1. Jun 21, 2013

### Xiliavus

Hey everyone just curious to how i would go finding the appropriate power needed to move a 20-25kg device around. I want it to be roughly walking pace so dont need a large motor or anything like that but could someone help as to find out approximately how much force, torque etc i would need to move it. I just need a push in the general direction i know most physics regarding newtons laws and all that just finding torques and stuff like that im getting a little confused. If anyone could help it would be great.

2. Jun 21, 2013

### Staff: Mentor

Coefficient of friction?
I would start with force and power. Torque on the wheels depends on the wheel size.

How do you plan to control the motion of that object?

3. Jun 21, 2013

### jack action

What you really need to know is how much power you need. There are basically 2 performance criteria you need to identify:
1. How fast do you want to go (defined mostly by aerodynamic characteristics)
2. How fast you want to accelerate (defined mostly by the mass)
This acceleration simulator will help you. The theory is at the bottom of the page.

For example, with the simplified version, by putting 0.1 hp of wheel power on a 25 kg RWD vehicle and a 0.7 tire friction coefficient, you get a vehicle that can reach a maximum speed of 29 km/h. Reading the lower left graph, it could reach 5 km/h in 0.57 s.

You can put drag coefficient and weight distribution characteristics of your own to get more accurate results.

4. Jun 22, 2013

### Staff: Mentor

I would expect that air resistance is negligible, and 29km/h is certainly not "walking pace".
Rolling resistance of the tires is the big unknown factor here.

5. Jun 22, 2013

### jack action

No, but if you want to have an acceptable acceleration to get to the "walking pace" speed, you will obviously be able to achieve a higher-than-needed top speed. For example, a typical car could easily have enough of 30 hp to reach 100 km/h, which is the usual maximum speed on most highways. But you would need around 60 s to get there, which is an unacceptable level of acceleration.

Rolling resistance is as negligible as aerodynamic drag. The big factor in this case is the acceleration needed, which I assume should be at least 0.2 - 0.3 g. It would represents an equivalent force about 10 times the rolling resistance and aerodynamic drag combined.

6. Jun 22, 2013

### OmCheeto

I agree. It's not even worth discussing unless you know the rolling resistance.

I always thought my driveway was completely flat, as the previous 6 cars I've owned wouldn't go anywhere if I left them in neutral and were not in gear.
When I bought my new truck and tried that, it rolled away.
I was like; "Come back here Truck*!"

If the rolling resistance hadn't stopped it, I imagine that it would have rolled for miles (given a perfectly flat surface) as I calculate the deceleration at the ~0.5 m/s speed it obtained to be ~0.0001 m/s2 due to aerodynamic drag.

ps. Given the vehicle mass of 1600 kg and a velocity of 0.5 m/s, the kinetic energy of the vehicle was about 200 joules, which indicates my driveway had an elevation change of ~ 1/2 inch. If it obtained its velocity over 2 seconds, it might be possible to figure out the equivalent torque required to accelerate the vehicle to that speed on a flat surface.

-------------------------
* My sister asked me what I named my truck after I bought it. I said; "Truck"

7. Jun 22, 2013

### Staff: Mentor

I'm not sure, and I do not have a 4-wheeled trolley to test it. How far does a trolley roll on rough surfaces (a street, sidewalks, ...) if we let it go with 5km/h? More than 1-2m? If it falls over, rolling resistance is always more important than acceleration.

5km/h = 1.4m/s, 0.2g looks reasonable - it gives 0.7s and 50cm to accelerate if that acceleration can be maintained the whole time. Compared to instananeous acceleration, the trolley is 50cm behind, easy to adjust for a human.

8. Jun 22, 2013

### jack action

Rolling resistance is defined as:

Rr = CRrmg

Where CRr is the Rolling resistance coefficient. The inertia is defined as:

F = ma

Or, if the acceleration is in g's:

F = aGmg

It is easy to see that if we assume aG = 0.2, then an equivalent CRr would also be 0.2 such that the two resistances compare. But even according to wikipedia, that corresponds to a tire rolling on sand. A more realistic value (tire on concrete) would be 0.01 to 0.015 (It might even be smaller than that if it is a harder type of wheel or tire), which is at least 10 to 20 times less than the 0.2 g acceleration, thus my earlier statement.

9. Jun 22, 2013

### Staff: Mentor

Hmm... language fail on my side. In German, "Trolley" is mainly used for this type of suitcase. The given weight range fits perfectly, so I assumed it would be something similar, and forgot all the other trolleys in English.
With larger wheels, rolling resistance will stay reasonable.

10. Jun 23, 2013

### OmCheeto

I don't see a language fail, as trolleys all have wheels as a common denominator.

I did an experiment in my driveway yesterday, and determined that the coefficient of rolling friction for my 280 gram volleyball is 0.14

-----------------------
volleyball rolling resistance experiment
initial height -- 0.79 meteres
distance traveled -- 5.5 meteres
vb mass -- 0.28 kg
mgh = 0.28kg*9.8m/s^2*0.79m=2.1 joules
*
F￼ is the rolling resistance force
￼Crr is the dimensionless rolling resistance coefficient or coefficient of rolling friction (CRF), and
￼N is the normal force, the force perpendicular to the surface on which the wheel is rolling.
*
N=mg cos(theta)
cos(zero)=1
N=mg
*
N = 0.28 kg * 9.8 m/s^2 = 2.7 newtons
*
f=ma
w=pe=f*d
pe=2.12 joules
f=pe/d = 2.1 j / 5.5 m = 0.39 newtons
*
F/N=Crr= 0.39/2.7 = 0.14

ps. Wo ist der OP gegangen?

11. Jun 24, 2013

### Staff: Mentor

They all have wheels, but the size is important.
Gute Frage (I have no idea)