Rating hills - percentages??



huhenio

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Jul 19, 2005
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I am a little of a mapping enthusiast and I want to make a map of my area with hill rating. I can read a topographic map just fine, but I am aware that there is a system for rating hills in percentile figures (3%, 6%, n%, whatever percent%).

What is the formula to use so I can give a correct rating to my very hilly stomping grounds?

Thanks in advance!!:confused::D
 
huhenio said:
I am a little of a mapping enthusiast and I want to make a map of my area with hill rating. I can read a topographic map just fine, but I am aware that there is a system for rating hills in percentile figures (3%, 6%, n%, whatever percent%).

What is the formula to use so I can give a correct rating to my very hilly stomping grounds?

Thanks in advance!!:confused::D
It's about the number of verticle feet in a certain amount of distance, 100 yards or meters I believe is the distance.
 
House said:
It's about the number of verticle feet in a certain amount of distance, 100 yards or meters I believe is the distance.
so is distance yards/feet?

1232 yards/200 feet = 6.16 % :confused:
 
huhenio said:
so is distance yards/feet?

1232 yards/200 feet = 6.16 % :confused:
The denominator is not set in stone. The "100" simply makes it easy to convert to %:

a 12% climb is one that rises vertically 12 feet for every 100 feet you traverse horizontally, so (12/100)*100 = 12%

likewise (42/350)*100 = 12%. (Height travelled vertically/distance horizontally)*100 is the formula you are after.
 
FrankBattle said:
The denominator is not set in stone. The "100" simply makes it easy to convert to %:

a 12% climb is one that rises vertically 12 feet for every 100 feet you traverse horizontally, so (12/100)*100 = 12%

likewise (42/350)*100 = 12%. (Height travelled vertically/distance horizontally)*100 is the formula you are after.
gotcha ... .thanks!!
 
As long as the units cancel out, it does not matter which units are used.
4 inches/100 inches = 4 feet/100 feet = 4 meters/100 meters = 4%
 
RickF said:
As long as the units cancel out, it does not matter which units are used.
4 inches/100 inches = 4 feet/100 feet = 4 meters/100 meters = 4%
so 160 feet/ 5800 feet = 0.0275 times 100 = 2.7[size=+1][/size]
 
huhenio said:
so 160 feet/ 5800 feet = 0.0275 times 100 = 2.7
Right, but that assumes it's just straight up with no undulations. You hear folks talk about average rating also. So, you could have one section at 4% and another pancake flat and you'd have a 2% average grade.

Crude example, but you get the idea. Many hills have varying grades. You could calculate 2% on paper, but that probably only counts the starting and end points and may ignore the 2 huge humps in the middle, for example.
 
I know .... but what other official method there is to asess the gradient percentage.

Today I counted 9 humps on a 1 mile hill. That has to be difficult to measure, but an overall every 200 feet has to be rather good for pointing out difficulty.

I cannot imagine what a 18% gradient for miles and miles has to be like on a bike.
 
Without getting into an extended discussion of variational calculus, accuracy is going to depend on the distance between measurements. If you have a topo map, you can go just go along your route and measure the distance along the road between each elevation line. Then load all of the points into a spreadsheet to get the grade profiles and total elevation gain. Depending on the scale and spacing of the elevation lines, it should work pretty well.
 
artmichalek said:
Without getting into an extended discussion of variational calculus, accuracy is going to depend on the distance between measurements. If you have a topo map, you can go just go along your route and measure the distance along the road between each elevation line. Then load all of the points into a spreadsheet to get the grade profiles and total elevation gain. Depending on the scale and spacing of the elevation lines, it should work pretty well.
Exactly. TopoUSA does a decent job of this, for readily available software. I don't know how much it is but friends have it and it works pretty well.
 
FrankBattle said:
Exactly. TopoUSA does a decent job of this, for readily available software. I don't know how much it is but friends have it and it works pretty well.
It runs $50-100 depending on the coverage. I haven't used it, so I don't know if that's a good deal. USGS topo maps are available for free at www.terraserver.microsoft.com. I've been playing around with Google Earth a bit lately. It's pretty cool, but it sucks bandwidth like crazy.
 
The topozone and terraserver are great sites - sometimes I spend way too much time on there trying to compare the Alps to the Appalachains (silly me!!!)

I've found that the numbers on the topo- maps can throw you a bit ... I've scouted several climbs online that looked more menacing than they played out - sometimes the more innocent looking climb on a map ends up being a better training route.

Most mountain-ish roads don't go straight up the mountains (too bad for cyclists!)- so when I scout a climb online, I look to see the angle that the road goes toward the incline. Most roads move somewhat across the grade, and the construction of the road may mean that the crews even dug down into the hill to build the road and lessen the difficulty of the grade. This isn't always evident from the map's numbers either.

I found from my experience that a steep, shorter climb can give some nice training results - what's more is that I found some fairly steep climbs in residential areas with wide roads where it's nice and safe- and the road aims straight up the hill.
 
I am using Google Earth - metric settings work great for calculations - on my neighbourhood.

It also makes a good 3D image of the terrain - with satelite pictures as "skin" - so to visualize the incline on the given road that you might want to tackle.

Since this are rolling hills, they are more or less a mile long and more or less 200 feet high, making them a 5 or 8 percent hills. That being said, there are sections that are far steeper than that, and sections of the same hill that are flatter than that. It ends up averaging the forementioned numbers. But there are 100/ 200 meter sections where I get anywhere from 10% to 20%!!!

Good training though ... It feels great to know that those sections not only look steep but they are actually steep!!

I thank you all for your imput, I will get back to my map thingy
 
OK, this always confused me too. I'm an engineer, so I apologize if my trig is a bit confusing.
If you take a section of an uphill, it can be imagined as a right triangle. We travel up the hypoteneuse, then the other two legs of the triangle would be the vertical height traveled, and the other would be the horizontal distance traveled. Therefore, if you were traveling on perfectly flat ground, the angle of the triangle would be zero degrees, so the horizontal distance would be the actual distance traveled.
So when you say that the formula is vertical height divided by horizontal distance traveled, I get confused as to what horizontal distance means. Is that the horizontal leg of the aforementioned triangle, or is that the hypoteneuse (i.e. the distance you actually traveled up that section).
I apologize if this is confusing.
 
crb189 said:
OK, this always confused me too. I'm an engineer, so I apologize if my trig is a bit confusing.
If you take a section of an uphill, it can be imagined as a right triangle. We travel up the hypoteneuse, then the other two legs of the triangle would be the vertical height traveled, and the other would be the horizontal distance traveled. Therefore, if you were traveling on perfectly flat ground, the angle of the triangle would be zero degrees, so the horizontal distance would be the actual distance traveled.
So when you say that the formula is vertical height divided by horizontal distance traveled, I get confused as to what horizontal distance means. Is that the horizontal leg of the aforementioned triangle, or is that the hypoteneuse (i.e. the distance you actually traveled up that section).
I apologize if this is confusing.
Technically, it should be rise over run, so it would be the horizontal leg of the triangle; however, when you are talking about 4%, or even 15% inclines, you are probably still within the error of measurement for both distance traveled and altitude gained if you use the distance traveled (hypoteneuse) as the denominator. Not many roads have 45 degree (100%) inclines, where the error would be a factor of 1.414213562373...

I love rails to trails. No incline on a standard US railroad bed is greater than 4%.
 
RickF said:
Technically, it should be rise over run, so it would be the horizontal leg of the triangle; however, when you are talking about 4%, or even 15% inclines, you are probably still within the error of measurement for both distance traveled and altitude gained if you use the distance traveled (hypoteneuse) as the denominator. Not many roads have 45 degree (100%) inclines, where the error would be a factor of 1.414213562373...

I love rails to trails. No incline on a standard US railroad bed is greater than 4%.


IOW, on a pure 6% climb for 1km, you'd climb 60 meters vertical and travel 1000 meters horizontal, but a whopping 1001.8 meters along the hypoteneuse (the road). The error is too small to bother with trig.
 
DiabloScott said:
IOW, on a pure 6% climb for 1km, you'd climb 60 meters vertical and travel 1000 meters horizontal, but a whopping 1001.8 meters along the hypoteneuse (the road). The error is too small to bother with trig.


Thanks for the explaination. I never ever considered small angle approximations. So then the slope of the hill (rise over run) is the percent grade. Thanks!
 
Is it true that all major high ways and interstate highways in the US are required have a max gradiant of no more than 6%? I have heard this for years from fellow cyclists. They say it's some kind of[size=-1] US Department of Transportation, Federal Highway Administration requirement.[/size]