Currently I'm doing a personal project, but I've some doubts about the calculation process. Well, I've a platform (2.30m length x 0.32m width). This platform is supported by 4 wheels with a diameter equals 120mm. This platform loads a motorbike. The maximum mass (platform + motorbike) that it will move is about 400kg. The surfaces contact are rubber with concrete (coefficient of friction 0.45). Rolling coefficient I think is 0.02). I'm uncertain wether to put 2 driven wheels or 4. The platform must travel a distance of 7 m. For the process I propose a speed of 0.1 m / s. Well, How can I calculate the required torque to move this platform with a motorbike?
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2 Answers
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Let's get a few things out of the way first.
Torque or force have nothing to do with the speed, they interact with angular momentum or acceleration.
Also if the travel path is level you will be fine with just two driven wheels
To figure the force needed let's say we apply enough force so that at the end of the 7m the speed changes from zero to 0.1m/s' as you mention and then the brakes engage.
$ x=\frac{v^2}{2a} \ ,\ a=\frac{v^2}{2x}$
So $a=\frac{0.1^2}{2*7} =0.01/14=0.0007$
$ F=ma =400kg*9.8*0.0007=2.8N\quad 2.8*1.02 =2.86\ \text{after the friction } $
this force should be divided by two for individual wheel and converted to torque applied at radius of 0.06m.
$ 2.86/(2*0.06)=24Nm \ \text{torqe/wheel} $
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$\begingroup$ Weight (plaform + motorbike)=4905 N;Radius of wheel=0,06m:Desired top speed = 0,1m/s;Desired acceleration time = 3s;Maximum incline angle=5 degree:mu(rubber-concret)=0,45;Crr = 0,02;Total Tractive Effort (TTE): TTE = RR + GR + FA where: RR - Force necessary to overcome rolling resistance GR - Force required to climb a grade FA - Force required to accelerate to final velocity RR = 4905 x 0,02 = 98,1 N GR = 4905 x sin 5º = 427,5 N FA = 4905 x 0,1 / (9,81 x 3) = 16,67 N TTE = 542,27 N Tw(wheel torque) = TTE x Rw (radius wheel) x RF (resistance factor) Tw = 542,27 x 0,06 x 1,1 = 35,79 Nm $\endgroup$– JoanCommented Aug 20, 2019 at 8:32
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I put an answer but it don't reads correctly.. I put there:
My calculations are: Total Weight (plaform + motorbike) = 4905 N Radius of wheel = 0,06 m Desired top speed = 0,1 m/s Desired acceleration time = 3 s Maximum incline angle = 5 degree mu (rubber-concret) = 0,45 Crr = 0,02
To determine the Total Tractive Effort (TTE): TTE = RR + GR + FA where: RR - Force necessary to overcome rolling resistance GR - Force required to climb a grade FA - Force required to accelerate to final velocity
RR = 4905 x 0,02 = 98,1 N GR = 4905 x sin 5º = 427,5 N FA = 4905 x 0,1 / (9,81 x 3) = 16,67 N TTE = 542,27 N
Tw (wheel torque) = TTE x Rw (radius wheel) x RF (resistance factor) Tw = 542,27 x 0,06 x 1,1 = 35,79 Nm
Are there correct? Thanks