General Cycling Discussion - What is the steepest hill you been up?

The steepest hill I've ever been up is in Brunswick, Maryland. Althoung it is about 100 yards long, this thing must have a 21-23 degree gradient, I can barely get my car up it... and I don't know how I got my bike up it.

Once in Chicago, I started to ride straight up the side of Sears Tower. I would have made it to the top, too, except there were a lot of pigeons doing foul things on the side of the tower I chose for my assault.

Actually the toughest slope I ever did was on a ride up into the mountains outside of Salzburg, Austria. If I'd have known the angle was important to you I would have taken a protractor.

Not really... I mean who needs the weight of a protractor?

I've never been to Europe, but I bet some of those climbs have to be pure hell to get up. As for the protractor, I agree. Campy or Shimano could make some serious money if they came out with a carbon fiber one ;).

White Road in Spokane WA, about 1 to 1 1//2 miles mostly 20% with a pitch near top at 22%, YUK!!!

Ride Flatter
Pat

I live just half a mile away from this hill....steepest in SoCal......The famous Fargo Street Hill Climb takes place every March here. Stoker-less tandem riders do well on this climb.

http://img.groundspeak.com/cache/40054_300.jpg

Wohoo George...
I think with this kind of hill, it will not take long to break something...
chain, sprokets, spokes, strip the crank... etc... Only mountain bikes are somewhat meant to climb something like this. :)

Originally posted by pat5319
White Road in Spokane WA, about 1 to 1 1//2 miles mostly 20% with a pitch near top at 22%, YUK!!!

Ride Flatter
Pat I've been up White Road, and I thought I was going to tear a muscle in my chest! :eek: And if I recall correctly, that was with my ol' Fuji that had the 39 x 28 low gear, too. I don't think I've climbed anything steeper except off-road.

I've been up McEwen many times. They had the college finals there a few months ago. Kids 1/3 my age were installing XTR rear deraliurs and cassettes on for it.
Then there is the last 1000' at the top of Mt.Diablo that
would still be a killer if it were the first 1000'.

Officially 22% for 1.2km (Trees Road, Tallebudgera, Queensland). Unofficially there were some that were steeper in Brisbane's 10 hills ride a few years ago. I'm told some of them were 30-35%! :eek:

This is the steepest I have ridden.
http://www.multimap.com/map/browse.cgi?X=558000&Y=101000&width=500&height=300&client=public&gride=&gridn=&keepicon=true&coordsys=gb&addr1=&addr2=&addr3=&pc=&advanced=&scale=25000&up.x=18&up.y=5

It is much steeper than any alpine road Ive ever seen, just goes straight up. I was busting a gut with 28/28, and going down is pretty frightening.
There is a town in New Zealand with a grid road structure, just planted down into a steep vally by some civil servant working in London. The roads go straight up the vally sides at horricious gradients.

MichaelW, that Might be Dunedin where the wrold's steepest street is. (Baldwin St)

Brendon

1 in 2, but only for a few yards.

Originally posted by NZLcyclist
MichaelW, that Might be Dunedin where the wrold's steepest street is. (Baldwin St)

I've heard about that. It's one of my ambitions to test myself on that street when I eventually get around to touring in New Zealand. Isn't the gradient of that 45% or something?

Mmm, I rode up a 21% gradient at a time trial meet last year, damn lactic acid :(

Originally posted by George
The famous Fargo Street Hill Climb takes place every March here.

Yes, I have heard about that event from one of its veterans, Jim Baross.

When I lived in Los Angeles, my friend's 19% driveway in Benedict Canyon was the steepest short-distance climb I ever tackled. The longest, toughest climb was probably Tuna Canyon (Malibu hills, just north of Topanga), which is about 3.5 miles / 6km, with an average grade of about 12%.

My steepest climb was about 1 mile long with a gradient of 25%. I did it once , and I'll never do it again!

Mine has been Hogpen Gap on the Six Gap Century.
http://www.dahlonega.org/sixgapcentury/

This is what is on the website
"The toughest climb is Hogpen Gap. The Hogpen Gap climb averages 7% for 7 miles, with sections as steep as 15%."

That was Plenty for me!! But I will probably do it again this Year.
:D :cry:

I think I'm confused about the formula for calculating a grade percentage. I had thought it might be altitude gained divided by distance travelled, but that can't be right because if you travel 1 mile and climb 1 mile in altitude that would be a 45% climb, but the math I used wouldn't come out that way.

So, if you know how far you traveled and how high you climbed, what is the correct formula?

Bob

I had always thought that a flat plane was 0% grade, a 90 degree wall (straight up & down) for example, would theoretically be 100% grade. A 45 degree hill would be 50% grade, and a 22 1/2 degree hill would be 25% grade. Does that make sense?
The Fargo Street Hill climb which I mentioned earlier, is 30% grade, & a quarter mile long.

There is this one section of road on Santa Barbara Ca that I cannot remember the name of, (I think it was Jimeno Rd) that was about 75-80 yards long, that was so steep cars would burn rubber going up. I would take that road almost daily to head into the mountains, and as I cranked on the pedals every stroke would make the front wheel of the bike bounce about an inch off the road. I don't know what percent the grade.

I had always thought that a flat plane was 0% grade, a 90 degree wall (straight up & down) for example, would theoretically be 100% grade. A 45 degree hill would be 50% grade, and a 22 1/2 degree hill would be 25% grade. Does that make sense?

The grade calculation is the altitude change divided by distance traveled.

A level road is a 0% grade, but a 45 degree hill would be a 100% grade. An example - if your altitude changes 1000 feet over a distance traveled of 4000 feet, you would divide 1000 by 4000 and get .25, which would equal a 25% grade.

wow.... a 25% grade is only 14 degrees!!! An angle that looks so shallow on paper feels like a beast on the pavement!

Thanks for the Calculation. So that means that Hogpen has some sections that everyone here is talking about. I thought that climb was a bear, So It should be in line with everyone elses, The Key difference is that this climb Starts at Mile 63 and goes for 7 miles Then you still have 2 more Gaps to climb to reach the Century Mark. No wonder it was the hardest thing I have ever done!!!

Thanks, NW NJ Biker.......your formula makes much more sense....

The highest number I have seen on a sign is 16%, on the lower slopes on Passo Costalunga back in 1998.

If that sign was right, I have probably been up short stretches of 20+ %.

The steepest road over a longer distance is probably a forced detour due to construction work on Passo Costalunga in 2001. I haven't found a profile of that road (haven't even seen it on a map...), but I wouldn't be surprised if it averages 12% for 10 kilometers. Fortunately it was towards the end of a wonderfully hilly Alpine tour, which made it possible to climb it even with fully loaded touring bikes.

The steepest confirmed roads I have been up is Mont Ventoux (Bedoin side, 22k @ 7,1% (with 10k @ 10%) and Colle Agnello (Italian side, 22k @ 6,5% (with the last 9 k to the 2744m summit just below 10%). Both done during the same Alpine tour in 2001.

Unfortunately, we missed the 10k @ 11,5% Monte Zoncolan (with its six kilometre stretch at 15%) due to time constraints.

I am lucky to have one of Sweden's best hills nearby, but 2.2k @ 6% is nothing compared to the Alpine giants.

/Csson

Baker Street, San Francisco, near the Presidio.
What gradient? All I know is the sidewalk had steps.

Originally posted by NW NJ Biker
The grade calculation is the altitude change divided by distance traveled.

A level road is a 0% grade, but a 45 degree hill would be a 100% grade. An example - if your altitude changes 1000 feet over a distance traveled of 4000 feet, you would divide 1000 by 4000 and get .25, which would equal a 25% grade.

Hmmm, I guess degree would be more what I'd want rather than percentage then.

Is there a direct relation ship? I mean, if 45 degrees = 100%, then would 22.5 degrees = 50%, etc? If so, then I mentally figure out comparisons easy enough. If not, I'd need a formula and I don't know what it would be.

Bob

I know that road BentRox, I was in San fransisco with my family years ago and we drove down it and that was almost scarry I could not imagine even trying to ride up it.

Well I once walked up the main street in Haworth, Yorkshire
(Bronte Country). I know it doesn't sound like much but
as I recall the cobles were set at a 45 degree angle to the
hill so the horses could gain purchase.
Has to be the steepest hill I've ever seen even tho
it wasn't long. I'd hate to ride that one, makes me think
of the Koppenberg
see attached photo, which is copied from

cyclingnews.com (www.cyclingnews.com)

Originally posted by TheRCF
Hmmm, I guess degree would be more what I'd want rather than percentage then.

Is there a direct relation ship? I mean, if 45 degrees = 100%, then would 22.5 degrees = 50%, etc? If so, then I mentally figure out comparisons easy enough. If not, I'd need a formula and I don't know what it would be.

Bob

nope.... you have to use trigonometry to figure out the angle....

specifically, find the arctan of the ratio of rise over run.

so for the 100% grade, rise over run = 1.... arctan 1=45degrees

Mount Washington has to be the longest and steepest I've ridden.

Average 12% grade for 12kms, with the last kicker at 23%.

Originally posted by Chris L
I've heard about that. It's one of my ambitions to test myself on that street when I eventually get around to touring in New Zealand. Isn't the gradient of that 45% or something?

It's 38%. I drove up in a Mazda Familia hire car. It made it up without too much hassle.

The steepest hill I've done at this stage is Bumble Hill on the Central Coast. They used to do it in the Commonwealth Bank Classic. There's a couple sections which are like walls.

Originally posted by deliriou5
nope.... you have to use trigonometry to figure out the angle....

specifically, find the arctan of the ratio of rise over run.

so for the 100% grade, rise over run = 1.... arctan 1=45degrees

uh-oh. Never took trig. Seems like there should be a simple formula though. I'm picturing a right angle triangle where I know the vertical length (altitude) and I know what I think is called the hypoteneuse (the diagonal part - the distance I've actually ridden). I don't know the horizontal line of the triangle (it would be a bit shorter than the hypotenuse), but there should be an easy formula for that. And once all three distances are known, there should be a formula that gives all the angles. We know one would be 90 degrees. The other two would have to total 90 degrees, but I don't know what the formulas are.

Bob

Originally posted by bentrox!
Baker Street, San Francisco, near the Presidio.
What gradient? All I know is the sidewalk had steps.

We lived over on Sacramento Street, just down the street from the Presidio. I know exactly which hill you are talking about. It's a mother to just walk up the sucker. I couldn't even think about riding my bike up that hill.

Take care.

Steepest road climbs here are , I'd have to guess at 20% or more, but they are only short, 100m to 120m in elevation.

Someone tell me if this makes sense for determining both the percent grade and the degree angle of a hill. And let me know if there is a simpler way of doing this.

First, we are dealing with a right angle triangle. The slanted portion (hypotenuse) is the distance you actually travel going up. The vertical portion is the altitude gained. The horizontal portion would be how far you would have traveled if you were on flat ground (something less than the diagonal distance climbing).

I found a formula that said that adding the square of the two sides making the right angle (the horizontal and vertical lines) would give a number equal to the square of the diagonal:

"a" squared + "b" squared = "c" squared (where c is the diagonal line).

Since we would know how far we rode up ("c") and one way or another can find the altitude (we'll call this "a"), it is the horizontal distance we have to calculate which I did with this:

Get the answer of "c" squared minus "a" squared. Now take the square root of that answer. The answer is the vaue of "b" which was our unknow (horizontal line).

Now that I had the three values for the sides of the triangle, I needed to get either the degree angle between the diagonal and horizontal or get the percent value. Someone elsewhere said that to get the percent, you divided the vertical by the horizontal (a/b). Is that correct?

Assuming it is correct, I then took the value I just figured for the horizontal and divided it into the vertical figure, hopefully establishing the percentage of the grade. For example, an altitude change of one mile divided by a horizontal distance of 1 mile would give the answer of "1". In percentage terms, that would be 100%.

Then, to get the degree, I made some assumptions. Since a climb of 45 degrees is considered a 100% grade, I mulitplied the result by 45. Thus a grade percent (in decimal form) of 0.05 would be a hill of 2.25 degrees.

Or I could be totally wrong.

Bob

It happens to be a hill near my grandfather's home in North York (Toronto). I don't know how to tell the % grade but if I had a protractor at the bottom of the road the angle of inclination would be about 50degrees so do the math on that. The hill is a good 450metres long.

On Slickrock in Moab baby! Without a doubt!

If you've ever ridden slickrock, you know what i mean, and the kind of an "up" you can climb. It takes a few hours of playing around on the rock and gaining confidence to attempt the climbs that others before you have done. ...and you know that they have climbed it by the tire marks on the "trail." Unbelievable traction on that stuff.

Originally posted by WorldIRC
It happens to be a hill near my grandfather's home in North York (Toronto). I don't know how to tell the % grade but if I had a protractor at the bottom of the road the angle of inclination would be about 50degrees so do the math on that. The hill is a good 450metres long.

Well, if a 45 degree hill = 100% as some have said, wouldn't that mean a 50 degree hill would be 110%? Geez, this gives me a headache because that just doesn't seem to make sense. I would think a hill going straight up would be a 100% hill (and, of course, a 90 degree hill).

Bob

Originally posted by TheRCF


So, if you know how far you traveled and how high you climbed, what is the correct formula?

Bob

From Southern Illinois University:

http://www.siu.edu/~ilbmp/img26.jpg

One picture = one thousand words.

Originally posted by doctorspin
From Southern Illinois University:

http://www.siu.edu/~ilbmp/img26.jpg

One picture = one thousand words.

Actually, that picture doesn't deal with the information as I gave it because it gives the altitude and the horizontal distance. What I had to work with was the altitude and diagonal distance.

However, it does confirm that if you know the altitude and horizontal distance you divide the altitude by the horizontal to get the percentage (in decimal format). Since I think I have a formula to calculate that horizontal distance from the other two, that's good enough.

But it still seems strange that a 45 degree angle would be equal to 100%.

Bob

Originally posted by TheRCF
Then, to get the degree, I made some assumptions. Since a climb of 45 degrees is considered a 100% grade, I mulitplied the result by 45. Thus a grade percent (in decimal form) of 0.05 would be a hill of 2.25 degrees.

Or I could be totally wrong.

Bob

You're totally wrong ;)

.... by that reasoning, if you doubled the angle to 90 degrees, it would be a 200% grade.... but that can't be since 200% means 2 feet of elevation per 1 foot of horizontal distance.... that's only an angle of 63.4 degrees.

In fact, a 90 degree slope would have an INFINITE perecent grade
an 89 degree slope would have a 5700% grade
an 89.9 degree slope would have a 57200% grade
an 89.99 degree slope would have a 572900% grade

.... you get my point ;)

believe me.. you NEED trigonometry to figure out this one.

I'll post up a graph of percent grade vs angle in just a second :)

Here's a graph that shows the percent grade of all angles from 1 to 89 degrees..... notice how percent grade shoots up to infinity as it approaches 90 degrees.

Now here's a more useful graph showing angles more commonly seen in nature :) Notice more importantly that there isn't a direct, linear correlation between percent grade and degree measure... which was implied that the formula that you were trying to use.

Originally posted by WorldIRC
It happens to be a hill near my grandfather's home in North York (Toronto). I don't know how to tell the % grade but if I had a protractor at the bottom of the road the angle of inclination would be about 50degrees so do the math on that. The hill is a good 450metres long.

Hoggs Hollow Brian?

the steepest climb that I did was at Hamilton embankment, Ontario I thinks its. 10 degrees and about 20 percent grade. or maybe higher, this is the place where the World Road Cycling championship will be held this October 2003

Yeah, last night I was doing figures in my head and realized many unrealistic results!

> but that can't be since 200% means 2 feet of elevation per 1 foot of horizontal distance.... that's only an angle of 63.4 degrees. <

Wait a minute. I must be missing something again. What do you get if you divide the altitude change by the horizontal distance?

Can you give a formula I can plug into a spreadsheet that would allow me to get the DEGREE of slope angle?

And then a formula for getting the percentage?

I'm missing something - probably getting simple things confused.

Bob

Originally posted by Scooby Snax
Hoggs Hollow Brian?

No. Cummer and Bayview down in the valley. On Cummer, there is a HUGE hill. That is easily 50-55 degrees.

Originally posted by TheRCF
..it still seems strange that a 45 degree angle would be equal to 100%.

Your over-thinking this one, Bob. Slope is percent. Angle is degree. Think unit rise per unit run and then multiply by 100 - that is percentage slope.

That's all it is...honest! 1 unit horizontally per 1 unit vertically x 100 is a 100% slope (an angle of 45 degrees, if you must think this way.)

Yes, technically on a climb the traveled distance would be measured along the slope (or diagonal as you called it,) not horizontally. If you're calculating ballistic trajectories you certainly have to do real trig, but when bicycling, the difference in distance between that on the slope and that on the horizontal is just too insignificant to affect the overall result.

So stick with rise/run times 100 and bring peace to your well-being.

Mr. Wizzard

> Your over-thinking this one, Bob. <

I tend to do that (grin). Hey, sometimes it even works!

> Slope is percent. Angle is degree. Think unit rise per unit run and then multiply by 100 - that is percentage slope. <

So, using a different example, a really steep slope where you rise two feet, but only travel horizontally one foot would be a 200% slope?

If I have that right, then that is the easy math one. But how do you get the degree angle?

> Yes, technically on a climb the traveled distance would be measured along the slope (or diagonal as you called it,) not horizontally. If you're calculating ballistic trajectories you certainly have to do real trig, but when bicycling, the difference in distance between that on the slope and that on the horizontal is just too insignificant to affect the overall result. <

True, it seems to be a small difference. But I think I have a formula that will give me the horizontal distance if I have the other two figures, so I might as well use the better figure.

> So stick with rise/run times 100 and bring peace to your well-being. <

Ah, but I really would like to know the degree angle, although from what I can tell from the posted chart, for the hills we would deal with, the two numbers seem to be pretty close to the same, so maybe it doesn't matter that much.

Bob