Bit geeky, but hoping someone with better maths than me can sort this out.

Yesterday, for the first time, I rode down a slope steep enough that I couldn’t stay in the saddle. The force on my trailing foot was sufficient that I was bodily hoisted up in the air. And dumped, unceremoniously.

On previous occasions, I’ve ridden down lesser slopes, where I’ve maintained control, but have thought that for any combination of wheel size and crank length, there must be an angle limit that can’t be exceeded using body weight alone (in other words you’d have to physically lock yourself in the saddle using your arm).

When you’re riding down a slope, your trailing foot is on the short end of a lever, formed from the radius of the wheel and the crank. My initial thought is that the ratio of crank to wheel radius has to be no less than the sin of the slope angle. At zero degrees (sin=0), there’s no component along the ‘slope’ (because it’s level) hence a crank length of zero is sufficient to resist. At 90 degrees (sin=1) the crank would have to be the same length as the radius of the wheel.

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.

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.

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.

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!!

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.

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?

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.

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.

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

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?

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

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.

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.

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

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??

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

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.

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.

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?

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

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.

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)