Hi there!
I am building a small robot and it needs to climb a hill at 30 degrees.
I already have a motor and i wanted to check if it was sufficient enough to do the job, however do not know what formula for torque to use as every website has a different method....
Mass = 0.3kg
Torque of e motor= .0.6kg/cm
wheel radius = 2cm
Angle of incline = 30 degrees
from this how can i calculate if its sufficient to climb the hill or at what speed it would climb at?
Thank you very much!
as every website has a different method
Take the results from several websites and average them, throwing out the obvious outliers, and tell us what you get!
(I get 0.03 Newton-meters just to overcome the force of gravity).
Note: this makes no sense
Torque of e motor= .0.6kg/cm
Can you tell me how you are doing your calculations please? By different results on each website i mean each one uses different parameters and now i am very confused!
0.6kg/cm = 0.06n/m of torque
Thank you!
Please figure out your units first.
Torque = force times distance, and is measured in Newton times meters, and sometimes improperly stated as Kg times cm. Division is not involved.
At 30 degrees the height climbed is 0.5 times the distance travelled along the slope, so
for a mass of 0.3kg the force down the slope is 0.5 times the weight, ie 0.5 x 9.8 x 0.3 = 1.47N (assuming negligible friction)
The radius of the wheel is 0.02m, so the rolling torque = force x radius = 1.47 x 0.02 = 29.4mNm (milli newton-metres).
Allowing for friction lets say that 50mNm is the minimum needed to ensure uphill progress.
Torque of e motor= .0.6kg/cm
Perhaps you mean 0.6 kgf-cm (killogram force x cm)? Silly units, lets convert to SI:
0.6kgf = 5.9N. torque = 5.9 x 0.01 = 59mNm
So I reckon that's enough.
In mechanics always work in SI units, it saves so much faff and chance for error.
In reality friction is a major factor, so you normally allow a big margin for
friction, perhaps a factor of two.
And torque is a product of force and distance, not a ratio. You can also think of
torque as energy per radian.
0.6kg/cm = 0.06n/m of torque
Again NO! The symbol for the newton is "N". Torque is force times distance,
not force over distance. kilogram is a unit of mass, not force. You can use kgf,
but that's assuming you know the local gravity (ie your values will be wrong
on the moon!)
When i say 0.06n/m i mean to say 0.06 newton- meters, Sorry!
Thank you very much for clarifying everything Mark !
Much appreciated
Last question...
To factor in rolling resistance and acceleration is this right?
Rolling resistance = mg x friction coefficient
Acceleration force = m x speed/time taken
? Thank you
No, you confuse sliding friction with rolling resistance - rolling resistance is a complicated
function of gears, tyre losses, etc, and tends to increase with speed or with square of speed.
Sliding friction is simpler and the force is constant and proportional to contact force as your
equation indicates. Sliding friction is only appropriate to wheels if skidding!
For acceleration you need to worry both about the linear acceleration of the whole vehicle
and the angular accleration of wheels/gears/motors.
Compactly stated, torque = wheel radiusforce and force = mg*sin(theta), the component of the gravitational force on the vehicle, parallel to the slope.
That is the torque required at the axle of the wheel to just overcome the gravitational force. In stating the motor torque, have you taken gearing and motor voltage into account?