tilt torque produces a yaw movement (not a torque). Yaw torque produces a tilt movement. In a steady turn, the rider opposes any tilt movement with a tilt torque, which eliminates any tilt precession, and forces the wheel to only turn about the yaw (vertical) axis.

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On 7/27/2022 at 12:04 AM, mrelwood said:

Sure, when you participate in a contest to spend as much time as possible in a 2m x 2m area, twisting is what you need to do. But it isn't useful in any actual on-road riding situation.

Your roads don't have traffic signals? Mine do, and when I don't have a green light or a walk sign, I tend to slow to a crawl 10-20 feet back and then inch forward until the signal changes. Also when I catch up to pedestrians on narrow or busy trails.

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On 7/27/2022 at 12:21 AM, rcgldr said:

Video example of a fully (3d) articulated hub. I set the link to where a full scale helicopter is getting ready to taxi (rotor plane forward) and then take off (rotor plane horizontal). The plane of the rotor changes angle while the wheels remain on the ground and the main shaft is nearly vertical. At the first scene transition, the plane of the rotor is angled much further forwards so the helicopter can taxi. Then at the next transition where lift off begins, the plane of the rotor is nearly horizontal. As the helicopter lifts off, the body pitches up, main shaft pitches back, and the plane of the rotor is adjusted to return it to horizontal. Since the hub allows the plane of the rotor to move independently of the main shaft, the hub only applies a linear tension force in the direction of rotor lift to the top of the main shaft, sort of like a mass suspended by a short rod with a hinged connector (the hub) at the top.

The main shaft tilted forward with respect to the fuselage centerline. That keeps the fuse closer to level during forward flight. You'll notice that when the heli lifts off vertically, the nose is high and the fuse has a clearly visible rear tilt.

There is some flex in the head, we've been over that already (the blades themselves flex too, btw). But the whole point of cyclic pitch is to make the fuselage rotate in the pitch and roll axes, so "independently" seems like vastly understating how closely the rotor axis is coupled to the main shaft axis.

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5 hours ago, techyiam said:

Why a movement and not a torque? 

I should have clarified, there's no torque in an idealized case in a steady state when there is nothing to oppose the precession, such as an attached mass. If there is an attached mass, the initial reaction is in the direction of torque, and a torque in the direction of precession exerted on the attached mass, eventually transitioning into a precession movement, at which point the attached mass is rotating at a constant rate, so no more precession related torque (assuming there's no friction opposing precession).

The Wiki article on counter-steering states that for some motorcycle at 40 mph, with some amount out outwards steering torque applied, the initial gyroscopic reaction contributes about 12% of the initial tilting torque, while out-tracking (lateral friction forces on the ground pushing outwards at the contact patches) contributes about 88% of the initial tilting torque. However once the leaning has started, the rider and|or trail effect steer the front tire inwards, which initially would result in an outwards gyroscopic reaction tilting torque

For an EUC, the rider uses a tilting torque to tilt an EUC to induce camber effect and prevent any tilt movement related to camber effect yaw torque of the EUC, and the EUC transitions into a steady state turn where there is no tilt movement and only a yaw movement. I don't know how quantify how much rider torque is required to compensate for camber effect yaw torque.

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5 hours ago, NSFW said:

The main shaft tilted forward with respect to the fuselage centerline. That keeps the fuse closer to level during forward flight. You'll notice that when the heli lifts off vertically, the nose is high and the fuse has a clearly visible rear tilt.

There is some flex in the head, we've been over that already (the blades themselves flex too, btw). But the whole point of cyclic pitch is to make the fuselage rotate in the pitch and roll axes, so "independently" seems like vastly understating how closely the rotor axis is coupled to the main shaft axis.

One purpose of a fully articulated hub is to eliminate the hub from having to produce a torque onto the main shaft to rotate the entire mass of the helicopter body. The hub exerts a linear force in the direction of the axis of rotation of the main rotor to the top of the main shaft, while the mass of the body of the helicopter is in effect suspended by the main shaft, with the top of the main shaft pulled in the direction of linear force that the hub exerts onto the top of the main shaft. Sort of like a ball suspended by a short string, where a linear force is applied to the top of the short string.

The main shaft of a full scale helicopter is fixed with respect to the helicopter body. On a helicopter with wheels, the hub allows the rotor to be angled forwards for taxi, and horizontal for takeoff.

 

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This helicopter tangent is just too painful to continue. We're both quite confident in our "understandings" about how rotor heads function, and yet what we believe couldn't be more different.

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4 hours ago, NSFW said:

This helicopter tangent is just too painful to continue. We're both quite confident in our "understandings" about how rotor heads function, and yet what we believe couldn't be more different.

From the Wikipedia article: "In a fully articulated rotor system, each rotor blade is attached to the rotor hub through a series of hinges that let the blade move independently of the others. ". This also means each rotor blade can move independently of the main shaft. For a two bladed rotor, an option is a tetering hub that allows the rotor blades to tilt as if a single unit, independently of the main shaft.

https://en.wikipedia.org/wiki/Helicopter_rotor#Fully_articulated

Animated example of fully articulate hub and a tetering hub:

 

 

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On 7/29/2022 at 5:09 AM, mrelwood said:

I often have trouble following the terms yaw and roll since they don't have good translations in our language. But based on the images and gifs that Google showed me for "yaw roll pitch", roll is the sideways tilt that the user forces on the wheel, and yaw is the resulting rotation/turn to left and right that the camber effect (or gyroscopic precession) creates.

The Finnish translation for yaw is "kääntyminen" and for roll "kallistuminen". Pitch would be translated as "nyökkääminen".

On 7/29/2022 at 5:09 AM, mrelwood said:

So, I shot a video on the test I mentioned.

Thank you for the video. Gyroscopic precession has many times been considered to be too weak to be meaningful for EUC riding. This video showed that it might have more meaning than thought before. I'm certain camber effect alone is not enough explain autocorrection of the wheel we have seen on the trick ride videos and experienced ourselves. In combination with gyroscopic precession it could make it. The gyroscopic precession is speed related. This means autocorrection would also be speed related. It would only work at certain speed range. Below the lower speed limit of this range the gyroscopic precession is too weak to make corrections and over the higher speed limit it would overcorrect. This matches very well to our riding experiences. We need to have a certain speed to be stable and at the higher speeds we get speed wobbles. Based on this thinking, the key against speed wobbles is to learn to tame wheels overcorrection.

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32 minutes ago, Eucner said:

Gyroscopic precession has many times been considered to be too weak to be meaningful for EUC riding. This video showed that it might have more meaning than thought before.

Perhaps the release of euc's like the Inmotion V13 will bring precession in the limelight.

I see in your profile, you only list the V8F. I am wondering whether you have ridden bigger wheels that have gyro effects, especially at above 50 or 60 km/h? 

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19 hours ago, rcgldr said:

For an EUC, the rider uses a tilting torque to tilt an EUC to induce camber effect and prevent any tilt movement related to camber effect yaw torque of the EUC, and the EUC transitions into a steady state turn where there is no tilt movement and only a yaw movement. I don't know how quantify how much rider torque is required to compensate for camber effect yaw torque.

I believe should you ride an euc with gyro effects, you will be surprised how different it corners than your V8F. A stock V12 has some gyro effects, but euc's like the Sherman, Abrams, and Monster Pro would have more.

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3 hours ago, Eucner said:

I'm certain camber effect alone is not enough explain auto-correction of the wheel we have seen on the trick ride videos and experienced ourselves. In combination with gyroscopic precession it could make it.

My V8F becomes stable and auto-correcting (for small imbalances) at 6 to 8 mph, too slow for precession to have much effect.

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2 hours ago, techyiam said:

I believe should you ride an euc with gyro effects, you will be surprised how different it corners than your V8F. A stock V12 has some gyro effects, but euc's like the Sherman, Abrams, and Monster Pro would have more.

I wonder how EUC girl does so well in handling the Begode Hero (80 lbs), which weighs more than her?

https://www.youtube.com/watch?v=joQ1ee3tkSU

As posted before, on my motorcycle, I don't get much of a sense of gryo effects (resistance to counter-steering toque applied to handlebars) at speeds below 50 mph, but an EUC wheel weighs more than my motorcycle front wheel, dual disk brakes and tire. Based on a poll I recall, only a low percentage of riders routinely ride faster than 35 mph. I weigh 187 lbs, and don't trust my V8F above 18 mph, and I mostly ride at 12 to 15 mph on the local bike trails, so I have no sense of what gyro effect would feel like.

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15 hours ago, rcgldr said:

I wonder how EUC girl does so well in handling the Begode Hero (80 lbs), which weighs more than her?

She is an exceptional talent, no doubt. In one of her earlier videos, her dad got her to ride the Monster Pro, which was almost twice her weight, and don't forget the size of the Monster Pro. And her dad wanted her to free-mount too.

Even though her dad said in the Hero video that the Hero is kind of top heavy, for me, I didn't find the Hero top heavy when compared to other suspension wheels such as the Master or S22. Alhough the Hero's 80 lbs is heavier than the Sherman's 77 lbs, the Hero actually feels more agile than the Sherman.

What I find interesting was her participation in racing with adult women. For acceleration, braking, and cornering, mass matter on an euc. These powerful wheels that they race with usually have gyro effects. Having said that when she get into her teen years, she is going to be a phenomenon. 

 

15 hours ago, rcgldr said:

Based on a poll I recall, only a low percentage of riders routinely ride faster than 35 mph.

Should you typically make turns by nudging your inner outside leg against the side of your V8F, you will be in a rude surprise when you attempt to make a turn for the first time on those big and heavy euc's, even at speeds as low as 20 or so mph. Basically, you will find that the euc won't turn as you thought it would. Your body won't instinctively know how to effectively get enough leverage to tilt the wheel.

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