Why would it make you work harder if you incline a stationary exercise bike? What's the physics involved? Assuming all other factors remain constant and only the incline changes, why would you burn more calories?
$\begingroup$
$\endgroup$
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
2
A stationary exercise bicycle presents a load to your legs which is entirely frictional and is generated either by a drag brake acting on the wheel, a fan that stirs up air, or an electromagnet that induces eddy currents in the wheel rim.
None of these friction sources has anything to do with gravity; they would all work just fine in deep space- and therefore the load on your legs is independent of the direction of the gravity vector.
This means that tilting the stationary exercise bike one way or another will have no effect at all on the difficulty of turning the pedals on it.
answered Jul 19, 2021 at 6:58
-
$\begingroup$ So if I hang a bike upside down from the ceiling and try to ride it, it's the same effort as sitting on the floor upright??? $\endgroup$ Commented Jul 20, 2021 at 18:24
-
$\begingroup$ yes, plus the effort required to hang onto it!!! $\endgroup$ Commented Jul 20, 2021 at 21:31
$\begingroup$
$\endgroup$
7
When tilting the bike machine, you are adding an additional gradient that you must now ride against. This increases the amount of power your body needs to output to maintain the same speed.
It might seem counterintuitive, but remember that the floor underneath the exercise bike keeps moving, while the incline remains, as if you were actually riding up a hill.
If you are riding on a flat surface, the total power output is $$P= F_R\cdot v$$ which is simply the power needed to overcome the resistance of the machine so that you are riding with a speed $v$.
If you were to tilt the machine, an additional factor comes into this equation, and that is the power required to overcome the gradient. Assuming you maintain the same velocity in each case, the power you require is now $$P=(F+m g\sin\theta)\cdot v = F\cdot v + m g\sin\theta\cdot v$$ where $\theta$ is the angle of the incline from the horizontal. In other words, now you are riding against the resistance of the machine plus a gradient when the bike is inclined.
-
$\begingroup$ So F is the force with which I press against the pedals, m is my body mass, g is gravitational acceleration, and theta the incline in degrees, correct? $\endgroup$ Commented Jul 20, 2021 at 18:30
-
-
$\begingroup$ So I incline the bike 10 Deg. I press on pedals with 50lbs force, turn the pedals at 10mph. I weigh 315lbs (that's the mg), sin(10deg) = .1736. P = 50(10) + 513(.1736)(10) = 500 + 546.84 = 1046.84. So by inclining the bike 10 deg, I need more than double the power output to keep pedaling at 10mph at the same resistance as when level, correct? Does that sound right??? $\endgroup$ Commented Jul 21, 2021 at 18:29
-
$\begingroup$ Looks about right but you have written $513(.1736)(10)$ instead of $315(.1736)(10)$. According to your calculation, the $500$ is the amount of power output you need without the incline. You then add the $546.84$ part which is the contribution due to adding the incline of $10^o$. I have not checked your calculation, but if it is correct, adding the incline doubles the amount of work/time as compared to level ground. $\endgroup$– joseph hCommented Jul 21, 2021 at 20:57
-
$\begingroup$ Apoligies for my fat finger. Thanks for confirming! $\endgroup$ Commented Jul 21, 2021 at 22:10
$\begingroup$
$\endgroup$
1
You would need to counter your gravitational potential energy when on an incline so you would need to work harder. While on a stationary bike there is no change in gravitational potential energy. Essentially on a inclined bike you have to work against gravity pulling you down.
answered Jul 19, 2021 at 4:24
-
$\begingroup$ I have to work against gravity pulling me down when it's level. How would I determine how much harder I'd work if inclined? $\endgroup$ Commented Jul 20, 2021 at 18:25
$\begingroup$
$\endgroup$
3
To the best of my knowledge, it would not. Perhaps pedalling would become more awkward due to the incline but in terms of the mechanics/physics of the bike it would be the same.
Most stationary trainers have a resistance setting to deal with the difficulty of the incline. But this is not really your question.
I can’t imagine any significant mechanism that would make an incline burn more energy. Is there some mechanism you are thinking of when asking this question?
-
$\begingroup$ Yes, a Harvard Health Publication article indicated the following: On a flat surface, a stationary bike burns about 391 calories per 30 minutes for a 155-pound rider, if you work vigorously. If you use an incline, however, you’ll burn more calories because you must work harder. -- I want to know how to determine how much more calories I'd burn because of a 10 degree incline up. $\endgroup$ Commented Jul 20, 2021 at 18:23
-
$\begingroup$ Very interesting, could you please link the article? I wonder if they have a system setup so that if stationary bike is inclined, it will automatically increase the resistance. $\endgroup$ Commented Jul 20, 2021 at 18:34
-
1$\begingroup$ Harvard link: ttps://www.health.harvard.edu/diet-and-weight-loss/calories-burned-in-30-minutes-of-leisure-and-routine-activities Link to article which indicates using incline takes more effort. livehealthy.chron.com/incline-bikes-vs-bicycles-10261.html $\endgroup$ Commented Jul 21, 2021 at 18:39