Unicycling Theory

23 degrees is 1:2.5 or 40% isn't it?

For a sustained gradient, without a brake, it'd be hard to ride
anything much steeper than that. Obviously for short slopes you can
spin out, and if you've got a brake you're only limited by tyre grip.

Joe

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iridemymuni said:

is it a theory or a hypothesis :S.

23 degrees is acutally quite steep. people will see a 23 degree angle
and say its like 40 or something.

Well, it's a proposed description, explanation, or model of the manner
of interaction of a set of natural phenomena, capable of predicting
future occurrences or observations of the same kind, and capable of
being tested through experiment or otherwise falsified through
empirical observation, so I guess you could call it a theory.

And yes, there's definitely a 'Woah dude, thought I was going to die'
factor in estimating slope angles.

Chz
John

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In order to truely deturmine this I believe some mid-level college
physics/engineering are needed. You would need to deturmine all of the
forces that are acting on the uni.

I would start by including the weight of the rider and how it is
distributed between the two pedals and the seat. Also the answer
should be different for a stationary uni than for a moving uni.... Of
coarse crank length and wheel size are important factors.

Sounds like a great science project for a student!!

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icon said:

Now, my wheel is 26" and my crank 150mm (why, oh why, must we mix
units?). Allowing for tyre, that's probably radius 380mm for a ratio of
.4ish. By my reckoning the theoretical maximum angle would be 23-odd
degrees. That sounds quite shallow, but I think people tend to
over-estimate slope angles. On the other hand, I might be wrong. And
yes, I appreciate that there are technique issues not addresses here -
allowing the uni to accelerate, holding the saddle (allowing for an
effective 'over-body-weight' force, riding partially across the slope
and probably others I don't know about.

Note that 26" refers to an approximation of the outside diameter of the
tire, not the size of the rim. So you are overestimating the radius;
it's probably more like 350-360mm on a 26x3" tire.

I am certain that I have ridden down slopes much greater than 23
degrees. For that matter, on roads I have ridden up slopes that steep.
I have ridden down slops where portions of the slope were not visible
from the top of the slope.

The difficulty in calculating a theoretical maximum is that a unicycle
is not a railroad. There are two major effects which increase the
unicyclist's ability to descend steep slopes. One is that a unicycle
naturally "tacks" across the face of a slope, and this effect becomes
greater on steeper slopes. The other is that you don't necessarily
need to apply back pressure to go down a hill; for sufficiently short
sections, you can just spin through them and regain control when you
get to the bottom.

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Smeltz said:

In order to truely deturmine this I believe some mid-level college
physics/engineering are needed. You would need to deturmine all of the
forces that are acting on the uni.

I would start by including the weight of the rider and how it is
distributed between the two pedals and the seat. Also the answer
should be different for a stationary uni than for a moving uni.... Of
coarse crank length and wheel size are important factors.

Sounds like a great science project for a student!!

I'm excluding the distribution by taking the limiting condition of the
rider's entire weight being borne on the trailing pedal. (I think) the
rider's weight is neither here nor there, since it's a common factor to
both parts of the 'equation'. I grant you that 'pedals level' is the
optimum position for resisting the tendancy of the wheel to rotate -
indeed when the pedals are vertical, there's briefly no resistance to
rotation. But then, thinking about it, is the other part of the lever
actually the radius of the wheel, or is it the horizontal distance by
which the wheel's contact point on the slope trails the axle?

Is there a mathematician in the house?

J

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To put that 23 degrees into perspective, the steepest public road in the
world is supposedly Baldwin St in Dunedin,NZ, which at it's steepest
point is 19 degrees, although some people say Hardknott Pass in the UK,
is 1 in 2.5 in part, or 21 degrees. Baldwin St is rideable up on a 20"
(although by Ken Looi), but I dunno if anyone has tried it on a 26".

I've ridden down a road signposted 1:4, on the Schlumpf in high gear,
but I sure as hell wasn't keeping it at a constant speed, and spun out
and lost it after the second bend even with pulling up on the saddle.

I think I've ridden down steeper stuff on the muni, but only when it
was a short steep section that I could spin through, I presume here
we're talking about sustained descents, because we all know you can
spin out down stuff. I don't think I've seen anyone ride up anything
approaching 23 degrees without side hopping up it.

Joe

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joemarshall said:

To put that 23 degrees into perspective, the steepest public road in the
world is supposedly Baldwin St in Dunedin,NZ, which at it's steepest
point is 19 degrees, although some people say Hardknott Pass in the UK,
is 1 in 2.5 in part, or 21 degrees. Baldwin St is rideable up on a 20"
(although by Ken Looi), but I dunno if anyone has tried it on a 26".

Sorry, you're right, I'm thinking of 23%, not 23 degrees. I haven't
ridden up anything at 23 degrees, though I've probably ridden down
stuff that steep.

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I've ridden UP Hardknot on a bike, but I reckon it'd be pretty scarey
downwards on a bike. It's signposted 1:3, but there could be parts of
it that are steeper. There are a couple of roads around Widemouth Bay
in Cornwall that are signposted 1:3, and those certainly have some evil
steep bits on the inside of bends. I don't think I could ride either
of them on my muni (26x3 with 150s) - there's a half-mile long road on
my commute that's between 1:5 and 1:6 most of the way up, and I can
ride up (on a good day) and down that, but I don't think I'd feel in
control going down anything steeper (obviously apart from short
slopes). I'm too chicken to spin out downhill - I ride steep hills
slowly, which is probably harder on the legs but easier on the face 🙂

Rob

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icon said:

Is there a mathematician in the house?

I've recently graduated with a degree in physics and, if anything, have
learned that there are some situations in which one's first-order
approximation is no where near what occurs in reality. We could argue
(and many have on this forum) about which formula is the best to
represent the physical situation and likely only achieve a crude
approximation; even then, we'd be at a loss as to which numbers we'd
plug into the formula.

In addition to the phenomena mentioned by yourself and tholub, there's
also that of slippage. I, too, have been down slopes significantly
greater than 23 degrees and have been able to stay in the saddle with a
combination of sliding, tacking, handle pressure and acceleration.
There are obviously many variables left out of your approximation that
renders it virtually meaningless.

Even if you're looking for a simple approximation, what are you going
to do with the number once you get it? Lay down a protractor and take
a grade measurement every time to come across a new hill? :p

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hmm ... its so simple ... if the force pushing back on the rear crank is
greater than that of your body weight you are going to come off the
saddle, that is if you want to stay on the front pedal ... but then
momentum of of the spinning crank combined w/ centripital force must
come into calculation
and thats going perfectly straight down the hill 😕

wheel size would change it too, because the torque on the axle would
change; crank size would do the same (correct me if I'm wrong)

there are wayy too many variables, forget it; it doesn't matter because
skill and technique has a lot to do w/ it ... i have gotten down about
a 70* slope (it was a dune and I was sliding)

i didn't even include wheel slippage or tire spring or any of that
stuff

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skrobo said:

hmm ... its so simple ... if the force pushing back on the rear crank is
greater than that of your body weight you are going to come off the
saddle

This is why you need to hold the saddle. If you ever get to ride a 36"
Schlumpf in geared mode you will know this as you have to hold on
during acceleration. :-)

Roger

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skrobo said:

but then momentum of of the spinning crank combined w/ centripital force
must come into calculation and thats going perfectly straight down the
hill

what you say makes no sense...

the crank is rigidly attached to the wheel, its motion hasn't much to
do with the dynamics of this problem. the crank does not have a
momentum separate from that of the wheel; the crank affects the wheel's
moment of inertia.

centripital forces keep the crank from flying off the axle; the
direction of this force is always towards the axle, not straight down
the hill.

check out 'hyperphysics' site on circular motion'
(http://hyperphysics.phy-astr.gsu.edu/hbase/circ.html#circ) for an
explanation of basic rotational concepts...

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i should have explained my self differently

the torque on the wheel is diff than the amount of torque than the
crank, but it's directly proportional correct?

Quoted message said:

centripital forces keep the crank from flying off the axle; the
direction of this force is always towards the axle, not straight down
the hill.

umm, that doesn't make sense, isn't the force in this situation exactly
90* from the axle... ?

umm maybe this isn't as complicated as we are thinking

isnt it if potential energy ( the cicular factor of the wheel...or does
that have anything to do w/ it?, the slope degree, and weight)

if potential energy is greater than the weight/ holding strength of the
rider's hand holding theirself to the seat and the multiplication of
that by the mechanical diasadvantage that the crank/wheel size gives
us... you fall 🙂
did I get it right??

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-stephen-
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BLAH we took the wrong approach ... there is a triangle made by your
unicycle from the wheel to the pedal (we will go w/ a perfect
horizontal stance) and the axle ... it is a right triangle if we had a
perfect tire and perfect balance and coordination we could go down a
slope w/ about 65*

the sides of the triangle depend on the specific unicycle

the only factor deciding how steep of a slope we can go down is wheel
slippage

my triangle is incorrect I know ... its just an example ... the
ultimate way to go down the hill would have the unicycle slightly
twisted ... wheel touching on the side and pedal touching the ground

I give up ... ohh well i tried

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skrobo said:

the torque on the wheel is diff than the amount of torque than the
crank, but it's directly proportional correct?

Torque is rotational force. For a conventional unicycle, the wheel and
cranks are fixed together, so they make the same rotational movement
(when the wheel turns 90 degrees, so do the cranks). So both wheel and
cranks experience the same torque.

You're likely talking about force, and tangential force at that.

skrobo said:

isn't the force in this situation exactly 90* from the axle

Before you start calling out angles, you need to establish a frame of
reference. The axle is a radially symmetric object; you can rotate an
axle any amount of degrees and it will be nearly identical to an
unrotated axle. The significant forces in this problem are
gravitational (pointed towards the center of the earth) and
anti-gravitiational (pointed away from the center of the earth)... the
pedal exerts its greatest force on the rider in the anti-gravitational
direction.

skrobo said:

umm maybe this isn't as complicated as we are thinking ...
isnt it if potential energy ( the cicular factor of the wheel...or does
that have anything to do w/ it?, the slope degree, and weight) ... if
potential energy is greater than the weight/ holding strength of the
rider's hand holding theirself to the seat and the multiplication of
that by the mechanical diasadvantage that the crank/wheel size gives
us... you fall 🙂
did I get it right??

this isn't as complicated as you're making it out to be... that last
paragraph sounds like absolute nonsense. have you taken any physics
classes yet?

skrobo said:

I give up ...

me too... perhaps they should be teaching you kids intelligent design
after all... (just kidding)

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maestro8 said:

I've recently graduated with a degree in physics and, if anything, have
learned that there are some situations in which one's first-order
approximation is no where near what occurs in reality. We could argue
(and many have on this forum) about which formula is the best to
represent the physical situation and likely only achieve a crude
approximation; even then, we'd be at a loss as to which numbers we'd
plug into the formula.

In addition to the phenomena mentioned by yourself and tholub, there's
also that of slippage. I, too, have been down slopes significantly
greater than 23 degrees and have been able to stay in the saddle with a
combination of sliding, tacking, handle pressure and acceleration.
There are obviously many variables left out of your approximation that
renders it virtually meaningless.

Even if you're looking for a simple approximation, what are you going
to do with the number once you get it? Lay down a protractor and take
a grade measurement every time to come across a new hill? :p

Well, I tend to differ with you on this one Maestro. One thing I have
learned is that there is no point adding extra possibilities to
intentionally make a simply expressed problem difficult. RTFQ and
then ATFQ ! So K.I.S.S. and don't overengineer the problem.
Why add slippage, acceleration, tacking etc etc to the problem?
Reduce the problem to essential bare bones as per the post and it
really becomes very simple indeed, save for one factor not mentioned.

So for the laymen in here:

Adding in tyre slippage is way over the level at which Icon was asking
the question. Icon was on the right mathematical track in paragraph 4,
appropriate for the level at which the question was posed, and for
simple non slippy slopes it would give a pretty good answer too, *for a
stationary unicycle*.

However there is one major factor that does need adding. Once the
cranks are not horizontal they are less effective in keeping speed
constant (or in holding the unicycle stationary on the slope).

It is impossible to ride down a path having a uniform slope, at a
uniform speed. Speed varies throughout each revolution of the wheel.
To keep a constant overall speed , the moments when the pedals are near
horizontal have to be used to actually slow the unicycle down, because
in those times when the pedals are near vertical, you will not be able
to prevent some acceleration down the slope. So the maximum angle as
calculated in Icon's simple sin formula is inaccurate , in that it
would only apply to a stationary still stand situation.

You can stillstand on a much steeper slope than that which you can ride
down without increasing your overall speed.

Nao

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Naomi said:

You can stillstand on a much steeper slope than that which you can ride
down without increasing your overall speed.
Nao

that said, you could turn sideways and as long as the pedal doesn't hit
the slope on impact you can jump down a slope that has over 45* of
slope 🙂 (this is where you need wheel slippage factor)

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tholub said:

I am certain that I have ridden down slopes much greater than 23
degrees.

How did you measure the angle and how steep where those slopes?

Klaas Bil - Newsgroup Addict
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The factors cancel eachother out beautifully and you end up with a
maximum slope that is proportional to wheel to crank ratio.
That is assuming that the tyre doesn't ever slip, that you start from a
standstill on the slope and that your unicycle is weightless.

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