What do you mean?

Do you mean the average slope (or grade) of only the actual uphill bits? Ie, how steep is the road while you are going up?

If so, are you trying to answer the question via map data / elevation profile, or via data collected by a cycling computer (GPS watch, altimeter, etc)?

How precise are you trying to be?

In short, I don't think there is one "right" answer. Well, there is....but, you are unlikely to have the data needed to do it "right". So, it come down to how wrong are you willing to be? Which gets back to the purpose of the question. Are you planning/deciding if you should try a route (too steep or not enough?), or looking for bragging rights? ...or?

I mean like in the below example, 10 km ride generally uphill, from sea level to 1000m height - - - but with declines on the way up

Hello All :)

What's the right way to calculate the average gradient of an uphill route that includes parts with negative slope on the way up ?

Thanks!!

What part of "average" don't you understand? ;-)

I think that 1,000m / (total ride run, including declines) is both right and simple.
it takes into account the fact that in the declines one 'runs' but not contributing to gaining of height (the opposite)

I know it's just joking, but you go ride Mt. Mitchell. A 1/2mi and a 2mi descent midway alters the average from like 4.3% up to a little over 5% if you take those out.

It goes from "only 4%" to realistically a more difficult grade you're spending most of your time going up. Still not Alpine Europe or the Rockies, but still.

Yes, it is simple. And if the question you are asking is what is the average slope of the entire ride, then it is "right".

If the question is how steep is the road while I'm climbing, then your approach will UNDER-estimate the uphill slope. The degree of error will depend on how much descending actually occurs. In your example:

1000 / 10000 = 10% average grade

What if those two descents are 100 m each? In that case, you climb 1000m, descent 200m, then climb 200m. So, you've climbed 1200m in that 10km. That's a 12% average grade. But, you also were only climbing for 80% of the total distance, or 8km. So, 1200 / 8000 = 15% grade for the uphill segments. So, when you are actually on your bike and going uphill the difficulty level you feel will be the result of a 15% grade---that's considerably different from 10%.

So, what you're saying is you want to know what the average is of ONLY the uphill portions...

So, what you're saying is you want to know what the average is of ONLY the uphill portions...

NNAG: Non-negative average gradient.

WAG: Weighted average gradient, where weighting could be based on time, TWAG if like.

NNWAG: Non-negative weighted average gradient.


The possibilities are endless :D


As for plain old average gradient....

Average Gradient = Rise / Run

Rise = (Altitude end - Altitude start)

Run = [(Distance travelled along road)^2 - (Rise)^2 ]^0.5

e.g.
Altitude start = 150 m
Altitude end = 483 m
Distance travelled along road = 3680 m

Gradient = (483-150) / ([(3680)^2 - (483-150)^2]^0.5) = 0.09086 or 9.09%

It's a reasonable approximation for most angles of inclination encountered on roads to simply use the distance traveled along the road as the Run value (since there's not much difference between the values of the sine and the tangent at small angles of incline). It'll very slightly underestimate gradient. For very steep inclines you really need to calculate the horizontal run though.

e.g. (483-150)/3680 = 0.09048 or 9.05%

Can you elaborate on what the "run" calculation is taking into account and why it is needed??? I'm not following at all why the second, simpler calc of 9.05% is not factually correct if we are calc-ing "average grade".

Because gradient is defined as rise/run. The distance traveled along the road is longer than the run. Distance traveled is the same as the hypotenuse of a right angled triangle, while rise and run are the vertical and horizontal distances respectively.
As long as there is a non-zero angle of inclination, then Run < Distance traveled.