Traveldave

I have a couple of rides that I frequent here in Atlanta. One is the Silver Comet Trail, which is a rails-to-trails conversion and is therefore pretty flat. The other is Stone Mountain, which has 5 or 7 mile laps and is somewhat hilly.

It seems to me that if I do 30 miles (out and back) on the trail or 30 miles worth of laps at Stone Mountain, I would have the same net vertical gain (zero) and thus my effort should be about the same. However, that's not the case. 30 miles on Silver Comet is both easier and faster than Stone Mountain.

I've got some thoughts on why this may be, but I figured I'd put this out there for some scientific discussion by the roadies.

forum*rider

tail wind?

better paved roads?

Maybe it's physcological. You think that Silver Comet is easier and therefore it is for you.

55/Rad

Simple math I believe.

Lets say you average 20 mph on the trail.

On the mountain loop, let's say you climb the hills at 10 mph. In order to average the same 20 mph overall as on the trail, you need to spend an equal amount of time at 30 mph as you do at 10 mph in order to make up the difference. Most people go down the hills at a much faster rate but they don't make up the deficit created in the climb simply because they don't spend as much time going downhill.

I hope that's clear - my junior high algebra teaching is turning in her grave.

55/Rad

sm266

what he said.

zensuit

Effort wouldn't be the same, either. If you do hill repeats, the effect on your system is pretty draining. To ride "equal" effort on a flatter trail would require taking yourself to the same heart rate you get to climbing...and that's usually pretty high. And it's unlikely that you would push yourself that hard unless you were time trialing competitively. You are required to exert yourself to get up a hill, since you can ease a little on the flats, you usually do...human nature.

bac

Given your logic, riding up, and down Mt. Everest (hey, I've got my studded tires on!) would be the same effort as riding on the flats for the same distance. Yes, the net vertical gain is zero, but you'd have done thousands of feet of climbing - that's the number that's important.

BlazingPedals

Simple thought experiment. Do a 20 mile out-and-back route, uphill all the way out. If you average 10 mph on the climb and take an hour to do the first 10 miles, how fast do you have to go after the turnaround, in order to average 20 mph overall? (hint: it's not 30 mph.)

55/Rad

You would have to magically disappear with your bike from the top of the hill and reappear at the finish line in the same moment - just to average 20 mph on this ride.

55/Rad

brent_dube

I don't think this is really about time or mph average. It's about energy used over a certain distance and time.

Isn't that like saying doing a 100 mile flat century requires the same effort as a 100 mile century loop with 3000 meters of climbing? It doesn't work that way.

blah, you explained it for me

recursive

Dave:

If you come off a long descent, and go right into a climb of the same height, your momentum will wear off pretty quickly. This is because of air resistance. If it wasn't for air resistance, you could hit ridiculous speeds of hundreds of mph given long enough descents. This speed would act as stored energy, and be spent moving you up the following climb. But since air resisantance increases proportional to the square of your speed, there is a limit to the speed you can reach on the descent. Even though it seems you're going fast, the air is sapping the energy that would push you up the next climb. In fact, if you were riding in a vacuum somehow, I think your theory of equal effort might be correct.

downhill!

To use the law of Conservation of Energy you would have to not pedal at all on the decent. This will get you close to using the same amount of energy on a flat ride vs. hill ride. However, by expending all of your energy on the climbs you do all of the work in less time, meaning you expend more power. This makes you more tired. In addition to that, wind resistance increases exponentially with speed, so the downhill portion of the ride will be much less efficient than the flats or uphill. This will cause you to have to pedal some on the descent to keep your speed up.

Avalanche325

You are confusing NET work of the verticle component, which is 0 since you are starting and finishing at the same point, with ENERGY EXPENDED.

You have to push the bike up, gravity will bring it down. So when cycling, you add all of ups together, for the work the YOU (the prime mover) is performing.

clausen

That wouldn't work. Even if the flats =0 energy expense and going up hill are a - energy expense you would have to get a + energy going downhill. Thats impossible.