Can a cornering motorcycle go faster if the rider puts a knee down?

[sophiecentaur]

Fun quiz question -- Why do Moto GP riders sometimes put their leg out under heavy braking into corners?

https://www.redbull.com/us-en/the-history-of-motogp-leg-dangle
View attachment 245454

Answer: Naked Fear!

 

[DannoXYZ]

hahahahahhahhhh!

I like this one:

Others have suggested that its enthusiastic embrace is more the result of a copycat club, the clown prince of Grand Prix racing demonstrating his massive global influence via some sort of meta joke — the racing equivalent to convincing the world the proper way to eat a Snickers is with a knife and fork.

Rossi, the eternal jokester, is trying to see how outrageous he can make his copycats dance!

 

[berkeman]

Rossi, the eternal jokester, is trying to see how outrageous he can make his copycats dance!

LOL, yeah. The first time I saw him do it on TV during a race, it looked like he was doing it just to keep his legs relaxed. He did it at a few places on the track during breaking, and I thought to myself, "Okay, looks kind of dorkey, but if it helps to keep you relaxed whatever." I guess there may be (or may not be) more to it...

 

[DannoXYZ]

New Olympic sport: Synchronized Leg Dangling!

 

[bobob]

Summary: Does the rider benefit from putting a knee down by being able to take the corner faster?

I remember this problem from my freshman physics class ~38 years ago. As I recall it was explained to us how a motorcycle rider could actually take a corner faster by putting his inside knee down on the ground. We see this in motorcycle racing where the riders have a knee pad that let's them place their knee on the ground, effectively sliding it along the pavement as they lean into turns..

...
So the idea is that, if the rider puts his inside knee down on the ground while in the turn, that corresponding amount of weight is removed from the apparent centrifugal force. So there's a little less centrifugal force so that allow the rider to go a little bit faster then when his knee was not sliding on the ground.

That is not quite how that works. The reason that riders put tgheir knees down (or possibly a little short of actually touching the ground) is that they use their body position to gauge their lean angle so they don't exceed the leangle possible for the bike. The knee actually doesn't function to assist in the turn apart from it's position contributing to getting the rest of your body into the right position. In order to get the maximum lean the rider actually shifts his entire body weight by moving somewhat back in the seat, which places a bit more weight over the rear tire with his body hanging slightly off the bike over the knee that is down with his chest and head pointed in the direction of the turn. Then the key thing to pulling off the turn while leaned way over is to countersteer, i.e., the front wheel is turned in the opposite direction the bike is turning. If you watch a video of motogp riders and you can see a bike from the front, you will notice that front wheel is NOT turned in the direction of the turn.

The onnly real purpose the knee serves beyond gauging the amount of lean and helping to obtain the right body position is to try to save your ass using brute force to keep the bike from falling over if you misjudge the lean angle.Actually, if you just look at the photos above, you can see that the front wheel is not pointed in the direction of the turn.

 

[Elroch]

For sure not the high speed banked turns like Daytona. I've read that the g-forces in those banked turns are high enough that the riders have to rest their helmets on a small pad on the tank, because they cannot hold their heads up against those strong g-forces. Yikes!

This cannot be strictly true. The sideways component of the force is of the order of 1.5 g, because it is limited by the friction of the tyres on the road. See https://www.f1technical.net/forum/viewtopic.php?t=20627 for data for F1. The support would make it a lot more comfortable and less tiring on the neck muscles.

 

[berkeman]

This cannot be strictly true. The sideways component of the force is of the order of 1.5 g, because it is limited by the friction of the tyres on the road. See https://www.f1technical.net/forum/viewtopic.php?t=20627 for data for F1. The support would make it a lot more comfortable and less tiring on the neck muscles.

Did you take the banking into account? That is what makes the difference...

 

[Elroch]

Did you take the banking into account? That is what makes the difference...

Short answer: not correctly! Long answer leaves me skeptical for a different reason.
For a car, in the frame where the road is horizontal, you can think of the component of the centrifugal force that is perpendicular to the road as being an increment k * g to gravity so g is replaced by g * (1 + k), and forces are scaled up with this pseudogravity in a fictitious frame where the road is horizontal.

So sideways acceleration experienced by the head of a car driver can be as high as g * (1 + k) * mu, where mu is the coefficient of friction. This is definitely relevant to neck strain for car drivers on banked tracks.

But for a bike, the vertical axis of the bike needs to be aligned with the total acceleration in a frame of reference that is rigidly attached to the bike. Otherwise it falls over - it only has stability in one direction, not two like the car. So the additional force is solely in the vertical axis of the bike's frame of reference (the vertical direction of its frame). My first thought was that this mean no lateral neck strain, but the rider's head is pointing forwards, so his head is being pulled forwards from his point of view, and is subject to that enhanced gravity, g * (1 + k).
I really can't see where the rider can rest his head (see image), so I think he just has to build his neck muscles if this is a problem.

 

[DannoXYZ]

Note that friction is not what generates traction forces on tyres. This was resolved decades ago when friction-model would only allow dragsters to reach 200mph maximum. We now know that's not true.

For example, F=Nu is higher for rubber+steel than it is for rubber+asphalt. Yet going around corners on asphalt has much more grip than sliding over sewer-lids.

 

[DannoXYZ]

So sideways acceleration experienced by the head of a car driver can be as high as g * (1 + k) * mu, where mu is the coefficient of friction. This is definitely relevant to neck strain for car drivers on banked tracks.

But for a bike, the vertical axis of the bike needs to be aligned with the total acceleration in a frame of reference that is rigidly attached to the bike. Otherwise it falls over - it only has stability in one direction, not two like the car.

Don't forget that there's actually two forces on cornering bike. Downwards gravity and lateral centripetal force.

This is similar to if you hung an apple from rear-view mirror of car via string. String would have 1G when car goes in straight line. Under 1G of cornering, string would experience 2G total. Note that apple and string are no longer vertical, but hanging at 45-degrees, aiming towards outside of turn.

Now, hang that same apple from bike's... rear-view mirror. It too would have 1G of load when bike goes in straight line. Under 1G of cornering, string would experience 2G of load. This is supported by monitoring suspension-travel and under steady-state cornering of +1G, suspension-compression is much higher than during upright-riding. On many street-bikes, suspension is completely compressed and locked at full-travel. Thus street-bikes taken to track must have upgraded suspension with much stiffer springs.

 

[Elroch]

Yes, but you can combine gravity and centrifugal force (the fictitious force in the non-intertial frame of reference is the relevant one), as I did earlier. A consequence is that the angle of the road changes.

For a bike, this force is along the line between the tyre's contact with the road and the centre of gravity - roughly the line on the tyre in your diagram. This is necessary (given minor simplifying approximations) for stability in the frame that moves with the frame of the bike (excuse pun).
Friction in this context is a complex summary of the phenomenon of a moving tyre moving along a road with a resultant force acting on it. It retains an easily understood meaning: the tangent of the angle at which a bike can tilt is the same as this quantity if the centre of mass is kept in line with the tyre. It determines the point at which a tyre slides, based on the relationship between the two components of force (perpendicular to the road and parallel to it).

(I am ignoring small corrections such as that due to the rather small angle between the planes of the two tyres).

 

[DannoXYZ]

Yes, extra centripetal force is pushing inwards on contact-patch and bike wants to flip over that patch to outside. It's an extra force in addition to gravity.

This is also static model, snapshot in time. Inertia wants to stand bike up and have bike go in straight line tangent to curve. Extra force and energy is needed to lean bike over and make it go around curve. Entire time in curve, I'm fighting to keep it leaned over. If I relax, it actually stands up, not fall over. Many videos of racers falling off their bikes and it stands up and goes straight by itself. I'm actually countresteering more than lean-angle just to force bike to stay leaned over.

This is balance of opposing forces, very dynamic configuration. You need to adjust your model to account for why suspension compresses twice as much under cornering compared to riding upright. Where is this extra force coming from to compress suspension?

 

[DannoXYZ]

A more familiar model to understand this is to imagine you're standing upright and someone slides sideways into your foot. Obviously they'd kick your feet out from under you and you'd fall over them.

But... what if you saw them coming? You can lean away from them at certain angle and not fall over when hit. As they impacted your foot, your entire body (if rigid and leaning away) would experience more than its regular 1G of force. Your head, even though not being impacted directly by sliding person, would be accelerated from its resting position and experience more than 1G. Obviously if they hit you at 10-m/sec and had enough energy to accelerate you to 10-m/sec in 1-sec, you would have to lean away at 45-degree angle and your head would experience 2G.

 

[Elroch]

Do you accept the point that if you pick your frame moving with the bike, so that the plane through the points of contact and the centre of gravity is considered as "vertical", then in this frame, the artificial centrifugal force (which results from the fact that the frame is rotating) and the gravity add to give a single perceived force in this "vertical line"?

The reason this viewpoint is useful is it is close to what the rider experiences. They experience something very close to an increased gravitational force which stays in the same direction relative to them and the ground becomes steeply sloping to the side. The effect of banking is to cancel out this slope in the ground. This is modified slightly by the choice of riders to shift their centre of gravity to the side, which means the tilt of the bike is lower (as the plane between the centre of gravity and the contact points is at an angle to the centre plane of the bike) . As far as I can see, this choice does not make it possible to go faster, as long as the tyre works over the full range of angles, which it should. It would be interesting to study whether riders really get any physical advantage out of changing their position on the bike, or whether (surprisingly) it just feels better.

Here I implicitly assume the rider is moving in a constant circle at a constant speed, but it is a very useful approximation.

 

[DannoXYZ]

Do you accept the point that if you pick your frame moving with the bike, so that the plane through the points of contact and the centre of gravity is considered as "vertical", then in this frame, the artificial centrifugal force (which results from the fact that the frame is rotating) and the gravity add to give a single perceived force in this "vertical line"?

The reason this viewpoint is useful is it is close to what the rider experiences. They experience something very close to an increased gravitational force which stays in the same direction relative to them and the ground becomes steeply sloping to the side. The effect of banking is to cancel out this slope in the ground. This is modified slightly by the choice of riders to shift their centre of gravity to the side, which means the tilt of the bike is lower (as the plane between the centre of gravity and the contact points is at an angle to the centre plane of the bike) . As far as I can see, this choice does not make it possible to go faster, as long as the tyre works over the full range of angles, which it should. It would be interesting to study whether riders really get any physical advantage out of changing their position on the bike, or whether (surprisingly) it just feels better.

Here I implicitly assume the rider is moving in a constant circle at a constant speed, but it is a very useful approximation.

Yes, that model works, but rider still experiences 2G of forces. Banking only cancels out lean-angle, but it doesn't cancels out cornering forces. While you may be able to take 1G corner without leaning, there is still 1G of gravity and 1G of cornering. Just happens that cornering-forces are in-line with bike; it's pushing up on tyre instead of sideways on it. In cases of Wall-of-Death, there can be unlimited "cornering" traction since there's no leaning whatsoever.

This is especially apparent on 2-wheel vehicles that don't have as high cornering forces as motorcycles... say bicycles on velodrome. These are banked at either 20 or 30-degrees depending upon tightness of curve. At tighter 0.25km velodromes such as Hellyer Park or Encino, you really can feel the extra G-forces on your neck. On 0.33km velodromes like Cal State Dominguez Hills, curve isn't as tight, so less G-forces, even though bike still stays straight & upright relative to track surface.

There is an advantage when you change angle between effective CoG and contact patch. It allows for higher cornering forces with less lean angle and you don't run off edge of tyre. There IS an upper limit for lean-angles where you WILL roll off and crash. None of this idealistic "assuming unlimited traction" stuff not connected to reality.

This is where testing part of scientific process comes into play to verify hypotheses. Numerous real-world experiments have been conducted to verify:

1. there is not unlimited traction on tyres
2. there is a lean-angle limit beyond which you'll roll off edge of tyre and crash
3. when at this lean-angle limit, separating body & bike to maintain lean-angle, yet increase angle between effective CoG and contact-patch DOES allow for faster cornering without crashing.

This is typically what racers do in practice. They go faster and faster each lap and typically stay in-line with bike at lower-speeds. As you get up to lean-limit, you can feel bike starting to slide. Next laps around, you keep lean-angle, but hang off more. Suddenly, bike doesn't slide anymore. Then laps after that, you can continue with increasing speeds gradually again.

 

[Elroch]

You agree the 2g ( (1+k)*g in my post), as measured by an accelerometer on the rider say, is near enough along your blue line (presuming that goes through the centre of gravity).

 

[DannoXYZ]

The 2G of force is inline with effective CoG and contact patch regardless of where rider's spine is located. In this example, it's in between spine of rider and axis of bike.

 

[Elroch]

Yes, I made my post more precise as you were posting! The reason for referencing the rider's anatomy was that back somewhere in this discussion we were discussing a claim that a rider had to rest their head on something.
The tilt of the rider's head in your penultimate photo makes some of this force to the rider's left, but not enough to be a problem, IMO. I think that the reason riders tilt their heads this way is it makes the ground look more horizontal and easier to mentally process.

 

[DannoXYZ]

Yes, that was for banked tracks where it's possible to get even higher cornering-forces than on flat ground. So for those tracks, if you lean bike to same relative-angle to ground as on flat, you end up with even more than 2G of cornering, 3G is possible. It's the combined lateral and vertical forces pushing head towards tank. It does get tiring if you've got your head turned 45-degrees to look out of corner. Heck, I get that on velodrome curves and those don't generate more than 2G on the tighter ones. Nice thing about velodrome banking is you can continue pedaling while in corners, thus can generate higher Gs than on flat ground. Here's good guide to banked curves.

http://dynref.engr.illinois.edu/avb.html

 

[Chris S]

Dave had the answer, but it's implication was not understood. Transferring weight from the tires to the knee transfers the cycle's traction from good traction components (tires) to a poor traction component (plastic knee pad), thus reducing the total traction from 100% on good traction components (tires) to less than 100% now that some traction is on the knee. Although Dave said this increases the risk of a slide, and risk of sliding doesn't answer the question, the implication of Dave's answer is that less traction means slower turn speed maximum, thus the answer the question appears to be "no." [However, the other benefits of knee dragging outweigh the slower theoretical maximum speed.]

 

[berkeman]

I forgot to mention that aerodynamics becomes an issue at 150+ mph, an issue with overall drag and apparently an issue with upsetting then handling balance of the bike, so the riders stay tucked in at 150+ mph, even while turning.

At Laguna Seca, turn 1 is a high speed slight kink, no one is hanging off there. From the start/finish line, the riders don't hang off until the very tight turn 2

It looks like MotoGP is pushing the envelope even more now. Wow.

 

[rcgldr]

It looks like MotoGP is pushing the envelope even more now.

Part of that is the tires, but much of the drifting in MotoGP and Superbike is due to the rules now allowing computer assists, for both traction and wheelie related aspects. The sport bikes sold to the public have had similar assists for quite a while now. On a side note, I'm waiting to see how the new Ducati V4R (street version has 16,500 rpm redline, 220 hp (15,250 rpm) or 230+ hp (15,500 rpm) with the race exhaust) will do in Superbike class, since the power will be increased further still. The street version sells for $40,000, so wealthy riders and probably a lot of racing teams buying them.

 

[DaveC426913]

...computer assists, for both traction and wheelie related aspects.

Well, that makes those wheelies I see a lot less amazing.

Kind of the bike equivalent of musical auto-tuning.

 

[berkeman]

Well, that makes those wheelies I see a lot less amazing.

No, @rcgldr is referring to anti-wheelie software. The boys and girls in MotoGP don't need any computer assists in lifting their front wheel. LOL

https://i.pinimg.com/originals/a4/3a/d0/a43ad099dcb42e0b7916782ba53570fc.jpg

 

[John Archer]

What it seems many are missing, is the relationship between traction and suspension.
We hang off for several reasons,
#1 Keep more tire on the road
#2 Keep the bike as close to vertical as possible, as our suspension is designed for vertical movement.
At 90 degrees, our suspension does virtually nothing. Hanging off keeps the bike more upright, and helps the suspension do it's job more efficiently.

Imagine a bump in the pavement, say 1" tall (yes huge but for this conversation, not a big problem). The tires absorb some of this, but the suspension absorbs some as well. Let's say the tire moves 1/2" and the suspension then moves the other 1/2". Now lean the bike over at 45 degrees. we have a lot less sidewall for the tire too move, but let's ignore that and say it still absorbs (lack of a better term) 1/2" of the bump (it deforms 1/2"). the suspension needs to take the other 1/2"...but it is at a 45 degree angle. Does this mean it now needs to move 1 full inch to get 1/2" of vertical travel? Yes, of course there is some flex of the suspension components, but they are making them as stiff as possible (as flexing parts don't have much damping effect on movement).

So in the end, yes. You CAN corner faster by hanging off, because the suspension will work better, giving you more usable traction, and the tires will do the same (larger contact patch).

Dragging a knee is mostly to judge lean angle, but can also be used to save a crash where the front tire losses useful traction.NOTE for previous issues:
You don't hang off on the banking at Daytona because of drag. The banking nearly eliminates lateral acceleration as measured relative to the wheels.

You don't hang off at some super high speed corners, mainly because even at speed, you are not near your traction limits...or at least you don't hang off fully and drag a knee. You know you have the traction to make the corner, hanging off here again is only to help the suspension work (the amount of work the suspension does at very high speeds is incredible).

 

[cjl]

Don't forget that there's actually two forces on cornering bike. Downwards gravity and lateral centripetal force.
View attachment 245922

This is similar to if you hung an apple from rear-view mirror of car via string. String would have 1G when car goes in straight line. Under 1G of cornering, string would experience 2G total. Note that apple and string are no longer vertical, but hanging at 45-degrees, aiming towards outside of turn.

Now, hang that same apple from bike's... rear-view mirror. It too would have 1G of load when bike goes in straight line. Under 1G of cornering, string would experience 2G of load. This is supported by monitoring suspension-travel and under steady-state cornering of +1G, suspension-compression is much higher than during upright-riding. On many street-bikes, suspension is completely compressed and locked at full-travel. Thus street-bikes taken to track must have upgraded suspension with much stiffer springs.

No, this gives 1.41G of total force, not 2G. The centripetal and gravitational forces are at 90 degrees to each other, so when each is 1G, the combined effect is sqrt(2)G at 45 degrees, not 2G.

 

[Tom.G]

In case anyone wants some numbers to plug in:

Referencing the video in post #6 https://www.physicsforums.com/threa...he-rider-puts-a-knee-down.972994/post-6189354

In the turn at 2:10 the angle of the horizon is 47° from horizontal,
at 4:23 the horizon angle is 51.5° from horizontal.