What determines wheel zippiness?

[Mono]

On 12/15/2019 at 5:10 AM, Mono said:

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.

Next up: the influence of pedal height under acceleration (or breaking). The disclaimer first: I consider low pedals as a significant safety risk. Getting caught with the pedal or foot is a reasonably common and wicked reason for crashes. I personally require at the very least 13 cm ground clearance to be on a reasonably safe side.

 

When the force vector is vertical (green cases), the pedal height has zero effect on the torque (the white bar is unaffected by vertical shift). That was easy 

Work in progress: what happens under acceleration, that is, when the force vector is tilted (red cases)? We have 

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

and the leverage increases with increase tilt angle if the pedals are below the axle and increases even more the lower the pedals are.

This looks like a significant effect, which should be further quantified. For example, without any forward displacement, the thrust is

    sin(tilt angle) x (distance to axle / radius) x weight / cos(tilt angle)
    = tan(tilt angle) x (distance to axle / radius) x weight

where distance to axle is the (vertical) distance between axle and the point of attack and is negative if the axle were below the attack point. EDIT: the more general formula including forward displacement is now in the OP.

Lowering the pedals by 10% of the wheel radius (2cm on a 16" wheel) at a lean angle of 10º adds for an 80kg rider 1.4kg thrust. As a reference, a 2cm forward displacement of the CoG at a lean angle of 0º and otherwise same setting adds 8kg thrust.

EDIT: A forward displacement does not influence the effect of pedal lowering and vice versa.

Edited December 16, 2019 by Mono

[Aneta]

32 minutes ago, Mono said:

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.

My bad. Translating the concepts that I learned in a different language. Moment arm it is!

[Mono]

ooups, I found a (much) simpler expression for the forward thrust under a stationary lean as

   thrust [kg] = weight x (horizontal displacement + tan(lean) x vertical displacement) / wheel radius
                      = weight x (horizontal distance between axle and force vector) / wheel radius

where the displacements refer to the forward and down position of the point of attack on the pedal away from the axle. Whhaaat? wow  Checked it twice... still only 75% convinced.

EDIT: it seems to be correct:

 

Edited December 17, 2019 by Mono

[Aneta]

 

4 hours ago, Mono said:

What happens under acceleration, that is, when the force vector is tilted (red cases)? We have 

   thrust [kg] = weight / cos(tilt angle) x lever arm / outer wheel radius

and the lever arm increases with increase tilt angle if the pedals are below the axle and increases even more the lower the pedals are.

Shouldn't it be moment arm in that formula?

Edited December 15, 2019 by Aneta

[Mono]

2 minutes ago, Aneta said:

Shouldn't it be moment arm in that formula?

yes, it should, I was too quick to adopt the terminology which turned out to be wrong 

[Aneta]

4 hours ago, Mono said:

Without forward displacement, lowering the pedals by 10% of the wheel radius (2cm on a 16" wheel) at a lean angle of 10º adds for an 80kg rider 1.4kg thrust. As a reference, a 2cm forward displacement of the CoG at a lean angle of 0º and otherwise same setting adds 8kg thrust.

So, if a light rider like Petra has difficulty going up steep hills at constant speed (most common situation, and safer one, as accelerating up a steep hill is asking for a faceplant), then lowering the pedals is not a solution, but making pedals longer is?

[Mono]

On 12/15/2019 at 11:20 PM, Aneta said:

So, if a light rider like Petra has difficulty going up steep hills at constant speed (most common situation, and safer one, as accelerating up a steep hill is asking for a faceplant), then lowering the pedals is not a solution, but making pedals longer is?

yes, good catch, if the lean angle is zero (no acceleration, negligible airstream), the pedal height has zero(!) effect on the thrust. Not necessarily what I would have expected...

EDIT: however, the lean angle cannot be zero even at constant speed. Neglecting air resistance, the lean angle must be somewhat proportional to the forward thrust with which the wheel pushes the rider forward to keep up the speed.

Edited March 14 by Mono

[Aneta]

1 minute ago, Mono said:

yes, good catch, if the lean angle is zero (no acceleration, negligible airstream), the pedal height has zero(!) effect on the thrust. Not necessarily what I would have expected...

I guess, the opposite may be useful - raise the pedals, so that you can elongate them without them scraping the slope.

Now, that would be a useful feature if pedals can be quickly raised/lowered on their hangers without any tools, and elongated easily (perhaps, similar to how compact printer paper tray can be unfolded to double its size from folded size)!

[mrelwood]

 

13 hours ago, Mono said:

That's not happening by the wheel controller, I am pretty certain.

The wheel moving backwards is just what would happen if the controller actually turned completely off during the fall in the extreme example of a firmware delay. Not relevant for further thought, as no soft mode actually goes completely dead.

13 hours ago, Mono said:

I guess we should check how large this effect can actually be

That does sound challenging to me, as I understand the firmware response curves to be quite complicated and not easily measurable. To me the effect has been significant however, at least on wheels that are capable for a very hard pedal mode, such as the V10F and the MSX.

[Mono]

8 hours ago, mrelwood said:

That does sound challenging to me, as I understand the firmware response curves to be quite complicated and not easily measurable.

It should be simple to compute the upper bound of the effect size, because the effect comes from gravity (falling forward). I also mount the unpowered wheel regularly (haven't tried to roll it yet).

[Chriull]

On 12/14/2019 at 11:26 PM, Chriull said:

I started with an excel sheet to get some numbers with this force system.

Seems more finished by now - a point i got just now from @RockyTop's youtube video linked above is that on an incline the EUC itself already creates a torque which has to be countered...

I'm ?quite? sure by now if this is already considered in my following spreadsheet by "splitting the gravitational force" for inclines in the downward part, the part along the surface and the part perpendicular to the surface. The part along the surface is working against the EUC trying to accelerate.

 

KS16S
Wheel radius	r	0,21	m	21	cm
Half pedal length	w	0,105	m	10,5	cm
Distance pedal-axle	l	0,115	m	11,5	cm
EUC weight	mw	17	kg
Rider weight	mr	90	kg
	g	9,81	m/s²
Riders force attack point on pedal is the leftmost pedal tip, rider facing to the left
inclination angle	epsilon		0,35	rad	20,00	°
Pedal tilt (clockwise)	alpha		0,03	rad	2,00	°
length of line between force attack point on pedal to axle	|rp|	sqrt(l²+w²)	0,16	m
angle between line from force attack point on pedal to upmost pedal fixing(axle) and pedal(clockwise)	beta	arcsin(l/r)	0,83	rad	47,60	°
Gravitational Force of Rider on Pedal	|Fg|	mr*g	882,90	N
acceleration of rider	|a|		2,00	m/s²
"Inertia Force" of Rider on pedal by an acceleration a	|Fi|	mr*a	180,00	N
angel of "intertia Force"/acceleration to horizontal (clockwise)	delta	epsilon (==inclination)	0,35	rad	20,00	°
resulting force of rider on pedal	|Fr|	sqrt( (Fg+Fi*sin(delta))²+(Fi*cos(delta))² )	959,49
angel between Fr and rp	gamma	beta + pi/2 - alpha	2,37	rad	135,60	°
Torque on pedal	M	|rpxFr|=|rp|*|FR|*sin(gamma)	104,54	Nm
accelerating force of wheel on surface	|Fa|	M/r	497,79	N	50,74	kg
Weight force wheel and rider on surface	|Fgs|	g*(mw+mr)	1.049,67	N	107,00	kg
Perpendicular part to surface	|Fgsp|	|Fgs|*cos(epsilon)	986,37	N	100,55	kg
Part along surface	|Fgss|	|Fgs|*sin(epsilon)	359,01	N	36,60	kg
resulting forward acceleration of rider and wheel	a	(|Fa|-|Fgss|)/(mr+mw)	1,30	m/s²	0,13	g
Force Scale	0,001

 

The libre office file is attached here - hope the formulas and considerations are somewhat valid

EUC acceleration.ods

[Mono]

On 12/17/2019 at 2:02 PM, Chriull said:

Seems more finished by now - a point i got just now from @RockyTop's youtube video linked above is that on an incline the EUC itself already creates a torque which has to be countered...

I figure that it should be sufficient to just add the (negative) thrust that the slope creates on the entire system in the end. That seems to be the only change necessary. I figured out this baffling simplification of the generated torque and thrust in the meanwhile. It is the length of the blue line (times rider weight divided by wheel radius):

Edited November 16, 2022 by Mono

[Chriull]

Some graphs derived from the above sheet:

Both are for a 20° incline - first shows the wheel acceleration (for an initial rider acceleration of 0) versus pedal tilt angle if for applying the riders weight force on the pedal tips (10,5 cm from center)

This second one is again for a 20° incline with 0° pedal tilt. This time the wheels acceleration is show versus the "Force attack point" - where the rider applies his weight force measured from the center of the pedal in m. The first point is about 8,5 cm from the center were the wheel accelerates forward up to the full (half) pedal length of 10,5 cm. That's imo the essence of @RockyTop's linked youtube video - the steeper the incline, the smaller the usable part of the pedal front for forward acceleration-

[Chriull]

8 minutes ago, Mono said:

I figure that it should be sufficient to just add the thrust that the slope creates on the entire system in the end.

That's imo what i did - and seems to be a valid way.

Quote

I figured out this in the meanwhile:

You have nicely shown all the angles and vectors in the diagramm! ... i thought of this too, but was too lazy

Ps.: I although tried to iterate the above formula set by putting the calculated acceleration from the motor torque into the field for "acceleration of rider" (for the riders force vector) and the system quite fast reaches a stable acceleration value. (at least with two examples i tried...)

Edited December 17, 2019 by Chriull

[Mono]

1 hour ago, Chriull said:

Some graphs derived from the above sheet:

Both are for a 20° incline - first shows the wheel acceleration (for an initial rider acceleration of 0) versus pedal tilt angle if for applying the riders weight force on the pedal tips (10,5 cm from center)

I don't understand this one  so I would need some clarifications:

• A positive tilt angle means forward tilt?
• Where is the CoG of the rider and how does it change under the tilt? I assume it should change vertically, does it also change horizontally?

I assume we agree that forward tilt moves the pedal tip backwards. Now we have two possibilities:

1. the point of attack (or support) moves backwards too, leading to an additional forward lean. 
2. the point of attack stays horizontally at the same place, hence it moves forward on the pedal which presumably happens when the rider tries to keep balance under a pedal dip.

I guess the latter is not possible in the above model when the attack point is already on the pedal tip. It would be interesting to see the difference though.

11 minutes ago, Chriull said:

That's imo the essence of @RockyTop's linked youtube video - the steeper the incline, the smaller the usable part of the pedal front for forward acceleration-

Or in other words: the more thrust we want to generate, the less space we have left in front of the support point on the pedal.

[Chriull]

6 minutes ago, Mono said:

I don't understand this one  so I would need some clarifications:

• A positive tilt angle means forward tilt?

No. Backwards. My "model" assumes rider facing left, tilt angle is on the riders side counted clockwise.

6 minutes ago, Mono said:

• Where is the CoG of the rider and how does it change under the tilt? I assume it should change vertically, does it also change horizontally?

The riders cog is straight above the pedal tip. Assuming for each pedal tilt angle a starting condition of zero acceleration. Leading by the torque to the acceleration in the graph by motor.

6 minutes ago, Mono said:

I assume we agree that forward tilt moves the pedal tip backwards. Now we have two possibilities:

No moves - just static points shown.

6 minutes ago, Mono said:

1.  

I guess the latter is not possible in the above model when the attack point is already on the pedal tip. It would be interesting to see the difference though.

One could start with "smaller" pedals having the point of attack nearer to the center and then increase it.

[Mono]

23 minutes ago, Chriull said:

The riders cog is straight above the pedal tip.

OK, but then the thrust is just proportional to the horizontal distance of the tip to the wheel axle. And what we see is some zoom into a sinus curve (the horizontal distance as a function of the pedal angle), or am I missing something?

Edited December 17, 2019 by Mono

[Chriull]

13 minutes ago, Mono said:

OK, but then the thrust is just proportional to the horizontal distance of the tip to the wheel axle.

 

Here the two extreme situations one time +5° pedal tilt, the second -5° pedal tilt:

As one sees, the moment arm (horizontal distance between weight force and axle) changes. (From 11,5 cm to 9,46 cm)

13 minutes ago, Mono said:

And what we see is some zoom into a sinus curve (the horizontal distance as a function of the pedal angle), or am I missing something?

Yes - more or less. The final computations to acceleration are (in this case) just some factors.

 

 

Edited December 17, 2019 by Chriull

[Mono]

3 minutes ago, Chriull said:

As one sees, the moment arm (horizontal distance between weight force and axle) changes. (From 11,5 cm to 9,46 cm)

Of course, still, AFAICS the moment arm is just the cosine of the pedal angle assuming that 0º means the tip is horizontal level with the axle, or the sinus of the angle if 0º means that the tip is vertically under the axle.

[Chriull]

Just now, Mono said:

Of course, still, AFAICS the moment arm is just the cosine of the pedal angle assuming that 0º means the tip is horizontal level with the axle, or the sinus of the angle if 0º means that the tip is vertically under the axle.

Yes.

Since in the beginning was @Aneta's point with the the pedal tilt change for climbing inclines, this was just a try to see how much different pedal tilt calibrations could change available acceleration/force (here about +/- 50%) on inclines.

On level surfaces the effect is much less (relatively seen)

[Mono]

13 minutes ago, Chriull said:

On level surfaces the effect is much less (relatively seen)

I see, I believe the effect should be weaker under acceleration when the rider needs to lean. Under leaning, lowering the pedal improves the moment arm.

Edited December 17, 2019 by Mono

[Chriull]

8 minutes ago, Mono said:

I see, I believe the effect should be weaker under acceleration when the rider needs to lean. Under leaning, lowering the pedal improves the moment arm.

Didn't look at the effect of the lean - the difference is imo that there is no gravitational component along the surface getting subtractef from the forward force. So there should be no difference in absolute numbers but the realtive change gets much bigger?

[mrelwood]

5 hours ago, Chriull said:

Since in the beginning was @Aneta's point with the the pedal tilt change for climbing inclines, this was just a try to see how much different pedal tilt calibrations could change available acceleration/force (here about +/- 50%) on inclines.

I’m not sure if calculating the maximum possible thrust describes or explains the real life difference of various pedal calibrations though. By the same mechanism, wouldn’t a simply more powerful wheel always look like the best climber, while we know by now that that’s not the case? It’s obvious that the rider can’t balance the CoG very close to the tip of the pedals, so other variables may rise to be even more relevant.

It would be interesting to calculate the required angle at the rider’s ankles for both calibration scenarios, assuming somewhat bent knees for an even closer representation of real life scenarios.

Edited December 18, 2019 by mrelwood

[Chriull]

6 minutes ago, mrelwood said:

I’m not sure if calculating the maximum possible thrust describes or explains the real life difference of various pedal calibrations though.

There are for sure other points, too. Like leaning forward on an already downwards tilted pedal us very "unnatural".

This just showed that with 5° upward tilt 3 times the acceleration could be reached as with 5° downwards tilted pedals. Or otherwise - with 5° upward tilt less lean/"bringing the cog forward" is needed for the same acceleration.

6 minutes ago, mrelwood said:

By the same mechanism, wouldn’t a simply more powerful wheel always look like the best climber, while we know by now that that’s not the case?

This considerations are just based on wheel geometry - no motor powers are regarded. Its just assumed that the motor delivers enough power.

 

[mrelwood]

5 hours ago, Chriull said:

There are for sure other points, too. Like leaning forward on an already downwards tilted pedal us very "unnatural".

Me and @Mike Sacristan seem to disagree. Pedals tilting forward makes the natural no-effort stance (ankles at 90•) tilt forward as well. Just like a tilt-back requires more effort to push through (despite the pedals moving forward). I think this is due to the ankle angle.

5 hours ago, Chriull said:

This just showed that with 5° upward tilt 3 times the acceleration could be reached as with 5° downwards tilted pedals.

Three times?! By moving the CoG just from 8.5cm to 10.5cm before the axle? Sounds incredible.

5 hours ago, Chriull said:

Or otherwise - with 5° upward tilt less lean/"bringing the cog forward" is needed for the same acceleration.

Less lean? This I don’t understand either. The center point, the no-acceleration position of the CoG is still the same, directly above the axle. For the effectively longer pedal, to reach the tip of the pedal, doesn’t it require a lean that moves the CoG (or point of force contact) further from the center? Just like longer pedals allow for a further lean, but still require the same amount of lean for the same displacement?