# What forces the maximum speed of a cyclist on a steep climb?

I know that on a flat road drag is the main force, I have to provide as much force as the drag (and the friction in the drive train) produces. I can also see that the faster I ride, the harder it gets and at speeds over 35 km/h that if I bend down to the handlebars, I can ride faster. But what about steep climbs? On a 15% climb I can barely ride at 5-6 km/h and the drag doesn't seem to be noticeable (I don't seem to go any faster when there's a moderate tailwind). I need to produce more force to combat gravity, but that does not depend on my speed - why can't I ride on the same climb at e.g. 9 km/h speed?

On a steep climb we can ignore drag and friction. If you wish to sustain higher velocity, you need to invest the same amount of energy in a shorter time, i.e. you need more power. This energy goes into your gravitational potential energy, given by $$E_p = mgh,$$ where $$m$$ is your mass (together with your bicycle), $$g$$ is gravitational acceleration and $$h$$ is height.
If your velocity is $$v$$ and angle of inclination you climb is $$\alpha$$, then your vertical component of velocity is $$v_z=\frac{d h}{d t}=v \sin\alpha.$$ In order to sustain this, the required power is $$P = \frac{d E_p}{d t} = mgv_z = mgv\sin\alpha.$$
You can see that the required power is linearly proportional to velocity. If the maximum power you can sustain is $$P$$, the top speed you can sustain is given by: $$v_{max}=\frac P{mg\sin\alpha}$$