Comments on: The Question of Hills

By: JohnB

JohnB — Sun, 07 Sep 2008 06:22:09 +0000

Regarding your statement about walking becoming a competitive option, I would say that’s true only if you are not carrying a heavy load, especially in a trailer. In that case, as I have experienced several times in the last few months with over 50 pounds of freight, I may be standing on the pedals in granny gear to climb the hill with the load, but if I get off and try pushing it uphill instead, it’s much harder still. I guess from a physics point of view you could say that at least in a heavy load-bearing situation, your gears really are doing a lot of work for you. Which is encouraging!

By: Mark Stosberg

Mark Stosberg — Sat, 23 Aug 2008 22:07:39 +0000

Craig, Thanks for the reply. I double checked the math and you are right.
However, my experience remains that going up hills doesn’t lower my overall average speed as much as I expect it to. This could perhaps be because the momentum of the downhill has a greater affect than any additional recovery required once I’ve reached the crest after an uphill.
For example, Consider a course that is one mile uphill followed one mile downhill and then one mile of flat. For some distance on the flats I’ll be able to continue to put the downhill momentum to work as my speed tapers back town a “normal” speed for flat ground.
In the rolling hills of Indiana, that also means that some amount of the downhill momentum is going into making climbing the raise immediately following it easier.
Of course, I’m using intuition again and you are using physics equations, so I’m prepared to be wrong about the reality again.

By: Craig

Craig — Sat, 23 Aug 2008 18:16:18 +0000

The math in this post is incorrect. Average speed is not the average of your climbing and descending speeds but is instead the total distance (climbing and descending) divided by total time. This difference will give you a radically different average speed but one that is correct.
For example, if you climb a hill for X miles at a pace of 3 MPH, then it is impossible to achieve an average speed for both climbing and descending of 6 MPH or greater. This is because even if you descend at the speed of light — not recommended on most bikes! — you merely double your total distance traveled to 2X, and doubling the distance merely doubles the average speed.
In the example in the post, climbing a hill for X miles at 3 MPH and descending at 21 MPH yields the following:
T = D / V (velocity)
D-up = X miles (given)
V-up = 3 MPH (given)
T-up = X miles / 3 MPH
D-down = X miles (given)
V-down = 21 MPH (given)
T-down = X miles / 21 MPH
T-total = (X miles / 3 MPH) + (X miles / 21 MPH)
D-total = 2X miles (given)
V-total = D-total / T-total
V-total = 2X miles / ((X miles / 3 MPH) + (X miles / 21 MPH))
V-total = 5.25 MPH
So in the above example, going up a hill at 3 MPH and down at 21 MPH gives an average speed of 5.25 MPH. This is not as encouraging of a result — sorry!
Two comments:
(1) Try this for yourself. Go up a hill at 3 MPH and down at 21 MPH. Measure how far you went in total up and down the hill as well as how much time elapsed. Then find a flat stretch the same distance and go 5.25 MPH. You should find that it takes the same time.
(2) An intuitive way to understand the math is that you spend a lot of time going up the hill slowly but zip down the hill quickly. Therefore, the uphill portion penalizes your average speed much more than the downhill.