Calculating motor force based on acceleration required 1

[makterna]

I am inventing a product which can in short be descibed as a 24V electrical vehicle which will transport itself along a special track, and I am unsure how to calculate how much torque my DC motor will need.

Car total weight: 100 kilograms
Required acceleration: 0.8 m/s per second (or 0.06 G)
Maximum uphill slope: 5%

Is this the correct formula to use: 6 Nm = 100kg * 0.06G

That can't be right..?

Do I also need to calculate tire friction or would that be marginal?

The speed will be less than 10 m/s (36 km/h).

I have searched for an online calculator page for these things but found none. I would appreciate any help with this embarrasingly simple question.

 

[AJone]

Force (in Newtons)=100kg* .8m/s^2=80N

-To convert into Torque, you need to take into account radius of wheel.


-This doesn't take into account the slope, however. To accelerate this fast uphill, you have to take into account the compenent of gravity pulling you down the hill.

See how far you can go with these clues and post your results.

 

[makterna]

Thanks for your help.

Weight: 100 kg
Required acceleration: +0.8 m/s per sec
Wheel radius: 0.04 m
Wheel width 0.04 m
Tire friction coefficient: 0.3
Uphill slope: 5%
Gravity acceleration constant: 9.82

Force needed for acceleration: 100*0.8 = 80 N
Force needed to conquer tire friction: 100*0.03 = 3 N
Force needed for moving upwards: 100*9.82*0.05 = 49.1 N

Torque required = (80+3+49.1) * 0.04 = 5.284
...or (49.1+3)*0.04 = 2,084 Nm if it is not required to accelerate when traveling uphill.

Air resistance has not been taken into consideration.

Have I made any mistakes now?

 

[AJone]

It's not correct to compute "Force needed for moving upwards" by multiplying by percent of the slope. You realy need to use trigonometry. However, in this case the answer works out to be pretty close. ( Consider a 100% slope which is only 45 degrees. 100*9.82*1.00 =982, but in reality only a percentage of the weight is pulling down the slope)

A tire friciton coefficent of .3 would probably be a static friction value and not appropriate at this point in the calculations. It would be used to verify the tires were sticky enough to transmit the required force to the track.

A fricition coefficent of around .03 may be a realistic value for rolling friction. In this case you would multiply .03 by the component of weight pushing normal to the ground. This value would then be the force required to conquer friction. This value will change depending on the slope.

Your method for computing torque appears to be appropriate. And the number you get for required torque above are probably within 15% of what is the minimum required.

As you note this is an idealized calculation and doesn't take into account alot of real world parameters and considerations such air resistance, recommended operating range of the motor, etc.

I hope this is enough information for the design to move forward.