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Electric unicycle and bicycle dynamics - gyro effects on steering


rcgldr

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On 8/1/2022 at 1:01 PM, mrelwood said:

I can't see how gyroscopic forces would enter into play other than actually dampening the tendencies to wobble.

Left tilt -> left yaw by precession -> right tilt (starting from nonzero tilt positioning) by misalignment of wheel axis and velocity vector (or centrifugal force, if you like) -> right yaw by precession -> (after crossing some symmetry axis) left tilt by (the opposite) misalignment of wheel axis and velocity vector -> left yaw by precession -> ...

I can't possible see how we would determine whether or not there is enough damping in this procession. As the new (opposite) tilt force gets in effect only after the wheel yaw crosses the symmetry axis and the involved forces increase with increasing speed, it looks like a good candidate for a speed wobble mechanism. The camber effect seems a worse candidate, because it doesn't switch the yaw until after zero tilt is crossed, which feels like too late to catch the wheel. And it doesn't increase with speed.

Thinking about it for a couple of minutes, I would even say the above mechanism for a speed wobble is the best I have come across so far (in some six odd years). Nice.

Edited by Mono
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1 hour ago, techyiam said:

By Gyro effects, I mean resistance force opposing my inner lower leg, as I nudge the side of the wheel with my leg. The wheel wiil not lean easily.

But does it return to vertical easier? Gyroscopic forces resist getting back to vertical just as much as they resist the initial tilt. If there is more resistance at the initial tilt, those are not gyroscopic forces.

1 hour ago, techyiam said:

If I did the exact same thing under the exact same conditions on my T3, it would have leaned with no resistance.

The Z10 is a great example of a wheel that weighs about the same as an 18XL, yet has a much wider tire. Gyroscopic forces are similiar with both wheels, yet the Z10 resists tilting after 10km/h as if it was precisely engineered to do that and nothing else. The 18XL is easy to tilt still at 40km/h.

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20 minutes ago, mrelwood said:

Gyroscopic forces are similiar with both wheels

Because? The motor/wheel unit of both euc's have the same moment of inertia? At the same spindle speed, they need to have the same angular momentum. 

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But the distribution of the mass on the periphery is different too—you have to integrate the forces of the different axes of yaw rotation across the width of the tire. Does the thought still hold in the limit case of a tire with zero width?

Edited by Tawpie
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On 8/5/2022 at 7:24 AM, Tawpie said:

Does the thought still hold in the limit case of a tire with zero width?

There would be no camber effect on a tire with zero width. In the case of bikes (bicycles), mathematical models based on zero width tires don't match reality in terms of capsize speed, where if a bike is moving at or faster than capsize speed, it will be unstable and tend to fall inwards at a very slow rate, and the calculated capsize speed is a fraction of the actual speed where a real bike would transition from self-correction to vertical to tending to hold a lean angle. In one case, a mathematical model for a 10 speed type bike with zero width tires showed a capsize speed at 18+ kph, while the actual bike was extremely stable at 30 kph on a very large treadmill test.  On motorcycles, the tendency to hold a lean angle rather than return to vertical starts to happen around 100 mph. For a riders that always counter-steer, this isn't an issue, and the only difference is at lower speeds, a rider has to use some outwards counter-steering pressure on the handlebars to hold a lean angle at moderate speeds to prevent the bike's self-correction from attempting to return to vertical, but no pressure is needed to hold a lean angle at high speeds.

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On 8/4/2022 at 11:44 PM, rcgldr said:

If you can prevent the precession in this manner, does the force it takes to tilt the wheel increase or decrease?

 

On 8/5/2022 at 12:42 AM, techyiam said:

Obviously, the force it takes increases. And I confirmed it.

Not so obvious, as shown in this video that I linked to before. An 8 lb gyro supported at one spinning at several thousand rpm gets an assisted start to precess, but then a small peg is placed in a hole to block the other end of the gryo, stopping the precession, and the gyro bounces back and immediately falls down as if it was not spinning (as if near zero angular momentum). Prior to this, the force from gravity was resisted by precession, but once the precession was blocked, the gyro just fell down, with little resistance to the pull of gravity:

https://www.youtube.com/watch?v=0L2YAU-jmcE&t=3104s

 

Edited by rcgldr
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25 minutes ago, rcgldr said:

Not so obvious, as shown in this video that I linked to before. An 8 lb gyro supported at one spinning at several thousand rpm gets an assisted start to precess, but then a small peg is placed in a hole to block the other end of the gryo, stopping the precession, and the gyro bounces back and immediately falls down as if it was not spinning (as if near zero angular momentum).

To me, it did exactly what it was suppose to do. No angular velocity of precession input, no torque output. 

But your question I replied to asked a different question. And my reply was correct. You need to get a bicycle wheel to experiment with, and then you can clearly see why it behaves the way it does.

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On 8/4/2022 at 1:18 PM, Eucner said:

V8F - I do believe in cone effect and gyroscopic precession. 

My V8F weighs 34.8 lbs, and the motor+wheel+tire weigh less than 17 lbs. I weigh 187 lbs before gear. I mostly ride at 12 to 15 mph (18 mph max), and I don't feel any significant angular momentum resistance to tilt steering, even with my legs clear of the upper pads, and just using pedal pressure to tilt the V8F side to side. 

The front wheel, brake discs, and tire on my 2001 Hayabusa motorcycle weigh about 25 lbs, and I don't feel any significant resistance to counter-steering inputs on the handlebars at 35 mph or less, and it just feels like I'm moving the handlebars. At 55 mph or faster, the sensation changes to where it feels like I'm applying counter-steering force with little movement of the handle bars. 

The rate of yaw from precession could be less than or greater than the camber effect rate of yaw. If less than, then it's opposing the camber effect yaw, acting as a damper. If greater than, it's assisting the camber effect. I still haven't found a source to quantify how much yaw torque precession could generate to oppose or assist camber effect yaw. Precession is mostly a movement reaction to a torque (a force at one point results in a displacement at a later point). In experiments where precession is completely blocked, there is virtually no precession related torque.

Edited by rcgldr
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37 minutes ago, rcgldr said:

I mostly ride at 12 to 15 mph (18 mph max), and I don't feel any significant angular momentum resistance to tilt steering

Ahhhhh. Speed, mass, diameter. For me, on my 16X the "really doesn't want to tilt by weight alone" effect starts becoming noticeable at about 16-18 mph. On the MTen, there's no (safe) speed that feels like it's resisting my efforts to tilt it, it's scary floppy even at 14-16 mph. So many variables.

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On 8/4/2022 at 6:37 PM, mrelwood said:

Precession only yaws when the roll angle increases, so during the  initiation of a turn. Once a stable roll angle has been achieved, precession stops yawing.

That's just nonsense.

Let's start with the trivial and obvious example of how things work without precession:

If there a torque (not motion) about the roll axis, that roll torque produces a rotation about the roll axis.

And then, the same scenario with a spinning object that exhibits precession:

If there is a torque about the roll axis, that roll torque produces a rotation about the yaw axis.

(Assuming of course, that the spinning object in question in spinning on the pitch axis, and ECU wheels do.)

You can produce a continuous yaw motion by applying a continuous roll torque.

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On 8/4/2022 at 4:10 AM, Eucner said:

Yes, when angle is not chancing, there no gyroscopic precession.

 

On 8/4/2022 at 4:11 AM, rcgldr said:

Precession requires an unbalanced inwards torque to yaw inwards. 

 

46 minutes ago, NSFW said:

If there a torque (not motion) about the roll axis, that roll torque produces a rotation about the roll axis.

This was mentioned before in prior comments. Since the rider exerts an outwards reactive centrifugal force at the center of the pedals, which is above the contact patch, this results in an outwards roll torque on the EUC that the rider must compensate for with an inwards roll torque (requiring the rider to lean a bit more inwards). In a steady turn, these roll torques cancel (the EUC tilt angle remains constant), so I'm not sure there's any precession response if there is no net torque on the EUC.

The other possibility is the camber effect yaw torque producing an outwards precession roll response, but the rider counters this, holding the EUC tilt angle constant. I don't know how to quantify any precession related roll torque due to the rider preventing a precession response. In the case of an 8 lb gyro spinning at several thousand rpm, supported at one end, precessing about 1 revolution per second, a thin peg is pushed into a hole that catches the other end of the gyro, stopping the precession and the gyro bounces back and drops down as if there's near zero angular momentum (near zero precession response).

https://www.youtube.com/watch?v=0L2YAU-jmcE&t=3104s

On my 34.8 lb V8F, the motor+wheel+tire weigh less than 17 lbs, and since I mostly ride a 12 to 15 mph, there's not much angular momentum. My 2001 Suzuki Hayabusa motorcycle front wheel, brake discs, and tire weighs about 25 lbs, and I don't get any sense of angular momentum resistance to counter-steering inputs on the handlebars at 35 mph or less. At 55+ mph, the sensation is a counter-steering torque applied to the handlebars with little change in steering angle, where the angular momentum is significant.

As for the prior comments about helicopters with fully articulated or teetering hubs that allow the rotor to tilt independently of the main shaft, here is animated video explanation of that I posted before in case you missed the earlier one.

 

Edited by rcgldr
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54 minutes ago, rcgldr said:

This was mentioned before in prior comments. Since the rider exerts an outwards reactive centrifugal force at the center of the pedals, which is above the contact patch, this results in an outwards roll torque on the EUC that the rider must compensate for with an inwards roll torque (requiring the rider to lean a bit more inwards). In a steady turn, these roll torques cancel (the EUC tilt angle remains constant), so I'm not sure there's any precession response if there is no net torque on the EUC.

 

 

Yes, if the roll torques are balanced, there will be no yaw.

We all agree that the forces can be balanced. You seem to think that the forces must be balanced, I don't see why.

If the rider wants yaw, nothing stops the ride from applying imbalanced roll torque, which (due to precession) will result in yaw.

Edited by NSFW
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3 hours ago, NSFW said:

Yes, if the roll torques are balanced, there will be no yaw.

We all agree that the forces can be balanced. You seem to think that the forces must be balanced, I don't see why.

Because there is no angular velocity at the roll axis during a steady turn. Are you saying that the forces on the roll axis can be imbalanced yet the roll angle would remain steady? How would that be possible?

I suggest for you to do the lift test as well and examine how the precession presents itself: Lift the wheel and let the motor spin forwards. Then start tilting the wheel to the side. Examine how the yaw by precession is related to the roll angle change. If you can create a situation where the roll angle is stable but precession keeps yawing the wheel, I'd love to see that on video!

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21 hours ago, mrelwood said:

Are you saying that the forces on the roll axis can be imbalanced yet the roll angle would remain steady? How would that be possible?

Yes, that's what I'm saying, because that's how precession works.

When a helicopter pilot wants to pitch forward, they command cyclic pitch that increases lift on one side of the helicopter, and decreases pitch on the other side. Naively you'd think that would produce roll, but due to precession it causes the rotor (and then the rest of the helicopter) to pitch forward.

If you have a friend with an RC helicopter, ask them for a demonstration. Or check youtube, surely someone has posted a video about this by now.

21 hours ago, mrelwood said:

I suggest for you to do the lift test as well and examine how the precession presents itself: Lift the wheel and let the motor spin forwards. Then start tilting the wheel to the side. Examine how the yaw by precession is related to the roll angle change. If you can create a situation where the roll angle is stable but precession keeps yawing the wheel, I'd love to see that on video!

I'd love to do the lift test, I really would. Unfortunately, my King Song S22 apparently has blown its MOSFETs. :( 

You'd need a gimbal to demonstrate that properly given the degrees of freedom involved, but I'm pretty sure it would look a lot like this:

Granted, that's an animation, but there's only so much searching I'm willing to do to on your behalf.

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On second thought, I was willing to search a little more. This video starts with a demonstration of the same phenomenon shown above.

I'll admit I haven't watched beyond the first 20 seconds or so.

Quote

 

 

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4 hours ago, NSFW said:

When a helicopter pilot wants to pitch forward, they command cyclic pitch that increases lift on one side of the helicopter, and decreases pitch on the other side. Naively you'd think that would produce roll, but due to precession it causes the rotor (and then the rest of the helicopter) to pitch forward. If you have a friend with an RC helicopter, ask them for a demonstration.

Some don't consider helicopter rotor response to be precession, just similar to precession. If you consider what each blade does, it's easier to explain, that maximum deflection of a rotor blade occurs as the lift force transitions through zero force. For full scale helicopters, the delay is about 90 degrees, so cyclic controls are advanced 90 degrees (close enough). For some RC helicopters, there's less delay, as little as 75 degrees, requiring mechanical changes and|or transmitter programming changes.

Almost all articles about precession describe a delayed and unimpeded movement response to a torque. The only reference I've found that quantifies the amount of precession related torque generated if that precession is impeded is the WIki article on counter-steering, with only one source as a reference. The wiki article mentions the torque generated by a 25 lb "wheel" (wheel, brake discs, tire) at 50 mph, being steered at 1 degree per second (no mention of the torque required do do this), generating only ~ 2.6 foot pounds of torque, versus the ~ 22 foot pounds of torque generated by the tires initial out-tracking response to being counter-steered.

 

Edited by rcgldr
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I've owned several rc helis over the years, with rotors from about one foot to five feet in diameter, and all had swash plates set up at 90 degrees with no electronic correction. I don't know what would cause that to vary but I've never encountered it.

I have no idea what you mean by "maximum deflection of a rotor blade occurs as the list force transitions through zero force."

If hovering, and you want roll, maximum pitch will happen as the blade passes over the tail boom or the canopy.

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36 minutes ago, NSFW said:

I've owned several rc helis over the years, with rotors from about one foot to five feet in diameter, and all had swash plates set up at 90 degrees with no electronic correction. I don't know what would cause that to vary but I've never encountered it.

I have no idea what you mean by "maximum deflection of a rotor blade occurs as the list force transitions through zero force."

On some models the "phasing" can be as little at 75 degrees. From a thread at helifreak: "Phasing is a set variance on every rotation. It is the variance from max blade pitch to max blade tilt. It does not rotate around the heli per rotor cycle (or control would be impossible). If phasing is 75 degrees instead of 90, you need to configure your TX or your flybarless controller to adjust the cyclic inputs to be 15 degrees later than normal."

Maximum blade deflection|tilt is easier to understand on a hub with a flapping hinge. Say the goal is to pitch forwards, and the phasing is 90 degrees, then maximum blade pitch and the associated lift occurs on the left and right sides of the helicopter, and maximum blade deflection and zero pitch and zero lift occur 90 degrees later, at the front and back of the helicopter.

Edited by rcgldr
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Yes, I know what phasing is. I've just never needed anything buy 90 degrees. My only guess is that it might be useful at very low rotor speeds, where gyroscopic effects get small. I never even noticed it during autorotations though, and I think it would be pretty obvious if the cyclic behavior was off-axis when coming out of an inverted autorotation.

I guess you were talking about spanwise deflection. If so then yes, of course. If cyclic pitch change is greatest over the tail and canopy, then it's necessarily zero to the left and right.

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  • 1 year later...
Posted (edited)
On 8/9/2022 at 9:04 PM, NSFW said:

I've owned several rc helis over the years, with rotors from about one foot to five feet in diameter, and all had swash plates set up at 90 degrees with no electronic correction. I don't know what would cause that to vary but I've never encountered it.

I have no idea what you mean by "maximum deflection of a rotor blade occurs as the lift force transitions through zero force."

That should have been blade pitch, not force, and only applies when there is zero collective. When setting up swash plate without the rotor moving, and with zero collective, then the maximum deflection points will occur at the points where the rotor blade pitch is zero, and 90 degrees ahead of that will be the point where the rotor blades have maximum positive and negative pitch.

For 90 degree advance the maximum blade pitch occurs 90 degrees ahead of the desired rotor pitch or roll. Assume rotor spins counter-clockwise as viewed from above, and the goal is to pitch forwards. The maximum lift will occur on the left, the minimum lift will occur on the blade on the right, the lift at the front and back will be the same and about the average lift between left and right blades.

 

 

 

Edited by rcgldr
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Posted (edited)
On 8/8/2022 at 11:32 PM, mrelwood said:

Are you saying that the forces on the roll axis can be imbalanced yet the roll angle would remain steady? How would that be possible?

 

On 8/9/2022 at 9:04 PM, NSFW said:

Yes, that's what I'm saying, because that's how precession works.

The external torque is imbalanced, but precession will generate a counter-torque in reaction to the torque that induces precession, and in the case of a steady precession, the internal and external torques cancel, so no net torque if the internal torque is included. The counter-torque is proportional to the rate of precession. Assume a torque about a roll axis induces precession about the yaw axis. If the rate of yaw is increased, the counter-torque is increased, and the gyro experiences angular acceleration in the opposite direction of the external roll torque. If the rate of yaw is decreased, the gyro experiences angular acceleration in the same direction of the external roll torque. If the yaw is stopped or prevented, the gyro experiences angular acceleration in the same direction of the external roll torque as if the gyro was not spinning. 

Consider the case of a gyroscope supported at the end of one axis. At the moment the other axis end is released, gravity exerts a roll torque, the gyro initially drops a bit, then starts to precess. If there is some mechanism to dampen nutation, eventually the gryo precesses at a constant rate. The precession about the yaw axis coexists with a roll torque that exactly opposes the roll torque due to gravity, so no imbalance in roll torques, no rotation about the roll axis, only the precession rotation about the yaw axis.

If the precession is stopped, then there is no torque opposing the torque due to gravity, and the gyro just yields to gravity and falls:

"For the setup in the images: the 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

Also mentioned in this video. The gyro is counter-balanced, so it precesses slowly and falls slowly.

 

 

Edited by rcgldr
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