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Gradient vs Incline Discussion Revisited...(11 posts)

Gradient vs Incline Discussion Revisited...funknuggets
Dec 30, 2002 11:49 AM
I know this comes up from time to time, but DougSloan's recent ride report indicated the last 1/2 mile of one climb was at 18%. Im confused. Im looking at the schedule of climbs for the 2003 TDF and the steepest climb... the Côte de Larrau in Stage 16 is 2.4 Km at 10.5%. Im a bit confused. How are these measured?

Are these measured as if they travelled in a straight line and they measure the elevation at the beginning and then at the end, and then measure the total distance travelled on the road and treat it as the hypotenuse and then figure the angle? At one point, he climbed 1500 feet in 6 miles, right? So, you would convert miles to feet and then you could find the distance for the base by squaring the sides, subtracting the squared rise distance from the squared distance traveled and then getting the square root of the remainder, then setting as a ratio... and coming up with a particular degree angle of this climb of 4.07%... or a percent of grade of just over 7%. Which is used the degree or % of grade?

Because, just for fun, the Col de Tourmalet is reported at 17.1 km at 7.4%.

What do they use for the tour, the degree, or percent of gradient? And from a climb perspective, did DougSloan just ride the equavalent of the plateau de Bonascre (uphill finish of next year's stage 13 9.1km at 7.2%)?

Thanks.

Chris
google search...C-40
Dec 30, 2002 12:22 PM
The web often has all the info you could ask for.

A google search under "percent grade" produced this:

http://www.howstuffworks.com/question380.htm
but if DougSloan...funknuggets
Dec 30, 2002 1:10 PM
I guess my main question is what they use as the hypotenuse in the Tour de France calculations. The website you provided said the rise over the run, using the hypotenuse as if a straight line was drawn from the bottom to the top, which did not explain, if the road had 20 switchbacks as opposed to 3, your distance travelled or hypotenuse will be much different. Either way, I just want to know explicitly to know whether DougSloan is a stud or not... haha.

Chris
of course I'm a studDougSloan
Dec 30, 2002 2:34 PM
There are several things to consider.

First, percentage is nothing more than rise over run (as ridden). If a hill is 2500 feet over 6 miles (31680 feet), then the AVERAGE grade is 7.9%.

I'm not aware of many hills that have a constant grade. So, parts will be steeper than others. Some people loosely refer to a hill's steepness by it's maximum grades here and there. So, while the hill in question is about 8% over all, and parts are 18%, even in that 18% 1/2 mile section parts will be flatter than others. Heck, though, hit a switchback, and depending upon where you are in the turn, it could be well over 25% for a while.

If we are demanding completely precise references to where we ride, I doubt many could report figures that rigorously. However, reporting over all figures and some steep parts within gives a pretty good idea of what you face: the total climb and the worst parts.

Note that in my post I said, "It goes from 2300 feet to 4800 feet in 6 miles, and finishes you off at the top with 18% grades for 1/2 mile." Does that literally mean that the last 1/2 mile is exactly 18% the entire time? Heck, I don't know. However, I have spent an enormous amount of time running every hill around here, and in fact, every ride I've ever done, through Delorme Topo USA and have studied topo maps for all parts of my routes. Then, I pay attention to what's going on when I ride, so that when I'm having to stand and grunt out a climb when geared at 35 gear inches, I can tell you pretty close what that grade is. I've done this a lot. I'm sure I'm not the only one, either.

The average grade reported by the Tour or whatever pro race is likely merely the net climb over the run (as the road goes). That can be very misleading as to the real difficulty, like the Angliru in the Vuelta. While the average for the Angliru was something like 10%, which is very doable, with parts at 25% those sections can destroy you and more readily separate riders, even if the 25% sections are pretty short.

All that said, I'm certain that most everyone here, given the right gearing, could climb any hill I've ever done. Even a 20% hill can be climbed with a 30 inch gear fairly readily. The speed at which this is done is another thing.

So, what was your question?

Doug
L'Anglirucyclopathic
Dec 31, 2002 9:35 AM
avg grade for climb was 9.6% for entire run with steepest part reported at 23.6%. There was a mile at steepest with avg grade ~18%. Nothing to sweat about ;)

CP

PS btw do not take DeLorme data for gospel. Sometimes it is very accurate (I had centuries where diff btw reported data was less then 4%). BUT the biggest disappointment was when I ran through it Skyline drive. It came with 17,500' northbound / 19,000' southbound, when the park service reports it at 10,600/11,200.
L'Angliru - one bad@ss hillDougSloan
Dec 31, 2002 10:00 AM
There is no doubt that is one tough hill, and thrown in at the end of 100 miles or so in the rain on stage 15 -- well, that has to reeeeallly hurt.

I realize Topo can be off. Lots of times, the routing does not actually follow the road closely, so it may think the road dips down a little valley or rises over a hill where the road does not even go. In a "micro" sense, I wouldn't trust it without a strong reality check.

Some areas you can check, though. If you know the base peak elevations for a hill, as well as the run (according to your bike computer), then it's pretty easy to compare.

I saw a little bubble device that gauges incline that you mount on your handlebar in Velonews. Can't find it on the internet, though. That would be fun. I'd bet that you'd find that lots of hills were not what you thought.

Nonetheless, the bottom line is that all this yearning for accuracy really doesn't get us much of value. Who really cares if a hill is 18% or 20%? If we're in the ball park, that's all that matters.

Doug
Switchbacks have nothing to do with it!Kerry
Dec 30, 2002 4:56 PM
You take the vertical distance gained and divide by the distance traveled on the road. If you wanted to be more precise (anal?) you could project the pure horizontal distance of the road and divide that into the vertical rise, but the difference in the % grade would be in the fourth significant figure for any "normal" road grade. IOW, grade of 100m rise in 1 km distance could be either 10% using road distance traveled, or 10.05% using projected horizontal distance. Even a 20% grade would only be 20.4% if you used the shorter projected horizontal distance. Try this with the Pythagorean theorem, the math is REALLY simple.
Switchbacks do matter...HillRepeater
Dec 30, 2002 7:04 PM
Since the 'run' is the distance as ridden, and switchbacks greatly increase that distance as opposed the a straight accent - it impacts the average grade.

Say you've got a cone shaped hill that's 200' high with a radius at the base of 1000'. Build a road straight up it, and you'll get an average grade of 200/1000 or 20%. Now, change the road to one with switchbacks - to get to the top of the hill, you know have to travel over 2000' of road, reducing the average grade to 10%. That's the whole purpose of switchbacks in the first place - to reduce the grade.
the point of the switchbacks...DougSloan
Dec 30, 2002 8:28 PM
was that when you are actually on a switchback turn, the inside is much steeper than the outside; that's all; like this:



Doug
You missed the pointKerry
Dec 31, 2002 4:24 PM
The questioner asked whether switchbacks made a difference in grade. You're working the numbers backward. My point was that it is vertical rise over distance traveled. Of course switchbacks make the distance traveled longer - that is done to reduce the grade. The calculation of grade doesn't care how the distance traveled is engineered into the road (long traverses or zig zag of switchbacks straight up the hill). The questioner implied that the existance of switchbacks would somehow change the calculations, which it does not. The calculation is vertical rise divided by distance traveled, expressed in %.
re: Gradient vs Incline Discussion Revisited...CHRoadie
Dec 30, 2002 2:29 PM
It's very hard but not impossible to ride up a 7% average road for an extended period of time. Most of us just don't do it at 15mph like Lance.

You'll usually see phrases like "average grade", which means that they took the altitude difference between the start and stop of the road they were measuring and divided by the distance travelled during the climb (as opposed to the "crow flying" distance).