It would be difficult to ride down a 10% grade and hold 200 watts. For me that would be approximately 53.5 mph or about 11 minutes.

Uphill 200 watts at 10% grade for me would be about 6 mph. For 10 miles that would be 1.66 hours or 1:40:00

So 1:51 overall for 20 miles

You're right that time taken to complete the route would vary substantially because drag losses are exponential with increasing speed.

But yes, average gradient is as simple as rise over run. It is a "state function," which means that all you need to do is compare the start and end points and find the gradient between those two points, and you have your answer, no matter what the many miles of roads do between those two points.

Average gradient is pretty useless for determining the difficulty of a ride, or anything at all, as you're probably figuring out now.

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Yep. Down at 52.5 mph and up at 6 mph is an "average" of 29.75 mph.

But 20 miles in 1:51 is an average of 10.8 mph. What gives?

For anything in series, you need a harmonic mean, not an arithmetic mean. You could also get there by averaging the minutes per mile instead of the miles per hour. For the harmonic mean you take the mean of the inverse and then flip it back over.

Point is, regardless of the physics or physiology, the slow sections slow you down more than the fast sections speed you up. If the down speeds you up 30% and the up slows you down 30%, they don't cancel out, you'll be 20% slower than just maintaining the same speed.

Flatter is faster.

Well, except for all downhill.

How long does it take to ride 10 miles at 6 mph?

Perhaps that's in jest and part of your point?

Actually, what the "many miles do between those two points" does matter. I used to live on a mountain which had two roads up: one on the front which was 3 miles and the other on the back which was 2 miles. The vertical diff was about 900 ft so going up the front was about a 5.7% avg gradient vs the back which was about 8.5%.

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If you’re starting and finishing at the same place your average is 0. I suppose you could calculate something like standard deviation from 0 of the gradient. That would probably be really useful, though I suppose just considering elevation gain per mile is already doing that.

Average gradient would just be the difference in the altitude between the start and finish divided by the length of the ride.

(end_alt - start_alt)/ride_length = ave_grad

So if you started a 10km ride at sea level and ended it at 100m then the average gradient was 1%.

If you then turned around and rode the 10km back to the start at sea level the average gradient becomes 0%....which is why the metric is kinda useless. All loop courses are 0% average gradient by definition.

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Another way to look at it is average absolute gradient, so a 2% uphill and a 2% downhill both count as 2% (not one as 2% and the other as -2%).

For a loop course, take the vertical climbed on the ride and double it then divide by the ride length. This gives the average grade of the road for the ride, independent of riding direction.

(2*ride_vert)/ride_length = |ave_grad| note: |ave_grad| is not the same as taking the abs value of the value from the first equation

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No.

Imagine a hill steep enough that it cuts your speed in half compared with riding on the flats. And then imagine doing an out-and-back ride on that hill, versus a ride of the same distance on pancake-flat ground. On the hilly ride, by the time you reach the top of the hill, you would have done exactly twice the distance on the flat ride... which means that the flat ride would be finishing at that moment! So no matter how fast your descent is, the hilly ride ends up slower.

Average gradient is somewhat useful for a climb, especially if it's fairly consistent. But it's utterly meaningless for a route, as others have suggested.
Distance and elevation is a more sensible way to describe a route. It doesn't tell you how steep the climbs and descents are but it gives you a good idea of the impact they will have on the ride regardless, since a short steep climb of a given elevation gain, is somewhat equivalent to a longer shallower climb of the same gain in terms of total effort and time - they're not equal, to be clear, but it gives you some sense of scale.

Any condition that slows your speed, whether headwind or climbing, will not be compensated by the equivalent tailwind or descent. There are a number of reasons for this, but rather than explain why, it may be best for you to think through a few examples to understand this. For example, if the first half of a route is uphill and is steep enough to halve your speed compared to riding on the flat, you would have to complete the second half instantaneously to manage the same time for the distance as if it were flat. Clearly this is impossible, as it is for any hill steeper than this. There is no magic happening as you reduce the steepness that makes the descent suddenly adequate to compensate once the climb speed rises above half the flat speed. The critical thing to spot is that the slow speed is happening for a much larger proportion of the ride than the high speed and has far more impact on the average than a much shorter stint at high speed. This is even more the case with wind, since the amount of headwind you experience is actually far greater than the amount of tailwind, since you spend far more time in it. Therefore wind is far more negative than positive on a looped course. The same is not quite true for elevation since a given amount of elevation equates to the same energy regardless how long it takes you to supply or harvest it.

Watch these videos, and then pick up the pieces when your mind explodes. All of these tracks have the same average gradient.

https://www.youtube.com/watch?v=_GJujClGYJQ

All, I'm going to give a shot at what the poster might be looking for..........versus what the math would dictate.

Often I see group rides posted as "feet per mile" where they take the total elevation gain you would see in Strava or your gps for a ride and divide it by the total distance in the ride.

Yes, if you end where you started your "net" is zero. However, that isn't useful for determining how long a ride may take you or what pace you or your group might expect over that route.

So, a flattish route might be 30ft per mile. Rolling might be 30 to 70 per mile. Hills might be 70 to 120 per mile. Mountains might be over 120ft per mile...........when you do that calculation.

For the original question, you won't output the same power up and down unless the gradients are very shallow and there are no corners or stops. So that immediately tosses the calculation.

To answer the question for practicality of planning a bike ride: yes total gain divided by total distance. Even the Tour de France has an "unspoken" guide of how they say a route for the day is flat, hilly, or mountainous. It generally deals with that calculation.

The result should be obvious. The greater dips result in a higher average speed. Take the limiting case where the "shallow dip" is zero. The ball wouldn't even move.

Doesn't relate to bike riding though, because:

a) We aren't coasting
b) Aero drag dominates

Note that the balls don't actually achieve the same height when they get to the other side. This is because the faster ball has lost more energy to friction (mostly aero drag I'd guess). It's a shame he didn't make a "fast" track with a simple slope like the "slow" one, and let them go for several laps. It would have shown the "slow" ball eventually overtaking the "fast" one.

It's not about how many miles you do against a given force (gravity or similarly wind), it is about the time spent working against the force. Even when you're doing a loop and get the tail wind for as many miles as you got the head wind, it will take you longer to do the headwind section than the tailwind section. So your watts are working against the wind for longer than your watts are working with the wind. The higher the average gradient (meters climbed / km's travelled) or the higher the wind speed the slower your average speed will be on a given number of watts because you will have been fighting the force of gravity or the force of the wind for longer than you took advantage of it.

The exact climb time would depend on CDA/position/wind but it is possible to calculate an approximate answer to your hypothetical climb and descent times with a fixed 200w power (let's assume 175lb rider and 15lbs bike) using this calculator: https://www.cyclingapps.net/calculators/climb-speed-calculator/ Model takes into account drag and gravity effects......here goes:


10mile ascent 10% = 124mins
10mile descent -10% = 10.5mins (no corners 😂)
20mile flat 0% = 77mins
Difference = 134.5mins - 77mins (57.5 mins) due to the climb or an additional 75% over flat 0% speed

So in your example it would take approx 1.75x longer to ride the 10mile up and down than to ride 20miles on the flat.

bw Alex @ fastfitness.tips

The OP asked that we ignore drag, etc. If we do the answer is yes, the hilly route would take the same time as a flat one -- but with zero drag, both routes would be completed very quickly.

That is incorrect. Ignoring drag, there is still the force of gravity being applied on the individual and the longer amount of time it takes to go up the hill than down it means gravity ends up being more of a negative than a positive and does not balance out. Depending on the course, ie: if there are a lot of short climbs and short descents, the impact can be mitigated because momentum will play more of a factor. The OPs example of a 10mile uphill then downhill there is no momentum effect and would be the slowest. If the course profile was 0.5 miles at 10% then 0.5 miles at -10% repeated 10 times, it would be slightly faster. 0.2 miles at 10% then 0.2 miles at -10% repeated 25 times would be slightly faster than the 0.5 example.

#going to become a classic SlowTwitch thread?

10% grade
75 Kg Bike+rider
CRR .008
CDA .250 (typical for me on road bike)

200 watts gets me 5.6 mph with
2.4 watts for wind resistance
14.7 watts from tire drag
183.3 watts for lifting the bike and rider on the climb

If you ignored drag from drive train losses, rolling resistance and air in this scenario 200 watts gets me:
75 Kg Bike+rider
CRR = 0
CDA=0
6.1 mph

One thing with shallower or shorter hills that are more repetitive than large climbs people forget is that you will probably wind up having to pedal down the hill possibly.

With big/long climbs, yeah, you don't get that "momentum" help up the next side of a hill. But with short/repetitive hills you likely never reach the "free" ungodly fast speeds a mountain descent allows.

So, even if you save your energy and don't pedal down short hills and get some "free" gain up the next one.......you never reached nearly the speed you would going down a big mountain.

Around here, on the 1x TT bike I'm often going downhill in the 54/11 at 200w at like 35mph. Mountain passes people go well into 40, 45, 50 mph.

Realistically I'd say I get about 7 to 10 feet "free" elevation from momentum on short repetitive hills.

This also assumes your repetitive hills begin exactly where the descent ends. Not always the case. Often a false flat, or flat area at the bottoms before you go up.

So I don't think this momentum thing plays as much a roll as people make out. You can find places it makes a big difference, but I don't feel I enjoy that much. 90% of the time there is a big curve or stop sign at the bottom of a hill negating any "free" momentum gain.

Agreed. And even with a perfectly designed rolling course to optimize momentum if you're keeping even watts it would still be slower than a flat course.

Not true. See my earlier post about harmonic mean.

Yes, even without drag or rolling resistance, the time will depend on the elevation profile (see brachistochrone problem).

Frankly, this is a bit of an ill-posed problem which invites many-flavored answers, but the gist is that you want some metric that represents the amount of climbing effort.

In general, total energy is all that matters. You can compare rides of any length or effort distribution by total energy expenditure, and anything else is biological or psychological.