Hi everyone, an interesting question came up today during my ride with a friend. Do you burn the same amount of calories when you do a 20 mile ride in an hour, compared to a 20 mile ride in half an hour. This assumes that both rides are under the same conditions. Flat, same route, same headwind/tailwind, etc.
My answer initally was yes because I thought that it would require the same amount of watts to move yourself and the bicycle 20 miles. But I don't know much about power meteres, or cycling power in general. My understanding is that you do burn a lot of calories in the first ride because of a higher speed, but since you ride less amount of time it should equal ride 2.
Hi everyone, an interesting question came up today during my ride with a friend. Do you burn the same amount of calories when you do a 20 mile ride in an hour, compared to a 20 mile ride in half an hour. ...
No, it takes more energy and thus more calories to cover the distance faster. Same way you burn through more gas to drive your car faster. Take a look at this chart of estimated calories burned while cycling: nutristrategy.comcycling.htmOpen ↗ Compare the estimate for 20 mph to the estimate for 10 mph. It takes a lot more calories to ride twice as fast even if you ride for half the time.
You mean in the sense of overcoming air resistance?
Assuming a dead flat road, then yeah the faster speed (20 v 10 mph) for half the time would be 2-3 times more calories.
Actually, I meant in the physiological sense, but you're right that more work is even being done by the faster rider by pushing against the greater resistance for the same distance.
If the power required to drive your bicycle twice as fast was twice as much, then the total calories would be the same (i.e., double the spend rate, but with half the total time). Except it isn't a linear dependence.
To go twice as fast requires more than twice the power, so the calorie expenditure is higher when you complete the same distance in less time.
If the power required to drive your bicycle twice as fast was twice as much, then the total calories would be the same (i.e., double the spend rate, but with half the total time). Except it isn't a linear dependence.
To go twice as fast requires more than twice the power, so the calorie expenditure is higher when you complete the same distance in less time.
Doubling speed takes from four to six times more power over the range of practical cycling speeds, the lower end for lower speeds (double speed from 5 to 10 mph) and the higher end for higher speeds (double speed from 15 to 30 mph).
This is due to the higher contribution of rolling resistance (linear with speed) at lower speeds. At very high speeds (think motorcycle) doubling speed requires almost an 8X increase in power.
Isn't that what I just said - it isn't a linear dependence.
And it isn't rolling resistance that's the big item, it's aerodynamic drag.
The entire equation of motion is nonlinear; however, there is a linear component (rolling resistance) that is pronounced at slow speeds. At 10 mph, 50% of the required energy is to overcome rolling resistance.
If anyone is interested, there are some models here to play around with. You can see for yourself what the major loss components are as a function of speed, and you can figure out the calorie expenditure.
