KevinF

It's a rather complicated way of doing it, but it is most definitely valid. He's just calculating the "real" run (i.e., horizontal distance) covered by the climb, and using that. As I mentioned in my earlier post, the difference between the road distance (5280 feet in the original example) and the "real run" (5256 feet) is basically insignifigant (less then 0.5%). From a math perspective though, this is probably a more accurate method.

DiabloScott

It's done that way because roads are designed by civil engineers looking at maps with elevations and horizontal distances in feet, and checked by surveyors who look through instruments that measure horizontal distances, not runs.

Mouserue

lvleph

I think I need to add this again. It does not matter if you use actual distance travelled or the horizontal distance travelled and here is why:

Let "a" be the angle of your hill, "z" the hypotenuse (road), "x" the adjacent (run) , and "y" the opposite (rise).

For small "a"
sin(a)~tan(a), because cos(a)~1; If cos(a)~1 it follows that x~z.

And so in this case one can use the hypotenuse (z) instead of the adjacent (x).QED


BTW, the error will not change based on the length of the climb only the % gradient (except that the gradient changes more over longer climbs).

Let's assume a 30% grade (gnarly hill). Assuming the grade was figured out using rise over run (y/x), we have

y/x=0.3
y=0.3x

z^2=x^2+ y^2
substituting 0.3x for y we have
z^2=x^2+(0.3x)^2
z^2=1.09x^2
z=1.04x
x=0.96z

%grade=y/0.96z
0.96*%grade=y/z

so we are looking at a whopping 4% error even on a 30% grade. Not that big of a deal in my opinion.
So using the estimate we get 29% versus the actual of 30%.

2Tired2Shift

All these other posters are wrong. It's actually the probability, in percent, that you will vomit during the climb. That is all.

merlinextraligh

^^^^
Incorrect. For example Brasstown is only 21%, but the hurling odds are much higher.

SDRider

There is a road with 10% grade in Palm Springs running up to the tram station and all I know is I wouldn't want to ride up it on my bike. Coming down in a vehicle I had to have it in second gear to keep from riding the brakes all the way down.

lvleph

How long is it? If it is more than 1 mile it sounds like a great hill to climb. You should force yourself to do it. I love and hate hills.

CastIron

My cyclometer tells me it's best guess. How it gets there means little. Here's my interpretation:
2% False flat. Vexing at times.
5% Shift a few gears
10% Gotta work it.
15% This hurts.
20% Don't puke...really, just don't puke...oh gawd...
25% WTF? No chairlift?

dddave

there's a sweet hill near my parents' house with a 32% grade.

KevinF

+10 ! The only thing to add is that somewhere around 20% is when you need to start worrying about keeping your front wheel from popping an unintentional wheelie.

mrbubbles

Is there a tool where you can measure slope gradient at a point by placing it on the ground?

redfooj

inclinometer

in geology we use a brunton

ExMachina

Yes. As the angle of a hill increases past 45 degrees, the weight of an object will exceed itself.

Furthermore, as the % gradient approaches infinity, the weight of the object also approaches infinity.

Special Relativity explains this by the fact that bikes going up steeper and steeper hils are approaching the speed of light.

That's why good climbers age more slowly than the rest of us.

[Moderator: this should be a sticky, IMHO]

SDRider

It's at least a mile...maybe 2. I've only been on that tram once and it was years before I started cycling again. I've heard that the tram is the second steepest tram in the world and goes from about 2600' above sea level to over 8500' feet above sea level with an average gradient of 26 degrees and a maximum gradient of 42 degrees.

I'll have to check out the road next time I'm in Palm Springs. I'll make sure it's during the winter though because it's 110-115 degrees there right now (I did some riding there a few weeks ago).

Mach42

Maybe my sarcasm detector is broken, but this is true for energy expenditure.

At 45 degrees or a slope of 100 percent, one must use enough energy to lift themself 1 foot to travel 1 foot horizontally, hence 100 percent energy. At 90 degrees, or a slope of infinity, you can put infinite energy into lifting yourself, but not travel a foot horizontally.

For a slope of 2:1, or 200 percent, you'll have to lift yourself twice the distance upwards for the distance traveled forwards.

I think that it makes sense.

DiabloScott

No it doesn't. Your statements about distance are correct but your statements about energy are nonsense.




By the way - this was my post #666 so I'm feeling especially ornery, no offense.

Edit with copy of my profile as proof below:

DiabloScott
It's MY mountain
Last Activity: 07-18-07 06:27 PM
Viewing Thread Could someone explain % in a climb @ 06:27 PM
Add DiabloScott to Your Buddy List Add DiabloScott to Your Ignore List
Signaturehttps://diabloscott.blogspot.com/

Forum InfoContact Info

urbanknight

Do you ride this on your road bike?!!!

ExMachina

Not sure what (or 'Watt' ) "100% energy" is...but the gradient number has nothing to do with weight, energy, or even physics for that matter.

Gradients are a convention for communicating angles in 'real world' terms. For example, what if I asked you to pave a road at an incline of 10 degrees? That means nothing to you, the foreman. What you and your crew want to know is that along x-meters of road bed, the road needs to rise y-meters. THAT *means* something and you can go to your topo-map and see exactly what needs to be done to make the road rise @ 10-degrees.

redfooj

all you should care about is the total change in height , not how you get to that height (long distance up a shallow slope or short distance up a steep slope)

fetch

i visited a site from a link on BF a few months back that was like a mash-up of a map+google maps feature to figure out the grade of a hill.


anyone happen to have that link? PM it to me if possible

shogun17

ok. 0-5% gradual.

5-10% reasonable-steep

10-15% very steep

15+% mind boggling.