Thanks for the continued input. I found Gauss' description of the reason precesion occurs particularly vivid & useful, I now feel like I understand it personally rather than just expecting it to happen "because the book said" & "because it always does".
I have also been running round the house holding a spinning unicycle wheel & I can really feel the expected forces & appreciate how they could assist the continued turn. I tried it with the cranks still on at first but they upset the balance too much to get the full appreciation.
My perception now is that the moment that enables a unicyclist to make a smooth continuous turn is principally provided by the riders centre of gravity relative to the tyre's contact point with the ground. This moment is also added to by the gyroscopic precesion of the rotating wheel.
/\/\
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Guest repliedRe: How do you turn corners?
On Sat, 12 Oct 2002 18:14:08 0500, gauss
<gauss.cfog0@timelimit.unicyclist.com> wrote:
>I am not for sure yet, but looking at the equation it seems that a
>decrease in the mass of the wheel means that there is a smaller required
>force to make the same precession. As I think you pointed out earlier,
>the relationship is linear, meaning that the limiting case of a massless
>wheel would require no force to make a precession. This seems sort of
>intuitive.
I figured that out too, for a coin, or a wheel only. You argued that
the polar momentum of the unicycle and rider is quite small; yet it's
of course not negligible. And then a rider can increase it by
stretching out his arms right? And even further if he would hold
weights. Would he then need to lean more to make the same turn
especially if the wheel is light?
It still is a bit elusive to me, but maybe we should stop the
discussion here, I think we've drifted offtopic for too long.
Klaas Bil
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Guest repliedRe: How do you turn corners?
Thanks, the spinning ball analogy was a useful mental image. I'm not
sure I understand the turningthroughleaning argument fully. I mean,
I think I understand the logics and the physics as you explained them
(in daily instalments), but I still find it hard to believe that the
mass of the wheel is involved in that. Would a uni with a heavier
wheel turn more at the same lean?? For now, I think I'd better just
accept it though.
Klaas Bil
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Guest repliedRe: How do you turn corners?
> Your formulae and my gut feeling agree that the twist tendency is
> proportional to the wheel mass. So if I have a lighter wheel the whole
> "turning through leaning" works different, i.e. less. And with a
> weightless wheel it wouldn't work at all unless I miss something. That
> is still counterintuitive to me.
Keep in mind there's no such thing as a weightless wheel. Intuition may
fail you with impossible things.
>
> I'll ask my colleague engineers what physical explanation they offer
> as to why a coin that you throw to roll in a straight line on a
> horizontal plane, but which (coin) is tilted as opposed to being
> vertical, starts to describe a circle as opposed to continuing in a
> straight line while falling at the same time. And if this is due to
> the procession, then try imagine a weightless coin with an additional
> point mass at its centre. There the procession explanation wouldn't
> work, would it? And still it would ride a circle, wouldn't it? (I'm
> not sure, maybe my intuition fails in this hypothetical thing.)
> I think that my failure to understand this coin behavior is core to
> my problem.
Precession. Once again, no such thing as a weightless coin. It's not a
useful mental image.
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Well countersteering was the key concept for me. Thanks to those who mentioned this idea. I read it yesterday and today while practicing I remembered the idea. I used to love countersteering turns on my BMX. In practice today I applied this idea and my poorly performed 3m diameter circles turned into 1.5 meter circles without pedal hits.
A countersteering turn is one in which the wheel is steering toward the outside of the circle. The greater the lean of the wheel into the turn, the greater the angle of countersteering required to overcome the lean. The best place to see this in action is any motorcycle race. These are getting easier to find on TV every day. Note how upright the wheel is compared to the frame of the cycle.
I countersteered by twisting my body in the direction opposite of the turning direction. Since a countersteer allows you to accellerate through the turn, the effect is a smother, sustained turn. I was able to get 3 or so circles of 1.5m diameter. Another advantage is that, although the unicycle may be leaning more, the angle of lean of the wheel is less, keeping the pedal from hitting the ground. Note that I currently have 150mm cranks on a 20".
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Hey, Klaas. I think I understand your coin situation. I think the reason is sort of a combination of all these things. I think if the coin is thrown slightly tilted, then when its centroid is further to the side than the contact patch, it would begin to fall to the side. Since it is rolling, the falling begins a precession that counters the fall and begins making a circle. But, I couldn't make a coin do the behaviour you described so I may be misunderstanding, or uncordinated, or both. I see why this massless situation messes with your intuition. I thought a while about it and came up with this: As you look down on a rider on a unicycle, the polar moment of inertia is small (the inertia resisting rotation about what we have been calling the y axis)(we weigh a lot (some more than others), but our mass is all pretty much very close to the axis). The angular momentum of a turning wheel is probably pretty significant (spin a wheel and try to tilt it without letting it precess, it will fight with you pretty hard, even for light wheels) these two things sort of work together to make a pretty significant effect.
I am not for sure yet, but looking at the equation it seems that a decrease in the mass of the wheel means that there is a smaller required force to make the same precession. As I think you pointed out earlier, the relationship is linear, meaning that the limiting case of a massless wheel would require no force to make a precession. This seems sort of intuitive.
gauss
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Guest repliedRe: How do you turn corners?
On Sat, 12 Oct 2002 12:03:12 0500, gauss
<gauss.cf7bb@timelimit.unicyclist.com> wrote:
> try the experiment again, but this time, spin the wheel and begin
>turning your self around in place while the tire is spinning. You will
>see once again the the uni tries to twist to turn into the circle.
I tried it and it works the way you described (though the previous
experiment seemed more to the point). The uni twists and since I keep
the seat at a constant height, the uni lifts itself up. The wheel
looses angular velocity in the process (I guess mostly by converting
angular momentum to potential energy i.e. the lift) so I could only do
the experiment in a transient fashion. If the wheel was driven I could
check if the uni stays in the "lifted" state as long as I turn it.
Your formulae and my gut feeling agree that the twist tendency is
proportional to the wheel mass. So if I have a lighter wheel the whole
"turning through leaning" works different, i.e. less. And with a
weightless wheel it wouldn't work at all unless I miss something. That
is still counterintuive to me.
I'll ask my colleague engineers what physical explanation they offer
as to why a coin that you throw to roll in a straight line on a
horizontal plane, but which (coin) is tilted as opposed to being
vertical, starts to describe a circle as opposed to continuing in a
straight line while falling at the same time. And if this is due to
the procession, then try imagine a weightless coin with an additional
point mass at its centre. There the procession explanation wouldn't
work, would it? And still it would ride a circle, wouldn't it? (I'm
not sure, maybe my intuition fails in this hypothetical thing.)
I think that my failure to understand this coin behaviour is core to
my problem.
>You can sustain a conversation indefinately using only four words:
>Yes, no, dude, guess.
Or these: lean, force, moment, circle. :)
Klaas Bil
I posted only a single copy of this message.
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You are right in assuming that a massless wheel would not exhibit the behaviour that I mentioned. These are kinetic equations and deal with the momentum. Also then you can't apply forces to massless things.
I often too find it difficult to sort out physical meaning from just equations and had to go through this several times before I got a handle on it. Try to picture a ball on a string that you are swining in a circle in a vertical plane. Imagine giving the ball a whack directly to the side while it is spinning. the spinning doesn't cease but sort of changes direction. Think of how it would. Now instead of a ball think of it as continuous like a wheel. You will have each little piece responding to now not a force applied to one point the instant it passes by, but to the whole continuum as it passes, and each little piece is trying to respond. This is the way that I understand it. Hope this helps.
gauss
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Klaas, you are forgetting that you no longer have an inertial reference frame when you ride in a circle. Although you are riding in a circle at a constant"speed," you are not riding at a constant "velocity". That is, there is always an acceleration towards the center of the circle which is keeps the force in the lean direction (couple about the x axis) . try the experiment again, but this time, spin the wheel and begin turning your self around in place while the tire is spinning. You will see once again the the uni tries to twist to turn into the circle.
gauss
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Guest repliedRe: How do you turn corners?
It remains a bit elusive to me. Although I've been trained in vector
notation long ago, I haven't followed all of your logic. But I think I
understand from it that the moment causing a rotation in y is
proportional to the mass of the rotating wheel. That would mean that
if I would have a hypothetical wheel with zero mass, then lean would
not cause the unicycle to turn. It may be so but it is against my gut
feeling.
>You can do an experiment yourself.
>Spin a unicycle wheel and hold it off the ground by the seat. Try to
>make it "lean" and you will see is twist into a turn in your hands.
I did the experiment and noted that the unicycle only tended to turn
in my hands as long as I increased the lean. Having a constant lean,
there was no turning tendency. This is different from riding a bend,
where a "constant" lean and a "constant" speed keep me riding in a
circle forever.
Klaas Bil
I posted only a single copy of this message.
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I too gave the matter some more thought. Of course we can't think of a rolling unicycle with static equations, as there is a certain amount of momentum. Moreover, the angular momentum of the wheel ends up being very important. I am using a computer on which I can't make jpegs, so you are free of diagrams this time, but you may draw your own from this. If you don't understand what a vector is then maybe just skip to the last paragraph and hold off on the usual replies to this post about how some idealized situation doesn't apply to the real world.
Imagine a disk spining about the z axis. If you apply a moment about the x axis, That is you lean your unicycle. The result is not a rotation about the x axis, but rather a precesion about the y axis. The reasons for this are pretty complicated, I don't think I can explain it well, as I don't understand it well, but...
think of a point mass moving with a constant speed in the xzplane (but not constant velocity). The partical has a momentum G =m v . I will try to use bold print to denote a vector quantity. If a force normal to this motion is applied it will cause a change in the momentum d G= d (m v ) in the direction of the force F. Newton's second law states that F dt = d G.
Here it gets a little complicated without diagrams, but you know the vector F is perpendicular to mv. So th adding them together gives a new vector describing the new direction of motion for the particle (that is velocity +change in velocity is the new velocity). Now we are expecting a precesion about the y axis, so let angle dtheta be the angle between the original velocity and the new velocity. the tan(dtheta)= Fdt/mv. If dtheta is small, (that is to say the change in momentum is small, then tan(dtheta)=dtheta. Solving for F we get F= mvdtheta. dtheta is an angle in the xz plane. so it is a revolution around the y axis. lets call it omega* j. Remember j is the unit vector in the y direction. omega is the rotation velocity in this direction (dtheta means how fast does the angle theta change which is velocity). so finally we have:
F = m omega x v
where "x" is cross product. Which makes sense considering the directions o f these vectors.
That was a logical case for point masses. This can be extended to rotations to get the same sort of thing considering newtons law as Moment = a change in angular momentum (M=dH) So now the original motion is a fast rotation about the z axis( instead of a motion of a point) a moument is applied about the x axis and the result from the cross product relation is a rotation in y.
I'm guessing that if you followed all this, particularly without diagrams then you already know what I am saying, maybe this was a refresher. If you didn't I will refer you to the Meriam text on dynamics where I snagged this, although any dynamic text will explain this phenomena one way or another. You can do an experiment yourself. Spin a unicycle wheel and hold it off the ground by the seat. Try to make it "lean" and you will see is twist into a turn in your hands.
gauss
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Guest repliedRe: How do you turn corners?
On Tue, 8 Oct 2002 18:03:02 0500, mike.hinson
<mike.hinson.c8930@timelimit.unicyclist.com> wrote:
>The conclusion of this long rambling claim is that it should be
>impossible to make a smooth continuous turn while decelerating. Even if
>Joe's "cone shaped contact patch" principle was helping, the forces
>described by Gauss would be working in the opposite direction (while
>decelerating).
If the forces work in the opposite direction, then you're still
turning, but just in the other direction. That sounds counterintuitive
though: I lean to the left while decelerating, hence I turn to the
right. Recipe for a quick fall I should say. Something seems wrong
with this argument.
I gave the turningthroughleaning concept some more thought. I now
think you wouldn't need a constant lean (and therefore a constant
accelleration). Unicycling is about dynamical balance, so there will
be a bit of forward and then backward lean all the time, in sync with
pedal motion. Maybe we inconsciously use those leans to derive a
momentum to turn?
Klaas Bil
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Gauss' answer sounds right to me for the major forces involved.
Klaas has got me thinking too & I am trying to understand where the moment comes from when riding down hill.
I suspect that the well established principal of leaning back while descending a hill may really only mean leaning back to the extent that the riders centre of gravity is only slightly forward of the point the wheel contacts the ground. (phew).
I would go further and say that true leaning backwards (meaning to locate the riders centre of gravity behind the wheel's point of contact with the ground while moving forwards) is only going to occur while decelerating.
The conclusion of this long rambling claim is that it should be impossible to make a smooth continuous turn while decelerating. Even if Joe's "cone shaped contact patch" principle was helping, the forces described by Gauss would be working in the opposite direction (while decelerating).
I don't think I ever go fast enough to decelerate for long enough to test this out, perhaps someone would care to give it a go please.
Thanks all,
/\/\
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Guest repliedRe: How do you turn corners?
On Mon, 7 Oct 2002 22:08:52 0500, gauss
<gauss.c6pzn@timelimit.unicyclist.com> wrote:
>Now, in the lower picture, the
>unicyclist shifts his weight foward. There is still the force pulling
>in, but now he is ahead of the contact patch. So there is a distance
>between the reaction force and there is a resulting torque making the
>uni twist. It is a little more complicated than this, but that is
>pretty much it. If anyone wants to fight about it, let me know, I'll
>give you my email.
Gauss, for want of your email I post here. Besides, mike.hinson and
others may be interested enough to continue some discussion here.
In the situation as in your last picture, there is indeed a moment in
the horizontal plane causing the unicycle to rotate around its
(almost) vertical axis. However, isn't there now also a moment that
tends to rotate the unicycle around a horizontal axis, i.e. to fall
forward? You can probably imagine why, or else it would be apparent
from a side view.
The only ways I can see to stop falling forward would be for the rider
to constantly accellerate (not your typical way of turning though), or
to have the forward lean only temporary, i.e. to initiate the turn but
then stop the forward lean. In the latter case the vertical rotation
(i.e. the turn) would maybe continue because of rotational inertia?
What do you think?
Klaas Bil
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Guest repliedRe: How do you turn corners?
"gauss" <gauss.c6pzn@timelimit.unicyclist.com> wrote in message
news:gauss.c6pzn@timelimit.unicyclist.com...
>
> I stayed out of this a while, but now i'll come in. I thought Klaas was
> going to hit what I was thinking as he set it up the same way. That is,
> how will a unicycle spontaneously turn when no in plane torque is
> applied?
As well as all the things people have said, there's another effect going on
too.
When you tilt the unicycle, the outside edge of the contact patch shape is
at a smaller distance from the centre of the wheel than the inside edge. For
the whole wheel to spin at the same rpm the smaller radius circle must move
a shorter distance, which makes the wheel turn towards that side.
The extreme of this effect is if you put a cone down on its side and roll
it, it will turn round in circles, despite not being balanced in the way a
wheel is.
Joe
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