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Old 2002-10-03, 03:06 PM   #16
mike.hinson
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GILD wrote:
> mike, i'm curious, why the need for 'smoother' turns?

Two reasons:
1). In my ignorance I assumed that if I followed tips to ride "smoother" I would also be riding in a more controlled way and find it easier to stay in my balance envelope.

2). I was trying to define the part of cornering that my academic question referred to, that is to say the ongoing cornering, rather than the initial (often jerky for me) start of the corner.

/\/\
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Old 2002-10-03, 04:53 PM   #17
Mikefule
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This bit is easy to understand in an intuitive way if not mathematically. Spin a coin on its axis and as it slows down it will flatten out so that the contact point between the edge of the coin and the table 'moves' around the coin. Roll an old tyre along the ground and see what happens as it slows and starts to fall to one side. The tyre is turning around its own curvature. Push your uni slowly and turn it and see what happens.

On a curve you are in a similar position to a body in orbit. Your momentum from the straight will tend to push you away from the centre of the corner (keep you going straight; make the uni stand up), and your lean will fight against that. On a smooth turn, the forces are balanced for longer; on a sharp turn, there are sudden imbalances and corrections.

But how do you start the turn? That was a good question. Intuitively, you'd think Newton's laws would say that you can't move your centre of mass sideways without something to push against. That is right. However, you can move part of your mass, by moving it agains the rest of your mass. So, you can start the uni turning by twisting your lower body, or by pushing the pedal in a certain way, by using your main body mass as something to react against.

You only need to start a slight lean of the wheel which then rolls out from under you a bit, and the centre of support (tyre contact patch) is no longer under the centre of mass. Then gravity starts to contribute (sometimes more than we would choose ) and you need to accelerate to counteract the fall. You accelerate out of the turn, or steer the wheel back under your mass, whichever way you look at it.

So, to start the turn, part of you pushes against another part of you. Legs go one way, body goes slightly the other. To continue the turn, the natural curve of the wheel does the job. To end the turn, either countersteer or accelerate.

In real life, of course, it either works or it doesn't.

As for turning smoothly... an excellent objective. For muni or road riding, you often need to follow a very specific path. When playing or performing, sudden sharp turns are more fun.
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Old 2002-10-04, 01:50 PM   #18
tats
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do you guys smack your pedles if you lean in too much in a quick turn!

how can i avoid doing it
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Old 2002-10-04, 06:20 PM   #19
john_childs
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Quote:
Originally posted by tats
do you guys smack your pedles if you lean in too much in a quick turn!

how can i avoid doing it
I can smack my pedals on the ground when doing a sharp turn on my 20" freestyle uni with 125mm cranks. Shorter cranks on the 20" would help avoid that. With 24" and larger wheels it is difficult to hit your pedals on flat ground when doing a sharp turn. I assume you're having pedal strike problems on a 20"?
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Old 2002-10-04, 10:44 PM   #20
mike.hinson
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I am still on the trail of the maths & physics behind unicycling. I have found this link which has mathematical explanations for counter steering on bicycles.

http://socrates.berkeley.edu/~fajans.../bicycles.html

I don't yet understand the maths, perhaps there is someone here who could review it and extract any of the aspects that apply to unicycling? (please).

Roger Davies pointed me at the author of this paper "1988 Johnson, R.C. Unicycles and bifurcations, American J. of Physics, volume 66, no.7, 589-92"

I have also found the author requesting information about unicycling here: http://www.unicycling.org/unicycling...mail/0305.html

But I have not been able to locate the paper or get a response from his e-mail address.

Perhaps I will have to content myself with the image of a coin rolling across the desk & turning in tighter circles as it runs out of speed. I may not understand why it does it, but it intuitively seems right & reasonable to expect it to happen.

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Old 2002-10-05, 06:02 AM   #21
fred
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Quote:
Originally posted by mike.hinson

Roger Davies pointed me at the author of this paper "1988 Johnson, R.C. Unicycles and bifurcations, American J. of Physics, volume 66, no.7, 589-92"
....
But I have not been able to locate the paper ...
I am searching for the same kind of material. The above paper can be bought here:
http://tinyurl.com/1svf

Please let us know if you get a response from the author.

Have fun,
Fred
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Old 2002-10-06, 08:53 PM   #22
Roger Davies
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Re: How do you turn corners?

I have talked to him on the phone some time ago and he seamed a very affable
guy. I was working on him to actually have a go at unicycling... but he is
a really tall guy (Jamie size) and at the time I didn't have a tall enough
unicycle for him.... if anyone does talk/mail him remind him he needs to put
his theory into practice at Shakespeare Hall on Tuesday nights. ;-)

Roger
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----- Original Message -----
From: "fred" <fred.c1duz@timelimit.unicyclist.com>
Newsgroups: rec.sport.unicycling
To: <rsu@unicycling.org>
Sent: Saturday, October 05, 2002 7:02 AM
Subject: Re: How do you turn corners?


>
> mike.hinson wrote:
> > *
> > Roger Davies pointed me at the author of this paper "1988 Johnson,
> > R.C. Unicycles and bifurcations, American J. of Physics, volume 66,
> > no.7, 589-92"
> > ....
> > But I have not been able to locate the paper ...
> > *

>
> I am searching for the same kind of material. The above paper can be
> *bought* here:
> http://tinyurl.com/1svf
>
> Please let us know if you get a response from the author.
>
> Have fun,
> Fred
>
>
> --
> fred - Uni Gwaihir
> ------------------------------------------------------------------------
> fred's Profile: http://www.unicyclist.com/profile/478
> View this thread: http://www.unicyclist.com/thread/20698
>
>

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>



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Old 2002-10-08, 12:55 AM   #23
jerryg
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Although my understanding of the theory is miniscule, and I have tried to work some math in my head while riding -well arithmetic really -but I was chewing gum at the time!... Anyway, I like to think smooth turns are accomplished (at faster speeds) by leaning slightly foreward of the balance point and slightly into the turn. Leaning backward during a turn usually makes the turn more sharp and jerky. Understand leaning forward causes speed to increase, and leaning backward decreases speed, but it seems to help smooth turning. I've also noticed, in my trail riding, that only the wheel needs to be leaned into the turn. Sharp turns between close trees can be negotiated while leaning the wheel (and rest of muni) into the turn, while the rider (me) leans away from the turn. It's great fun, but bumping trees at speed leaves bruises on my shoulders. -Perhaps I should lean back a little.
I'm sure that thinking about it too much while riding does not help.
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Old 2002-10-08, 03:08 AM   #24
gauss
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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?
I think you can get this by looking at the wheel's contact patch versus your center of mass. Think of a unicycle riding straight down the road with the rider centered up on it. If he leans over without turning he falls. I attached an image that shows why. The left half shows this situation when viewed from behind. The red block is your center of mass. the black block is the tire. when riding straight there is your force down and because of newton's third law the road pushes up on the tire with the same force and there is a balance. Immediately to the right of that there is a leaning unicycle. Here the center of mass has moved away from the contact patch. the forces are still equal but now there is what is called a "moment". A moment is when two equal and opposite forces are applied but there is a distance between them. I included what this disatnce is in this case, but that isn't important. A moment is torque and makes things twist, and is the force times the disance between them (think of torque in your car. That is why the units are ft lbs it is the force times the distance). Anyway, so the second two pictures are viewed from above. The top one is again a lean. there is a force pulling out and since the contact patch is below, there is a torque tipping the unicycle. 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. I'll not be responsible for another "wheel puzzle" thread of doom.
-gauss
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Old 2002-10-08, 08:09 AM   #25
tats
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Quote:
Originally posted by john_childs

I can smack my pedals on the ground when doing a sharp turn on my 20" freestyle uni with 125mm cranks. Shorter cranks on the 20" would help avoid that. With 24" and larger wheels it is difficult to hit your pedals on flat ground when doing a sharp turn. I assume you're having pedal strike problems on a 20"?
yeah so its not just me riding badly other people do it to
actually i think ive done it a couple of times on my 24" but never got it back up

oh well ta for that any way
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Old 2002-10-08, 10:43 AM   #26
Joe Marshall
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Re: 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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Old 2002-10-08, 10:03 PM   #27
Klaas Bil
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Re: 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 e-mail 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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Old 2002-10-08, 11:03 PM   #28
mike.hinson
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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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Old 2002-10-09, 10:48 PM   #29
Klaas Bil
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Re: 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 turning-through-leaning 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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Old 2002-10-10, 02:49 AM   #30
gauss
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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 x-zplane (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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