I have wondered for a long time: Why are the gradiants on hills referred to in percents instead of degrees? Degrees seem to me to be much more descriptive. So, will someone please give me the formula for translating percent into degrees? For example, a 10% grade will be at what angle in degrees? Please enlighten me.
I have a pi/30 radian hill right outside my door. Makes for a tough last mile of the ride.
OK, that’s a little too much math for me. What is the % grade, and what is that in degrees?
I believe 10% percent grade means 10 feet vertical change for 100 feet horizontal change. If that is correct, then the translation to degrees would be 10% multiplied by 90 degrees (100% being a vertical wall) equals 9 degrees. Correct?
(100% being a vertical wall) equals 9 degrees. Correct?
Not quite. A 100% grade is 90 degrees.
(rise/run)*100 = percent grade
So a 10% grade would be a 1ft rise over 10ft.
Or the 10% grade is actually the tangent of the true angle. I.E. 10% is equal to .1 and 5.710 degrees has a tangent of .1…assuming my math is right.
~Matt
I think it is just the elevation gain divided by distance covered.
ie you climb 1000m over 20k
1000m/20000m = 0.05 = 5% slope.
Yes, 10% of 90 equals 9. You misread my sentence.
Rise over run would suggest the linear distance covered, I think it is the hypotanus distance not the linear distance. It would be a lot of work for people to calculate the actual linear distance covered for all these stages.
In 5 words, percent grade is
“Rise over Run times 100”
or
(rise/run)*100 = percent grade
Where both the rise and the run must be in the same units - typically feet. So, for a rise of 120 ft over a distance traveled of 0.9 miles, it would be
(100 * 120) / (0.9*5280) = 2.52 %
Now to turn that into degrees, you’re gonna have to use calculus. You get to use words like “sine,” “cosine” and “tangent”
The steepness of a road is generally measured in % grade, which in mathematical terms is the slope, or TANGENT of the angle, measured from the horizontal. This is the ratio of elevation change per horizontal distance traveled, often called “rise over run”. Typically a road that rises 1-in-10, is otherwise called 10% grade.
Measuring the distance along the surface of the road instead of horizontally gives practically the same result for most road gradients. The distance along the road surface gives the SINE of the angle in contrast to the horizontal distance that gives the TANGENT. For practical purposes SINE equals TANGENT for small angles (up to ten degrees or so). For instance, a 20% grade (11.3 degrees), whereas measuring along the road surface gives a 19.6% grade.
The slope of a road is more useful than its angle because it gives a direct way to assess the effort required to move forward against the grade, whereas the angle in degrees does not readily reveal this information. A 5% grade requires a forward force of approximately 5% of the vehicle weight (above and beyond the force it takes to travel similarly on flat ground). A 15% grade requires a propulsion force of approximately 15% of the vehicle weight.
Although the angle may be more easily visualized, it does not convert easily to effort without a calculator. For instance a 20% grade is an 11.3 degree angle and is a steep and difficult gradient. The relationship between angle and slope is non linear becoming 100% (1:1) at a 45 degree angle. In contrast, the SINE of 45 degrees is 70.7% while the SINE of 90 degrees (straight up) is 100% for which the slope
(TANGENT) is infinity (or undefined).
The most accurate way to measure this without a precision inclinometer, is to use a level, a one meter long bar and a metric ruler. Resting one end of the rod (held level) on the road at a representative spot, measuring the distance down to the road at the other end in centimeters gives the percent grade directly. Using a carpenters level and a one meter long rectangular bar can give accurate readings to a couple of tenths of a percent.
And if you think I didn’t just Google all that up, well, thanks for the compliment.
Not quite. A 100% grade is 90 degrees.
No. A 100% grade is 45 degrees. Like others have said, it’s simply vertical distance over horizontal distance. The slope of a 45 degree line (for example, the line y=x) is 1, and therefore it has a grade of 100%. Draw one on paper and you’d think you could cycle up it.
-Colin
Slope gradient is measured as rise over run, so a 100% slope is 45 degrees (horizontal and vertical distance the same). For slopes less than about 20 degrees, a reasonable approximation of the formula you seek is:
(rise/run)*57.3
If you want to be exact, the angle is the arc tangent of the rise over run - you can use this calculator and enter the slope (raw value, not the “percentage”, eg. .10 for a 10% grade):
http://iws.ccccd.edu/jwilson/atancalculator.html
Here’s a table, showing that the degree of slope is a fairly unimpressive number, hard to make objective comparisons with, until you get to a pretty steep hill, maybe that’s why it is not used much. There’s a big difference between, say, 3.5% and 5.2%, but both are “2 something degrees”, so, counter to your assertion, degrees are not really more descriptive. That, and maybe because it’s not linear, so it’s less useful for road engineers.
grade degrees
1% .57 degrees
2% 1.14 degrees
3% 1.72 degrees
4% 2.29 degrees
5% 2.86 degrees
6% 3.4 degrees
7% 4 degrees
8% 4.57 degrees
9% 5.14 degrees
10% 5.7 degrees
15% 8.5 degrees
20% 11.3 degrees
Thanks for setting me straight! I deserve a “bad grade” on that question!
Thanks for all the somewhat confusing answers. It seems I’m not theonly one who was confused. I think I’ll just go with Skip’s formula and table.
That table is so depressing. You’d think an 11 degree slope would be almost nothing, but try riding up it! Anyone happen to know how steep typical ski trails are? For example, what would be the range for a double diamond trail somewhere in Utah or Colorado?
Now to turn that into degrees, you’re gonna have to use calculus. You get to use words like “sine,” “cosine” and “tangent”
And if you think I didn’t just Google all that up, well, thanks for the compliment.
I think this is trigonometry, not calculus.
Now to turn that into degrees, you’re gonna have to use calculus. You get to use words like “sine,” “cosine” and “tangent”
It is Geometry or Trig at the most.
Here’s Cunego climbing at 31% (17 degrees) by my pixel ruler estimate, probably a bit steeper since the camera isn’t level:
http://www.cyclingnews.com/photos/2005/giro05/?id=teampres/13
He makes it look so easy 
.
Distilling the confusion, the right formula is:
angle = arcsin(rise/run)
If you have a sufficiently geeky calculator, just push the “asin” button. But Skip’s formula is plenty accurate – to better than a tenth of a degree.
I knew it was from High School, which was many years and many … well, let’s just say it was many years ago.