# How do you calculate the best gear reduction for max speed?

I know that a gear reduction would decrease the RPM if the motor ran at constant speed, but in reality its speed is a function of the load.

I first tried turning a wheel with a direct drive system, but the motor started getting warm (burning off the lubricant inside) and was spinning much slower than without a load because it didn't have the torque required. I know that a gear reduction would cause less stress (torque) on the motor, which in turn, would allow it to [possibly] go faster. I am essentially looking for the highest speed I can get, so how would I determine the proper gear reduction ratio to use?

• You would need to inspect the RPM to efficiency graph Jul 15, 2015 at 15:24
• @ratchetfreak is the graph specific to the motor? If it is, I do not have it. This (comingsoon.radioshack.com/…) is the motor I used. Is there a way I can make the graph given the few specs listed? Jul 15, 2015 at 15:59
• why do you want to run the motor at its max speed? If you do that, you'll have to maximize the gearbox ratio (to get minimum torque at the motor shaft), to levels that would practically not work. Jul 16, 2015 at 9:03
• The motor can direct drive, but it gets very hot and the wheel spins slowly, so I think it would work with a 1:2 or 2:3 ratio. Jul 17, 2015 at 18:44

What you really need to do is measure the power required to turn the wheel at various RPMs. Then, match the power output of your motor to this data. So, for example, if your motor generates 1 HP, see what RPM value corresponds to 1 HP of required power (maybe leave yourself some margin and use 0.9 HP instead), then set your gear ratio accordingly to get these RPMs.

The hard part is, of course, measuring the power at different RPMs. You may have to do this by trial and error. You already know that direct drive (1:1 ratio) is too fast, so maybe try something like 3:2.

Measuring the rotational speed of the wheel is easy with a non-contact optical tachometer: I bought one a few years ago for less than \$50. The RPMs that the motor is rated to is usually stamped on the motor somewhere. Once you measure the maximum RPMs that you can turn the wheel at (based on the power of your motor), the drive ratio is just the rated motor RPM divided by the measured wheel RPM.

• How would I go about doing this? I am not sure how to count the RPMs unless I used an opto interrupter or similar (which would be over complicated). Once I do get the RPMs and HP, what calculation do I do to find the needed ratio? Jul 15, 2015 at 16:55
• @electricviolin the page you cite gives motor speed at load at various voltages. That can be turned into a rough speed/power /load graph. By measuring current at a given voltage for your load you can see how your load compares to their load. Go from there. Ask more if needed. Jul 16, 2015 at 8:22
• So I should put an ampmeter between the motor and supply and check the current it draws? I am only using 12V for this project... Please explain further Jul 17, 2015 at 12:47
• @russellmcmahon Jul 17, 2015 at 13:05
• @carlton would a 1:2 ratio work? I am not sure if that makes the torque half is stressful on the motor, but if it does, it may work best. Jul 17, 2015 at 18:46

If you have

• the power and torque curves (Assume the following power/torque curve in this example): Figure : engine power and torque curves for a car ICE engine of about 150 hp (source carthrottle.com )

• the diameter of the wheel (assume $$d_w= 0.7[m]$$ then the procedure is the following:

## procedure

1. you need to find the RPM at the maximum Power.

E.g in this example its 110 kW at 5500 rpm.

1. Find the top speed by equating the maximum power to the aerodynamic losses

The top speed is theoretically reached at the maximum power output, and the highest gear.

The top speed can be approximated by equating the car engine power and the losses (i.e. drag) using the following equation:

$$P = \color{red}{F_D} \cdot u = \color{red}{\frac{C_D}{2} A \rho u^2} \cdot u$$

so the top theoretical (see below why this is stressed) speed that engine can reach is:

$$u = \sqrt{\frac{2 P}{C_D A \rho }}$$

where:

• $$C_D$$ is the drag coefficient assume 0.35 for this example
• $$A$$ is the frontal cross-sectional area (Assume 2 $$m^2$$)
• $$\rho$$ is the air density 1.225 $$kg/m^3$$

For these value the theoretical top speed is approximately 228.75 kph (63 m/s)

1. Calculate the rpm of the wheel for the top speed

The rpm of the wheel $$n_W$$ are equal to:

$$u = 2\cdot \pi\cdot n_{w} \frac{d_w}{2}\Rightarrow n_{w} = \frac{u }{d_w\pi}$$ where:

• $$u$$ is the top speed in m/s
• $$n_w$$ is the rotational speed of the wheel in rpm
• $$d_w$$ is diameter of the wheel in meters

The result for the example is : $$n_w = 275.92[rpm]$$

1. Calculate the drivetrain ratio

The drivetrain ratio will be given by

$$i= \frac{n_{eng}}{n_w}$$ where:

• $$n_{eng}$$ are the rpm of the crankshaft when the engine is producing max power (i.e. when top speed is reached)
• $$n_{w}$$ are the rpm of the wheel at top speed.

for this example:

$$i= \frac{5500}{275.92}= 19.932$$

If you know the differential ratio and its e.g. $$i_d=4$$, then from the crankshaft to the differential the overall gear ratio for the top speed should be $$i_{ts}= \frac{i}{i_d}$$