Physics question about lean angle

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CTK
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Physics question about lean angle

Post by CTK »

There are two forces acting on a motorcycle tire contact patch- the normal force of gravity, and then whatever accelerative forces are on the bike acting in a shearing manner. OK. Correct me if I'm wrong, but in a lean, the cotangent of the ensuing lean angle = the ratio of the horizontal force acting on the tire vs gravity (the vertical force)- not tangent as lean angle is measured from vertical.

OK.

So based on that and these reported max lean angles, do these max G estimations sound right?

Image

Scooter = 0.83g
Street bike = 1.2g???
Supersport = 1.4g??!?!??
SBK= 1.88g?!?!?!?!???????????????
MotoGP = 2gs!?!?@@?@?!@!?@?!??????

I know it sounds pretty crazy but I remember seeing the simulation sheet for Argentina and they had estimations of ~1.3-1.5gs through corners and into braking zones

Did I miss something?

*EDIT* Could the missing .5gs on the MotoGP bike be lost to heat?

Hanuman
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Re: Physics question about lean angle

Post by Hanuman »

Where's mu?

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Geonerd
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Re: Physics question about lean angle

Post by Geonerd »

Yea, those numbers look right.

It's a straight tangent function where Tan (lean angle) gives you the lateral acceleration generated. Figure the vertical leg of the triangle will always be 1, so Tan gives the length of the other leg, or G.

In a situation where the rider is hanging off, using the actual lean angle is not correct. Instead, you want to find the angle (from vertical) between the tire contact and the center of mass of the bike+rider. Since this angle is higher than the pure lean angle, a hanging-off rider can make more lateral G than the lean angle suggests.

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bobkarol
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Re: Physics question about lean angle

Post by bobkarol »

In order to know what the correct lean angle is, we have to know how the lean angle is measured. If it is measured via a gyro-accelerometer affixed to the chassis, then the actual lean angle will be greater than measured, due to the unsensed angle added by the rider hanging off. However, if the lean angle is calculated from gps-measured distances over a constant increments of time, then the posted lean angle will be correct, as the Cg of the bike being measured includes the rider's position. The telemetry system installed on my bike utilizes the latter system. If I hang off further while following the same cornering radius, my speed is greater. The bike covers greater gps-measured distance in the same increment of time, thus the computer calculates a greater actual lean angle.
Last edited by bobkarol on Fri Jul 06, 2018 5:38 pm, edited 1 time in total.

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bobkarol
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Re: Physics question about lean angle

Post by bobkarol »

It might be interesting to know how GPS bike telemetry works: GPS defines the position of the combined Cg of the bike/rider at three points along the bike's line in small increments of time. Given the change in angle of the bike's direction over that measured distance, a standard dynamics algorithm calculates average speed, then instantaneous cornering radius, and from that the corresponding lean angle of the bike's combined Cg. GPS telemetry is also used to calculate the bike's rate of acceleration or braking, or g-force, over those three points. That's how the leaning and force diagram you see on-screen during a motoGP race is produced.

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Mikesbytes
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Re: Physics question about lean angle

Post by Mikesbytes »

That brings up the possibility that the rider can slightly influence the electronics by how much they sit off the bike slightly impacting the measured lean angle
My signature isn't particularly interesting

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bobkarol
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Re: Physics question about lean angle

Post by bobkarol »

Hanuman, above, asked about the effect of mu: Mu would be the combination of tire composition (for conformity to track surface and durability), braking load thus contact area, and track surface condition (moisture, texture, temperature). The actual limit of Mu for any set of condtions can only be determined by performing a locked wheel (tire) drag test to measure the maximum rate of deceleration. Racers find the limits of Mu by feel while cornering. When they feel the tire push or slide, they've exceeded Mu. Thus lean angle and braking force become their gauges of Mu's limits.

The maximum permissable lean angle at any instant is a function of Mu (grip) less any simultaneous braking or accelerating force. As cited by CTS, above, the maximum lean angle (64°) implied a maximum 2.0g cornering force. That is the measure of Mu that enables that specific lean angle. Any braking or accelerating force applied while leaning reduces the amount of Mu that is available for cornering. Lean angle has to be reduced in order to stay within the total available Mu.

As CTS further cites, the estimated force in Argentia was 1.3-1.5 g's through corners and into braking zones. This implies riders also could have consumed a large portion of Mu by simultaneously braking in the corners, plus conserving a bit to maintain a margin of safety. Corner-braking required much greater force than the missing 0.5g that CTS pondered. For example, given 2.0g Mu, less a 10% margin of safety, it can be calculated that a rider would be braking at 1.0g while cornering at 1.5g's, or a 56° lean angle.

In order to maximize his cornering speed, the rider could fully release the brakes, eliminate his margin of safety (unused grip), and lean farther (63°) to consume the full 2.0g Mu of traction. Then again, a rider does not have to consume all 2.0g Mu while cornering. If the corner was a wide, flat-out corner, and the fastest line dictated a 56° lean angle (1.5g's), plus a moderate accelerating force (0.5g) to maintain the bike's high speed, his margin of safety would be 20%.

The possible combinations of lean angle and braking force are endless. They produce all the different lines through a corner that the riders use to seek an advantage.

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speeddog
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Re: Physics question about lean angle

Post by speeddog »

Geonerd wrote:Yea, those numbers look right.

It's a straight tangent function where Tan (lean angle) gives you the lateral acceleration generated. Figure the vertical leg of the triangle will always be 1, so Tan gives the length of the other leg, or G.

In a situation where the rider is hanging off, using the actual lean angle is not correct. Instead, you want to find the angle (from vertical) between the tire contact and the center of mass of the bike+rider. Since this angle is higher than the pure lean angle, a hanging-off rider can make more lateral G than the lean angle suggests.
This ^ is correct.

I am fairly certain the latest spec system does not allow GPS information to be used onboard.

Additionally, the angle of the tarmac is unknown, the more the track is banked, the larger the actual cornering 'g' for a given angle from vertical.

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bobkarol
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Re: Physics question about lean angle

Post by bobkarol »

Speedog writes, "the more the track is banked, the larger the actual corner g-force for a given angle from vertical."

It doesn't matter if the given angle of lean from vertical is on a horizontal or banked track. The effective cornering g-force and speed will be the same. The track banking serves a different purpose.

On a horizontal track, the lean angle relative to the surface will reach a limit where the cornering g-force that's generated exceeds the available traction limit, Mu. The tire slides out.

On a banked track, the bike can lean at an greater angle from vertical than would be possible on a horizontal track. But this greater lean angle is reduced relative to the angle of the banking.

As a result, the lateral g-force exerted on the track surface is less than the traction limit, Mu. The tire continues to grip, hence enable the bike's greater effective lean angle and speed.

Increase the bike's speed and angle of lean from vertical, as you increase the angle of the banking: Eventually, the track banking is nearly vertical, and the bike is going around leaned almost horizontally. The rider is now a carnival performer riding the Wall of Death.
Last edited by bobkarol on Wed Aug 14, 2019 5:56 pm, edited 3 times in total.

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bobkarol
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Re: Physics question about lean angle

Post by bobkarol »

Speedog also references Geonerd's original discussion of the prevailing error of using a bike's chassis-measured Cg, instead of the combined Cg of bike+rider (CCg), in order to determine the correct effective cornering lean angle that produces maximum lateral g-force.

This brings up an interesting question: In the absence of GPS tracking to provide the position/distance/time data needed to calculate an accurate answer, how can the effective cornering CCg be determined?

Basic motorcycle dynamics equations can easily churn acquired basic data: 1) knowledge of the traction limit Mu (instanteous g-force drag test in actual conditions), 2) g-force measurement of any simultaneous braking or accelerating, 3) speed data calculated from wheel rpm and tire size, or distance divided by time, and 4) track banking, if applicable.

Initially, the calculations will produce the CCg effective cornering lean angle. This result includes any degree the rider is hanging off (or not).

Despite different bike Cg and CCg lean angles, their effective cornering lean angle is actually the same. The difference between the chassis-measured Cg lean angle and the data-calculated CCg lean angle represents the degrees that the rider is hanging off.

So, even without the benefit of GPS data, we could learn how much a rider contributes to his bike's speed by hanging off. That might be interesting to know, especially when he's totally hung out, dragging this elbow through a corner.

This isn't the end of the effective race-bike cornering lean angle problem, however: The effect a sliding rear tire has on the leaning CCg also needs to be considered.

MotoGP riders drag and slide the rear wheel into and out of corners, because it produces appreciably increased cornering g-force, hence speed.

How this happens is a different thread.

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speeddog
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Re: Physics question about lean angle

Post by speeddog »

bobkarol wrote:Speedog writes, "the more the track is banked, the larger the actual corner g-force for a given angle from vertical."

~~~SNIP~~~
Yeah, I stuffed that one. :lol:

Indeed, they can calculate the actual lean angles relative to the track, as they do get intermittent track positions from the track splits markers.
IIRC, there's more than the 4 sectors we see on the video.
Given an accurate track profile of banking vs position map, they can calculate track position from wheel rpm and lean angle.
Tire slippage on acceleration or braking will corrupt that calculation, but the banking changes quite gradually so not likely a big error.

Acceleration/deceleration provided by the tire steals away from the cornering capacity of the tire (traction circle)

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speeddog
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Re: Physics question about lean angle

Post by speeddog »

bobkarol,

I replied to your PM, but it seems to be stuck in my 'outbox'.
I've sent a message via alternate channels to our wizard, see what I'm doing wrong....

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bobkarol
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Re: Physics question about lean angle

Post by bobkarol »

I previously concluded, "The difference between the chassis-measured Cg lean angle and the CCg data-calculated lean angle represents the degrees that the rider is hanging off."

I forgot to mention that the chassis-measured Cg lean angle includes any sliding the rear wheel does, which the CCg data would also include. But that's not the significant part of my oversight ––read error:

The CCg is a combination of the bike Cg and the inward Cg of the rider hanging off (RCg). That means the rider is hanging off farther than the CCg data indicates.

How far, isn't really knowable from the data we can collect. It might be twice as far, if we ingore the portion of the lean angle contributed by the rear wheel sliding. An equation for the approximation would be RCg = 2 (CCg – Cg).

I reckon the true way to answer the original question, "how much does a rider hanging off add to tthe bike's speed?" is to acquire specific rider Cg data. We could do that by fastening an accelerometer to the rider's leathers where his belly button is located and track his shifting position.

Enough, I think. I suspect my lean angle blather is more than enough for most readers. My apologies for not adding the numbers.
To quote a brief acquaintance, who told me, "I get a headache when I don't get paid for doing it."

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