One Wheel Drive car

[Adam Farabaugh]

It's true that the path would be shorter, but why would a 20" wide motorcycle be faster than a 20" wide car?

Transients make the difference I think; motorcycle has to shift CoG way too much in a slalom to be fast.

EDIT: Fundamentally a quasi-static lap sim is a worse approximation for a motorbike than a four-wheel car. QSLS assume that yaw dynamics are fast, which I think is not true on a motorcycle. Anybody know what order of a system motorcycle yaw is? Shifting your body to turn is something like jerk (3rd derivative) or snap (4th derivative) input I think, but not really sure.

[Z]

There is the potential for some good Vehicle Dynamics discussions starting here...

I won't say much more about Charles Taylor's One-Wheeler because it involves too much gory Classical Mechanics (hint, hint ... anagram!), and CM seems to be a forbidden subject in the racecar world.

But good to see that Jonny and Adam are using Sims to help decide whether a two-wheeler bike is faster than a four-wheeler car. But how good are these Sims???

Some miscellaneous thoughts on simulating bike vs car.

1. The bike should be faster "through" the slaloms because it does not have to move its CG sideways AT ALL! Remember that the FSAE cones are quite short, so in plan-view the bike's CG can trace a straight path OVER the cones, with only the tyreprints and wheels "side-stepping" the cones in an S-shaped path. The taller the bike, the easier this is (the reason is found in the end-view...).

2. Big subject here, but... If you are riding a bike in a straight-line..., and you want to turn leftwards..., then you MUST FIRST STEER THE FRONT-WHEEL RIGHTWARD! Oh yes ... much debate on this one over the years...

3. The above "set-up phase" for a bike entering a corner is NOT necessary with a car. So cars generally do sudden "transients" much better than bikes.

4. (For Adam.) Rigid Body Dynamics, epitomised in Euler's RBD-Equations, only requires consideration of the second derivative of position. Specifically, you only have to know about Forces causing changes to Momenta (P). So, since P = m.V and F => dP/dt, you only have to know up to "acceleration". Higher derivatives, such as "jerk", are only of use with squashy bodies (eg. the difference between your body being hit by a high velocity bullet, or a heavier but slower moving rock, both of which have the same P).

5. Though many people have tried, no one (that I know) has ever developed a good aero-DOWNFORCE package for racing motorbikes. Good aero-DF is the major differentiator in most forms of car circuit-racing, including (IMO) FS/FSAE. So, since "AERO-ABOVE-ALL!!!", I will put my money on the cars here.

Z

[Adam Farabaugh]

Z, that's not true. It really does depend on what order derivative of position your control input is.

Take for example a quadrotor, electric motor dynamics are fast enough that you can assume infinitely fast thrust changes, but the input is at the 4th derivative! To move horizontally, you first have to roll the vehicle so that the thrust vector will give you horizontal acceleration, and since you actuate roll moment, your control input is the 4th derivative of position. So the only trajectory a quadrotor can follow exactly is one with zero snap at the beginning and the end.
http://ieeexplore.ieee.org/stamp/sta...number=5980409

I believe steering a motorcycle is similar. You actuate some sort of roll moment which in turn allows the bike to exert a yaw moment against the ground. You can't turn without leaning the bike. I probably have some details mixed up but I am pretty sure you do not directly actuate yaw moment when driving a motorbike! This is the reason that bikes' transients are slower than a car, in a car you do actuate yaw moment directly (assuming tire transient dynamics are fast, which compared to the timescale of rolling a bike/car, they are).

[Z]

Z, that's not true.

Adam,

I see two possibilities here. Either you soon get a Nobel prize for proving Classical Mechanics wrong, or we are misunderstanding each other.

I reckon the latter is more likely. So, to clarify my point, understanding the Dynamics of massive particles (as per Newton) or massive Rigid Bodies (as per Euler) only requires consideration of "derivatives of position of the masses" up to the 2nd, namely acceleration. I suspect your view (eg. in your second paragraph above) is based on some sort of "4th order control equations" that are not a direct representation of the relevant "Dynamics".

That is, your above sort of "cogitatio caeca", or "alphabet soup thinking" as I call it, makes it very easy to make ANY physical phenomena look as if it is modelled by ANY "Nth" order equation (ie. just keep stirring the soup!). Sadly, much as I would like to learn more about your point of view, I am hindered from doing so by the paywall blocking your linked paper. And since my experience is that 99.999...% of all modern papers are junk, and I am not even allowed to browse the paper to see if it is worth it (as was possible in any Library in the olden-days!), I am not going to pay. Ahh... progre$$!!!
~~~o0o~~~

Getting A Bike To Turn A Corner.
========================
Indeed, a bike does "Yaw" its way around a leftward corner by the rider "steering" the handlebars leftward. And for a correctly set-up bike in the middle of a steady-state leftward corner, the rider must keep applying this leftward couple to the handlebars (*). Contrary to popular opinion (at least IMO ), most of the road-to-tyre cornering forces ("Fy"s) come from "slip-angle" behaviour, rather than from "camber thrust". So at the "tyreprint" level the behaviour is very similar to that of a car.

But at the "whole vehicle" level there is obviously a huge difference. A bike has no "trackwidth", so obviously no possibility of resisting a Roll-Couple (not R-"Moment"!!!) via "Lateral-Load-Transfer". So the bike/rider's CG MUST lean into the corner in order to "equilibriate" the FBD.

BUT (!!!), BIG QUESTION: How does the bike rider, initially travelling in a straight-line, START the leftward lean that is necessary to get them around a leftward corner?

A word of warning to those of you starting a family. It is around the time that little kiddies learn to ride pushbikes that they realise that Dad is NOT that infallible, God-like, figure who knows everything. Thinks the kiddie, "Either he's got NO idea about this bike-riding thing, or he is trying to KILL me!" (Hint: DO NOT tell them "If you want to go left, then first you have to lean left...")

(* BTW, in my younger days I used to be able to ride a pushbike more-or-less indefinitely without touching the handlebars. So, with hand-on-hips for that cool look. In this case steering is done partly by foot-pedal forces, but mostly by bum-cheeks clenching the saddle and flicking the rear-half of frame so as to control the front-half of frame (= front-wheel + forks + handlebar). Mass-distributions and interconnecting geometry of the two half-frames makes a big difference to ease of control.)

Z

[zhangxiaobao]

I think what Adam means is that in some cases all derivatives up to the n-th (in his case of a quadrotor it's up to the 3rd) must be continuous.

[stever95]

I'd put my life savings on the FSAE car as well for what it's worth.

[Zac C]

I'd put my life savings on the FSAE car as well for what it's worth.

depends on the FSAE car. I've been to FSAE test sessions where a golf-kart had FTD until after lunch.

I'm not sure a bike or motorcycle would do very well on a typical FSAE course though. It'd do really well in the slaloms and any kind of straight away or big sweeping turn. It wouldn't handle the transitions between linked turns very well and the tighter turns would be a problem. With slow speed turns you have to do a bit more steering of the bike to get through without wrecking.

[Jay Lawrence]

For the 2-wheel doubters,
youtube this: /watch?v=fXWVYtsf43Y

Admittedly, the skill required to get around a FSAE track quickly in a FSAE car is vastly lower so as far as the 'weekend autocross racer' goes it's not exactly comparable, but it's interesting to see the capabilities.