# Electric Longboard Gear Ratio

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1. Sep 26, 2016

### Gurfin321

I would like to know what gear ratio i would need for my electric longboard. The goal is to run the longboard at 30 - 40 km/h at max power. Me and the longboards weight is approximately 65 kg. The brushless motor i am running has an rpm of approximately 18000 rpm (as it is an airplane engine) and a torque of 1,24 Nm. I would like to be able to go up a 10 degrees incline at 10 - 20 km/h. The longboard wheel is 75 mm in diameter. My question to you is: "How big a gear ratio should i use, and achieve said goals?". I would like to know all formels and equations, and also all the variables.

//Josef

2. Sep 26, 2016

### Staff: Mentor

Welcome to the PF.

Is that the kind of electric motor that is routinely used in commercial battery-powered longboards? Such a high RPM motor (which is useful for RC aircraft would seem to be not well suited to a lower-RPM higher-torque application like a longboard.

3. Sep 26, 2016

### Gurfin321

No it is not, however i can if needed switch motor, as long as it will suit my needs, budget is not a grate problem. I would like to get it as cheap as possible, while stil remaining top quality.

4. Sep 26, 2016

### Staff: Mentor

From an engineering perspective, the losses in such a high gear ratio will cost you power and battery life. It would be much better to find out what type of motor is used on commercial longboards, and see how inexpensively you can buy one (like on eBay or similar). Do you have access to a commercial battery-powered longboard that you can check the motor type/brand/size? If not, I have a good friend who commutes to Stanford on his battery-powered longboard. I could ask him...

5. Sep 26, 2016

### Gurfin321

I do not have access to an electric longboard, so it would be greatly appreciated if you could ask him for advise. Thanks!

6. Sep 26, 2016

### Staff: Mentor

Will do.

7. Sep 26, 2016

### Staff: Mentor

Here is his reply. Can you follow up on the Internet info about the board and motor?

8. Sep 26, 2016

### Gurfin321

9. Sep 26, 2016

### jack action

Quick analysis:

Possible top speed $v_{max}$ in (m/s):
$$v_{max} = \sqrt[3]{\frac{P_{max}}{0.5\rho C_DA}}$$
Where:
• $P_{max}$ = maximum motor power (2200 W, from your source);
• $\rho$ = air density (1.225 kg/m³);
• $C_D$ = drag coefficient (http://www.taylors.edu.my/EURECA/2014/downloads/02.pdf [Broken] $\approx$ 1.00);
• $A$ = frontal area (standing human $\approx$ 0.9 m²).
This gives 15.86 m/s or about 57 km/h.

Power required to go 15 km/h (= 4.17 m/s) on a 10° incline:
$$\begin{split} P &= 0.5\rho C_D A v^3 + (mg\sin\theta) v \\ P &= 0.5 (1.225) (1.00) (0.9) (4.17)^3 + ((65)(9.81)\sin(10)) (4.17) \\ P &= 502\ W \end{split}$$
The gear ratio needed (assuming there is sufficient power at the motor rpm):
$$GR = \frac{rpm_m}{rpm_w} = \frac{\pi}{30}\frac{rpm_m r}{v}$$
Where:
• $rpm_m$ is the motor rpm (rpm);
• $rpm_w$ is the wheel rpm (rpm);
• $v$ is the speed (m/s);
• $r$ is the wheel radius (m).
Say you have a 70 mm wheel (= 0.035 m radius), then if you want to reach 40 km/h (= 11.1 m/s) when the motor is at 18 000 rpm, then you need a gear ratio of 5.94:1. But you may not need to reach that rpm since you have enough power to reach 57 km/h.

Of course, you need to make sure the motor can produce the required power at every speed (i.e. considering actual rpm motor at that speed), in every condition you expect (i.e. incline), with the gear ratio selected.

Last edited by a moderator: May 8, 2017
10. Sep 27, 2016

### Gurfin321

THANK you! Great help, really got stuck with this one.

Where did you learn all this, i would like to know the source as it might come in handy to solve other practical problems.
Now me and my buddy are gonna build an awesome longboard.
Again Thanks!

//Josef

11. Sep 27, 2016