If you used the difference in height over the actual distance traveled instead of the horizontal distance, then 100 % would indeed be vertical. I.e. instead of using sin(a)/cos(a) you could just use sin(a) and for small angles the values are close but the later one approaches 1 while the former one diverges as you approach 90°. I would probably have picked the later one but the former one is probably somewhat easier to work with in practice, e.g. for construction work.

> sin(a)/cos(a)

Also called tan(a) ;)

The other variant was actually meant to read sin(a)/1 to highlight the ratio in both cases but I abandoned that halfway through because I thought it makes the hypotenuse look special while it really is not in the sin(a)/cos(a) variant where the more natural description would be tan(a)/1 but that kind of ruins the simple association with the side lengths of a right triangle. I had another idea how to formulate it that would not favor one of the variants but as far as I can tell the necessary distinction between the two catheti that exists in German does not exist in English and so I left it as it is.

That's right. It's simple to remember once you get the hang of it. 100% grade: every foot forward is one foot of climb, or every meter forward is one meter of rise. 10% grade: ten feet forward, one foot up, or ten meters forward, one meter up. 110% grade: one foot forward, 1.1 feet up, or one meter forward, 1.1 meters up. 6% grades are steep enough to earn warning signs, if you pay attention to signage in the mountains.

Sonora Pass, for example, has sections of 26% grade. That sounds immensely steep (and is, for driving), but it sounds more reasonable when you say "every 50 feet forward, you've climbed 13 feet." Solve the triangle, and that's 14.574°. If you've been over the Sonora Pass, you know that it feels like you're in a rocket climbing vertically at a couple points, and you might be surprised to learn that's only 14.6° of grade.

22nd Street in SF is over 31%, IIRC, and is one of the steepest streets I can think of.

In my experience percent grade is usually used as a measure when people want to make something sound really steep.

E.g. "this ski slope has a gradient of 90%!!!"

That would, in fact, be an extremely steep slope. An 80% grade is steep enough that if you fall you do not stop until you reach the bottom.

http://www.foxnews.com/travel/2017/01/30/5-most-dangerous-sk...

More:

> In general, beginner slopes (green circle) are between 6% and 25%. Intermediate slopes (blue square) are between 25% and 40%. Difficult slopes (black diamond) are 40% and up. ...

https://en.wikipedia.org/wiki/Piste

That is very true. But my point still stands. People hear a percentage number and assume the max is 100.

Also generally with ski slopes the quoted gradient is some tiny portion of the run. Top to bottom the actual gradient is much less.

That is what an ice axe would be for, I presume.

The nice thing about it from an engineering perspective is that it doesn't distort at the ends of a scale.

A 10% grade over a horizontal distance of X will gain exactly twice as much height as a 5% grade over the same distance.

Percent grade is standard for road steepness at least in the US, e.g., 6% which requires effort for a loaded truck to climb.

So a 90° angle is... infinity%?

It’s related to a ratio. A ~6 degree slope would have you dropping one meter for every ten meters you move forward: 10%. 45 degrees means one meter for every meter forward: 100%. Keep plotting that out and it approaches infinity as you reach 90 degrees (or just ask, “for every meter forward, how many up/down?” if it is 90 degrees.)

The problem is that "forward" is ambiguous. You can measure it relative to your direction of travel, or perpendicular to the gravitational vector. For small gradients these are more or less the same, but the steeper the slope the more these two measures diverge.

Fortunately, as 110% it's not ambiguous anymore.

With slope in degrees and grade in percent,

100*tan(slope) = grade%

slope = arctan(grade%/100)

so 110% is about 47.7 degrees, and tan(90 degrees) is infinitely large.

Precisely. Math makes it clear to understand and settle down the discussion, thanks ;)

It would be, yes, basically.

If you have a 90° angle that would be called a “wall” not a “slope”.

If you plotted the altitude with a function f(x), it's just the derivative f'(x) expressed in percent.