There are two types of courses that I am considering for a 10 km trial:

  1. Completely flat. 5 km out and 5 km back.
  2. 5 km uphill at a grade of 1%, and 5 km downhill on a grade of 1%.

Assuming you start and finish at the same place and the elevation increase/decrease is zeroed out, is one of these courses faster than the other? Or are they both the same? Would a lighter rider benefit from the 1% uphill gradient?

  • 3
    If you're organising a TT event and planning the route, then pick the more interesting. Consider spectator access, emergency ambulance access, and risks from traffic and weather. IE if there's a lot of fallen leaves in Autumn, consider that a negative. Also consider picking the route with lower overall traffic, or whether you can get the local authority to close the road.
    – Criggie
    Oct 21, 2020 at 10:34
  • 1
    If you're planning training routes, why not both? One must practice climbing and descending, and while 1% is not exactly steep its a start.
    – Criggie
    Oct 21, 2020 at 10:35
  • 4
    Sure this isn't a physics class problem? :)
    – Armand
    Oct 21, 2020 at 18:49
  • 2
    @Armand The idealized physics problem has the opposite result! Due to the realities of drag and friction, going down and up is less efficient on a bike. But in a frictionless "physics world", if two objects start with the same speed, one that goes down a hill and then back up will have an equal or higher speed at all times compared to one that moves over flat ground. Oct 22, 2020 at 20:56
  • Another minor diff, In a perfect physics world, are you measuring distance travelled horizontally, or distance rolled? a 5 km flat distance would be 50 metres high at 1% (which is 0.57 degrees) and a total rolled distance of 5,000.25 metres. Even a 10% gradient only results in 25 metres further to roll. More fun at calculator.net/triangle-calculator.html
    – Criggie
    Oct 24, 2020 at 23:18

6 Answers 6


The problem is that average speed depends on time, not distance.

If you go the first 5km distance at 20km/h and the other 5km at 40km/h your total time is 22.5 minutes for an average speed of 26.67km/h (not 30km/h as you might have thought).

Drag also increases with velocity squared, so all that precious potential energy you’ve gained by riding up the mountain will be lost to drag at high speed long before you’ve reached the foot of the mountain.

So generally it’s more important to be fast on the uphill parts. In the above example, increasing your uphill speed by 10% (e.g. by losing body weight and getting a lighter bicycle or by increasing your power output by 10%) would improve your overall time and average speed by 6.5%. Increasing the downhill speed by 10% would only improve the overall time by 3%. To increase the downhill speed by 10% would also require more than a 10% increase in power output.

That being said, even at 20km/h you’d still have considerable aerodynamic losses, so improving aerodynamics is always a good idea.

  • 4
    Lots of important points, but you don't seem to address the actual question. (I'm aware OP accepted your answer, but for the future it would be good to have.) Oct 22, 2020 at 11:55
  • @saaru I'm not quite sure if Michael meant that - possibly would have been better in a comment like this, and let Michael edit that in. Or put it in an answer of your own.
    – Criggie
    Oct 22, 2020 at 21:55
  • 1
    Oh sorry, I though Michael had to approve the edit of their answer and would reject if it was not to their liking. I'll rollback and put the link in this comment. @Michael, if you want you can add it for a more in depth explanation of speed averaging. Oct 22, 2020 at 22:06

It is pretty obvious. Just imagine the hill rose a bit, say to 5%. And even more, to 10%. Going across a mountain is never fast. Mountainous race stages are slow. The total meters needed to be climbed over the stage count, not just the different between the start and the finish.

Do remember that the average speed is not averaged over the distance but over time. So if you spent more time in the slower section, it counts more. And you do, because you are, well, slower...

There is an old riddle. You have one hour to go to a certain place 10 km away and back. So you plan to go 20 km/h. But it was uphill on the way there and you only went half the planned speed. How fast do you have to go on the way back to get back in time?

I also notice you ask about a light rider. The more there is meters to be climbed per every kilometer of the course, the better for a light rider although the time trial discipline has its specifics and specialist riders.

  • 3
    For those who want the solution to the riddle: you can't get back in time. Half the planned speed of 20 km/h is 10 km/h. So you spent your entire hour you had for the entire trip just on the trip away, and you have no time left to get back.
    – Nzall
    Oct 22, 2020 at 7:39

Fast Fitness Tips recently addressed a similar question: whether an out-and-back time trial would be faster with still air, or into a headwind for half, tailwind for half, and how to pace the windy TT. Key quote:

sadly the gain from the tailwind is not as large as the losses in the headwind and as a result there is a net loss in speed whenever its windy on a looped course or an out and back.

I believe the same would hold true for the course with the climb.

On level ground, the more powerful rider will have the advantage. On hills, rider weight isn't so much an issue as the rider's power:weight ratio. A rider who weighs 75 kg and develops 300 W should be evenly matched against a rider who weighs 50 kg and develops 200 W (although bike weight will be relatively constant, so the heavier rider might actually have a better net power:weight ratio). But this only comes into play on steeper hills—1% is close to flat, so the more powerful rider still has the advantage. Between two riders of equal power, sure, the lighter rider will have an advantage, but on a 1% grade it's not huge.

  • 2
    The problem with this argument is that even if you do get all the advantage from the 5 km/h wind and go 25 km/h one way and 35 km/h the other way, you still lose (yes, meaning lose time). It is not necessary to have any losses whatsoever. It is more obvious when it is 10 km/h one way and 50 km/h the other way. Oct 21, 2020 at 7:54
  • @VladimirF: You lose time, but if the others lose more time you still win the race.
    – Michael
    Oct 21, 2020 at 7:56
  • @Michael As in any other race, I do not see a point. Oct 21, 2020 at 7:57
  • Well, part of the question was "is one of these courses faster than the other"
    – ojs
    Oct 21, 2020 at 8:14
  • 2
    Yes and no - the headwind/tailwind factor is because the power lost to air drag scales with the cube of speed, but climbing power scales linearly with speed. However at constant effort this same effect means the effective headwind (even in still air) is greater on the downhill segment, and the reduction on the uphill doesn't make up for it
    – Chris H
    Oct 21, 2020 at 9:25

Here's an empirical example. My local 10 mile course has a little bit of a climb (6-11% for 500m) at about half way, and then after doubling back comes down the same stretch of road. It's not a hilly course overall, but has a few other bumps as well. My average speed when I finish the descent is never as high as when I start the climb (whether or not you include the false flat at the bottom). This is true regardless of the wind direction. The prevailing wind is at your back on the climb, but I've done it climbing into the wind. That means more headwind over the whole course, and more work (or more likely a lower average speed) before the climb.


Generally it's true that hilly courses are slower. This is because the aerodynamic drag force is proportional to the square of the apparent wind, and a hilly course will have slow parts (uphill) and fast parts (downhill). The squared velocity term means the total energy required to complete the course in a given time increases when the speed is more variable.

However, 1% is not much of a grade at all. It's not enough to say with certainly whether one course is faster than the other when considering other factors which may vary between the two.

Does the out-and-back course require braking to turn around? Re-accelerating the bike requires work.

Which way is the wind blowing? If the hilly course has even a gentle tailwind the whole way it will likely be faster than the out-and-back course which will have a headwind for half of it.

Does one course have more turns than the other? Turns, like hills, make the speed more variable. They also require the rider to assume a less aerodynamic position or risk a crash. If the turns require braking, that requires even more energy to overcome.

Steve Gribble has a nice calculator and also some detailed explanation of the physics involved. Let's put some numbers on it by assuming the cyclist will maintain a constant 250 watts, and otherwise using the defaults for weight and aerodynamics from the calculator.

On a completely flat, windless course, 250 watts will achieve 22.65 mph, for a completion time of 8:13.8.

On a 1% incline, 250 watts yields 19.85 mph, completing the first half of the course in 4:41.73. The second half with a -1% grade is completed at 25.50 mph in 3:39.31. The total time is 8:21.04.

These estimated times don't account for the initial acceleration, but since that's pretty similar between the courses we can ignore that for comparison.

Thus the hilly course is 7.44 seconds slower in highly idealized conditions. In practice, an experienced cyclist will use a more efficient pacing strategy for the hilly course, which is to put in a little more power on the uphill portion and a little less power downhill, which will further reduce the time difference.

This is not a big difference, easily less significant than the other variables previously described.

Weight is also not very significant. Running the hilly example again, but with a rider 5 kg heavier, the uphill time is 19.73 mph and downhill 25.54 mph, for completion times of 4:43.44 and 3:38.96 respectively, for a total of 8:22.41. That's only 1.37 seconds slower than the lighter rider. The heavier rider completes the climb in more time but descends faster, but this means a more variable speed and thus higher aerodynamic losses.

But in practice, a heavier rider is also able to produce more power, and this gap will be less.

We can continue to run numbers, but I believe the point has been made: 1% is a very slight grade, with only minor effects. If you had said 10%, we could definitively say the course will be slower, and it will favor smaller, lighter riders which tend to have a higher power to weight ratio. But at 1% aerodynamic effects are still dominant by a significant margin, and as such neither the completion times nor the optimal rider type are very much different from a flat course.


Assume a 40% incline - it will take at least an hour to get up that (plus about 2 minutes to come down). You ought to be able to do much better on a flat course. ;)

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.