It's very common for physics to make various simplifying assumptions.
Assuming unlimited friction doesn't render it meaningless, though - it just means the x-axis continues on indefinitely, i.e. there is no upper limit to cornering speed (except ground clearance for the required lean angle, which we can also assume isn't a problem, just for fun).
The limit of grip is usually expressed in terms of g, so that implies a maximum useful lean angle for that tyre - i.e. a maximum attainable speed for a given cornering radius.
Example: given a tyre with lateral grip limit of 1.2 g, the maximum speed a bike so equipped could take the 40' radius corner at is about 38 ft/s, 26 mph - it would do so with the C.o.G at an angle of 50°. [Sanity check: tan(50°) = 1.2]
For fun, a speed of 10 000 mph around that 80' circle requires a lean angle of 89.97° and generates nearly 1700 g.
Clearly, lean angles of 65° are not meaningless, as it's routinely achieved in MotoGP (sticky tyres), so the graph is clipped at a sensible point, I'd say, despite the notion of "unlimited friction".
Another sanity check, below. Right in the middle of the table you can see 33 ft radius, 22 mph and 1.0 g generated; very similar to the above 40 ft, 26 mph and 1.2 g.
https://en.wikipedia.org/wiki/Circular_motion
Measuring pure lateral g is very hard to do on a bike, precisely because they lean. That means the only cornering force "felt" by the bike is actually straight down into its wheels at the angle it's leaning, so the accelerometer cannot accurately distinguish between a corner and a dip in the track. You have to infer lean angle from the forward speed and the rate of change of heading (many IMUs measure on rotational axes; GPS or magnetic methods work, too - it's preferable to use all three), using an equation of the form that generated the above graph.
Last edited by IndelibleInk; 06-27-2019 at 04:50 PM. Reason: More better pictures
2009 Shiver (White) | UK
It is meaningless if the question is what is the maximum G's attainable. The value of the model is only that it allows you to estimate lateral G's from a measured lean angle - hence my question about lean angle measurement.
Again, we have not established the measurement reference. I can attest from my professional experience that a properly designed 6DOF IMU will have no trouble resolving an accurate lean angle in spite of the coupled forces of gravity and lean. However, it is very likely that lean angle is 'zeroed' on level ground, and when MotoGP tells us that the rider achieved 62deg for 1/10th of a second, they did not subtract the 7 degrees of banking (or whatever it may be) for that particular corner on that track.
So yes, MotoGP's report of 65 degrees may well be meaningless. They have a vested interest in impressing us.
The difference between 1.0 and 1.2 lateral G's is the difference between a Mustang and a Ferrari. So not really sane.
Finally, to throw into question your entire thesis -> That lean angle can be deployed as a proxy to measure lateral G's:
Note that a very high instantaneous lean angle can be achieved in a pure slide, where very little lateral cornering force is generated (see Pacejka model 'Magic Formula curve: https://en.wikipedia.org/wiki/Hans_B._Pacejka).
That is to say, that for an instant, a rider can achieve a very deep lean angle, while hardly turning,thereby generating very little lateral G.
An illustrative video:
More Alan Birtwistle, just because he is awesome and I love Thunder Multimedia videos:
The best way to compute this equitably would be to either:
1) Go out an run the skidpad test that cars do.
2) If you had a speed and position plot of a flat (unbanked) turn on the racetrack, we could compute a good approximate of the cornering force based on the velocity and rate of change in direction (motorcycle heading).
I don't have any of this data (nor the time to collect it right now) - but in any case, for the OP:
A good rider on a good motorcycle will typically be able to pull between 0.8 and 1.2gs in any direction.
Use a gyroscope.
2009 Shiver (White) | UK
Gs don't matter, traction matters.
What the fuck is the point of traction if you're not using it to accelerate?
2009 Shiver (White) | UK
Because you can't accelerate without traction. You can't turn either. Without traction you can't even sit on the seat.
And after you're done acellerating you can't stop without traction.
Last edited by CapoEVT; 06-29-2019 at 10:24 AM.
You know that g = 9.8 m/s/s or 33 ft/s/s? That means it's a measure of acceleration. [specifically the acceleration due to gravity at sea level]
Turning is acceleration. Acceleration is a change in velocity - velocity changes when its direction changes, even if its magnitude is the same. [velocity is a vector quantity]
Stopping is acceleration. It's just negative with respect to the "forward" direction. [vectors again]
Your body is accelerated, partly by the seat, once the bike has been accelerated - in any direction.
Where a net force is being applied, whether by traction or otherwise, there will be acceleration. Newton: F = ma; F = mg is an object's "weight". For this reason, we find it convenient to express acceleration in terms of gravity - i.e. "g-force". Perhaps it's poorly named.
Finally, (to repeat myself) tyre traction limits are commonly expressed in terms of g. Including in Pacejka's work, linked above.
2009 Shiver (White) | UK