I calculate ~4.72:1
Spreadsheet doesn’t paste in …

It’s attached.

I used the back EMF constant to compute the motor speed from the applied Voltage minus the I*R voltage drop.

The ratio of the motor speed and the load speed (given) defines a gear ratio.

Given the power at the load, compute the torque on the load side.
Reflect this torque back to the motor side (1/N).

Given the motor side torque, use the motor torque constant to compute the motor current. Here I broke the algebraic loop, by using a separate cell.
The motor current then changes the IR Voltage drop.

For my previous post, I neglected gear/belt efficiency.
Was that the 80% you were referring to?
If it is, I used 100% - not realistic.

I am attaching a revised spreadsheet - I believe I have resolved the error.
I have included the Calculations you provided.
At 100% efficiency, the motor mechanical power and the fan mechanical power are identical.

Along the way, I added an efficiency factor for the gear/belt system.
80% => N = ~4.678, Mechanical motor power is ~25 watts, Electrical motor power is ~30 Watts.

In an ideal motor, voltage vs. free speed is a linear relationship and voltage vs. torque is also linear (therefore voltage vs. mechanical power is quadratic).

Maxima 5.27.0 http://maxima.sourceforge.net
using Lisp GNU Common Lisp (GCL) GCL 2.6.8 (a.k.a. GCL)
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
freeSpeed: (motorVolts/specVolts)*freeSpeedSpec$
stallTorque: (motorVolts/specVolts)*stallTorqueSpec$
motorSpeed: fanSpeed*gearRatio$
motorTorque: stallTorque*(1-motorSpeed/freeSpeed)$
motorPower: motorTorque*motorSpeed$
fanPower: motorPower*powerEfficiency$
ev(fanPower,
motorVolts=7.8,
fanSpeed=2500*%pi/30,
gearRatio=[4.65324181,4.678],
powerEfficiency=0.8,
specVolts=12,
freeSpeedSpec=18730*%pi/30,
stallTorqueSpec=100.54*0.00706155
)$ float(%);
[20.00000014978008, 17.80773091738802]

Whoops, divided free speed by motor speed, that’s not a gear ratio… Okay, that should be a ~4.618 speed up. With the motor spinning at 541.4 RPM. And then melting down.

Also, my bonus problem is actually a lot easier, since there’s no quadratics involved and only one solution.

I’m assuming that you’re going to give the motor 12V now…

As the power efficiency is constant, and the power required to run the fan is a monotonically increasing function of fan speed, the maximum speed you can spin the fan is such that the motor is running at maximum power.

Max power out of the motor at 12V is 348.1W*, which is 13.92 times as much as in the original problem (20W/80%=25W). As the fan power requirement is proportional to the cube of the speed**, the fan is running 2.406 times as fast as the original problem, or 6,014 RPM. At max power output, the motor is running at 6087 RPM*. So the required gearing/belting is a 1.012:1 reduction. In practice, 1:1.

These numbers from the previous post.

** From Kevin, but I’ll buy it because energy is proportional to mass times speed squared and mass per unit time is proportional to speed, making power proportional to speed cubed.

It’s not just a suggestion, it’s the law. I had reason to discover this when a client asked about upsizing a fan motor on a system someone else designed.

Your math is right for the RPMs at 12V, but you calculated your ratio at the 7.8V free speed/peak power. I realized I didn’t specify 12V or 7.8V, so I ran numbers for both:

@12V:
Motor Power = 347W
Fan Power = 347W * 80% = 277.6W
Power Ratio = 277.6W / 20W = 13.88
Speed Ratio = 13.88^(1/3) (cube root) = 2.403
Max Power Speed = 2500 * 2.403 = 6007.5 RPM
Gear Ratio = (18700 RPM / 2) / 6007.5 RPM = 3.113
@7.8V
Motor Power = 347W * (7.8/12)^2 = 146.6W
Fan Power = 146.6W * 80% = 117.3W
Power Ratio = 117.3W / 20W = 5.864
Speed Ratio = 5.864^(1/3) = 1.803
Max Power Speed = 2500 * 1.803 = 4508 RPM
Gear Ratio = (18700 * 7.8 / 12 / 2) / 4508 = 1.348