Simple question about percentage of incline:

So in the just finished Tour de France in some of the last stages in the mountains, they talked about a hill climb of 9 percent. I assume horizontal, neither going up against gravity or downwards, would be zero percent climb, I expect. But what is 100 percent? Straight up, that is to say, 90 degrees from horizontal?

If so, then 50 percent would be 45 degrees? 5 percent = 4.5 degrees? 1 percent=0.9 degree? 1 percent = 54 minutes? 9 percent climb = 8.1 degrees?

If so, just what is the utility in that where they could just say, it's an 8 degree climb?

Are we supposed to be so dumb we can't figure out what 90 degrees is?

It's the ratio of vertical distance to horizontal distance.

So, for example, if you ride up a 10% grade for 10 miles, you climbed a total of 1 vertical mile.

Originally posted by Anthem
It's the ratio of vertical distance to horizontal distance.

So, for example, if you ride up a 10% grade for 10 miles, you climbed a total of 1 vertical mile.

you would have to use a bit of trig if you were measuring distance travelled as i believe the ratio you want is (vertical change)/(horizontal change)

so the flat is 0% and a 45 degree incline is 100% (not 50😵

a cliff face has infinite % incline!

in the UK the 1 in X system is used which I believe is more user friendly for drivers

Nice little chart from the wikipedia entry on Grade (road):

http://en.wikipedia.org/wiki/File:Grades_degrees.svg

Originally posted by wolfgang59
you would have to use a bit of trig if you were measuring distance travelled as i believe the ratio you want is (vertical change)/(horizontal change)

No, he is correct, it is the ratio of elevation change to distance traveled.

Originally posted by forkedknight
Nice little chart from the wikipedia entry on Grade (road):

http://en.wikipedia.org/wiki/File:Grades_degrees.svg

Thanks, that makes it a lot clearer. So it is an asymptote approaching 90 degrees.

So 9 percent is how many km up for 1 km of road? Just 9 percent of 1 km? 90 meters?

Originally posted by forkedknight
No, he is correct, it is the ratio of elevation change to distance traveled.

Going to that wiki link that you recommended shows that 100% is 45 degrees, which confirms that the ratio in question is 'vertical change':'horizontal change' and not 'vertical change' : 'distance travelled'

Your definition would be more useful but it is not correct.

Originally posted by sonhouse
Thanks, that makes it a lot clearer. So it is an asymptote approaching 90 degrees.

So 9 percent is how many km up for 1 km of road? Just 9 percent of 1 km? 90 meters?

Yes - approximately.

To get the exact figure you need to use some tans & cotans or else Pythagoras. (see previous post)

I guess for gradients less than 20% you can use distance travelled instead of horizontal distance travelled to get a good approx.

Ah, yes, I looked it up, and I was misinformed. Thank you Wolfgang59.

Originally posted by wolfgang59
Going to that wiki link that you recommended shows that 100% is 45 degrees, which confirms that the ratio in question is 'vertical change':'horizontal change' and not 'vertical change' : 'distance travelled'

Your definition would be more useful but it is not correct.

But his statement, driving 10 miles up a 10 % climb puts you 1 mile in the air is wrong because 'driving 10 miles' is actually the hypotenuse of the resultant triangle? So that ten mile drive would actually be covering somewhat less distance of horizontal travel?

I guess you would have to do the a squared plus b squared deal to get the actual distances right, eh.