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Say the same 5mile ride going at 5 miles per hour versus 10, 15 20 etc. Even though the faster speed means shorter ride does it also always result in more calories burned?

I have the same commute looking to up the exercise somehow.

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  • Next question is where do those calories come from, and is it the right source for your exercise goal? There is solid science behind need for a lot of low/medium effort, long distance training. Often the best way to up the exercise is simply take the long way. – mattnz May 21 at 22:58
  • @mattnz: I’d assume that for a given distance, riding it at higher intensity (i.e. faster) will always result in more “gains”. Of course the question is if it’s appropriate from a training volume and injury risk point of view. – Michael May 22 at 6:56
  • Have you ever driven a car and had to refuel? – user2705196 May 22 at 20:30
  • Is walking an option ? – Dan K May 23 at 11:10
  • @Michael speed is not the same thing as intensity. there are other factors like wind. – emory May 23 at 14:45
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All other things being equal, yes. The increase in effort more than makes up for the decrease in ride time.

Consider two scenarios where one rides 20 km: first at 25 km/h, second at 30 km/h. In the first case, riding at 25 km/h takes 96 W of power, in the second, riding at 30 km/h takes 150 W according to this web page (using my own stats).

 96 W * 20 km / 25 km/h = 276 kJ over 48 minutes  
150 W * 20 km / 30 km/h = 360 kJ over 40 minutes

Thanks to a lucky mathematical coincidence, kilojoules of work is almost exactly equal to calories burned. Some estimating tools treat calories burned as 10% above kJ of work.

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    But going faster does not (necessarily) imply burning more calories. One can go faster because they are in an aerodynamic "tuck" that lets them burn fewer calories. – Daniel R Hicks May 21 at 20:06
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    @DanielRHicks Right. Which is why I led off with "All other things being equal…" – Adam Rice May 21 at 20:09
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    "Thanks to a lucky mathematical coincidence, kilojoules of work is almost exactly equal to [kilo]calories burned" Do you have more info about this claim? There's a factor 4.18 between a cal and a Joule. Do you consider the human efficiency to be ~25%? – Eric Duminil May 22 at 10:47
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    @EricDuminil 25% is not precise but it's close enough to be useful. – user_1818839 May 22 at 12:54
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Speaking purely in calories and power, this is just a small elaboration on Adam's answer: the power required to maintain a certain speed against aerodynamic drag is proportionate to the cube of that speed. The time to complete a distance is inversely proportional to speed. Thus, you will always burn more total calories if you increase your speed. The fact I asserted about power and aerodynamic drag is available in the calculator that Adam linked.

I am oversimplifying a bit, since I'm ignoring elevation change and rolling resistance. The power to overcome rolling resistance (that is the power consumed to overcome the friction between tires and ground) is linearly proportional to your speed. However, I believe that the speed where aerodynamic drag is greater than rolling resistance is quite low, possibly in the region of 10mph. Whatever the case, the total power required to hold a certain speed does not increase linearly, thus giving you your answer.

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