Torque and Speed of a DC Motor with Pulleys and Wheels

Question

I want to know how to calculate the speed of an electric skateboard

If i have that situation(from the picture below) i will need an acceleration=0.16m/s^2?

and a force F~=15N?

So i will need a power output Po=150W to move that skateboard with a velocity=10m/s ??? Can i calculate the torque that i need with this T=r*F ??

where r=radious of the wheel

and F i think that's the force that i need

if yes,i will need a torque of 0,036m*15=0.54Nm?

So,in that case,if i have a motor that has a torque=0.54Nm with a power output=150W will move the skate with a speed of 10m/s? With what i'm wrong??

I can calculate the torque of the motor with this

T = (I * V * E *60) / (rpm * 2pi)?

So in my case if i have a motor 270kv 1665W with Continuous current: 80A with 60%efficiency with a Lipo Battery 22.2V 10.4Ah 30C, I will get:

Power Input =22.2*10.4 =230.88W

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Power Output =230.88*0.6=138.52W

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RPM=270*22.2=5994RPM

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Torque=(10Ah*22.2V*60%ef*60)/(5994RPM*2*3.14)=7.992/37.642,32=0.21Nm?

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So what speed the board will have??

With what i'm wrong?Please explain

How the diameter of the wheel affect the speed and acceleration?

if i have 72mm wheels,so 0.22m circumference with 5994 rpm will have 0.22*5994=1.318m/min? so 21.96m/s?? if i'm not wrong,how the mass affect that speed? and the not enought torque?

How the diameter of the pulley A and B and the number of teeths and the distance between the teeths affect the torque and speed?

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How can i calculate the output torque and speed of the pulley A ??

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With what i'm wrong,how i can calculate what i need please be patient i never take physics/engineering class i just searched on internet don't be annoyed with my calculus

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Update: if i have torque=rm(v/t) i will get v=(timetorque)/(rm),in my case v=(60*0.21)/(0.036*90)=3.88m/s?? It's that right? If i get more torque with the pulleys from the motor to the wheel i can get something like 7m/s?

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Ok Update2: i have 2 gears ,one with 27T other with 14T ,i founded that the ratio is 27/14=1.92 so the torque is 0.21*1.92=0.4Nm? and the speed will be 5994/1.92=3121RPM?

If yes,i will have v=(timetorque)/(rm) V= 7.4m/s? i can't understand what i'm doing wrong..

Maybe if i think that i have enought torque to move the board,and if the wheel rotate with 3121rpm and 72mm diameter with 0.22m circumference,it will make 0.22m each rpm so 3121*0.22=686.62m/min=11.44m/s. What is the true ??#Please help !#

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Update 3: if i will use a 36T pulley with a 14T pulley on the motor with 138.52W power output,5994RPM,0.21Nm torque,70mm wheels

and i need a torque=0.54Nm

i will have 36/14=2.57 ratio ,so 0.21*2.57=0.54Nm torque and 5994/2.57=2332RPM

so each min i will make 0.22m(circumference of the wheel)*2332rpm=513m/min=8.55m/s??

Answer 1

Maximum speed is affected by air drag, rolling resistance and power of the motor.

The force pushing the skate is simple to calculate:

$$ f_a = k_m{ P \over v }$$

Where $P$ is the power output of the motor (including transmission losses) at the current skate speed $v$. With the right transmission you will achieve peak power at maximum speed.

Air drag force can be approximated to:

$$ f_d = k_dv^2$$

Where $k_d$ is the drag constant which you need to know, this depends on your body area, your shape, your height, density of the air and so on...

Then there is the wheels friction force which is highly variable and depends on many things:

$$f_f = f(v,m,typeOfTerrain) ??$$

Friction depends on your weight, the current speed, type and diameter of wheels, type of terrain, etc... It is a very complex phenomena.

To calculate max speed you will have to write the equation

$$ f_a = f_d + f_f$$

And solve for v. When the pushing force of the engine is equal to the losses you are at maximum speed.

But this is pretty much impossible without having a pretty good friction model for your skate wheels and also a given "standard" surface. Imagine trying to move with your skate on sand, you will be much slower than on a perfectly flat concrete surface...