The "best feeling" wheel (wheel zippiness and controlability)

[rcgldr]

1 hour ago, techyiam said:

if the pedals are mounted at axle height, once the COG is beyond the front pedal edge, the rider would lose balance and fall off the front.

If the center of mass is somewhat in front of the pedals, the normal reaction to avoid falling by pressing with the toes onto the front of the pedals would result in the EUC accelerating forwards and exerting a backwards torque onto the rider to re-balance the rider. In that video I linked to of Lando accelerating about 1/3 g on an 18XL, Lando is leaning about 22°. If the restriction was center of mass instead of center of force not being in front of the pedals, then Lando would only be able to lean about 8°. 

Whenever a rider is increasing lean angle, forwards | backwards | left | right, the rider is "falling" into the direction of lean and in the case of forwards | backwards also by the EUC decelerating | accelerating out from under the rider. Leaning to turn is done by counter-steering, from a vertical position, steer right to lean left and vice versa. Once leaned, steer more to lean less, steer less to lean more. Once a desired lean angle is achieved, the rider inputs then re-balance the rider via acceleration in the direction of lean.

Edited March 9 by rcgldr

[Rider1]

42 minutes ago, rcgldr said:

If the center of mass is somewhat in front of the pedals, the normal reaction to avoid falling by pressing with the toes onto the front of the pedals would result in the EUC accelerating forwards and exerting a backwards torque onto the rider to re-balance the rider. In that video I linked to of Lando accelerating about 1/3 g on an 18XL, Lando is leaning about 22°. If the restriction was center of mass instead of center of force not being in front of the pedals, then Lando would only be able to lean about 8°. 

Yes, I'm sorry that I oversimplified this, but the point of my comment was that the pedals being tilted forwards doesn't help with accelerating because you can't move your body beyond a certain point (let's put it that way), which doesn't change with pedal tilt (though it might even move backwards due to the pedals being located below the center of the tire, yet this rather small effect is most likely countered by the mass of the wheel moving forwards a little from the tilt). Also, air resistance would play a role, too.

But when using powerpads, a soft pedal mode will help you lean forwards quite a bit further since the pads on the front "move away from your legs", giving you more space to lean.

Btw, Lando was most likely squeezing the wheel really hard, giving him extra potential to lean forwards.

Edited March 9 by Rider1

[techyiam]

1 hour ago, rcgldr said:

If the center of mass is somewhat in front of the pedals, the normal reaction to avoid falling by pressing with the toes onto the front of the pedals would result in the EUC accelerating forwards

If a wheel is padless, and the pedals are mounted at the height of the axle, maximum acceleration is reached when the rider's COG is on the leading edge of the pedal.  Moving the COG further would not result in greater acceleration. The rider merely falls off.

[Rider1]

25 minutes ago, techyiam said:

If a wheel is padless, and the pedals are mounted at the height of the axle, maximum acceleration is reached when the rider's COG is on the leading edge of the pedal.  Moving the COG further would not result in greater acceleration. The rider merely falls off.

(I already wrote 3 different replies to this, changing it over and over again, due to realising new aspects when proofreading it, so make of the outcome what you will. And sorry btw if it was already clear what causes lower pedals to be better for zippiness, I didn't know, and might be wrong now, too, but here you go )

This could explain why lower pedals make accelerating easier, since the vector of the force that you put on the pedals isn't pointing straight down anymore when factoring in the force of the acceleration keeping you upright when leaning forward further than the frontmost point of the pedal (assuming there is enough friction so you don't slide off). When splitting it up in a vertical and horizontal component, the normal distance between the pedal and the wheel axis suddenly plays a role.

[techyiam]

21 minutes ago, Rider1 said:

what causes lower pedals to be better for zippiness,

When a wheel accelerates, the rider standing on its pedals accelerates with it.

The pedal will see a force equal to the mass of the rider multiplied by the rate of acceleration. (F=M x A)

If the pedals are mounted below the height of the axle, that force would create a moment about the axis of rotation.

This torque adds to the rider's input torque for acceleration.

The lower the pedals, the longer the moment arm.

 

[rcgldr]

On 3/9/2024 at 10:38 AM, techyiam said:

If a wheel is padless, and the pedals are mounted at the height of the axle, maximum acceleration is reached when the rider's COG is on the leading edge of the pedal.  Moving the COG further would not result in greater acceleration. The rider merely falls off.

"If a wheel is padless, and the pedals are mounted at the height of the axle", then the maximum acceleration occurs when the rider's center of force is on the leading edge of the pedals, not the rider's center of mass. If accelerating, the center of force is below and behind the center of mass at the pedals. This may be more clearly understood by considering that when turning, the rider's center of mass can be well beyond (inside) the inside edge of the inside pedal because the left | right centripetal acceleration results in the center of force being below and outside the center of mass, between the pedal edges, preventing the rider from falling inwards. Why should this be any different for forwards | backwards acceleration? The point here is that it's the center of  force that needs to be within the bounds of the pedals, not the center of mass. In the case of turning, the center of force has to pass through the contact patch of the tire (otherwise, the rider would be falling inwards or outwards, such as adjusting lean angle), while the center of mass can be well inside the contact patch.

Roger EUC's video of Dawn Champion on a V13 at about 30 to 33 mph:

https://www.youtube.com/watch?v=IyRxroEKHas&t=2638s

Wrong Way's video about tires and turning, the middle video of a Z10 (demonstrating it's 4 inch wide tire is more responsive to tilt than the other tires).

https://www.youtube.com/watch?v=NsXW4OKnmWc&t=313s

 

Edited March 11 by rcgldr

[alcatraz]

On 3/8/2024 at 11:21 PM, Rawnei said:

I see it more as problem solving, you have to define the problem you want to solve. What problem do you want to solve that power pads doesn't solve?

Good point. There are a few. 

Not being able to fold pedals. 

Not being able to get through narrow gaps when trollying.

Not being able to park the wheel in small spaces.

Not being able to change leg positions if your feet/legs get tired. 

Not being able to close your legs when riding seated.

...are what come to mind.

[Panzer04]

On 3/10/2024 at 6:39 AM, techyiam said:

When a wheel accelerates, the rider standing on its pedals accelerates with it.

The pedal will see a force equal to the mass of the rider multiplied by the rate of acceleration. (F=M x A)

If the pedals are mounted below the height of the axle, that force would create a moment about the axis of rotation.

This torque adds to the rider's input torque for acceleration.

The lower the pedals, the longer the moment arm.

 

This is an excellent point, and one I hadn't considered before. I'd generally neglected to consider pedal height as a factor in acceleration/braking, honestly, despite it being widely accepted that lower pedals make wheels feel more responsive. I suppose this would be exactly why - It goes a bit further too - The same logic would apply to the EUC itself (ie. if it carries weight high-up then that would apply another countertorque during acceleration).

Incoming wall of text + some quick math

Thought experiment: A wheel with the pedals at ground level (ignoring the ground :P) would experience a countertorque from the riders weight directly equivalent to that produced by the wheel to go forward - which means that you could totally ignore the rider weight when computing the pedal forces required to accelerate to a certain speed - because the torque required to accelerate the rider weight would also be completely produced just by the pushback of that weight at the pedal.

Of course, there's also the weight of the EUC to consider - and it seems the lower-slung (further away from the top) it can be made, the easier things will be to accelerate as well because this would also apply a countertorque, but in the direction resistant to the desires of the rider :(. Given how some wheels (*cough* master) carry their batteries rather high, along with a substantial chunk of metal holding the controller quite high, that probably adds up to quite a bit of resistance to acceleration. 15kg of weight an average of a wheel-diameter away from the axle would add (for a master), accelerating at 2m/s^2

15 * 0.25 * 2 = 7.5nm of countertorque

vs the forces required to accelerate at 2m/s^2 by the wheel, for an 80kg rider (ie. me)

(80 + 40) * 2 = 240N

To produce 240N at the ground, the wheel needs to produce:

T = 240 * 0.25 = 60Nm of torque.

There's definitely some more complexities to the math, but at a surface level the long story short certainly appears to be that the lower you can carry the weight, the less force required to initiate and hold an acceleration because part of the mass will then, by being accelerated, create an additional torque on the wheel. I guess the way to look at is that weight carried halfway between the axle and the ground contributes only half as much to the "apparent" weight of the whole system in terms of torques required to produce a force...

Naively my intuition is saying this:

• Weight at axle - 1x multiplier: Counts as itself for the purposes of the rider torque needed to accelerate
• Weight at (theoretical) ground level (-1r from axle) - 0x multiplier: It doesn't exist for the purposes of calculating torques for a given acceleration, although it would affect the torque required by the motor - it's just that you, as the rider, don't need to generate the torque to accelerate that weight.
• Weight at top of wheel (+1r from axle) - 2x multipler: It counts 2x for the rider torque needed to accelerate

And as the weight gets higher the multiplier is directly proportional to the distance from the axle.

So for example, on a Master:

• 80kg rider weight, approximately at axle level (for simplicity) = 80kg apparent
• Not sure how to count the motor - I'd guess it can be considered to *always* apply its weight as a countertorque - but most of it is centered close to the axle with a small radius, but guessing ~possibly 15kg (motor/tyre/rim) at +r/2 = 1.5*15 = 22.5kg apparent
• 15kg of wheel frame, batteries, etc. at 1r, 10kg at the axle, so 2x15 + 1x10 = 40kg

Adding all of this up gives an apparent weight against acceleration of 140kg, rather than 120kg as naive math would suggest (assuming all of the above are actually correct XD)

It's not a massive effect, but that's still ~15% additional torque required for a given acceleration, on par with (actually a little bigger) than the effect of choosing a 18" /12"rim wheel instead of a 20"/14" rim, compared with if the weight were lower-slung and mostly in-line with the axle. This is also in line with the torque calculation earlier which implies a roughly ~15% countertorque from 15kg above.

 

I think the above is what I was missing when comparing other wheels (in particular the S22) vs the master - I believe the S22 slings its weight lower than the Master, on top of weighing less. It might also have been the pedals were slightly lower after accounting for suspension sag (though I'm not certain on that front) - given even a couple of centimeters adds up to 3-4Nm of assistance that adds up fast as well.

 

 

[Panzer04]

I suppose it also explains why people like the EX30 so much - despite weighing more than the Master, it keeps its batteries low and wide, mitigating some of that weight, and I believe its pedals are also a bit lower.

Given all this, I guess the "ideal" design is to have low, wide batteries putting as much of the weight as possible as low as possible. Suspension makes this more awkward, because you'll often be sitting 50-100mm higher than you could be to give it room to slide. That being said, lots of wheels have taller designs that could easily be wider if they wanted to - just staying within the confines of the wheel at full suspension travel, and otherwise as as low as possible..

It also occurs to me that the above implies that the wheel size downsides could be mitigated if the weight is still kept below the (steadily rising) axle with bigger wheels. I suppose you will always suffer the inherent penalty of being able to apply the same torque via your feet vs increasing requirements from the wheel.. but if the pedals are low, then you don't need as much torque anyway..

Edited March 14 by Panzer04

[Rawnei]

People like the EX30 because they offset the weight with powerful controller and motor.

It's still a clumpsy wheel.

[noonewantstobepeterchris]

6 hours ago, Rawnei said:

People like the EX30 because they offset the weight with powerful controller and motor.

It's still a clumpsy wheel.

I agree that it is kind of clumsy wheel at low speed, and I personally would not take it on a trail.

However, the EX30 is still one of the fastest wheels on a straightaway. Sure, a wheel can have a high stop speed and still be clumsy and difficult to maneuver. Both the weight and weight distribution help with stability it at high speeds. The EX30 tends to place very high in races, which means that the maneuverability is at least acceptable.

[Planemo]

7 hours ago, Rawnei said:

It's still a clumpsy wheel.

I'm hoping to mitigate that a little with my recent fitment of -30mm hangers. If it's near to what a Sherman OG feels like I'll be super happy. I should know soon, but I think it might nearly get there. I don't think a stock EX30 is miles away from a Sherman as it is tbh despite the bit of extra weight high up (thick metal controller box, suspension linkages/shock).

[Rawnei]

32 minutes ago, noonewantstobepeterchris said:

I agree that it is kind of clumsy wheel at low speed, and I personally would not take it on a trail.

However, the EX30 is still one of the fastest wheels on a straightaway. Sure, a wheel can have a high stop speed and still be clumsy and difficult to maneuver. Both the weight and weight distribution help with stability it at high speeds. The EX30 tends to place very high in races, which means that the maneuverability is at least acceptable.

Exactly, people like this wheel because it's fast, that's it, not because it's agile or awesome at anything but going fast.

[Mono]

16 hours ago, Panzer04 said:

Thought experiment: A wheel with the pedals at ground level (ignoring the ground :P) would experience a countertorque from the riders weight directly equivalent to that produced by the wheel to go forward

Not quite, the distance which counts for the lever is the distance to the axle orthogonal to the (down) force from the rider, not the distance at the point of attack. Some slightly involved calculations suggest that when taking into account the lean angle of the rider too (instead of only the forward displacement), the distance that counts is the horizontal distance to the axle. Lower pedals only help because leaning increases this distance the quicker the lower the pedals are. The effect is quantified in this post (quote: lowering the pedals by 20% [of the wheel radius] at a 14º lean angle yields 5% x weight additional thrust):

 

Edited March 14 by Mono

[Mono]

On 3/9/2024 at 7:38 PM, techyiam said:

If a wheel is padless, and the pedals are mounted at the height of the axle, maximum acceleration is reached when the rider's COG is on the leading edge of the pedal.  Moving the COG further would not result in greater acceleration. The rider merely falls off.

In the stationary condition with maximal acceleration or maximal forward force, the COG of the rider is in front of the leading edge. The rider does not fall off, because the wheel constantly pushes from behind, so the force vector from the (COG of the) rider onto the pedal edge does not go straight down, but goes down+back.

[noonewantstobepeterchris]

29 minutes ago, Rawnei said:

Exactly, people like this wheel because it's fast, that's it, not because it's agile or awesome at anything but going fast.

No, it is agile at speed. That’s why it consistently places at races, which wouldn't happen if it wasn’t somewhat agile. The master pro and monster are fast but not agile.

[rcgldr]

On 3/9/2024 at 11:39 AM, techyiam said:

When a wheel accelerates, the rider standing on its pedals accelerates with it. The pedal will see a force equal to the mass of the rider multiplied by the rate of acceleration. (F=M x A) If the pedals are mounted below the height of the axle, that force would create a moment about the axis of rotation. This torque adds to the rider's input torque for acceleration. The lower the pedals, the longer the moment arm.

The pedals will also see a vertical force equal to weight of rider. Using my prior example, assume acceleration is 1/4 g, which translates into a 14° angle. The combined weight and reaction force to forward acceleration result in a line of force at 14°, below and behind the riders center of mass. If this line of force goes through the center of the axle, there is zero torque. The rider has to lean forwards by more than 14° so that the line of force is in front of the axle to generate the forwards torque needed for that acceleration. The torque is equal to that force times the distance perpendicular to the line of force to the center of the axle. Lower pedals don't change this, but that line of force intercepts the pedals further back if they are lower, and lower pedals allow the rider to generate more torque by leaning further forward before the line of force goes past the front of the pedals. I'm assuming hard mode, where the EUC remains vertical during acceleration. Soft mode would reduce the leverage since the pedals would be moved back as the EUC tilts forward.

The torque related leverage from the pedals is limited to somewhat more than half the length of the pedals, while the leverage from power pads is a much larger distance, from the pedals up to the power pads, allowing a rider to lean more (mostly by bending at the hips) to generate more torque.

Edited March 14 by rcgldr

[Panzer04]

5 hours ago, Mono said:

Not quite, the distance which counts for the lever is the distance to the axle orthogonal to the (down) force from the rider, not the distance at the point of attack. Some slightly involved calculations suggest that when taking into account the lean angle of the rider too (instead of only the forward displacement), the distance that counts is the horizontal distance to the axle. Lower pedals only help because leaning increases this distance the quicker the lower the pedals are. The effect is quantified in this post (quote: lowering the pedals by 20% [of the wheel radius] at a 14º lean angle yields 5% x weight additional thrust):

 

I agree - this is what I found in the first post, with big references to your calculations there.

This second thing is a separate, second, very noteworthy force component. If the wheel is accelerating, then as Techiyam suggests, it will experience another force consisting of the wheels CoG pushing backwards (with F=MA), with this applying another torque force depending on where this CoG is.

Then it also has to accelerate the rider, and since their only point of contact is at the pedal, the pedal is therefore being pushed back by the riders weight during acceleration. If the pedal is at axle-height, this force applies no net torques. Crucially, however, if the pedal is lower than the axle, this acceleration force applies a net torque to the wheel, which in some sense you can view as reducing the torque that the rider needs to apply in the conventional down direction you computed (for a given acceleration)

After all, this backwards force, below the pedal, is attempting to rotate the frame forwards about the axle, so the wheel must also counter it. If the pedals were super high, above the axle, it would work against you because the force now rotates the wheel backwards, reducing the amount of torque it needs to supply.

I think this is a super significant consideration, because it makes a big difference, in theory - on par with the effects of smaller wheels. In theory a well designed wheel, with all of it's weight placed low, could feel much easier to accelerate. I think it explains to some extent why I found a V13 and EX30 actually quite easy to push around, given their additional weight (and larger tyre, for the V13) - lower pedals and better-distributed EUC weight.

So assumingg I'm correct, the overall model has two major components:

- The rider's weight, applied vertically at a point along the pedal - the primary control input

- The weight of the EUC + rider applying a force in the opposite direction during acceleration. The lower both of these components, the less force required on the main control input for a given acceleration, as part of the torque is supplied by the mass itself being accelerated.

 

On further thought there's more to it - but I think there's an intuitive sense to this interpretation: consider a rod mounted with a free bearing in the middle. If you placed a weight on the bottom, then moved the bearing forwards the rod would rotate forwards at the top. If you placed a weight at the top, it would rotate backwards at the top. Apply this analogy to an EUC and you can see how these torques would help or hinder acceleration.

I don't know how to make sense of this with a rider that will lean to maintain stability in this situation and how that effects things, but clearly there must be a forward force applied at wherever the point of contact happens to be, right?

[Mono]

15 hours ago, Panzer04 said:

Then it also has to accelerate the rider, and since their only point of contact is at the pedal, the pedal is therefore being pushed back by the riders weight during acceleration. If the pedal is at axle-height, this force applies no net torques.

This doesn't sound right. Where and how the rider makes the contact with the wheel is irrelevant for the force balances, AFAICS. We always assume no relative movement between wheel and rider, but how this is achieved (and whether this can only be achieved with (power) pads) is a different question. The only relevant parameters are mass and CoG of rider+wheel, position of the axle, position of the contact patch and the gravitational acceleration. If these are given, everything is determined and the pedal height is irrelevant, AFAICS.

In my calculations, I separated the rider mass and CoG from those of the wheel (assuming a neutral wheel CoG) and added the wheel mass later as "efficiency" factor rider_weight / (wheel_weight + rider_weight). This was probably not a good idea and it ignores the effect of the CoG of the wheel weight not being in its axle (I think the calculations are exact when the wheel CoG is in the axle). I probably did it because I was in particular interested in the necessary rider displacement/lean (the less the zippier), but I could/should have gotten the forward displacement of the rider by separating rider and wheel mass only after the calculations.

15 hours ago, Panzer04 said:

So assumingg I'm correct, the overall model has two major components:

- The rider's weight, applied vertically at a point along the pedal - the primary control input

- The weight of the EUC + rider applying a force in the opposite direction during acceleration. The lower both of these components, the less force required on the main control input for a given acceleration, as part of the torque is supplied by the mass itself being accelerated.

 

AFAICS it depends only on the angle between CoG of rider+wheel and the tire contact patch, not (just) on the height of the CoG. For example, if the angle is 45º, we need to accelerate with one g (to exactly make up for gravity), no matter how high the CoG is. Specifically, for the lean angle to remain constant we must have

  forward_thrust = tan(lean_angle) x weight

where lean_angle is the angle of the overall CoG compared to the contact patch and weight is the overall weight (wheel+rider). If the CoG is in front of the contact patch, which it has to be under acceleration, lowering the CoG changes the angle and therefore means stronger acceleration to keep the rider from toppling over. This indeed implies that a top heavy wheel needs more lean to (strongly) accelerate. It also implies, as suggested and observed before, that tilting the wheel forward helps accelerating because it displaces its CoG forward w.r.t. the axle.

 

Edited March 15 by Mono

[Planemo]

It never ceases to amaze me just how much physics is actually going on with something as simple as a self balancing wheel. I'm almost convinced that the original designer didn't know the full math/physics behind it and simply attacked the problem from a mechanical viewpoint - getting it to work without actually knowing all the physics. I could be doing them a disservice by saying that but it's pretty damn involved when you actually break it down.