# Why does this fan with one blade missing rotates counterclockwise while running? The fan worked just fine until my friend tried to stop the spinning blades with her finger and knocked one off. Now it always rotates counterclockwise when running. Can someone explain in details why? Does this have something to do with the shape of the blades?

• Unbalanced fan generates huge vibration; which direction it moves or rotates depends mostly on strength of friction at each contact point on the table. . Aug 24, 2022 at 12:19
• @CarlWitthoft I don't think random contact friction with the table explains the consistent counterclockwise rotation.
– Eric
Aug 24, 2022 at 14:10
• When the unbalanced blade is going down, gravity helps you. when it goes up, gravity opposes you. So if it pushes less air when it goes up, that's a force to turn the fan. That says the fan should go counterclockwise if the blades spin clockwise when you look at them from the front. It should stop if you break off the opposite blade. I don't think the question should be closed either. Aug 24, 2022 at 14:13
• This is actually a good physics question. the unbalanced blade causes the whole fan to vibrate in a circular sense i.e., in the vertical and horizontal axes with a phase angle separating these components. this vibration causes the stiction between the fan base and the tabletop to come unstuck and depending on the phase angle, the fan base can shift position slightly while it is unstuck- and the fan then moves. this is the principle behind a vibratory bowl feeder used in assembly lines to sort, orient and present small parts to an assembly tool for automatic assembly. Aug 24, 2022 at 16:03
• When the unbalanced blade is on the left, it pushes air forward. When the unbalanced blade is on the right, it pushes air forward MORE. So there is more force pushing the fan backward when the unbalanced blade is on the right. It's a possibility. There could be some bigger force which happens to get the same result and which completely overshadows this one, though. Aug 24, 2022 at 16:10

From the video, it looks like that the axis of the fan rotor is not orthogonal to the vertical direction.

When a motor-driven rotor spins in a direction, a torque acts on it due to aerodynamic drag in the opposite direction, i.e. the torque can be represented by a vector orthogonal to the plane of the rotor, that can be approximately written as

$$\mathbf{M}^{aero} = - \dfrac{1}{2} \, C_T \rho R^3 |\mathbf\Omega| \mathbf{\Omega} = - \dfrac{1}{2} \, C_T \rho R^3 |\Omega| \Omega \,\mathbf{\hat{n}}$$ ,

being

• $$C_T$$ the torque coefficient depending on the parameters of the blades, of the rotor as a whole and non-dimensional coefficients describing the motion, like Reynolds number

• $$\rho$$ is the air density

• $$R$$ the radius of the rotor

• $$\mathbf{\Omega} = \Omega \, \mathbf{\hat{n}}$$ is the angular velocity of the rotor around its axis, following the right-hand rule for angular velocity. If we choose $$\mathbf{\hat{n}}$$ so that $$\Omega > 0$$, from what I get from the video, $$\mathbf{\hat{n}}$$ points slightly downward, so that we can write $$\mathbf{\hat{n}} \cdot \mathbf{\hat{z}} = \sin \alpha < 0$$, defining the angle $$\alpha$$ as the angle between $$\mathbf{\hat{n}}$$ and the horizontal plane. Now,

• if the axis of the fan is orthogonal to the vertical direction $$\mathbf{\hat{z}}$$, the rotor doesn't introduce any torque around the $$\mathbf{\hat{z}}$$-axis, since $$\mathbf{\hat{z}} \cdot \mathbf{\hat{n}} = 0$$ and this aerodynamic action is balanced by normal reactions exchanged by the table and the base of the fan

• if the axis of the fan is not horizontal (as it looks like from the video), the aerodynamic torque has a vertical component

$$M_z^{aero} = \mathbf{\hat{z}} \cdot \mathbf{M}^{aero} = - \dfrac{1}{2} \, C_T \rho R^3 |\Omega| \Omega \,\mathbf{\hat{n}} \cdot \mathbf{\hat{z}} = - \dfrac{1}{2} \, C_T \rho R^3 |\Omega| \Omega \sin \alpha > 0$$,

that makes the fan rotate around a vertical axis, in the counter-clockwise direction if we look at the fan from above, if the friction between the base and the table is not enough to balance the external action.

This process is independent from the missing blade, but only on the inclination of the rotor plane. You could test this explanation if you have a fan with all the blades, and play around with the direction of the axis of rotation. Even if you only have the broken one, you can change the orientation of the axis, so that it point upwards and see if the fan rotates around z-axis in the opposite direction.

Other contributions (I guess minor, but I currently have no time to do the calculations) could come from the unbalanced rotor.

• "This process is independent from the missing blade, but only on the inclination of the rotor plane." But when there was no missing blade, the fan doesn't rotate at all in spite of the same inclination.
– Eric
Aug 24, 2022 at 14:04
• Are you 100% sure? Have you tried with the full rotor? Lying the fan on the very same surface? (with rough surfaces friction may be enough to resist a small external torque). I still have to perform computation with an eccentric mass, and the "eccentric" aerodynamics. I'll be back in the next days Aug 24, 2022 at 14:10
• 100% sure. All other variables are equal, just the missing blade and it starts rotating! I also find Eli's answer reasonable, although I'm not sure about how $F_w$ comes to be.
– Eric
Aug 24, 2022 at 14:18
• Fw likely comes from lift on the blades. Maybe it's convenient to think at a "missing lift", coming from the missing blade. Anyway rotors are not trivial, since they act as a filter on the support: the missing lift rotates with the disk, I can't easily realize the influence on the fan. Some results may be counter-intuitive Aug 24, 2022 at 14:34
• I adjusted the rotor plane to make it orthogonal to vertical direction, the fan rotate the same. So we can rule out the inclination.
– Eric
Aug 24, 2022 at 15:07 the red points are the blades center of mass . the rotation about the y-axes ,cause a wind force $$~F_w~$$ towards the y-axes. the torque about the z-axes ,$$~\tau_z~$$ cause the ventilator to rotate .

with \begin{align*} \begin{bmatrix} \tau_{xi} \\ \tau_{yi} \\ \tau_{zi} \\ \end{bmatrix} =\begin{bmatrix} r_{xi} \\ r_{yi} \\ r_{zi} \\ \end{bmatrix}\times \begin{bmatrix} 0 \\ F_w \\ 0 \\ \end{bmatrix}\quad\Rightarrow \end{align*}

$$\tau_{zi}=r_{xi}\,F_w\quad\text{hence }\\ \tau_z=F_w\,\sum_{i=1}^n\,r_{xi}$$

the torque $$~\tau_z~$$ is zero only if $$~\sum_{i=1}^n\,r_{xi}=0~$$. obviously is for this ventilator not the case

• Why is $F_w$ towards the y-axis?
– Eric
Aug 24, 2022 at 14:08
• I remain unconvinced that the lack of a blade leads to this particular matrix operation. Any rotational effect in your diagram applies for all positions of the missing blade, leading to a net average zero torque. Aug 24, 2022 at 14:39
• @Eric this is what the ventilator dose , the rotation of the blades cause that you get air in your face ? the y-axis is arbitrary
– Eli
Aug 24, 2022 at 16:58
• @CarlWitthoft this is a simplification , but I don't think it is wrong .
– Eli
Aug 24, 2022 at 17:00
• So basically the rotation is just precession?
– Eric
Aug 25, 2022 at 3:59