A small incline angle θ on a still-air treadmill supplies the same power as the aerodynamic drag the athlete overcomes outdoors at race pace.
Find. The treadmill incline angle $\theta$ that reproduces the outdoor wind-resistance power at the same running speed.
Approach. Compute the race pace, then the aerodynamic drag force and the power it consumes outdoors. On a still-air treadmill there is no relative wind, so the equivalent extra power must instead come from climbing the incline; equate the two powers at the SAME speed (the speed cancels), leaving a direct equation for $\sin\theta$.
Outdoor aerodynamic drag and power.
$$F_d=C_D\left(\tfrac12\rho V^2\right)A=(1.1)\left(\tfrac12(1.19)(4.167)^2\right)(0.7)=\boxed{7.95\ \text{N}}$$
$$P_{wind}=F_dV=(7.95)(4.167)=33.1\ \text{W}$$
Treadmill incline power balance. On the incline, climbing at speed $V$ along a slope $\theta$ raises the athlete's height at rate $V\sin\theta$, so the extra power is $mgV\sin\theta$. Setting this equal to the outdoor wind power (the speed $V$ cancels from both sides):
$$mgV\sin\theta=F_dV\ \Rightarrow\ \sin\theta=\frac{F_d}{mg}=\frac{7.95}{(90)(9.81)}=9.01\times10^{-3}$$
$$\boxed{\theta=0.52^\circ\quad(\text{grade}\approx0.90\%)}$$
