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Computing hill gradient(8 posts)

Computing hill gradientride_hard
Aug 21, 2003 6:41 AM
How do you come up with a percentage when describing a hills gradient. I found one formula:

altitude difference (top - bottom) divided by the lenght of the hill x 100%.

Is there an easier less formal way?
That IS the easy way...PdxMark
Aug 21, 2003 6:57 AM
It is an approximation (a pretty accurate one, actually, for road grades) that lets you avoid the trigonometric calulation to get the "right" gradient.

Or you can get a bike computer that does it all for you...
slope = (y2-y1)/(x2-x1) * 100 to get % grade270bullet
Aug 21, 2003 7:00 AM
Simple wayjrescpa
Aug 21, 2003 8:17 AM
Here's how I do it - approx. Computer says .87 for a climb. Altimeter gained 400 feet from bottom to top. You need an altimeter to this this, by the way. .87 x 5,280 (feet in a mile) = 4594.

400/4594 = 8.7% average grade. Not exact but close enough.

I assume no liability for the accuracy of the above calculation. I took 3 semesters of Calculus 20 yrs ago and I got a c+ in Calc 3. These days I can't even help my daughter with her algebra equations. :)
Not quite....Alexx
Aug 21, 2003 10:10 AM
....since your equation seems to be using the surface distance, whereas a gradient is the rise over the horizontal run. Your equation will be further and further off the steeper the hill gets.

The only decent way to approximate the gradient with surface distance is to use a Taylor Series. Of course, this method is far from trivial.

You can read more about it here:

russw19 "more on percent grade" 7/18/03 8:35pm
actually, Not quite.... ETHoward
Aug 21, 2003 11:21 AM
(Note: I am ET, not the name that appears. I'm using a friend's computer, as I'm having trouble logging on under my own name after a hard drive crash yesterday.)

Taylor's series is not the only way to go. In case my post given in the link isn't clear enough, I'll try and be more concise here. In previous discussions it has been pointed out that if one uses the computer distance (i.e., the hypotenuse of the triangle formed when heading up a hill, with the rise and run being the other two sides) instead of run in the denominator of rise/run, that the percent grade calculated is still close to actual. I pointed out last time that this is not so mysterious, and has to do with the fact that for reasonably small angles (which ours usually will be), the tangent (rise/run) and the sin (rise/computer distance) are close, as can be seen from the Taylor series. However, you don't need to use that. In fact, using only the most easily obtained info available, i.e., the distance (as displayed on the computer) and the rise (as obtained from an altimeter or road signs), when you get home you can find the answer exactly no matter how steep the grade. It is simply

tan[arcsin(rise/computer distance)].

For explanation, since the the rise over computer distance is the sin of the angle, the arcsin by definition is just the angle giving that sin. So now we have the angle. Taking the tangent of this angle gives rise over run.

As an example, suppose you find rise/computer distance to be .5. Well, arcsin(.5) = .523599 (this is in radians; as a check, you can multiply by 180/Pi to get 30 degrees, and we all know from high school that the sin of 30 degrees is 1/2). Then tan(.523599) = .57735, the percent grade. As a check, this is indeed 1/SQRT(3), or tan(30 degrees).

In case you're moaning and groaning, in Excel this is just


So when you get home from your ride, replace the .5 with whatever you get for rise/computer distance and get the exact answer, with no additional error (i.e., other than that of your own measurements).
Only anal retentive mathematicians would do thisContinental
Aug 21, 2003 11:57 AM
For all practical purposes at actual grades encountered you get the virtually the same answer using rise/computer distance. If your rise over computer distance gives a 30% grade, going through all this nonsense gives you 31.4% grade. At lower grades the difference is even less. If rise over computer distance gives a 20% grade, the more exact method gives you 20.4%. If you're climbing 57% grades as shown in the example above, then trig is the least of your problems.

This is an example of why engineers exist. We are lazy, sloppy, and get things done to improve your life. Mathematicians are neat and exact but we'd be cold and hungry in caves if we took them too seriously
Just one small thing Mr Engineer........Synchronicity
Aug 21, 2003 7:30 PM
............PREPARE the flame-thrower:

If you're building a tunnel, digging from both ends and you neglect to use the most accurate calculations available to you out, of sheer laziness as you put it, you could find that your tunnel won't meet in the middle!! Wouldn't you look stupid then?!

How do you know what the error in the calculation is going to be until you've calculated it, eh? Saying "oh it's close enough" AFTER someone else has done the work just isn't good enough.
If you really are an engineer, I hope I never have the misfortune of using (or standing on anything) that you've /helped/ to "engineer".

Never mind the Maths & Engineering,
If it weren't for Materials Scientists there'd be no materials good enough to build your skyscrapers, your bikes, and your spaceshuttles, etc...........

Honestly, some of you engineers are so up-yourselves. You think yours is the only profession that the world couldn't do without.