This is a very tame/gradual climb, isn’t it? Can anyone put into words or describe what a climb like this might look or feel like?
Thanks
that would be pretty gradua,l take the 500 ft dividd by the 3 miles in feet and you have your grade, which is less than 1
% grade
.
which is less than 1
% grade
Are you sure that’s right? Grade = Y/X, X being the distance and Y being the rise so 500/15840 =.0315 or 3.15% grade. Still pretty gradual but not 1% gradual.
Edit to add; A 100% grade is 45 degree angle. Also for the OP the “Difficulty” is kinda associated to gearing I would suppose. For me with a somewhat typical road bike 10%+ grade is “Hard”, but I suck at climbing hills and have never been “Whip” thin 
~Matt
I get 3.1% grade, not steep, but still a hill, probably feel like pedaling into a medium strength wind.
i forgot to move my decimal. Can you tell I have been up too many hours already and it is only 7 am
http://www.gmap-pedometer.com/?r=1836263
This is probably the closest one to you that is similar if you want to see for
yourself (oak hill rd in Harvard).
2.2 miles, 350 ft (3.01%)
A “false flat” where you try to go 22 mph but can only manage 19.
For all those that have done Hawaii, I believe that it is just under 700ft in the last 5 miles to Havi. SO just about the same as the OP asked…I can tell you that on many of the days out there, it does not seem like a moderate, gradual climb, and we definately do not go 19mph up it. OF course there is typically a headwind, so that skews it, but don’t discount a climb just because the paper numbers look tame…I always thought that climb was a lot higher, but that is what the elevation charts say it is
I do 500 ft in less than one mile everyday on my commute to my house. Not too bad.
t
Never ridden that one in Harvard, but most of my rides go through Harvard, so I’ll check that one out sometime.
I get the gist of this climb, nothing too big of a deal. Doing a road race tomorrow which starts/ends with this hill, so I wanted to have an idea what to expect before I ride it for the first time.
Last year I did a race without much of a warmup that started with a very steep climb for a couple miles, so I basically cooked myself within the first 10 minutes. Don’t want to repeat that
I live on a 500’ hill that is 2 mi up and 3 mi down. The 2 mi is much harder than the 3.
“A 100% grade is 45 degree angle” Really? I had no idea.
That’s what I like about this place. I often learn things that I didn’t even know I didn’t know.
100%, means you’re going equal distance horzontal as you do vertical… hence the 45 deg answer…
Though I agree with the math that it’s a 3.1% grade, I’m confused by the notion that 100% is a 45-degree climb. Based on X/Y, wouldn’t 100% be straight up, or 90-degrees and a 50% grade be at 45-degrees (e.g, 1.5mile (or 7920ft) gain over 3-miles = 50%)?
Doing a road race tomorrow which starts/ends with this hill, so I wanted to have an idea what to expect before I ride it for the first time.
Oh, you should have said so earlier. That’s a completely different story.
It will be easy on the first lap, really hard on the second lap and a %$#Q%^@ near impossible on any subsequent laps!!
Though I agree with the math that it’s a 3.1% grade, I’m confused by the notion that 100% is a 45-degree climb. Based on X/Y, wouldn’t 100% be straight up, or 90-degrees and a 50% grade be at 45-degrees (e.g, 1.5mile (or 7920ft) gain over 3-miles = 50%)?
It depends on what you use for X, the length of the road or the horizontal distance traveled. For most of the grades we typically encounter the distance between the absolute horizontal distance and the actual road length is small enough that it doesn’t make much of a difference. But as the grade increases the discrepancy between the two measures becomes more apparent.
I’m with you, I have always assumed X was the length of the road itself, so the only way you could get a 100% grade would be to have 100’ of road that goes 100’ straight up. However, if you used horizontal distance, then a road going straight up would cover 0 feet of absolute horizontal distance, giving an infinite grade.
Your 1.5/3.0 mile example is correct. But look at the most simple case rolling 100 feet forward and gaining 10 feet elevation doing so.
(10/100) * 100 = 10% and atan(.1) = 5.7 deg
Then your example: 50 ft of gain (50/100) * 100 = 50% = and atan(.5) 26.5 deg
100 ft off elevation gain = (100/100) * 100 = 100% and atan(1) = 45 deg
If you were to go up more than you went forward, (which would suck, BTW) you have something like
(150/100) * 100 = 150% grade atan(1.5) = 56 deg . Time to buy a compact crank.
Doing a road race tomorrow which starts/ends with this hill, so I wanted to have an idea what to expect before I ride it for the first time.
Oh, you should have said so earlier. That’s a completely different story.
It will be easy on the first lap, really hard on the second lap and a %$#Q%^@ near impossible on any subsequent laps!!
Dude, that is a nearly perfect description (well, not near impossible, but fricken painful on the 3rd lap). Race went up the hill 3 times, first time cake, second time pretty hard, third time felt like a 3 mile sprint.
All in all, it was not steep at all, but there were a few sections that went to a significantly steeper grade.
From what I understood, the gradient of the road is usually expressed as the tangent of the angle in percentage format, i.e. the vertical rise divided by the horizontal component (and of course, the horizontal component is much harder to obtain than the actual road distance traveled, especially on a curvy road), so that is why a 45-degree hill is 100%, because its tangent is 1. But no cyclist really has any business heading up or going down a road that is more than 45 degree anyways, right (unless he’s a mountain biker)?