Could someone explain % in a climb
 

I'm very new to cycling, starting this year at 44 and riding for fitness mainly but hoping to do some centuries and other club type rides. I've always enjoyed the TDF for what I think it is, a VERY demanding test of endurance, strength and mental capacity.

I know the % is a relation to length of climb and altitude gain. I was wondering at what degree of angle they are climbing for that distance? Is an 8% climb a climb of 8 degrees? Does this question even make sense to anyone?...lol. I told you I was new.

vertical is 90deg or 100%.

Rise/Run = % of climb.

i.e. 50 feet vertical over 1000 feet is a 5% grade.

incorrect

as the angle approaches vertical then % grade approaches infinity.

100% is 45 degrees. rise/run

I'm pretty sure it's calculated with run being the length of the road not horizontal distance covered.

I understood it to be every 1ft elevation gain in 100ft distance travelled = your percentage. In other words 1ft elevation gain in 100ft distance travelled = 1% incline. 2ft elevation gain in 100ft distance travelled = 2% incline, etc.

Is this incorrect?

That's the correct answer.

Check this out . . . all you could want to know:

http://www.geocities.com/sidestreetluge/grade.html

It depends if you use the sine function (opposite over hypotenuse. elevation gain over actual distance traveled) or if you use the Tangent function (opposite over adjacent. elevation gain over horizontal distance.)

In real world cases they come out very close to each other but in extreme cases they begin to differ a lot.

Wikipedia has a pretty good explanation. I think it is more complicated. The answer previously given that a 45 degree angle is a 100% grade is correct.

http://en.wikipedia.org/wiki/Slope

See the section on calculating the slope of a road/railroad:

http://upload.wikimedia.org/math/e/7...31073cb9cb.png http://upload.wikimedia.org/math/1/f...9c0a7bfc9a.png

Wrong.

This is an interesteing tid-bit from your site (http://www.geocities.com/sidestreetluge/grade.html) I did not know...


Q) Why use percent of incline vs. degree of angle?

A) The reason for using percent of incline instead of angle, is that percent gives a direct way to assess the effort required to move forward against the grade, whereas the angle in degrees does not readily reveal this information. For example; a 5% grade requires a forward force equal to 5% of the weight of the
object (above and beyond the force it takes to overcome surface resistance on flat ground at the same
speed).

According to the link previously given, they say

I didn't know that, either (about % gradient being tied to the force necessary to climb the slope). It gives me a new perspective on the importance of resistance training.

Strength is good!

I started to add you can get a hell of a lot more complicated about it, but simply rise/run works pretty well for cycling purposes

no that just means they're close for normal grades.

I'm pretty sure that's not true.

Wow, I had absolutely no idea it was this complicated. Makes sense if you think about it though.

It is usually calculated with the run being the length of the road, simply to make the math a little easier. For the angles on any sort of "normal" road though -- it simply doesn't make any signifigant difference to the final answer.

It's simple geometry really. I usually use the pythagorean theorem to calculate the slope, IE on a 1 mile , 500 foot climb

a^2+b^2=c^2
a^2+250000=27878400
a=5256

rise/run = b/a = 500/5256 = ~0.095 or 9.5% gradient

I was under the impression that percent grade was synonymous with gradient not just conveniently close around 0 which means that it would be calculated that way and the rise/run method would be the convienent approximation.

25% means get off and walk!

gradient is most definitely not calculated this way.

Ok could you correct me then?

no I'm wrong.

It's a rather complicated way of doing it, but it is most definitely valid. He's just calculating the "real" run (i.e., horizontal distance) covered by the climb, and using that. As I mentioned in my earlier post, the difference between the road distance (5280 feet in the original example) and the "real run" (5256 feet) is basically insignifigant (less then 0.5%). From a math perspective though, this is probably a more accurate method.

It's done that way because roads are designed by civil engineers looking at maps with elevations and horizontal distances in feet, and checked by surveyors who look through instruments that measure horizontal distances, not runs.

I think I need to add this again. It does not matter if you use actual distance travelled or the horizontal distance travelled and here is why:

Let "a" be the angle of your hill, "z" the hypotenuse (road), "x" the adjacent (run) , and "y" the opposite (rise).

For small "a"
sin(a)~tan(a), because cos(a)~1; If cos(a)~1 it follows that x~z.

And so in this case one can use the hypotenuse (z) instead of the adjacent (x).QED


BTW, the error will not change based on the length of the climb only the % gradient (except that the gradient changes more over longer climbs).

Let's assume a 30% grade (gnarly hill). Assuming the grade was figured out using rise over run (y/x), we have

y/x=0.3
y=0.3x

z^2=x^2+ y^2
substituting 0.3x for y we have
z^2=x^2+(0.3x)^2
z^2=1.09x^2
z=1.04x
x=0.96z

%grade=y/0.96z
0.96*%grade=y/z

so we are looking at a whopping 4% error even on a 30% grade. Not that big of a deal in my opinion.
So using the estimate we get 29% versus the actual of 30%.

All these other posters are wrong. It's actually the probability, in percent, that you will vomit during the climb. That is all.