How to Get Work from a Variable-Inertia Flywheel 2022-01-0640

Suppose we have two identical variable-inertia flywheels and we connect them to the inputs of a differential. The output is connected to the driveline of a vehicle. There are several types of three-element mechanical differentials (e.g. ring-gear/carrier, epicyclic, etc.). The specific type of 3-element mechanical differential is inconsequential in the following analysis except to say there are two inputs (e.g. side gears) and one output (e.g. carrier/ring-gear). What’s important is simply the relationship - For example, using the notation ‘a’ for the first side gear and ‘b’ for the second side gear and ‘c’ for the carrier, then the relationship is: c=(a+b)/2. Understand that ‘a’, ‘b’, and ‘c’ can each be an input or an output.

Using the designation ‘omega’ (ω) then the relationship looks like this: ωc=(ωa+ωb)/2. So, we have one variable inertia flywheel (VIFa) and a second variable inertia flywheel (VIFb) connected to two side gears, a and b, and a vehicle driveline connected to the differential carrier, c.

For starters, we set the inertia of VIFa to its maximum inertia value, Ia_mx, and the inertia for VIFb to its minimum inertia value, Ib_mn. Then we will spin up (with some auxiliary power source) VIFa to a minimum initial velocity (ωa_mn) and, in a reverse direction, VIFb to its maximum velocity (ωb_mx). At this point, the ‘output’ to the vehicle driveline will be (ωa_mn-ωb_mx)/2. But, at these initial conditions we need for the driveline angular velocity to equal zero. This can only be done by incorporating speed-changing gear set between the flywheels and the differential inputs. This gear set can change the speed of either VIFa or VIFb or both. Let’s apply the ratio (r=ωa_mn/ωb_mx) equally to each differential input.

Let’s put some numeric values to this system: ωa_mn=10; ωb_mx=-30. If we want equal speed change for both flywheels as a percentage (i.e. equal ratio of change) then we can find the ratio (r) by taking the root: r=sqrt(ωb_mx/ωa_mn)=1.732. Thus, the initial velocity of each VIF at the input to the differential is: ωa_dif=r*ωa_mn=17.32 while ωb_dif=ωb_mx/r=-17.32.

What we have accomplished with these two gear sets applied to the differential inputs is that at the initial velocities for each VIF, the output of the differential is zero. The point of this is that when the vehicle is at a standstill, if we change the inertia setting of each flywheel, decreasing the inertia of VIFa (its inertia was initially at its maximum) while simultaneously increasing the inertia of VIFb (its initial inertia was set to its minimum value) we will cause the angular velocity of VIFa to increase (conserving momentum) while the negative angular velocity of VIFb will decrease (conserving momentum) and ,since VIFb is rotating in a reverse direction of VIFa, the torque produced by each flywheel as they change velocity will be in the same direction and applied to the differential output thus accelerating the vehicle.