- #1

- 5

- 0

Thanks in advance!

//Josef

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Automotive
- Thread starter Gurfin321
- Start date

- #1

- 5

- 0

Thanks in advance!

//Josef

- #2

berkeman

Mentor

- 60,894

- 11,280

Welcome to the PF.

Thanks in advance!

//Josef

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

- 5

- 0

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.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.

- #4

berkeman

Mentor

- 60,894

- 11,280

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...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.

- #5

- 5

- 0

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

- #6

berkeman

Mentor

- 60,894

- 11,280

Will do.

- #7

berkeman

Mentor

- 60,894

- 11,280

I'm in the Idaho backcountry biking between hot springs . My board is a Boosted Dual Plus. Not sure whose motor they are using, but there is a teardown on the net so you can find it. Battery is 99 Watt-Hours.

Cheers

- #8

- 5

- 0

I was not able to find boosted motor however, on some other longboard diys they used motor like these http://www.hobbyking.com/hobbyking/...sk3_5055_280kv_brushless_outrunner_motor.htmlHere is his reply. Can you follow up on the Internet info about the board and motor?

While my motor should be even more powerful! (Sorry swedish site) http://www.hobbex.se/sv/artiklar/rimfire-80-50-55-500-borstlos-elmotor.html

Will it work?

- #9

- 2,250

- 4,069

Quick analysis:

Possible top speed ##v_{max}## in (m/s):

[tex]v_{max} = \sqrt[3]{\frac{P_{max}}{0.5\rho C_DA}}[/tex]

Where:

Power required to go 15 km/h (= 4.17 m/s) on a 10° incline:

[tex]\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}

[/tex]

The gear ratio needed (assuming there is sufficient power at the motor rpm):

[tex]GR = \frac{rpm_m}{rpm_w} = \frac{\pi}{30}\frac{rpm_m r}{v}[/tex]

Where:

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.

Possible top speed ##v_{max}## in (m/s):

[tex]v_{max} = \sqrt[3]{\frac{P_{max}}{0.5\rho C_DA}}[/tex]

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).

Power required to go 15 km/h (= 4.17 m/s) on a 10° incline:

[tex]\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}

[/tex]

The gear ratio needed (assuming there is sufficient power at the motor rpm):

[tex]GR = \frac{rpm_m}{rpm_w} = \frac{\pi}{30}\frac{rpm_m r}{v}[/tex]

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).

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:

- #10

- 5

- 0

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

Possible top speed ##v_{max}## in (m/s):

[tex]v_{max} = \sqrt[3]{\frac{P_{max}}{0.5\rho C_DA}}[/tex]

Where:

This gives 15.86 m/s or about 57 km/h.

- ##P_{max}## = maximum motor power (2200 W, from your source);
- ##\rho## = air density (1.225 kg/m³);
- ##C_D## = drag coefficient (standing human ##\approx## 1.00);
- ##A## = frontal area (standing human ##\approx## 0.9 m²).

Power required to go 15 km/h (= 4.17 m/s) on a 10° incline:

[tex]\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}

[/tex]

The gear ratio needed (assuming there is sufficient power at the motor rpm):

[tex]GR = \frac{rpm_m}{rpm_w} = \frac{\pi}{30}\frac{rpm_m r}{v}[/tex]

Where:

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.

- ##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).

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.

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

- 2,250

- 4,069

Where did you learn all this, i would like to know the source as it might come in handy to solve other practical problems.

I've put it all in a website, theory and real world application.

The basic concepts you need to understand are Newton's 2nd law, aerodynamic drag, rolling resistance, friction and mechanical advantage. Mix well and pour.

Share: