| **road gradient** | villanum
*Aug 22, 2001 8:42 AM* | | can anyone give me an idea of how steep an 8% grade road is? i've read the formula and i understand how they get the number but i still don't know what an 8% hill looks like. anyone in the san diego area have an example (what grade is the torrey pines hill from the beach to the golf course )? |
| **Topozone.com** | Chen2
*Aug 22, 2001 9:00 AM* | | If you'll go to www.topozone.com you can enter the location you are interested in and view a topographic map with contoured elevations. This can be printed out and you can measure distances and read elevations. With some basic math you can calculate grade. (A long 8% grade can be tough, especially on a hot day.).
-Al |
| **Good old rise over run...** | RhodyRider
*Aug 22, 2001 9:01 AM* | | Get yourself a nice straight 2x4, the longer the better but at least 3 or 4 feet long. Carefully tape a bubble level to the mid point of the 2x4. Take the board, a decent tape measure, and a calculator to the hill in question. Place the board on the road surface and then lift up the trailing end until the level indicates...level. Measure the distance (at a right angle) from the end of the board down to the road surface. Divide that measurement by the length of the board. There's your percent gradient. Take several readings along the course of the incline and average them to get the most accurate picture. You may look like a chooch doing this, but it will satisfy your curiousity. And this is one more thing we can add to the "you may be a bike geek" list... |
| **Couldn't you approximate it as...** | PdxMark
*Aug 22, 2001 10:36 AM* | | OK, here's a math guess about how gradient could be approximated... using a bike computer with an altimeter. Let me know what you all think...
If X is the horizontal run, Y the vertical rise, and Z the inclined distance, you could calculate the X as:
X=(Z^2 -Y^2)^1/2 (just Pythagorian Theorem)
so gradient G in percent is:
G=100*Y/X
But here's the rub. For gradients under 10% (even 20%), the square of the rise is much smaller than the square of the inclined distance, so X will be fairly close to Z (for 10% grade, X is 0.995Z, for 20% grade, X is 98%Z) - so really, you can approximate gradient G as
G=100*Y/Z
So a 3 mile climb of 1000 feet is approximately 6.3 percent
Thoughts? |
| **316 vert. feet per road mile is 6%, or...** | PdxMark
*Aug 22, 2001 11:06 AM* | | Working backwards... it seems that a 6% grade gives vertical rise of 316 feet per inclined mile... with 316 feet as a guideline, you can approximate grade with your altimeter by dividing rise over a mile by 316 feet - or maybe easier, for every 52.5 feet you climb over a mile you get another % of gradient |
| **re: road gradient** | alex the engineer
*Aug 22, 2001 10:56 AM* | | You can define hills in either slope, radians, degrees, or gradient. Slope is rise/run, degrees are the angle from horizontal, radians are similar, with 180 degrees = pi, and gradient is the percentage rise compared to the run. In this case, you rise 8 feet for every 100 feet. This works out to an angle of 7.2 degrees, .126 radians, and a slope of .08. 8% is probably the maximum grade which you would ever see on an interstate highway, and would be considered a "moderately steep" hill. |
| **Torrey Pines** | UncleMoe
*Aug 22, 2001 11:20 AM* | | I'm not sure of the gradient of the Torrey Pines hill. I think the elevation gain for the outside (road) is 150-200 feet over 1.5-2 miles. Its not the grade that kills you, its the distance. Takes me about 12 minutes.
The inside route (the park road) is tougher at the bottom because you gain so much so quick, but easier at the top. The inside is also actually a shorter distance. However, it takes me about the same amount of time. The inside is probably 8%-10% grade. Almost like an intermediate trail on a ski slope. The outside is like a beginner slope.
I've ridden tougher hills, but Torrey is a pretty tough hill depending on what part of your ride you are in. Easy for the start of a ride, tough at the end. |
| **I envy those who ride around there** | Tig
*Aug 22, 2001 11:42 AM* | | I was in Del Mar last year on business and loved the climbs. I haven't seen so many people on bikes since the last local century! Great views and challenging climbs. Torry Pines was a big 'un. I have to either drive 90 minutes or ride over the Seabrook/Kemah or Houston ship channel bridges to get any climbing in! |
| **gradient/categories?** | girodebirdman
*Aug 22, 2001 12:01 PM* | | Assuming I am using the gradient formula of rise/run, and not the angle of the incline (I think the former is standard for cycling, but am not sure) then what defines a Category 4,3,2,1,HC? I just did a 7 mile climb at 6.3% yesterday, and was trying to get a feel for how steep it is comapred to some of the climbs in the tour. I though perhaps category 1. Last week, I did a climb of 9.9%, but it was only 4km long. What would this be rated under the climb classifications? Any guidelines or links? |
| **re: road gradient** | Ian
*Aug 23, 2001 5:45 AM* | | The easiest thing to do would be to purchase a cyclocomputer that tells you the grade. I have the Specialized Speedzone Pro which tells me the grade and total elevation gained. A new company, Ciclosport, also has one that does this. I am going to switch to this one because it also estimates your watt output. I now know the grade of my local hills and can stucture a workout to include climbing. Yesterday I did a 40 mile ride and climbed 2,340 feet, not bad for Florida. |
| |