# Why aren't average speeds computed over distance? [closed]

Average speed is computed so that average speed = total distance / total time. This means that if you go up a hill, which takes a long time, your average speed drops. This isn't regained on the downhill, since going down takes so much less time.

If you compute the average over distance instead (i.e., your speed every tenth of a mile), the uphill and downhill are equally weighted, since they have the same number of samples. This means that your average is closer to what your speed would be on flat ground.

Your speed over flat ground seems more important to me. It doesn't depend on how hilly your route is, and that way you could compare cyclists more easily. Why don't we compute the average speed over distance instead?

Clarification edit: I know that average speeds are computed with time because we care about the average speed -- it'd take me that same amount of time to travel a certain distance if I traveled my time-average speed.

I don't care about time anywhere here (minus speed) -- the definition of average speed doesn't bother me, I'm just wondering why this isn't a more widespread metric.

• Because that's the way it's done. Simple and easy to understand. You can always compute it the other way if you wish -- just learn how to program an Android. Commented Jun 15, 2013 at 20:41
• I don't understand what you mean by computing it over distance. Could you give an example of what that equation would look like? Commented Jun 15, 2013 at 21:07
• @jimirings One useful definition: calculate the average pace and then calculate its reciprocal Commented Jun 16, 2013 at 21:33
• As a tourist I want the standard scheme. If I've got 30 miles to go and I've been averaging 15 miles an hour I have a reasonable expectation of reaching my goal in about two hours. Using the OP's scheme I'd have no idea. Commented Nov 22, 2015 at 13:44
• What would be the use of your "average speed every 100m" metric? If I knew it, what would it tell me? What could I do with that value? I suspect the reason no-one calculates it is that the answer to my question is "nothing". There might be benefit to Sam's "time at speed" risk metric, but it would be a bit subtle to calculate because the relationship between risk and speed isn't simple.
– Móż
Commented Nov 25, 2015 at 20:22

This is an interesting point of view. Let's unpack this a bit.

Assume I have a ride that is 10 miles of flat, 10 miles of ascent, and 10 miles of descent.
On the flat I maintain a constant 20 miles/hour.
On the ascent I fall back to a constant 10 miles/hour.
On the descent I maintain a constant 30 miles/hour.

My average speed for this would be:
(10 miles + 10 miles + 10 miles) / (0.5 hour + 1.0 hour + .33 hour) = 16.39 miles/hour

By the proposed calculation I would have:
(100 tenths/mile * 20 miles/hour + 100 tenths/mile * 10 miles/hour + 100 tenths/mile * 30 miles/hour) / 300 tenths/mile = 20 miles/hour

If I had an average of 20 mph I should be able to complete the course in 1.5 hours. The trouble is that the course would actually take 1.83 hours to complete at the actual average of 16.39 mph. I know it hardly seems fair, since by all rights you did the vast majority of miles at 20+ mph. The caveat is that you spent by far the greatest amount of time at 10 mph.

• Can you always assume that your speed difference is the same for uphill and downhill? You can say that the force is always the same (mgsin(theta)), but there's also wind resistance (and other factors, I'm willing to bet) to factor in. Commented Jun 15, 2013 at 23:05
• @Scott - The speed difference could be the same, but the rider's power output could vary substantially in relation to a number of factors. The best metric is probably WTHarper's suggestion of work/distance or work/time. See the graph of resistance and velocity in their post. Commented Jun 15, 2013 at 23:23
• @Scott - You definitely cannot assume that the speed difference is the same going up vs down. Likely nowhere close. I may average 15 mph on the flat, hit 30 going down a given hill, without even pedaling, and go up that same hill at 6 mph. And consider that, as a hill gets steeper, the uphill speed will eventually reach zero (ie, the cyclist simply cannot climb that hill). But downhill speed can easily exceed 50 mph. Commented Jun 16, 2013 at 1:53
• This isn't worth adding another answer for, but speed is defined as the derivative of distance with respect to time: dx/dt. The average speed is almost the same formula: (end_x - start_x) / (end_t - start_t). No good can come from redefining well accepted mathematical definitions. Commented Jun 16, 2013 at 13:56

Your average speed is always going to be a measure of distance over a period of time. What I think you're trying to get at is accounting for your grade losses (i.e. riding up a hill), but it would be equally pertinent to include losses for wind resistance and friction as well. This would allow you to determine the amount of work per distance (or per time) for the trip, but calculating all of the variables which would verify your actual work is impractical for the average cyclist.

Products like Power Tap hubs allow you to generate torque and cadence data from your rides. That information would give you better insight into your speed with regards to sustained torque and cadence and it will also better illustrate where total resistance is increased or decreased (e.g. going up or down a hill.)

As far as taking average speed over a set distance, you're still averaging. It doesn't matter if your sample is 0.1 miles or 10 kilometers, it is still a measure of distance per time.

Here is some more information on calculating propulsion resistance. I don't remember how to do much calculus, but it is there for anyone better equipped.

• +1 for identifying average power as the measure that the poster is looking for. Commented Jun 16, 2013 at 13:49

The main issue with what you're talking about is that from a mathematical perspective, 'average' already has a strict meaning, and in this context nearly always means 'arithmetic mean'.

https://en.wikipedia.org/wiki/Average

If you want to talk about another measure of speed, you have to drop the term 'average speed' at the very least. What your method seems to be referring to would be something like the average of multiple constant interval average speeds, and as such two people both reporting their AoMCIAS could vary wildly, even if they traveled side-by-side.

For example, why pick tenths of a mile? There are infinite variants of division lengths unit choices such that two AoMCIAS numbers can't even be related (unless the distance has been mutually agreed upon before measurement collection). Good luck getting the U.S. to adopt the metric system or the rest of the world to adopt Imperial units, so at minimum you'd have U.S. bicyclists reporting it over 0.1 mile intervals and everyone else at, perhaps, 0.2 km intervals.

For that matter, why use length as the interval control? If you use a GPS, it's storing the data in semi-regular interval time units (depending on the unit and resolution setting), perhaps someone else tries to unify the U.S. with the rest of the world and proposes 10 second intervals as the new standard.

The result is that AoMCIAS speed numbers would not convey enough information in and of their own value, like averages do. You'd have to report them as "24.56 miles/hour AoMCIAS over 0.1 mile intervals", and that value would vary so wildly by interval choice that it could ONLY be compared to other AoMCIAS speeds with the exact same interval. There wouldn't be a static conversion that could be done, either, you would need to completely resample the intervals from raw data, if it was even available.

All of this is completely independent of it's relevance as a speed measure for bicycling (I have a mathematics degree, and don't time any of my bicycle rides). What I mean by this is: it's possible that you could devise a creative method such as AoMCIAS with an ideal interval distance or time such that the number reflects something more accurate about bicycling performance, and it may even be useful inside of the bicycling context (and likely ONLY the bicycling context). However, it will be of little to no value to anyone else, mathematically or even quick comparison-wise in colloquial speech. Two numbers could only really compared with equivalent intervals, and any ability to do neat things quickly in your head with such values would be relegated to special calculators, computer programs, and perhaps some genius savants.

• You cant't compare it - that's a good reason.
– Uooo
Commented Jun 19, 2013 at 8:25
• I think AoMCIAS = pace. That's what most people use when running. Commented Nov 29, 2021 at 9:43

By computing your average speed over a distance, you basically sample the distance (delta S) and measure the time (t_i) each time you reach the defined sampling distance. The formula then to compute your average speed would be:

Then the problem starts. By decreasing your sampling distance as previously proposed to increase your "accuracy", you will reduce the time difference (t_i - t_(i-1) ). Let say you decrease your sampling distance towards zero, your time difference will tend towards 0 to... Which lead to a mathematical problem of 0/0, which is undetermined... The only correct average speed you can get from this formula is by choosing your sampling distance to the complete distance of your ride. You have then only one sample (n=1) and t_0 is your start time and t_1 your end time.

But if you want to "mathematically" increase your average speed, then you can apply this formula and chose a sampling distance which corresponds to your wish.

1. Average speed is a useful measure for estimating your arrival time.
2. It's a simpler calculation (see other answers).
3. It is possible to estimate the effort involved in a ride, taking into account the elevation gained. See http://www.cptips.com/formul2.htm
4. Most competitive cycling sees winners as those who completed the course in the shortest time, which is related to average speed - distance/time. If you're going up and down the same large hill, it might take 30 minutes to climb and 5 minutes to descend. In terms of doing it in the shortest time, improving your climbing speed by 10% is going to make more difference than improving your descending speed by the same amount, saving 6 mins instead of 1 minute.

You can only accurately compare cyclists on the same course in the same conditions. Even then you'll find that some cyclists will be faster on a hilly course while others are faster on a flat course.

A statistic is a just a mathematical tool for summarizing data so as to answer a particular question (or set of questions).

The usual definition of average speed relates time and distance, and helps to answer questions which crop up regularly in practice: "how long will it take me to get home?" "can I make it to the café before it closes?" "do I need to take lights on this ride?"

It's not clear to me that there are any practical questions that your statistic helps to answer.

• It would at least likely match the algorithm used within Expresso exercise bikes when stating the speed of the "pacer". Commented Jun 21, 2013 at 1:03
• Exactly. When we have a set of different data we can simplify by replacing them all with the average and still get the same result in some calculation that we care about. If someone travels at various speeds on a route we can simplify by replacing each speed with the average speed and we still get the same duration.
– bdsl
Commented Jan 19, 2015 at 17:40

This is really a statistics / maths question rather than a bicycles question.

I think the sum you are proposing would be closer to the median speed than the mean (average). There are 3 key statistical measures that can all be useful:

Mean or average in the case of speed would be distance/time.

Let's say we take the average speed over every 1 minute over a 60 min ride, the median is the speed at which 30 of the samples are below the median and 30 are above.

The Mode is the most commonly occurring average speed.

Calculating the average speed after a tour is easy. You know the start time, the end time, and the length of your route. Therefore, you can find out the average speed.

Using "average over distance" approach: What if your track is so hilly that calculating every 10th of a mile is useless again? Then we would need to use every 100th of a mile. This will get more accurate, but will not be exact.

You could use a unit like "speed per meter". You would have to track the average speed of every meter then, which will be extremely difficult (even with assistance of a bike computer/smartphone). Still, every meter is an average speed, so, mathematically seen, it is still not exact. You would need the speed of an infinite small part of a meter to get the exact speed, which is impossible.

So, from my point of view, it is not done for two reasons:

1. Difficult to track
2. Mathematically incorrect

However, you can calculate anything the way you want. We will not tell anyone ;-)

• There are other, non-exacting ways to go about this, and they're used often in devices like GPSes. They receive and store point data - latitude, longitude, elevation and time in chunks as fine or course as your unit and settings specify. From this, you can get "speed between each point", and is one way they can use to calculate the total average, though the intervals are spaced by time units and not distance. For these devices it is both easy to track and mathematically as accurate as your device is calibrated to. Commented Jun 19, 2013 at 8:19
• @Ehryk `as accurate as your device is calibrated` - of course you can do that. But the device limits the accurateness (although it is still very exact). The question was `Why aren't average speeds computed over distance?` (in general), and this are the reasons I can think of. Nothing blocks you from calculating it differently, if you want to.
– Uooo
Commented Jun 19, 2013 at 8:22
• Agreed, I just wanted to add that the points you have apply more to human tracking, and aren't much of a factor to modern electronics. Commented Jun 19, 2013 at 8:25

You can. But to make the calculus work nicely, you need to invert it:

Consider `Is = 1 / s` where `s` is speed in meter per second.

If you go `100m @ 10m.s-1` and `100m @ 5m.s-1`, you mean speed is `sum(d) / sum(t / d)` for each segment, or `6.7m.s-1`

But your `Is` is `sum(d * Is) / sum(d)`, or `sum(d / s) / sum(d)`

Which is `1.5s.m-1`, or `(.1 + .2) / 2`

So as soon as you consider the number of seconds for a meter instead of the number of meter every second, it averages on distance.

But that does not make you go faster.

Indeed your average speed is total distance over total time

Regarding sampling this is where you go wrong:
"If you compute the average over distance instead (i.e., your speed every tenth of a mile), the uphill and downhill are equally weighted."
No they are not equally weighted. The denominator is time (not distance). You need to take even samples in time. You cannot sample on distance if you want to average the sample.

20 miles up the hill and 20 miles down the hill.
Up assume 10 mph and down assume 30 mph.

First total distance over total time
Total distance is 40 miles
Total time is 20 miles / 10 mph + 20 miles / 30 mph = 2 hour up + 2/3 hour down = 8/3 hour

Average speed = total distance over total time = 40 miles / 8/3 hours = 120 / 8 mph
= 15 mph

The average speed is not the 10 + 30 / 2 = 20 mph because spent more time at 10 mph. Spent 3 times as long at 10 mph compared to 30 mph.

If you sampled every mile then indeed you would get the wrong answer of 15 mph average.

But if you sampled every minute you would get the right answer.
Up the hill you would have 120 samples at 10 mph and down the hill you 40 samples at 30 mph.
(120 * 10) + (40 * 30) / 120 + 40 = 1200 + 1200 / 160 = 240 / 16 = 15.
If you want to average the sample then the sample needs to based on the denominator.

But it is easier to just use total time over total distance.

If you had a magic hill that was 10 miles up and 30 miles down then your average speed would be the average as you would spend the same amount of time going up as down.

You can use any number you want for up versus down the hill. 18 mph up and 20 mph down. The average speed will not be speed up + speed down / 2. Because you will spend more time at the lower speed.
Average speed is caluculation:
d is total distance
su is speed up and sd is speed down
total distance / total time
d / (d / 2*su + d / 2*sd)
d / ( sd*d/2*su*sd + su*d/2*sd*su )
d / ( (sd*d + su*d) / 2*sd*su ) d*2*sd*du / d(su + sd)
2*sd*su / (su + sd)
hill same distance up as down
average speed = 2*sd*su / (su + sd)
in statistics it is called the harmonic mean
try 10 and 30 and get 15
try 20 and 20 and get 20
try 18 and 20 and get 19.95

• Or just view it from the opposite angle: instead of speed (km/h), average the pace (~s/km). Commented Nov 29, 2021 at 19:55

Mainly because it doesn't make mathematical sense. Consider a vertical cliff (the problem is not really related to cycling) of say 1km.

Person A: climbs up and down in 4 hours total

Average 2km/4hours = 0.5km/h

Person B: climbs up in 10 hours carrying a parachute and down in 36 seconds.

Average (0.1km/h + 100km/h)/2 = 50km/h

It doesn't make any sense that the person who takes twice as long has 100 times the 'average' speed.

(It's also a reminder of why driving fast in a city doesn't get you to your destination much quicker).

The OP is right. There's really nothing special (other than convention) about averaging something over time versus averaging it over something else (like distance). If you computed "average speed over distance", you wouldn't necessarily get a lower average speed on a hilly course like you do when you calculate average speed over time. That would be great and useful for comparing various athletes or even yourself on different days or courses. Another great thing would be that your average speed over distance would not go down if you stopped! All of us would love that.

You would calculate average speed over distance like this: Sum of (instantaneous speed * roll-out distance) / total distance. On a bike with a magnet on the wheel, every time there is a pulse you just multiply your instantaneous speed by your roll-out and add the product to a running total. To get your average speed over distance at any point, just divide the running total by the distance you have covered to that point.

I think most people arguing against this don't get the concept. Trust me, once you get past the shock and disorientation of NOT averaging over time, it's really simple and useful. Too bad there aren't any devices that calculate average speed this way.

• Consider if I travel 10 miles at 10 miles per hour in one hour. Then the next day I travel 9 miles at 10 miles an hour, stop and stand for an hour, then do the last mile at 10 miles per hour. By the OP's method both of these were 10 mph runs, even though the second took twice as long as the first. Commented Nov 22, 2015 at 13:40
• @DanielRHicks Idle time is a separate question - spending a couple of minutes stopped at the traffic lights, does your stopwatch pause? Its a short recovery rest, but you still have to get back to speed on the green light.
– Criggie
Commented Nov 22, 2015 at 19:35
• @javaman The word "speed" has a definition of "distance / time" so when you use the word in other ways, it makes for confusion. What are the units of your average {speed}- metres/second still? Then its a speed. If your units are something else, what? On re-reading I still don't get it. Are you referring to splits where the window is not the entire trip, but shorter lengths like 100m and 1km ? The answer is still a speed in m/s
– Criggie
Commented Nov 22, 2015 at 19:40
• @Criggie - Suppose I didn't stop, but simply slowed to a crawl. Same difference, same absurdity. Commented Nov 22, 2015 at 19:54
• @Criggie we could just redefine "distance" in this case to mean what other people call time, and then it all makes sense.
– Móż
Commented Jun 21, 2016 at 3:58

Your speed over flat ground seems more important to me.

Or you want to ignore outliers, correct? The median is remarkably suited for that. Why not collect speed values and then compute the median?

The two methods, collecting the speeds via time or distance are just different sampling methods to estimate the distribution of speeds. One may give a better approximation of the speed distribution, but hopefully the median is invariant to this.

Computing the median has solid theoretical backing: OP would like to minimize the effect of outliers on the data. This is a feature of medians.

We can think of our speeds as a random variable X that changes according to terrain/weather/etc. In continuous time, thinking about the probability distribution function (PDF) of X implies that we collect data points at every instant.

But given some samples, how can estimate the PDF? Probabilists have considered this problem, and have developed maximum likelihood estimation and the Bayes estimator (which requires a prior on the PDF of your speeds X).

EDIT I have generated graphs that show the difference between time- and distance-based sampling.

With this data, we collect median values of

``````| Sampling method | Median | Mean  |
| --------------- | ------ | ----- |
| native          | 16.02  | 15.68 |
| time            | 15.74  | 15.36 |
| distance        | 16.33  | 16.02 |
``````
• "collect your data over time or distance -- in this case, I would image it doesn't matter" but you would be wrong to do so, for the reasons pointed out with your original proposal.
– Móż
Commented Jun 20, 2016 at 3:18
• This is the first time on any SE site that I've seen an answer with a negative score marked as correct! Commented Jun 20, 2016 at 8:18
• @andy256 note that the answer is from the person that asked the question.
– Móż
Commented Jun 21, 2016 at 2:52
• @Móż Yes, I belatedly spotted that. Since the answer is factually, mathematically incorrect, I don't see how we leave it as it is on a site that is supposed have reference quality information. I'll try flagging as low quality. Commented Jun 21, 2016 at 3:02
• @andy256 I have made it more clear that on typical routes, I believe the method to collect data shouldn't matter. The justification behind this answer is mathematically correct. Commented Jun 21, 2016 at 3:14