What determines wheel zippiness?

[Mono]

Inspired by this thread, I am trying to understand zippiness and acceleration. I thought the following figure should help.

 

The green and red arrows indicate forces from different rider positions. In summary, the lengths of the blue bars represent the generated thrust (they are the thrust when multiplied by rider weight and divided by the wheel radius) assuming the rider position is stationary w.r.t. the wheel. FTR, here the pedal has a length of 95% of the wheel radius and is inclined by 3º. The question was which forces/torque the rider can/does induce.

 

Green arrows: down force of the rider (weight) assuming constant speed (two scenarios, same weight). The CoG of the rider is vertically above the point of attack on the pedal. Forces and lean induced by the airflow are disregarded.

 

Red arrows: (down) force from the rider towards the wheel under acceleration (or considering air resistance), two scenarios, all with the same rider weight force component (assuming stationarity). The arrow depicts a weight force combined with a lean angle due to acceleration (and possibly air resistance) such that the rider is in balance. The lean angle is directly tied to acceleration which is tied to the torque. That is, not every arrow positioning is a possible stationary acceleration scenario (or rather: only a few are and the shown are not likely to fall into this category, though they look somewhat close). The effective leverage (AKA moment arm or lever arm) of the respective force is the smallest distance of the force vector line to the axle, the created torque is proportional to the moment arm (see also here). EDIT: The forward thrust computes to (as derived here):

 

    thrust [kg] = weight x blue bar / outer wheel radius

 

where the blue bar is the horizontal distance between the force vector and the axle. For example, 80 kg x 3 cm / 20 cm = 12 kg forward thrust. The factor 80 kg / 20 cm = 4 kg / cm is the characteristic constant for the given wheel and rider. A 16" wheel (radius 20 cm) with 80 kg rider delivers 4 kg thrust for each cm horizontal distance.

 

We can express the length of the blue bar in terms of the support point on the pedal and the position of the CoG of the rider (see here). For any stationary lean angle (the support point of the rider is not vertically under the CoG, but the lean does not change) we have

 

   thrust [kg] = weight x (forward displacement + tan(lean angle) x down displacement) / outer wheel radius 

 

where

• the forward displacement is the horizontal distance between the support point on the pedal and the axle (negative if the support point is behind the axle)
• the down displacement is the vertical distance between the support point on the pedal and the axle (negative if the pedal is above the axle)
• the lean angle is determined by the horizontal displacement of the CoG of the rider and the support point

Remark that we have approximately

angle      0º  3º  5.7º 11º 17º 27º 45º 
tan(angle) 0%  5%  10%  20% 30% 50% 100%

• The lower the pedals (relative to the wheel radius), the more additional thrust we get during acceleration. WIthout lean angle it has no effect. The effect seems of moderate size: lowering the pedals by 20% at a 14º lean angle yields 5% x weight additional thrust, the same effect as shifting the CoG by 5% of the wheel radius forward. With an 80kg rider and 20 cm radius, we get 1 kg thrust per 1 cm lowering at 14º lean angle.
• On a slope of 15º (or 27% = tan(15º)), we need to displace the weight by more than 26% = sin(15º) of the wheel radius to get more than 0.26 x weight forward thrust to start get going. The right green vector above the white diamond depicts the situation where the wheel is at standstill. Not quite coincidentally, the contact point happens to be shifted 26% forward in this case (see also this Youtube clip).
• The right red arrow produces more torque than the right green arrow, even though its point of attack on the pedal is further to the center and with the same rider weight. Again, this doesn't need to be a situation which a rider can possibly sustain.

 

Black arrows (EDIT: are meaningless): tangential torque force direction induced by the rider. If the pedal does not dip, the wheel must provide the respective counter torque. The larger the force angle between the colored and the black arrow (if they point into the same direction, the angle is zero), the less weight force can be converted into torque. At 90º, no torque is generated. The closer the point of attack is to the middle, the larger is the angle and the closer is the point to the axle, and both means less torque is generated. EDIT: it's simpler to deduce the effective leverage from the white bars.

 

Random thoughts:

• There is more to it than the simple inverted pendulum, as we can shift the "pendulum" on the pedal.
• Influence of rider weight: the input force to generate counter torque is proportional to the rider weight. The system acceleration is inversely proportional to system mass = rider + wheel mass. That is, the rider vs wheel weight induced efficiency should amount to the rider mass divided by the sum of rider and wheel mass, hence heavier riders have an advantage. An example with a 20kg wheel: a 60kg rider has 60/80=75% weight-related efficiency, a 100kg rider has 100/120=83% efficiency, which is 11% more than the 60kg rider. Not a huge difference, but children get an automatic power reduction for free 
• Influence of rider height: apparently none, only the lean angle matters. To get in an accelerating position, the necessary displacement of the wheel (in the opposite direction) is proportional to the CoG of the rider?
• EDIT: the influence of the pedal height: when the force vector tilts forward (positive lean angle, for example under acceleration), the resulting thrust from this lean is proportional to the lowering of the pedals (relative to the wheel radius).
• Pedal dipping changes the support point on the pedal and seems to increase the force angle between colored and black arrow hence reducing torque (which should be good as it allows the wheel to catch up). EDIT: pedal dipping should not have any effect in the static force analysis (i.e. if the rider follows the dip with the ankle to stay in balance). It moves the contact point more to the front but leads to a less extreme dorsiflexion of the ankle.
• Acceleration induces an additional rotational force on the pedal-shell system, if the CoG of the wheel is not on axle level. My hunch is that this force is not decisive, but we better use equations to find out . As a starting point: a V8 has 7kg head weight at wheel radius, which is small compared to the riders "head" weight. For acceleration (or slowing down), the wheel weight should be the determining factor, not it's distribution. However the stationary lean angle of the rider should depend on the weight distribution of the wheel? The rider needs to lean (slightly) more if the wheel is head heavy.

 

Edited March 14 by Mono

[Chriull]

6 hours ago, Mono said:

Red arrows

Shouldn't they be the sum of the corresponding black and green arrow? (With neglecting the wind drag) Just a thought - but it's still quite early here...

[Mono]

4 hours ago, Chriull said:

Shouldn't they be the sum of the corresponding black and green arrow? (With neglecting the wind drag) Just a thought - but it's still quite early here...

The colored arrows depict just different scenarios (two constant velocities, two accelerating situations). Black arrows depict the part of the weight induced force that is transformed into torque. Arrow lengths or thicknesses do not exactly reflect the amount of force, but the angle of the force. Maybe there is a different way to interpret the arrows, but that would have been unintended 

It may make sense to decompose, say, the green vector into the black torque vector plus a downwards pointing vector plus a forward pointing vector. The decomposition should be unique, if we know the length of the torque vector (which I assume is cos(angle) smaller than the original down force). I now wonder whether this gives us the associated acceleration.

[mrelwood]

1 hour ago, Chriull said:

Shouldn't they be the sum of the corresponding black and green arrow? (With neglecting the wind drag) Just a thought - but it's still quite early here...

I think the (marvellous!) picture doesn’t consider the strength of the forces as arrow lengths, just the directions. And in my understanding the green and red arrows are different situations, from a different point in time.

To my understanding everything in the picture is correct. Although, it is not able to consider time as a variable, which is pretty essential when considering acceleration (m/s^2).

 

Firmware

I believe the programmed response of the firmware to be crucial as well in determining wheel zippiness: The acceleration curve of the firmware decides how much the pedal-shell is allowed to tilt for a certain amount of power to be applied to the motor. Meaning, how quickly and effectively the wheel tries to counteract the rider induced forces.

If the wheel responds too fast, the rider will have trouble shifting the CoG even to the rightmost coloured arrow: The acceleration immediately pushes the CoG back closer to the axle. The rider tries to lean, but his/her efforts are nullified, and the resulting acceleration is minimal. Very unzippy wheel. Sluggishness feels even worse than on the MSX.

The other end of the spectrum is an acceleration curve that reacts too slowly or too little. The CoG reaches the coloured arrows too easily before acceleration. The rider feels as if the wheel can’t keep up, and must push down with the toes to find the balance between the amount of acceleration that is happening, and the CoG staying stable in one point. Does not introduce trust, as the rider feels that the available power is about to be exceeded (no matter if it actually is or isn’t), resulting in a faceplant. Very unzippy wheel. Spongyness feel even worse than the X3, MCM4, or any other vey old wheel.

In conclusion, the firmware must be programmed just right in order to riders with different riding styles consider the wheel feeling zippy. But that is indeed just one factor.

 

Wheel diameter

If the wheel diameter is increased, the following attributes change:

• The frontmost pedal edge is closer to the axle, in relation to the wheel edge. This makes the range of the tangential forces that can be applied (the black arrows) smaller.
• Pedal height. The distance from the axle, both relative and actual, increase.
• Relative rider height decreases. The difference is likely too small to bear much effect though.
• Tire contact point length increases.

(I will dwell deeper into these at a later time.)

[Mono]

3 hours ago, mrelwood said:

I think the (marvellous!) picture doesn’t consider the strength of the forces as arrow lengths, just the directions. And in my understanding the green and red arrows are different situations, from a different point in time.

Yes.

Quote

To my understanding everything in the picture is correct. Although, it is not able to consider time as a variable, which is pretty essential when considering acceleration (m/s^2).

True, it is just one point in time. The point of attack where the colored force hits the pedal may be (and is) constantly changing, as part of the rider controlling the wheel.

Quote

Firmware

I believe the programmed response of the firmware to be crucial as well in determining wheel zippiness: The acceleration curve of the firmware decides how much the pedal-shell is allowed to tilt for a certain amount of power to be applied to the motor. Meaning, how quickly and effectively the wheel tries to counteract the rider induced forces.

Do you agree that the only degree of freedom of the firmware response can be captured in the (temporary) change of the pedal tilt angle, and everything else is directly tied to this variable? Is there another independent additional variable I am missing? After all, as you wrote elsewhere, the only thing the wheel can do is either produce more torque or less torque which should be directly tied to a changing tilt angle (everything else being equal).

Quote

If the wheel responds too fast, the rider will have trouble shifting the CoG even to the rightmost coloured arrow: The acceleration immediately pushes the CoG back closer to the axle. The rider tries to lean, but his/her efforts are nullified, and the resulting acceleration is minimal. Very unzippy wheel. Sluggishness feels even worse than on the MSX.

I don't see that an immediate reaction of the wheel can prevent the rider from "leaning". Here is why: a simple but useful alternative conception/description of (forward) leaning is lifting/releasing the forefoot for a short period of time. This moves the point of attack on the pedal behind the axle and consequently drives the wheel behind the rider and hence the rider gets into a leaning position. If the wheel responds "instantaneously", I can achieve any lean angle "instantaneously" (well, as quickly as masses can move around).

Many beginners feel that they cannot accelerate uphill even though they try to push hard. This is however due to a mistaken feedback loop on the rider side rather than on the wheel side, as most of us find out with time. It feels quite counterintuitive to lift the forefoot to induce acceleration, when acceleration indeed only happens finally by pushing into the forefoot.

Edited June 10, 2019 by Mono

[mrelwood]

 

Quote

Do you agree that the only degree of freedom of the firmware response can be captured in the (temporary) change of the pedal tilt angle, and everything else is directly tied to this variable?

If I understood the question correctly, then yes. But taken into account the factor of time, the change in pedal tilt angle (ie. adjusting the amount of power applied to spin the wheel) does introduce several potential parameters:

• How much tilt allowed, or time let by until the initial response from the software. (Very, very little. Still, may exist.)
• The acceleration curve for a peaceful lean.
• Acceleration curve for a fast lean.
• Other acceleration curve variations based on the events of the past second.
• Maximum tilt allowed under any circumstance.
• Combination of all of the above and the firmware’s response to the required vs expected power requirement for each situation.
• Etc...

 

Quote

I don't see that an immediate reaction of the wheel can prevent the rider from "leaning". Here is why: a simple but useful alternative conception/description of (forward) leaning is lifting/releasing the forefoot for a short period of time.

That is indeed a relevant point. But what happens immediately after that is the CoG moves forward. The firmware can have a small window of allowable tilt, which the lifting of the forefoot may partially fit in. When the CoG then moves forward beyond the window, the firmware can try to keep the pedals as level as possible. This would initiate a fairly powerful acceleration even before the rider reaches the intended lean. And as you surely know, the wheels read the sensors and respond to the input hundreds of times per second.

Whenever I bring up the firmware into this topic of discussion, I think about the first experience I had with the V10F (perhaps the first firmware version). The behaviour was ridiculous, and it felt impossible to understand at first. I was literally not able to get into a proper leaning position. The pedals felt rock solid, and to a 16S rider (at that time) it felt almost as if there was a tiny but immediate tilt-back trying to stop me from going forward. Later on with an updated firmware the same wheel felt normal, and it now had a senseable softness to the riding mode. Meaning, I was able to feel some tilt allowed when starting to accelerate. The wheel now felt quite natural, and it responded to my commands in a predictable and sensible way. I was able to accelerate fast, for example, and to do so was effortless.

The only way I can explain my experience is a stupendously hard riding mode. And being that the behaviour was quite soon changed based on user feedback, tells me that I (and the owner of the wheel, at least to some extent) wasn’t the only one with said experience.

 

Quote

If the wheel responds "instantaneously", I can achieve any lean angle "instantaneously" (well, as quickly as masses can move around).

That is true. But as soon as you start putting pressure back on your forefoot to stop you from falling over, the CoG starts to move forward, and the firmware can decide how the wheel reacts. If the mode is hard enough to speed up the initiation of a lean as you described, it will just as rapidly try to cancel the forward lean as soon as you start to slow down the fall forward. Which I’m pretty certain does generally displace the CoG a lot more.

Quote

Many beginners feel that they cannot accelerate uphill even though they try to push hard.

I think a few of us will never forget the experienced downhill skier that wasn’t able to get his nee Gotway Tesla to go faster than 4km/h no matter how much he ”leaned”... As soon as we suggested that his ”lean” might actually be a skiing crouch and that he should instead push his hip forward, he stopped replying. And I like to think was happily cruising on his Tesla at whichever speed he wanted...

 

Btw @Mono, I love discussing and pondering this (or any) topic finally with a person of your level of understanding on physics, and your mature and humble attitude of curiousness! That is a first for me.

Edited June 10, 2019 by mrelwood

[Mono]

Weird that it is possible to make the controller so sensitive to essentially prevent fast acceleration. I think I also saw weird braking behavior on a V10 video and started to wonder about the physical limitations due to pedal height and lean angle. The above depiction however seems to suggest that this should not be a physical limitation.

Thanks for the compliment, it's indeed very nice to discuss and getting this opportunity to deepen our understanding (slowly but steadily).

[Chriull]

On 6/10/2019 at 11:26 AM, Mono said:

The colored arrows depict just different scenarios (two constant velocities, two accelerating situations). Black arrows depict the part of the weight induced force that is transformed into torque. Arrow lengths or thicknesses do not exactly reflect the amount of force,

Ok. But once i see something with forces and torques in a "stable" system, it's always yelling inside my brain the sum has to be zero and it won't stop until i have it...

On 6/9/2019 at 11:51 PM, Mono said:

There is more to it than the simple inverted pendulum, as shifting the pendulum has a relevant effect.

As your drawing show different scenarios of a stable system, this are "just" the solution/result of this special inverted pendulum(1) formula for one stable point over time.

The "?original/?real?" inverted pendulum afaik has one more "degree of freedom" - the inverted pendulum is "free mounted" on the axle, but the rider on the wheel is "more" fixed on the pedals. The rider determines the lean angle by himself, whereas the inverted pendulums angle is determined by the control algorithm of the system/inverted pendulum "formula".

Or are both systems covered under the same name?

On 6/10/2019 at 11:26 AM, Mono said:

but the angle of the force. Maybe there is a different way to interpret the arrows, but that would have been unintended 

It may make sense to decompose, say, the green vector into the black torque vector plus a downwards pointing vector plus a forward pointing vector. The decomposition should be unique, if we know the length of the torque vector (which I assume is cos(angle) smaller than the original down force). I now wonder whether this gives us the associated acceleration.

My thoughts so far regarding the decomposition ("summing up the vectors to zero")

-The torque at the tire gives a force along the floor (which is countered by the floor) and determines the acceleration of the system ( F = (m_rider+m_wheel)*a, F = M/r)

- Assuming that the wheel itself is "balanced" (rotational symmetric) it does not introduce (or "need") any torque, so the countertorque for this above M is delivered by the rider standing on the wheel and putting force on the pedal. So the tangential part (along the circle) of this force has to be F_pedal = -M/r1.

- The forces on the CoG of the rider are (The force with which the riders pushes on the pedal)

-- The gravitation force Fg = m_rider * g acting vertically down

-- By the acceleration a of the riders weight a force F_CoG = m_rider * a acts horizontally (parallel to the floor)

The difference of the Fg + F_CoG and F_pedal in horizontal direction ?has to be? F_diff_h = m_wheel * a, the vertical part of this difference will act together with the gravitational force on the wheel into the ground.

From my guts feeling this F_diff_h acts "somewhere" on the wheel and can be ?ignored? for this model - in the "worst case" it is involved with some torque in this model, so the torque on the pedal is not the same (negative) torque at the tire diameter but all three of them sum up to zero?

Hopefully some ?mechanical/structural engineer/physicist? can chime in here - should be no problem for them to show the right answer? ... or over the time the solution will evolve, too. Slowly ...

[Chriull]

As there is the term "zippiness" in the topic - i'll try to contribute some thoughts to this too. As just looking at the stable system like above is just valid for a constant acceleration and has nothing to do with zippiness, which i would understand as the needed effort to get a positive acceleration change per time intervall.

As first thought in regard to this after the above consideration(hypothesis) it should be mainly determined by the wheels geometry and the mechanical points @Mono already stated above.

The wheels firmware should have just some minor if any influence? The firmware can only determine the aggresiveness of the control algorithm (the weakness/stiffness) of the pedal. For a weak reacting firmware there will be a bigger pedal angle in the beginning (the wheel accelerating less), but then it has to accelerate "much" more to reach this stable system.

A "stiff" reacting firmware has a "more or less" constant acceleration and reaches this stable system much earlier.

There imho the main difference should be the "riders impression" of this different behaviour. As with a "weak" reacting firmware - if the rider takes to chance to lean into this bigger pedal angle, leading to even more acceleration or panics back which would lead to less acceleration, while with the stiffer firmware one has a more "controlled" leaning with immediate reaction?

As there are often reports of the different zippiness of the KS18XL and the MSX, which should/could be about comparable wheels there exact geometry would be interesting - to see if this could be a valid explantion.

[mrelwood]

 

Quote

The wheels firmware should have just some minor if any influence? The firmware can only determine the aggresiveness of the control algorithm (the weakness/stiffness) of the pedal. For a weak reacting firmware there will be a bigger pedal angle in the beginning (the wheel accelerating less), but then it has to accelerate "much" more to reach this stable system.

My experience with the V10F firmware might well be accompanied by the negative feedback loop that the wheel’s own high CoG causes. But as the wheel was not my own, I have tried it only for fairly short distances, almost removing the variable of getting accustomed to the wheel behaviour. Between the extremely hard first impression and the second fairly natural one, I only rode my own familiar wheel, at medium setting. The difference in the V10F behaviour was not minor. It was indeed huge.

The behaviour is well noticeable even on the MSX, although I don’t like the way the softer ride modes are implemented (they should’ve been made a lot more suitable for such a large wheel). For a specific amount of acceleration (and the increase rate up to the acceleration) the soft mode does assist the rider, so that a smaller leaning effort eventually transfers to a faster acceleration than on the hard mode.

 

Quote

There imho the main difference should be the "riders impression" of this different behaviour. As with a "weak" reacting firmware - if the rider takes to chance to lean into this bigger pedal angle, leading to even more acceleration or panics back which would lead to less acceleration, while with the stiffer firmware one has a more "controlled" leaning with immediate reaction?

Unless/until the term is specified a lot more precisely, ”zippyness” has no measurable parameter other than the rider’s impression. It is indeed up to the rider’s trust on the wheel. A beginner doesn’t trust any wheel, so he might well not notice a clear difference with different riding modes, and perhaps even not very much between different wheels.

 

Quote

As there are often reports of the different zippiness of the KS18XL and the MSX, which should/could be about comparable wheels there exact geometry would be interesting - to see if this could be a valid explantion.

The difference between a 2.5” and a 3” tire is not small. Marty Backe had measured the outer diameter of the tires, and I think it was 0.75”, iirc. Pedal height and the vertical CoG being others that affect.

And to sound like a broken record, even the hard riding mode on the XL is initially a good deal softer than any of the modes on the MSX.

[Mono]

12 hours ago, Chriull said:

Ok. But once i see something with forces and torques in a "stable" system, it's always yelling inside my brain the sum has to be zero and it won't stop until i have it...

True. I commented on the decomposition somewhere. If I am not mistaken zero sum only applies if nothing is accelerating or kept on moving. I think the decomposition in horizontal, vertical, and tangential force makes quite some sense, but I didn't give it a deep thought (yet).

Quote

As your drawing show different scenarios of a stable system, this are "just" the solution/result of this special inverted pendulum(1) formula for one stable point over time.

The more interesting part of the model to me was indeed to realize that the contact/support point can and does change/shift (which sets the model apart from the inverted pendulum). I couldn't do the animation 

Quote

The "?original/?real?" inverted pendulum afaik has one more "degree of freedom" - the inverted pendulum is "free mounted" on the axle, but the rider on the wheel is "more" fixed on the pedals. The rider determines the lean angle by himself, whereas the inverted pendulums angle is determined by the control algorithm of the system/inverted pendulum "formula".

I am not sure what you mean with "determines the lean angle by himself". To me it seems reasonable to assume that the rider can only(!) do two things:

1. shift the contact point on the pedal without changing the CoG (via push or release the forefoot using the ankle), thereby introducing a lean and
2. shift its CoG vertically (bent or stretch the knees or the hip, the latter could also introduce a rotation and move the wheel forward, but that's maybe a too advanced question), thereby reducing or increasing the weight force for a short period of time.

Of course riders can also weave or rotation their arms and do other movements that may have some effect, but to me it makes more sense to try to understand how the system works without those (for the time being). Do you see any other first-order-relevant input from the rider?

 

Edited June 13, 2019 by Mono

[Chriull]

52 minutes ago, Mono said:

If I am not mistaken zero sum only applies if nothing is accelerating or kept on moving

Ups. Yes, of course!

54 minutes ago, Mono said:

I am not sure what you mean with "determines the lean angle by himself".

The rider can choose the lean angle freely, the inverted pendulum's lean angle is controller by the "system".

 

1 hour ago, Mono said:

. I think the decomposition in horizontal, vertical, and tangential force makes quite some sense,

Divide and conquer

1 hour ago, Mono said:

but I didn't give it a deep thought (yet).

Refining this should open the possibility to do nice simulations. Even in excel...

1 hour ago, Mono said:

The more interesting part of the model to me was indeed to realize that the contact/support point can and does change/shift (which sets the model apart from the inverted pendulum).

This seems to be a nice fine tuning possibility for us riders, until the full lever by the pedal is used.

1 hour ago, Mono said:

I couldn't do the animation 

There should be enough tools available to make animations from prepared data files - but i never happened to get in touch with any of them. Matlab could be such a prog? Or all/many of the render programs should take prdefined paths for the objects?

[Mono]

7 hours ago, Chriull said:

The rider can choose the lean angle freely

The rider is heavily bounded by the physics in how they can "choose" (rather: influence) the lean angle. There is no easy way to displace a mass by choice, am I wrong? So the usual idea that we have in mind when saying "leaning" is more or less physically impossible.

The by far easiest way to influence the lean angle is, AFAICS, using the ankle joint as actuator (and thereby moving the wheel in a different place relative to the rider). The same is true when standing on the ground (the ground doesn't move but the contact point still does), unless one has a wall to push against to move ones CoG.

Edited June 13, 2019 by Mono

[Chriull]

17 hours ago, Mono said:

The rider is heavily bounded by the physics in how they can "choose" (rather: influence) the lean angle. There is no easy way to displace a mass by choice, am I wrong?

No - you are right. The tangential force (from the "torque") counters the riders applied gravitational force. But just the vertical part - if these are not equal, the pedal angle will change. That's the self balancing part, that the wheel adjusts the torque accordingly.

Then there is a "vertical" horizontal part left over, that "accelerates" the rider - determines his lean angle. But one could turn it the other way round, that the rider can control the acceleration by leaning. But yes - this lean is restricted and not free choosable.

What's now puzzling me, is that the "full counter torque" accelerates the wheel and rider against the floor...

And how and if this vertical horizontal difference is applying to the rider acceleration or is just lean support.

Once i need to take an example and draw the forces with real values - should bring some more knowledge ... Hopefully ....

Edited June 14, 2019 by Chriull

[Mono]

12 minutes ago, Chriull said:

Then there is a "vertical" part left over, that "accelerates" the rider - determines his lean angle.

I assume you mean horizontal left over instead of vertical here? Do you mean that the acceleration determines what is the stationary lean angle? (As I would still insist that accelerating the rider is not the decisive mechanism to change the lean angle.)

[Chriull]

4 minutes ago, Mono said:

I assume you mean horizontal left over instead of vertical here?

Yes. Mixed it up - also at the end of the post

4 minutes ago, Mono said:

Do you mean that the acceleration determines what is the stationary lean angle? (As I would still insist that accelerating the rider is not the decisive mechanism to change the lean angle.)

But the rider has to "outlean" the acceleration, or there would be a torque acting on the rider turning him backwards?

[Mono]

On 6/14/2019 at 12:26 PM, Chriull said:

But the rider has to "outlean" the acceleration

for each acceleration away from zero there is a stationary lean angle away from zero (I am avoiding the word stable here, because it is a stationary but, AFAICS, unstable point of the model).

On 6/14/2019 at 12:26 PM, Chriull said:

or there would be a torque acting on the rider turning him backwards?

The mechanisms which let the rider fall behind would be inertia (in an acceleration scenario) and wind forces, but I don't see that those are a torque. Gravity adds a down movement when the rider is not in the stationary lean angle (anymore).

[Chriull]

5 minutes ago, Mono said:

The mechanisms which let the rider fall behind would be inertia (in an acceleration scenario) and wind forces, but I don't see that those are a torque. Gravity adds a down movement when the rider is not in the stationary lean angle (anymore).

As the feets are (relative) "fixed" at the pedal. Every force(difference), as the mentioned inertia, wind drag, gravity creates a torque.

So with the "wrong" lean angle a rider will fall forward or be pushed backwards == turn around the fixpoint (feet at the pedal).

[Chriull]

New approach - took some time, but seems sound (at least to me for now  )

Once the rider puts his "gravitational force" a bit outside the center of the pedals a torque is created - namely the cross product of the vector of this gravitational force and the vector from the axle to the point on the pedal this gravitational force is applied.

This wheel has to create a countertorque to keep the pedal straight and this countertorque is acting as a force from the wheel to the floor accelerating the whole system.

So the force accelerating rider and wheel is only dependend on the wheel geometry, the mass of the rider and the point on the pedal the rider puts his weight on(1). And the acceleration as found in the previous topic results from this force and the riders and wheel weight.

The riders lean is "only" an "instinctive" reaction, so that we won't fall of the wheel - like one is riding in a train, just standing and not leaning somewhere or holding onto something with our hands. (In reality we will anticipate how we have to lean and adjust the pressure point on the pedal by leaning forward.)

The firmware could with this just influence the zippiness indirectly - if the wheel is to weak or too slow to counter our "pressure shift on the pedal" we would fall forward and overlean, so we will shift the pressure point forward slower to no endager ourselves -> the wheel is less zippy.

(1) Was imho posted here already some years ago from a member, describing the formula with the geometry and riders mass the maximum reachable acceleration?

[Chriull]

So for my KS16S this would lead to the following numbers (hopefully right - did not double check )

Wheel radius: ~21 cm

(shortest)Distance Pedal - Axle: 11,5 cm

Half Pedal length: 10,5

EUC weight: ~17 kg

Rider weight: ~90 kg

F gravitational = 90 * 9,81 = 883 N

Angle between "pedal tip" and vertical: alpha = atan( 10,5 / 11,5 )

Angle between "Torque Force" (normal on vector from axle to pedal tip) and F gravitational: 90°-alpha

Distance from pedal tip to axle: r1 = Sqrt(10.5^2 + 11.5^2) = 15 cm

Torque on pedal: M = r1 = Fg * cos (90° - alpha) * r1 = 92 Nm

Accelerating force on Wheel  F = M / wheel_radius = 441 N

Acceleration: a = F/(m_rider + m_wheel) ~4 m/s²

Edit: for calculating torque on pedal one should need cos alpha and not cos(90°-alpha)

Edited June 17, 2019 by Chriull

[Chriull]

Some more thoughts: one uses the from @Mono stated inertia to increase the pressure on the pedals. The "attack" angle could also get better.

And with the pedals always tilting a bit (especially in soft mode), the distance to the axle gets bigger, but angle gets "worse" -> eventually more torque on pedal?. And a better radius factor (leverage) is reached for more acceleration force.

[Chriull]

On 6/13/2019 at 9:50 AM, Mono said:

I couldn't do the animation

(Un)fortionately i found openSCAD and wasted some hours of my sparetime... 

https://i.imgur.com/oLsHDcI.mp4

This was calculated with 100 frames a second, but i did not manage to make this to the mpg - so with the 25 fps it's now a 4 times slow motion.

Looks like some heavy leaning - but the max acceleration reached is 6,9 m/s² with finally reaching 62 km/h over 43 meters within 5 seconds...

I let the rider put his "pressure" point from the middle to the front within 2.5 seconds and than back again.

Attached the ods (libre office) file with the calculations, second sheet to be exported as csv (maybe not working as with the german settings i did some replacement of the , to .) and the animation.scad file.

Had some quirks in the formulas  So here everything new with a real time movie:

 

A short description

I made the situation also a bit more realistic - the "pressure point" only goes forward until 5,5 cm within 1.5 seconds instead the full 10.5 cm (21 cm pedal length), so the acceleration stays in "normal" ranges. After this within 1 second the "pressure point" is shifted back again to the center.

The resulting force (black arrow) from motor torque on the pedal and the riders inertia and weight is exactly in line with the pressure point and wheels axle, so this should induce no torque on the pedals anymore?  ( i would not really put my hands into a fire for this, but could be true???)

Edited June 27, 2019 by Chriull

[hyperair]

@Chriull That's a really nice animation. Could you share the OpenSCAD code please?

[Chriull]

46 minutes ago, hyperair said:

@Chriull That's a really nice animation. Could you share the OpenSCAD code please?

I made by now a small c++ programm to generate the data (animate-data.scad) for the frames - the output from this program is then included in the openscad animation.

I am looking forward to implement pedal tilt as next step, then a "rider control loop" - lean angle is not immedeately adjusted perfectly in reality and the "wheel control loop" to experiment with PID settings.

animate.cpp animation.scad animate-data.scad

Edited July 4, 2019 by Chriull

[Mono]

On 6/26/2019 at 7:32 PM, Chriull said:

 

A short description

I made the situation also a bit more realistic - the "pressure point" only goes forward until 5,5 cm within 1.5 seconds instead the full 10.5 cm (21 cm pedal length), so the acceleration stays in "normal" ranges. After this within 1 second the "pressure point" is shifted back again to the center.

The resulting force (black arrow) from motor torque on the pedal and the riders inertia and weight is exactly in line with the pressure point and wheels axle, so this should induce no torque on the pedals anymore?  ( i would not really put my hands into a fire for this, but could be true???)

Very nice animation! I struggle to understand (the physical model of) what happens in the beginning. The behavior of the green ball falling forward suggests that the contact point was moved backwards, but I don't see that this happened, and I also do not see the wheel moving backwards as a result of this. In other words, how is the first acceleration of the system initiated?