On level terrain a bicyclist is faster than a runner.
On a steep slope a runner is faster than a bicyclist.
(That's why you see bicyclists pushing their bikes uphill.)
Can we calculate the angle where it becomes more advantageous to discard the bicycle and start running? If so, what is the formula?
Apparently we can calculate the steepest gradient for a bicycle. Rhett Allain (professor of physics at Southeastern Louisiana University) calculates a "a maximum incline of 38.7°" (80%) on a dry road. Cyclist Magazine writes, 60% is probably more realistic.
So the angle where cycling becomes less efficient than running must be at least a little less than that.
Let's assume the most basic model with "ideal" circumstances: a straight road with a constant incline, dry, asphalt, no wind, etc.
(As the question is probably already difficult enough, please don't artificially complicate it by introducing exceptional circumstances like nighttime, rain, hail, oncoming traffic, a flat tyre, alien attacks, etc.)
Best Answer
We can at least calculate the angle at which the bike can no longer make progress. In order to maintain any nonnegative vertical velocity, the distance-averaged output force applied by the cyclist to the ground via the rear wheel must equal the force of gravity down an inclined plane:
$\bar F_\text{out} = g(m_\text{man}+m_\text{bike})\sin(\theta)$
Let $D$ be twice the crank length of the bicycle pedal, that is, twice the distance from the pedal to the center of the front gear.
Let $L$ be the minimum displacement of the wheel along the slope per half revolution of the crank
$L = \frac{\pi}{2} \times \text {(Wheel diameter) (number of front teeth) / (number of rear teeth)}$
The ratio of output to input force for the bicycle is $R = \frac {\pi D}{2L} = F_\text{out}/F_\text{in}$
So at the point where $\bar F_\text{in} = \frac{g}{R}(m_\text{man}+m_\text{bike})\sin(\theta)$ exceeds the maximum force the athlete can supply to the pedal without falling off (or leaping off), the cyclist can no longer make progress. This must be less than $gm_\text{man}$, or the cyclist will jump off.$^1$
$\theta_\text{max} \lt \arcsin\left(\frac{Rm_\text{man}}{m_\text{man}+m_\text{bike}}\right)$
For a 70 kg man and a 10 kg bike, with a minimum $L$ of $25\pi$ cm and a $D$ of $34$ cm giving $R = 0.68$, that gives
$\theta_\text{max} \lt 37 ^\circ$
This is considerably higher slope than I'd consider possible to bicycle up, which probably reflects the fact that balancing your entire weight on one moving pedal of a slow-moving bicycle, just to keep from rolling down the hill, is much more difficult than just getting off and walking. (Not to mention slower, and likely to damage the chain or crank.)
Given Superman and a specially designed indestructible bike, any angle could be achieved, since the athlete could, hypothetically, pull up on the bike to counteract pushing down on the pedals and greatly exceed his own weight in thrust. However, such an athlete would be better suited to leaping the hill in a single bound.
1: I have assumed that the cyclist (except for Superman and his superbike) has no good way of pulling up on the bike in a normal cycling posture. It may be possible to exert more force on the crank by pulling up with the opposite foot, if the athlete is clipped in to the pedals. This would switch the problem from a question about maximum force to a question about the maximum lateral torque that can be applied without flipping the bike, which in turn is a question about how well and how far the athlete can alternate leaning back and forth to counterbalance. I think that puts the question out of reach of a first-principles approach.