What determines wheel zippiness?

[Mono]

1 hour ago, Chriull said:

If one cancels the restriction for the "acceleration force" to be horizontal it will also work for inclines.

Yes, it seems that the resulting "forward thrust" points necessarily in whatever direction the road designates as "forward", am I wrong?

[Mono]

One major insight I gathered from this thread was that I should make a distinction between forward weight displacement (of the CoG) and forward lean. I consider as lean here the angle between a vertical line and the line between the CoG of the rider and the (center) point of attack of the summary (down) force on the pedal. When the rider has a non-zero and non-changing forward lean angle, there must a horizontal force that prevents the rider from falling forward (usually wind and/or acceleration). The vertical component of the resulting vector that acts on the pedal is always the rider weight. Angle and vertical component determine the length of the resulting vector uniquely to be sec(angle) x weight = weight / cos(angle). Therefore we have

   thrust [kg] = weight / cos(lean angle) x effective leverage / outer wheel radius

where the effective leverage is the amount of displacement of the resulting vector from the wheel axle (orthogonal to the vector). Thoughts?

[Aneta]

21 hours ago, Mike Sacristan said:

 

 

This would require the backward tilt calibration to allow tolerance towards level.
However it did not work that way when you were on level ground but it did when you were on the incline.  Maybe you weren't going fast/hard enough on level ground.

Yes I am aware of that I was just trying to comfort you after the unlikely paradox you presented. 

So basically you are saying that what you did was impossible.
And that is 100% fine by me because that's the way I see it too. 

Impossibility for the wheel to detect an incline and trick with pedal tilt are two different things.

[Aneta]

1 hour ago, Mono said:

One major insight I gathered from this thread was that I should make a distinction between forward weight displacement (of the CoG) and forward lean. I consider as lean here the angle between a vertical line and the line between the CoG of the rider and the (center) point of attack of the summary (down) force on the pedal. When the rider has a non-zero and non-changing forward lean angle, there must a horizontal force that prevents the rider from falling forward (usually wind and/or acceleration). The vertical component of the resulting vector that acts on the pedal is always the rider weight. Angle and vertical component determine the length of the resulting vector uniquely to be sec(angle) x weight = weight / cos(angle). Therefore we have

   thrust [kg] = weight / cos(lean angle) x effective leverage / outer wheel radius

where the effective leverage is the amount of displacement of the resulting vector from the wheel axle (orthogonal to the vector). Thoughts?

I think it should be apparent weight (W'), not just weight, in that formula. For riding on level surface, W' = sqrt((mg)^2 + (ma + D)^2), where D is aerodynamic drag of the rider. For incline, it'll a bit more complicated since both ma and D will have vertical components.

PS. If rider is leaning into airflow, the angle of attack is no longer 90 degrees and there will be small lift as well, not just drag. Especially if "chooching". But still negligible, surely less than 1kg.

Edited December 14, 2019 by Aneta

[Chriull]

3 hours ago, Mono said:

Thoughts

Nothing concrete till now. I started with an excel sheet to get some numbers with this force system.

First results (not checked/verified) were for the KS16S geometry getting about 4m/s^2 acceleration by a ?90? kg rider standing on the pedal tip. With this 4m/s^2 acting on the rider the motor increases acceleration to about 4.5m/s^2. (Could all be easily beyond the KS16S limits...)

Interesting is the "positive feedback loop".(1)

Once the sheet is finished and checked it could be interesting to look further into this. Such a positive feedback loop should be definitely something making a wheel "zippy". 

Edit:(1) seems it has to be this way. If an additional torque "appears", the motor has to accelerate more

Edited December 14, 2019 by Chriull

[Mono]

2 hours ago, Aneta said:

I think it should be apparent weight (W'), not just weight, in that formula. For riding on level surface, W' = sqrt((mg)^2 + (ma + D)^2), where D is aerodynamic drag of the rider.

The idea was that when a stationary angle is given and we know the weight, we can decompose the vector into vertical and horizontal component and compute sqrt((mg)^2 + (ma + D)^2) without to know (ma + D), because there is no degree of freedom left for the size of the horizontal component. 1 / cos(lean angle) = sec(lean angle) = sqrt(1^2 + (horizontal_component/weight)^2) = W' / weight is the factor to get W' from the weight, or mg if you like, AFAICS. We can of course also compute the horizontal component from the angle and vertical component, which is tan(angle) x weight, I think.

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For incline, it'll a bit more complicated since both ma and D will have vertical components.

Good point. So we could compute how much heavier the rider becomes due to the vertical acceleration and then do the same as above. OK, we cannot, because we don't know the acceleration, that's what we want to compute, kind-of.

That goes to a bigger point that I don't know the angle/positions of the stationary points of the entire system. I guess I will leave it at that for the time being.

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PS. If rider is leaning into airflow, the angle of attack is no longer 90 degrees and there will be small lift as well, not just drag. Especially if "chooching". But still negligible, surely less than 1kg.

Same as above, we can account for lift (or downforce) from airflow in the vertical component first and solve from there, if a stationary angle is given.

Edited December 14, 2019 by Mono

[Chriull]

17 minutes ago, Mono said:

OK, we cannot, because we don't know the acceleration, that's what we want to compute, kind-of.

Yes. It's a dynamic system. Imo best just simulated for "interesting" cases.

As it is in reality - the rider starts with some cog shift/lean, the wheel reacts with some acceleration, the rider reacts again to stand this which forces a new wheel reaction, and so on...

With accelerations there could be no easy, static system existing.

[Aneta]

36 minutes ago, Mono said:

The idea was that when a stationary angle is given and we know the weight, we can decompose the vector into vertical and horizontal component and compute sqrt((mg)^2 + (ma + D)^2) without to know (ma + D), because there is no degree of freedom left for the size of the horizontal component. 1 / cos(lean angle) = sec(lean angle) = sqrt(1^2 + (horizontal_component/weight)^2) = W' / weight is the factor to get W' from the weight, or mg if you like, AFAICS. We can of course also compute the horizontal component from the angle and vertical component, which is tan(angle) x weight, I think.

Ah, I see, you're absolutely right, I misread your formula and was thinking that the cos(lean angle) was there to compute the torque arm of the force from the distance between center of pressure on the pedal and the center of the pedal directly under the axle. If we denote the latter as COP_shift, then effective leverage (or what I call torque arm, an is a standard concept in rotational dynamics) = COP_shift * cos(lean angle), or 

 thrust = weight * COP_shift / outer wheel radius

Thoughts?

[Aneta]

17 minutes ago, Chriull said:

As it is in reality - the rider starts with some cog shift/lean, the wheel reacts with some acceleration, the rider reacts again to stand this which forces a new wheel reaction, and so on...

So, it's like Achilles and the tortoise?

[Mono]

53 minutes ago, Aneta said:

the torque arm of the force from the distance between center of pressure on the pedal and the center of the pedal directly under the axle. If we denote the latter as COP_shift

I don't quite get what that is. What is "the center of the pedal directly under the axle" and why is that relevant? I figured that the calculations are entirely independent of the pedal (of course the pedal must be in the way of the force vector to catch the force, but otherwise it doesn't seem to make a difference where it is and how it is tilted).

I checked Wikipedia: what I called effective leverage (white bars in the figure) is rather called moment arm. Is that what you refer to as torque arm?

[Mono]

58 minutes ago, Chriull said:

Yes. It's a dynamic system. Imo best just simulated for "interesting" cases.

RIght, but AFAICS it is only stable if the rider has the right control strategy, which would need to be part of the simulation, not sure how else to find the interesting cases. I am not likely to investigate this, in particular if I don't see a chance to find some closed solution.

1 hour ago, Chriull said:

With accelerations there could be no easy, static system existing.

Agreed.

[Aneta]

6 minutes ago, Mono said:

I don't quite get what that is. What is "the center of the pedal directly under the axle" and why is that relevant? I figured that the calculations are entirely independent of the pedal (of course the pedal must be in the way of the force vector to catch the force, but otherwise it doesn't seem to make a difference where it is and how it is tilted).

I checked Wikipedia: what I called effective leverage (white bars in the figure) is rather called moment arm. Is that what you refer to as torque arm?

Yes, moment arm, or lever arm it is!

By COP shift I meant the distance between the center of pressure (where green or red force vectors intersect the pedal in your OP) and the intersection of pedal and thin grey vertical line going through the axle.

[Mono]

27 minutes ago, Aneta said:

By COP shift I meant the distance between the center of pressure (where green or red force vectors intersect the pedal in your OP) and the intersection of pedal and thin grey vertical line going through the axle.

Got it, but I still don't see why that is not some arbitrary value. I could turn the pedal around the point of attack and fix it into a different position which would change the COP shift without changing torque. Or I could move the red arrow with the pedal in direction of the arrow without changing the lever arm or torque but changing the COP shift.

[Aneta]

43 minutes ago, Mono said:

Got it, but I still don't see why that is not some arbitrary value. I could turn the pedal around the point of attack and fix it into a different position which would change the COP shift without changing torque. Or I could move the red arrow with the pedal in direction of the arrow without changing the lever arm or torque but changing the COP shift.

But the wheel tries to keep the pedal level no matter what? Are we assuming "infinitely hard pedal" mode in our discussions? If yes, then the only way to "tell" the motor to change the torque is to shift COP, by shifting our weight within our footprints (or pedal outline, whichever is smaller).

Edited December 15, 2019 by Aneta

[Mono]

3 minutes ago, Aneta said:

But the wheel tries to keep the pedal level no matter what? Are we assuming "infinitely hard pedal" mode in our discussions?

Yes. I was considering different fixed pedal geometries that should all lead to the exact same outcome (only the position of the force vector should matter) but don't compute to the same COP shift.

[Mono]

Inspired by this comment and by this video

I updated the original figure adding a 15º inclination line.

I started to figure that a major part of zippiness may be that small (forward-backward) weight displacements lead to large effects. I guess I am not the first one  Effect size per displacement is independent of the wheel power(!) but decreases with increasing wheel size. Let's look at some numbers.

The weight displacement to get a certain desired thrust is proportional to the wheel size (and the rider weight). It is as simple as this: if I weight 80kg and want 8kg forward thrust, which is 10% of my weight, I need to displace my weight by 10% of the wheel radius forward. Stupidly simple.

The same displacement on a 14"x2.125" wheel produces only 76% of the thrust on a 18"x2.5" wheel. In other words, on the bigger wheel we need a 32% (≈ 1 / 0.76 - 1) further weight displacement for the same thrust. 10kg added wheel weight reduces the effect of the thrust additionally by the factor 92 / 102 (assuming 12kg + 80kg to begin with). Combined, we need a 46% (≈ 1 / (0.76 x 0.9) - 1) further displacement on the bigger wheel, for example 4.4cm instead of 3cm to induce the same acceleration. All this does not depend at all on motor power or battery size or firmware tweaks or anything inside the wheel. Nice and simple and probably one relevant aspect of zippiness.

[Aneta]

26 minutes ago, Mono said:

Inspired by this comment and by this video

I updated the original figure adding a 15º inclination line.

I started to figure that a major part of zippiness may be that small (forward-backward) weight displacements lead to large effects. I guess I am not the first one  Effect size per displacement is independent of the wheel power(!) but decreases with increasing wheel size. Let's look at some numbers.

The weight displacement to get a certain desired thrust is proportional to the wheel size (and the rider weight). It is as simple as this: if I weight 80kg and want 8kg forward thrust, which is 10% of my weight, I need to displace my weight by 10% of the wheel radius forward. Stupidly simple.

The same displacement on a 14"x2.125" wheel produces only 76% of the thrust on a 18"x2.5" wheel. In other words, on the bigger wheel we need a 32% (≈ 1 / 0.76 - 1) further weight displacement for the same thrust. 10kg added wheel weight reduces the effect of the thrust additionally by the factor 92 / 102 (assuming 12kg + 80kg to begin with). Combined, we need a 46% (≈ 1 / (0.76 x 0.9) - 1) further displacement on the bigger wheel, for example 4.4cm instead of 3cm to induce the same acceleration. All this does not depend at all on motor power or battery size or firmware tweaks or anything inside the wheel. Nice and simple and probably one relevant aspect of zippiness.

Very good and simple, I like it!

And we achieve COP displacement by either placing our feet on pedals differently, or shifting our weight on toes or heels. If pedals are small and the wheel is big diameter, we will experience discomfort and tiring effort by either placing our feet too much forward for the incline (and having the edge of the pedal cut into our soles) or "standing on tiptoes" like a ballerina.

[mrelwood]

 

1 hour ago, Mono said:

Combined, we need a 46% (≈ 1 / (0.76 x 0.9) - 1) further displacement on the bigger wheel, for example 4.4cm instead of 3cm to induce the same acceleration. All this does not depend at all on motor power or battery size or firmware tweaks or anything inside the wheel.

Exactly! Now let’s go and make all the ”zippy motor wattage” people understand, since I need a bit of help with the quest... 

Although, while I believe what you wrote being the main factor by a large margin, the effect of the firmware can’t be left out. A suitably soft riding mode can help the rider in reaching a larger displacement with lesser effort.

If the wheel would stop balancing for 0.1 seconds just as you have started to lean forward, you’d start falling. The wheel migh even roll back a bit in a standing start. The ”fall” doesn’t require any added effort for it to accelerate and reach further displacement. During the 0.1 second downtime you have fallen sufficiently far so that when the motor kicks in, you have reached a large enough displacement for the wheel to accelerate fast to keep the balance.

The downtime would of course feel horrible for the rider, but a softer pedal mode has a bit of the same effect as the wheel ramps up the power only after the pedals have tilted slightly already.

[Mono]

11 hours ago, Aneta said:

By COP shift I meant the distance between the center of pressure (where green or red force vectors intersect the pedal in your OP) and the intersection of pedal and thin grey vertical line going through the axle.

If the force vector is tilted, this COP shift can be zero or even negative, while we still produce positive torque. That's why it does not look like a suitable measure to me.

[Mono]

9 hours ago, mrelwood said:

Exactly! Now let’s go and make all the ”zippy motor wattage” people understand, since I need a bit of help with the quest... 

Although, while I believe what you wrote being the main factor by a large margin, the effect of the firmware can’t be left out. A suitably soft riding mode can help the rider in reaching a larger displacement with lesser effort.

Yes, that's what I would like to understand better.

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If the wheel would stop balancing for 0.1 seconds just as you have started to lean forward, you’d start falling.

Interesting point. Pretty weird that a delayed reaction of the wheel shall increase its zippiness. I do see the mechanism of the idea though.

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The wheel migh even roll back a bit in a standing start.

That's not happening by the wheel controller, I am pretty certain. Of course a rider can do this actively, but that doesn't differentiate different wheels.

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The ”fall” doesn’t require any added effort for it to accelerate and reach further displacement. During the 0.1 second downtime you have fallen sufficiently far so that when the motor kicks in, you have reached a large enough displacement for the wheel to accelerate fast to keep the balance.

I guess we should check how large this effect can actually be and which parameters are involved. That doesn't look too difficult. For example, how long does it take to increase 1cm displacement to 1.5cm assuming no pedal support? (My first estimate suggests that 10mm become 10.1mm after 0.1s, but I might have messed up something in the calculation).

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The downtime would of course feel horrible for the rider,

yes

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but a softer pedal mode has a bit of the same effect as the wheel ramps up the power only after the pedals have tilted slightly already.

I still wonder whether there are other candidate explanations for how a softer mode improves the perception of zippiness.

Edited December 15, 2019 by Mono

[Chriull]

8 hours ago, mrelwood said:

but a softer pedal mode has a bit of the same effect as the wheel ramps up the power only after the pedals have tilted slightly already.

 

59 minutes ago, Mono said:

I still wonder whether there are other candidate explanations for how a softer mode improves the perception of zippiness.

A softer pedal mode, were the pedal is allowed to tilt forward some time imo maninly ease the leaning forward of the driver, as there is less support from the pedal.

After this delay the wheel has to deliver the full torque to at least keep the pedals at this angle or a bit more to turn them horizontal again.

So imho, beside the bit better leverage by the pedal tilt its the ease of leaning is increased (if one dares)

 

[Mono]

7 minutes ago, Chriull said:

the ease of leaning is increased

I don't understand what "ease of leaning" would mean, and/or what it physically relates to other than what @mrelwood suggested, that is, to my understanding, "automatically" falling forward due to lack of pedal support. The effect size of this "automatic" falling should be easily quantifiable as a function of starting lean angle and time without support?

[Chriull]

8 minutes ago, Mono said:

what it physically relates to other than what @mrelwood suggested, that is, to my understanding, "automatically" falling forward due to lack of pedal support

Nothing. I just "decleare" this falling forward as ease of leaning. What could have the effect of making the wheel zippier (mainly by firmware?)

[Mono]

22 minutes ago, Chriull said:

What could have the effect of making the wheel zippier (mainly by firmware?)

Agreed, that would be a firmware effect. I still don't know whether this effect is of perceivable size though. Pedal softness in itself of course is.

[Mono]

15 hours ago, Aneta said:

Yes, moment arm, or lever arm it is!

I just checked the terminology. Moment arm and lever arm are crucially not the same  The force is by definition perpendicular only to the moment arm.

Edited December 15, 2019 by Mono