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more on percent grade(6 posts)

more on percent gradeET
Jul 18, 2003 11:09 AM
I had a few comments to add to the percent grade thread that has now dropped off the page. (I had trouble logging in since it's been a while. BTW, hi, everyone!).

As was stated in that thread, percent grade is defined as rise over run, i.e. the vertical distance traveled over the horizontal distance traveled, or in math terms, the slope. It is not taken to be the angle formed (or equivalently, the percent of 90 degrees), although I wish it were, as that is somewhat easier to comprehend.

It was suggested that one could use the road distance (the hypotenuse of the triangle formed) as a proxy for run, which would be convenient since run is often unobtainable, whereas road distance can be obtained from your bike's computer readout. Doug Sloan gave a link to an interesting chart he constructed revealing that the error in using distance traveled instead of run is small, even for rather steep grades. I just wanted to elucidate on that a bit. It turns out it's not so mysterious or coincidental.

Rise over run is simply the tangent (side opposite over side adjacent) of the angle of the triangle formed in the ascent up the hill. Rise over road distance is merely the sin of the same angle. The reason the two are close is that for reasonably small values, sin and tan are close to each other. This can be seen by looking at the Taylor series expansion of the two:

sin(x) = x – (x^3)/3! + (x^5)/5! – x^7)/7! + .

tan(x) = x + (x^3)/3 + 2(x^5)/15 + 17(x^7)/315 + .

It is worth noting that this shows that when tan(x) (i.e., rise over run) is positive, it always exceeds sin(x) (i.e., rise over road distance), not surprising since tan(x) = sin(x)/cos(x) where cos(x) is less than 1. If you truncate each of the above series at any term, there are known formulas bounding the error in terms of the first term discarded, so we could get a nice upper bound on the difference between the two series as well if we wanted to. In any case, here's something very practical: even for steeper grades where the error between the two grows, given rise over road distance you can anyway find (well, when you get home) rise over run: rise/road distance = sin(x), so x = arcsin(sin x), and now that you have the angle x, you can find tan(x). An Excel spreadsheet has all the built-in functions (e.g. ASIN) to handle this.

As an example, suppose you have found rise/road distance to be .14834. Then arcsin (.14834) = .14889 (confirming the well-known adage that x and sin(x) are close in value for small x, something obvious from the Taylor series expansion). Then rise/run = tan(.14889) = .15, i.e. a 15% grade. The error using road distance instead of run is .15-.14834 = .00166. Incidentally, the angle .14889 translates into 8.53 degrees.

It should be pointed out, of course, that due to undulating road conditions, road distance will always be a bit longer than the actual hypotenuse of the triangle used in the calculations, so that adds a bit of error.
re: more on percent grade03Vortex
Jul 18, 2003 11:17 AM
Thnaks but too complicated for me. I think I will just suck it up and accept whatever the road has to throw at me. Sometimes, the less we know the better.
A fellow engineer?Alexx
Jul 18, 2003 11:59 AM
Or perhaps a math professor? Gee, the lat time I saw a Taylor series was either in Physics 3, or in Claculus 2.

From what I can remember, taking the series to 3 iterations chould be accurate enough, therefore:

sin(x) = x – (x^3)/3! + (x^5)/5! – x^7)/7!

tan(x) = x + (x^3)/3 + 2(x^5)/15 + 17(x^7)/315

is quite suficient.
no way!!ET
Jul 18, 2003 12:23 PM
Pure mathematician here. So if you need to know, say, how much milk a cow makes in an hour, well, let's see: assume it's a perfectly spherical cow...

Anyway, I now work as an actuary (gotta pay the bills, you know :-)).
Hey!ET_SoCal
Jul 18, 2003 3:08 PM
ET, you made me change my login a few years ago...
I use to be "ET" on mtbr.com, but now I've changed.

Bike content? How about I'm going to be purchasing my first road bike next month and I've decided on:
1) Steel frame
2) Triple crank (What can I say? I'm a dirty rider)
3) 105 components or better.
So what am I looking at? Something two or three years old for around 1K ?
Hey!russw19
Jul 18, 2003 7:35 PM
I can get you a new LeMond steel frame w/ 105 for that.

For a grand and a year or two old, you should be able to find an Ultegra bike with an 853 frame and some nice wheels. Shop around.. the deals are there, you may just need to find them. Also, '04 bikes are starting to come out.. if you LBS has 2003's in stock, they are gonna start to get ancy about moving them.

Russ