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Cornering Angle ?


Matthew Miller

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Matthew Miller

Here is a general question that I have. Here is a hypothetical scenario. Take a constant radius corner at 40 MPH following the correct racing line through the corner. Do all bikes corner at the same angle in this scenario or does frame geometry and fork rake, control the amount of angle in the corner. I have owned Honda, Suzuki, Kawasaki, Yamaha, Moto Guzzi and my current bike a BMW R1100RL. They have ranged from Sport to Touring to Sport touring. When I think about it, it seems that they all have cornered very similarly when it comes to leaning excluding ground clearance and tires. What do you think?

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I believe that the center of gravity is what is at issue, so bike weight, rider weight, rider position, and lean angle would all be factors. There are likely others, such as traction, tire profile, wheelbase length and rake.

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Joe Frickin' Friday

Think of the following three points in space:

 

-the rear tire contact patch

 

-the front tire contact patch

 

-the combined rider+bike center of mass

 

These three points define a plane, and that plane has a particular lean angle measured relative to the ground. Any rider+bike navigating around a particular curve at a particular speed will have the same lean angle as any other rider+bike on that same curve at that same speed.

 

The lean angle (measured from vertical) is ATAN(V^2/(R*32.2)), where R is the bike's speed (in feet per second) and R is the radius of the curve (in feet). So if you're on a cloverleaf interchange (R=200 feet) at 30MPH (44 ft/s), your lateral acceleration is 0.3G and your lean angle is 16.7 degrees from vertical, whether you're on a Harley or a Ducati.

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Asymmetrical

Hi Mitch -- Would you explain this a bit further? I am presuming that the center of mass varies from one bike mfr. to another due to frame geometry and weight placement, so wouldn't that change the lean angle from the 30 mph constant you show in this example as 16.7 degrees? Regards, Greg

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Think of the following three points in space:

 

-the rear tire contact patch

 

-the front tire contact patch

 

-the combined rider+bike center of mass

 

These three points define a plane, and that plane has a particular lean angle measured relative to the ground. Any rider+bike navigating around a particular curve at a particular speed will have the same lean angle as any other rider+bike on that same curve at that same speed.

 

The lean angle (measured from vertical) is ATAN(V^2/(R*32.2)), where R is the bike's speed (in feet per second) and R is the radius of the curve (in feet). So if you're on a cloverleaf interchange (R=200 feet) at 30MPH (44 ft/s), your lateral acceleration is 0.3G and your lean angle is 16.7 degrees from vertical, whether you're on a Harley or a Ducati.

 

So for a given curve and speed:

 

1. If, as on a dirt bike, I place my standing mass above the front wheel and lean the bike,

 

2. and as on street bike, I "kiss the mirrors" placing my seated mass low and inside, keeping the bike more upright,

 

You are saying the bike lean angle must be the same?

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The lean angle of the center of mass will be the same, if I'm reading Mitch's post correctly. Not that of the rider, or the bike, but the center of mass.

 

Jan, in your examples, you are using body position to move the center of mass around, relative to where it would be if you stayed on the seat with your spine in line with the lean angle of the bike frame. This produces different effects, hopefully effects that are useful in that riding situation.

 

For example, in the street situation, where you are kissing the mirrors, you are moving the COM to the inside of the bike, allowing the bike to stay more upright than it would be if you didn't KTM. This allows the suspension to work better, and keeps you on tread closer to the center of the tire. It also gives you more room to lean the bike manually, if the turn tightens up unexpectedly or if you have to dodge a piece of debris in your intended line.

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Joe Frickin' Friday
Hi Mitch -- Would you explain this a bit further? I am presuming that the center of mass varies from one bike mfr. to another due to frame geometry and weight placement, so wouldn't that change the lean angle from the 30 mph constant you show in this example as 16.7 degrees? Regards, Greg

 

To clarify, there are three centers of mass involved here:

 

1. the center of mass of the bike

2. the center of mass of the rider

3. the combined center of mass of bike + rider

 

If you draw a line in space between #1 and #2, #3 will be somewhere on that line - generally closer to #1, since the bike is usually quite a bit heavier than the rider. It is #3, the combined center of mass of bike + rider, that matters here; it is for that point that the lean angle is absolutely determined by speed and turning radius.

 

The bike's center of mass may be higher or lower from model to model, and it may also be more to the front or rear. It might even be off to one side, if you've loaded your luggage unevenly. But if you put this guy:

 

pocket-bike-500-1150665809m4tv44wtmk.jpg

 

and this guy:

 

Towering-Tall-Bikes-L.jpg

 

through the same turn at the same speed, they will both lean at the same angle, as measured to the combined center of mass of bike + rider.

 

So for a given curve and speed:

 

1. If, as on a dirt bike, I place my standing mass above the front wheel and lean the bike,

 

2. and as on street bike, I "kiss the mirrors" placing my seated mass low and inside, keeping the bike more upright,

 

You are saying the bike lean angle must be the same?

 

For a given speed and turning radius, the lean angle of the bike itself will be different if you move your body's center of mass off to one side. Heck, if you're leaning way off of a 20-pound bicycle, you can navigate around some pretty tight corners while keeping the bicycle itself absolutely vertical. But if in #1 and #2 you consider the lean angle of the combined center of mass of bike+rider, then that lean angle in both cases is the same.

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Asymmetrical

Thanks for the explanation. Naturally, if the rider adjusts weight, that changes everything about lean angle and I realized you were assuming the rider remains centered.

 

My intuitive reasoning was that it would seem that a lower center of mass would require less lean angle than a higher center of mass because the centrifugal forces would be at a higher 'altitude' due to the higher center of mass - therefore requiring more lean angle to overcome that and to maintain the perfect line.

 

I suppose that's why intuition is unreliable!

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Dave_zoom_zoom

Excellent illustrations Mitch! I to will take the same application and use it on two extremes to help clarify something in my own mind. :) It's clear to see the rider on the very tall bike must move a greater distance to achieve the required angle thus his bike will be slower to respond. This could mistakenly leave the rider thinking he has to lean further to make the same curve at the same speed as the guy on the mini bike.

 

Dave

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My intuitive reasoning was that it would seem that a lower center of mass would require less lean angle than a higher center of mass because the centrifugal forces would be at a higher 'altitude' due to the higher center of mass - therefore requiring more lean angle to overcome that and to maintain the perfect line.

A lower center of gravity makes a difference -- but only in how quickly you can get the bike leaned over, not what angle you end up at. A lower CG makes it easier to get the bike leaned over because less torque about the tire contact patch level is needed rotate the bike over (or for the same amount of torque (steering input) the bike rotates over faster than with a high CG), but as Mitch indicates once leaned over, the height of the CG is irrelevant because it must always be in the plane with the supporting structure (the contact patches) for there to be balance.

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My intuitive reasoning was that it would seem that a lower center of mass would require less lean angle than a higher center of mass because the centrifugal forces would be at a higher 'altitude' due to the higher center of mass - therefore requiring more lean angle to overcome that and to maintain the perfect line.

A lower center of gravity makes a difference -- but only in how quickly you can get the bike leaned over, not what angle you end up at. A lower CG makes it easier to get the bike leaned over because less torque about the tire contact patch level is needed rotate the bike over (or for the same amount of torque (steering input) the bike rotates over faster than with a high CG), but as Mitch indicates once leaned over, the height of the CG is irrelevant because it must always be in the plane with the supporting structure (the contact patches) for there to be balance.

 

The height of the CG has to matter with respect to the lean angle. If both a "regular" bike and Mitch's tall bike both weigh the same, what angle would the tall bike have to come down to match the CG of the regular bike?

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Joe Frickin' Friday
If both a "regular" bike and Mitch's tall bike both weigh the same, what angle would the tall bike have to come down to match the CG of the regular bike?

 

Same angle.

 

The following will be messy, but if you can make sense of it (and of my horrible handwriting), you'll see that height doesn't matter.

 

Here’s a vector diagram of a rider on a bike in a turn:

 

lean-angle-L.png

 

We are viewing bike+rider from the rear, and the bike is in a steady right hand turn with a radius of R. Bike and rider are moving at a steady speed of V. As described upthread, there are three centers of mass (CoM) that are of interest:

 

-the CoM of the bike

-the CoM of the rider

-the combined CoM of bike+rider

 

In this diagram, the rider is hanging off to the right side of the bike, so his CoM is way off to the right of the combined CoM, and the bike’s CoM is a little off to the left (it’s a heavy bike, which is why the combined CoM is much closer to the bike than to the rider).

 

Gravity exerts a downward force on the combined CoM. If the road suddenly disappeared, bike+rider would accelerate downward at 1 g. The classic formula for acceleration is F = m * a; in this case a is the acceleration due to gravity, equal to 1 g. So gravity must be pulling downward on bike+rider with a force of Fw = m*g.

 

The contact patches of the tires exert a force that can be broken down into horizontal and vertical components Fx and Fy, respectively. The vector sum of this force, Ftire, must pass through the combined CoM, because if it doesn’t, the lean angle will start changing.

 

A constant lean angle means the combined CoM is holding at a steady altitude above ground. For this to be true, the vertical force exerted by the tires, Fy, must be exactly equal to the weight of the bike:

 

Fy = Fw

or,

Fy = m*g

 

Since our rider is traveling at constant speed V through a turn of constant radius R, the lateral force from the tires, Fx, must exactly cause the lateral acceleration needed in order to travel on that path:

 

Fx = m * alateral

 

The equation for lateral acceleration in a turn is V2/R, so we can rewrite this as:

 

Fx = m * V2/R

 

Ftire is the vector sum of Fx and Fy. Placing the component vectors Fx and Fy head-to-tail – also as seen in the diagram – allows you to calculate the angle of Ftire. Since Ftire passes through the combined CoM as described earlier, that angle is also the lean angle, THETAlean, of the combined CoM. We can calculate that angle with trigonometry:

 

Fx/Fy = TAN( THETAlean )

 

THETAlean = ATAN( Fx / Fy )

 

Substituting the earlier equations for Fx and Fy, we get:

 

THETAlean = ATAN[ (m*V2/R) / (m*g) ]

 

the m appears on top and bottom, so they cancel each other out and can be removed, leaving this:

 

THETAlean = ATAN[ V2/(R*g) ]

 

where:

  • THETAlean is the lean angle as measured to the combined center of mass of rider+bike
  • V = the forward speed of bike+rider
  • R = the radius of curvature of the path through the turn
  • g = the acceleration due to gravity

If you use speed in feet per second and turn radius in feet, then g = 32.2 feet per second squared. If you use speed in meters per second and turn radius in meters then g = 9.81 meters per second squared.

 

If you hang your body to the inside of the turn, then the bike may be slightly closer to vertical, but the lean angle measured to the combined center of mass will still be given by the above equation.

 

Note that for a lateral acceleration of 1*g, the component vectors form a square (Fx = Fy) and your lean angle will be 45 degrees.

 

This math applies to any object traveling on a curved path. A ball being swung around on a string (here the angle of the string is what matters), an F-16 banking through a turn, a sprinter running around the curve of a track, or a motorcycle riding around a bend. All the same.

 

The important thing to note is that the height of the bike’s CoM does not appear anywhere in this analysis.

 

Hope I haven't screwed anything up...

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I stand corrected. Mitch and Mark are correct. I would like to mention one more time there is a difference (sometimes huge) between the CG lean angle and the actual lean angle of the bike, in a given turn.

 

Mitch, nice force vector diagram. :thumbsup:

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Hope I haven't screwed anything up...

Perfectly correct.

 

Put another way, when you run the force balance equations, both the weight (mass) and the height cancel out on both sides of the equation, so the height and amount of the CG doesn't matter.

 

Whether you are on a minibike or a 20 foot-tall circus bike, if you want to go around a turn with X radius ay Y speed, the lean angle of the CG will be exactly the same.

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Now for a somewhat related question.

 

GS riders stand on their pegs to "reduce the center of gravity", at least that's what I've been told, when I asked why they did that.

 

1. Is that statement true?

 

2. What difference would it make for a rider?

 

 

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Envying Bud

 

Actually that raises the riders coG & the combined coG BUT seeing as the rider isn't solidly attached to the bike when standing up it allows the bike & rider to react differently & somewhat independently.

 

Also, when riding off-road, or in certain difficult to traverse terrain, a GS rider (off-road rider) uses a LOT of cross control (leans against bike lean-in) to keep bike weight over the tire contact patch. If not cross controlled there is a greater chance of the tires washing out from under the bike & rider.

 

Standing also somewhat uncouples the rider from the bike so the bike is easier to flip around.

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Matthew Miller

Thanks everyone for the answers. unfortunately I think I need a physics or something lesson to understand the equations. Short answer seems to be all road bikes will lean the same angle unless the rider directly changes the angle with his body position.

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Thanks everyone for the answers. unfortunately I think I need a physics or something lesson to understand the equations. Short answer seems to be all road bikes will lean the same angle unless the rider directly changes the angle with his body position.

 

Evening Matthew

 

On paper yes, in the real world no.

 

Things like tire shape (how far off-center the tire contact patch ends up during & through the lean-in, tire widths (again how far the tire contact patch changes during lean-in, tire traction (how much tire side slip in the turn), tire air pressures, bike lateral center of mass location, front/rear tire tracking offset (not all bikes have the rear tire center exactly in-line behind the front tire), rider fore/aft position on the bike (effects side slip & weigh bias therefore contact patch shape), things I have missed.

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The lean angle of the center of mass will be the same, if I'm reading Mitch's post correctly. Not that of the rider, or the bike, but the center of mass.

 

Jan, in your examples, you are using body position to move the center of mass around, relative to where it would be if you stayed on the seat with your spine in line with the lean angle of the bike frame. This produces different effects, hopefully effects that are useful in that riding situation.

 

For example, in the street situation, where you are kissing the mirrors, you are moving the COM to the inside of the bike, allowing the bike to stay more upright than it would be if you didn't KTM. This allows the suspension to work better, and keeps you on tread closer to the center of the tire. It also gives you more room to lean the bike manually, if the turn tightens up unexpectedly or if you have to dodge a piece of debris in your intended line.

 

:thumbsup:

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Hope I haven't screwed anything up...

Perfectly correct.

 

Put another way, when you run the force balance equations, both the weight (mass) and the height cancel out on both sides of the equation, so the height and amount of the CG doesn't matter.

 

Whether you are on a minibike or a 20 foot-tall circus bike, if you want to go around a turn with X radius ay Y speed, the lean angle of the CG will be exactly the same.

 

We have sailed for a lot of years. I remember when our teenage son was at the top of the mast and Nancy and I were in the cockpit. He told us to stop moving around (we didn't even notice) as small movements side to side in the cockpit made the top of the mast swing in a large arc. I bet the guy on the 20 foot tall circus bike has the same sensation when leaned at the same angle as the guy on the pocket bike. :grin:

 

Being a flatlander, my lean sensor doesn't work the same as someone like Foot, who has no flat land close to his house on which to ride.

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Matthew Miller

There's something I understand as I have sailed small and large sailboats most of my life. Kind of like being sea sick, Go to the lowest part of the boat you can get to.

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Lone_RT_rider
The bike's center of mass may be higher or lower from model to model, and it may also be more to the front or rear. It might even be off to one side, if you've loaded your luggage unevenly. But if you put this guy:

 

pocket-bike-500-1150665809m4tv44wtmk.jpg

 

and this guy:

 

Towering-Tall-Bikes-L.jpg

 

through the same turn at the same speed, they will both lean at the same angle, as measured to the combined center of mass of bike + rider.

 

What about this guy?

 

MN%20trip%202005%20052-XL.jpg

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The bike's center of mass may be higher or lower from model to model, and it may also be more to the front or rear. It might even be off to one side, if you've loaded your luggage unevenly. But if you put this guy:

 

pocket-bike-500-1150665809m4tv44wtmk.jpg

 

and this guy:

 

Towering-Tall-Bikes-L.jpg

 

through the same turn at the same speed, they will both lean at the same angle, as measured to the combined center of mass of bike + rider.

 

What about this guy?

 

MN%20trip%202005%20052-XL.jpg

 

It appears that bike will not lean. :rofl:

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The bike's center of mass may be higher or lower from model to model, and it may also be more to the front or rear. It might even be off to one side, if you've loaded your luggage unevenly. But if you put this guy:

 

pocket-bike-500-1150665809m4tv44wtmk.jpg

 

and this guy:

 

Towering-Tall-Bikes-L.jpg

 

through the same turn at the same speed, they will both lean at the same angle, as measured to the combined center of mass of bike + rider.

 

What about this guy?

 

MN%20trip%202005%20052-XL.jpg

 

It appears that bike will not lean. :rofl:

 

Bud, with a bike like that you need to know RideSmart to ride it. No need to lean. Lol

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The bike's center of mass may be higher or lower from model to model, and it may also be more to the front or rear. It might even be off to one side, if you've loaded your luggage unevenly. But if you put this guy:

 

pocket-bike-500-1150665809m4tv44wtmk.jpg

 

and this guy:

 

Towering-Tall-Bikes-L.jpg

 

through the same turn at the same speed, they will both lean at the same angle, as measured to the combined center of mass of bike + rider.

 

What about this guy?

 

MN%20trip%202005%20052-XL.jpg

 

It appears that bike will not lean. :rofl:

 

Bud, with a bike like that you need to know RideSmart to ride it. No need to lean. Lol

 

Ya but............ The range can not be very far with that short electric cord. :rofl:

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