- 25 Jul '11 21:19So 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? - 25 Jul '11 22:26

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)*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.

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 - 25 Jul '11 23:43

Thanks, that makes it a lot clearer. So it is an asymptote approaching 90 degrees.*Originally posted by forkedknight***Nice little chart from the wikipedia entry on Grade (road):**

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

So 9 percent is how many km up for 1 km of road? Just 9 percent of 1 km? 90 meters? - 26 Jul '11 00:34

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

Your definition would be more useful but it is not correct. - 26 Jul '11 00:37

Yes - approximately.*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?

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. - 27 Jul '11 18:00

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?*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.

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