# General Cycling Discussion - Is it possible to ride up 45 deg slope?

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joe99
09-13-07, 06:06 AM
Is it theoretically possible to ride up a 45 degree slope (without relying on momentum)?

Assuming a coefficient of friction of 1.0, it would require virtually all the weight to be on the driven wheel so that the front wheel would be on the verge of lifting off the ground. Does this mean it would be impossible to apply any torque to the back wheel to actually move the bike without back flipping?

Regardless of what any theories might suggest, has anybody actually done it?

I recall many years ago a car show where a Land Rover was driven up a 45 degree ramp to demonstrate its hill climbing ability so maybe it is possible. (The ramp was probably made of expanded metal or something similar so the coefficient of friction could have been greater than 1)

eubi
09-13-07, 06:43 AM
As long as the CG of the bike and rider is ahead of the rear axle, yes. Just be careful when you are starting. It's the acceleration of the bike that will impart an additional torque adequate to flip it over backwards.

mayukawa
09-13-07, 12:32 PM
Possible, yes...but at 45 degrees, it'll be faster if I carry my bike than have my bike carry me.

Steve Hamlin
09-13-07, 12:50 PM
Would you count this. . .?

http://vids.myspace.com/index.cfm?fuseaction=vids.individual&videoid=15272932

(Not my video, sorry about the quality. . .)

I went 'round three times on a single rider version at the Tech Museum in Silicon Valley. That was quite enough. As I told the folks operating it, "This like a bicyclist's bad dream . . . !"

AEO
09-13-07, 05:20 PM
you mean like these, only with pedal power?

http://www.pro-hillclimbers.org/new_en8.jpg
http://www.pro-hillclimbers.org/KP%20WIDOW%2085.jpg

They're stock dirtbikes with super-long custom made rears (duh) so they don't tip over 45+ degree inclines. I think some of them go 60+ degrees

Nermal
09-13-07, 10:06 PM
I've gone up 90 degree slopes. 45 is nothing.

Un huh.

c0urt
09-13-07, 10:34 PM
yes i have before there is a hill near my house that is rated at somewhere around 40-60 degrees, I have done it before on a road bike no less.

you have to lean most of your weight over your handle bars and pray your rear doesnt break loose if you are clipped in. most sane people either zig zag back and forth up that kind of incline or hop off and walk.

ken cummings
09-13-07, 10:39 PM
That franken-motorbike is the right idea. As one poster said, center of mass ahead of center of rotation. And the COR is the point where the rear tire touches the ground; the point you rotate around as you flip over backwards. At 90 degrees you would need a trike and a built-in winch. Thats' hoisting, not riding.

njm
09-13-07, 10:49 PM
Is it theoretically possible to ride up a 45 degree slope (without relying on momentum)?

Assuming a coefficient of friction of 1.0, it would require virtually all the weight to be on the driven wheel so that the front wheel would be on the verge of lifting off the ground. Does this mean it would be impossible to apply any torque to the back wheel to actually move the bike without back flipping?

I suggest you do a calculation like the one in another thread in this forum (http://www.bikeforums.net/showthread.php?t=342815) to see what kind of power that would require at a reasonable speed, say 5 mph ~= 2 m/s.

Of course, it doesn't really address your question of tipping over versus staying upright, but I figure it would take the typical 75 kg (165 pound) rider about 1000 W on top of what he usually puts out while riding at that speed.

So I think the answer is that to sustain that rate of ascent, you have to really dial it up.

GRedner
09-16-07, 04:27 PM
yes i have before there is a hill near my house that is rated at somewhere around 40-60 degrees, I have done it before on a road bike no less. Are you sure it isn't 40%-60% grade instead? That would make a lot more sense. 60% grade is just over 30 degrees.

rm -rf
09-16-07, 04:52 PM
You need a bike like this one (http://www.pbase.com/coaster/image/58055849) used on the Fargo Street Hill Climb. Of course, that's "only" a 33% grade (45 degrees would be a 100% grade). Check out the 60+ tooth cog in the back, and tiny chainring. Gotta spin up the hill with this one!

Other riders used a tandem with one rider to get the long wheelbase.

It would be hard to balance on a 100% slope, since you would have to ride very slowly.

TomM
09-16-07, 06:35 PM
Un huh.

With the slope rotated 90 degrees.

oopfoo
09-16-07, 07:34 PM
It's obviously a joke. 90 degrees FROM THE VERTICAL.

markjenn
09-16-07, 08:06 PM
I think many of you are confusing percent-grade with degrees of slope. A 45-degree slope is a 100% grade, which is almost ridiculously steep - even the most steeply pitched roofs are seldom 45 degrees. A 45-degree slope is one in which you would probably not be able to arrest yourself if you fell off the bike - you'd simply tumble down the grade. 100% grade slopes are for mountain climbers with ropes and carbiners.

At about 20% grade, most riders start to have serious difficulty on anything but the briefest grades. I don't know about theoretical issues of CG and traction, but riding up a 100% grade for any length of time is out of the question.

- Mark

dynaryder
09-17-07, 07:43 AM
I know of a road that goes up a hill into a developement,and if it's not 45deg,it's pretty darn close. Not fun to climb,but doable.

KevinF
09-17-07, 03:57 PM
Are you sure it isn't 40%-60% grade instead? That would make a lot more sense. 60% grade is just over 30 degrees.

Considering the steepest street in the world (http://en.wikipedia.org/wiki/Baldwin_Street) is at a 35% grade (19 degrees), I think it's safe to say that there is no paved road that's anywhere close to 30 degrees.

KevinF
09-17-07, 04:07 PM
I think many of you are confusing percent-grade with degrees of slope. A 45-degree slope is a 100% grade, which is almost ridiculously steep - even the most steeply pitched roofs are seldom 45 degrees. A 45-degree slope is one in which you would probably not be able to arrest yourself if you fell off the bike - you'd simply tumble down the grade. 100% grade slopes are for mountain climbers with ropes and carbiners.

I'll agree that a 45-degree slope is heart-stopping steep... I've skied a couple pitches like that, and trust me, it is frightening to look down an extended grade like that. Falling is definitely a really, really bad idea, as you won't stop.

Hiking up a steep grade like that... It would depend on what's underfoot. Tuckerman and Huntingon Ravine (on Mt. Washington in New Hampshire) are that steep, and people hike them all the time without protection. Then again, both have thousands of big footholds. If there are lots of trees and vegetation to hold onto -- you're fairly safe. If it's an exposed granite boulder that you're trying to walk up -- get a rope. ;)

ivegotabike
09-17-07, 07:10 PM
i realy want to mesure some hills in columbia

ivegotabike
09-17-07, 07:12 PM
I dont see how everyone is freaking out about WALKING up 45 degree slopes, thats crazy, maybe you are all a bunch of old fogies

AEO
09-18-07, 08:24 PM
I dont see how everyone is freaking out about WALKING up 45 degree slopes, thats crazy, maybe you are all a bunch of old fogies

45 degree is like... triple black diamond in terms of difficulty for a ski slope... Black diamond goes up to 40 degrees. Now if you go over 40 degrees that falls in the category of extreme skiing.

45 slope, probably I'd tie a safety rope to myself when walking down or up it.

ivegotabike
09-19-07, 07:08 AM
Im not a skier. But i mean there is equal vertical and horizontal movement in a 45 degree slope. Its not that huge a deal.

chipcom
09-19-07, 08:16 AM
Im not a skier. But i mean there is equal vertical and horizontal movement in a 45 degree slope. Its not that huge a deal.

Cool, so you can show us where that 45 degree slope that you climb is, using Google Maps?

bac
09-19-07, 08:22 AM
Would you count this. . .?

http://vids.myspace.com/index.cfm?fuseaction=vids.individual&videoid=15272932

You made my knees hurt. :eek:

That's pretty cool though. I wanna do it!

StephenH
09-22-07, 05:45 PM
For practical purposes, the force you can exert on a pedal is not too much over your weight, which limits the amount of torgue you can generate in the pedals. Depending on your gearing, you could simply not have enough force in the pedals to move forward. (Envision doing this on a coaster-style bike). This could be overcome by by using different gearing from the norm. So you get into the question of "could it be done" versus "could I do it on my bike".

Similarly, with the problems with angles and weight shift- you COULD build or modify a frame specifically for this purpose, but whether an off-the-shelf bike would work is a different issue.

I was looking at some videos of the world's steepest roads that I found on here. I noticed on one, that when the guy went up it, he had to zigzag back and forth to do so.

As to the 45 degree slope in general... Any slope looks a lot steeper after you get on it than before. If you have a fairly smooth surface at a 45 degree slope, you could quite likely walk up it. But, one slip, and you're into an uncontrolled fall. You hear about mountain climbers and hikers "falling" and never "sliding/tumbling down a steep slope". That's because when they slide and tumble down a steep slope, it's simply counted as a fall. They do make special sticky climbing shoes that would help walk up this kind of slope.

One other point. A bicycle increases efficiency over walking by letting you go farther with the same amount of effort. On a very steep slope, practically all of that effort just goes into raising the weight, and that's true whether riding or walking. So even if you can ride up, it's not necessarily any faster than walking up (plus, you have to move your bike up the slope, too). Of course, coming back down is a good bit faster. I used to hike up a very steep access road near my house in Colorado, and on occasion, I have walked past mountain bikers who were pedaling up.

ivegotabike
09-22-07, 08:25 PM
Im not saying that i ride any 45 degree roads, im saying i would like some perspective.

StephenH
09-23-07, 09:38 AM
One way to approach it:
Assume a weightless frictionless bicycle.
Assume a person standing on the pedal, with crank arm horizontal, with 170 mm crank arm.
Assume a 26.00" OD tire.

Let's say the bike moves 1" along the slope. It will move 0.707" up. The rider standing on the pedal has to move at least 0.707" down relative to the bicycle. In 1", the bicycle wheel rotates 1"/13" = 1/13 radian (4.407 degrees). The pedal moves approximately 0.707" x 25.4mm/in /170mm = 0.1056 radians = 6.05 degrees. So you have to have gearing such that your pedals rotate 6.05/4.4 = 1.375 times each time your wheel rotates. IE, your rear sprocket has to be 1.375 times as big as your chain ring. This is for a weightless frictionless bike, and also neglects the average dropoff in force due to pedals not always being at the horizontal position. In real life, you'd need a ratio somewhat greater than that, maybe 50% or more higher.

ivegotabike
09-23-07, 12:04 PM
or maybe you just need to be able to exert a force greater than your own weight on the pedals...pulling?

markjenn
09-23-07, 12:44 PM
In practical terms, no, you can't ride up a 45-deg slope. It's absolutely ridiculous. I doubt anybody could ride up much more than a 30-deg slope and even that would be super tough.

In pure physics terms, it's an interesting problem - probably analogous to determining what the critical slope is - the slope where an object will just barely stay on the slope due to friction without sliding off.

Imagine a bicycle with its wheels locked sitting upright on a flat surface (ignore that the bike might fall over for now). Start tilting the surface slowly until the bicycle no longer can stay on the slope and slides off. This is the critical slope. As I recall, the critical slope is a very simple trig function of the coefficienct of friction. 45-degrees sounds about right as I think it is something like:

tan theta = coefficient of friction

tan 45 degress is 1.0.

All from memory, could very well be wrong and my math is too atophied to derive it.

- Mark

StephenH
09-23-07, 06:30 PM
Revised approach- still assume a weightless, frictionless bicycle, with a person standing on the pedal.
Assume the pedals make half a rotation. That means the weight of the rider has dropped 340mm relative to the bike. So assume that amount of motion also moves the bicycle 340mm vertically upward, or 481 mm along the slope. That corresponds to 481/(13"x25.4 mm/in) = 1.456 radians or 83 degrees of motion. Meaning the pedals must rotate 2.157 times for every time the wheel rotates. The difference between this and the first approach is that it accounts for the variation in torque due to rotation of the crank arm (using average torque instead of maximum torque).

"or maybe you just need to be able to exert a force greater than your own weight on the pedals...pulling?"

If you know what that force is, and if you know the weight of the bicycle, you can factor those in using the same method. Whether you can actually do this depends on the rider/frame geometry and the strength of the bike. You'll still wind up with a minimum gear ratio that you need to move up the slope, and it will still quite likely be greater than that used in normal bikes.

The coefficient of friction is the frictional force divided by the normal force. For a mass on a slope, the normal force is W x cosine of the slope angle, frictional force is W x sine of the slope angle, so the coefficient of friction becomes W x tangent of the slope angle, or 1.0 for 45 degrees. I would suppose this could be done, might require a rubbery surface. If all else fails, use a rack system like on rack railroads.

It should be possible to test this, if someone wants to slowly ride their mountain bike up one of those skateboard ramps until they stall.