Society of Robots  Robot Forum
Mechanics and Construction => Mechanics and Construction => Topic started by: elimenohpee on November 24, 2010, 01:38:07 PM

I read through the robot dynamics tutorial, but it didn't seem to suffice for my calculation.
These are the parameters I've worked out.
mass = 2kg
wheel radius = 0.0381m
max velocity robot needs to move = 0.31 m/s
rpm = 90
acceleration to max velocity = 1.3m/s^2
I need 2 driving motors based on these specifications, but I'm not sure I'm calculating it correctly. I was assuming the force the motor would have to push would be weight, but this results in a large torque
torque = force * distance = mass * acceleration * distance = 0.748 Nm or 0.374 Nm per motor.
How does the acceleration of the motor play into this calculation? I assumed it would be a sum of gravity and the acceleration of the motor, and when the motor reaches the max velocity, the motor acceleration falls out leaving only the acceleration due to gravity.
Those torque values just seem a little to large to after searching around for motors, can anyone validate or correct my calculation?
Thanks so much in advance.

Did you try using the RMF calculator?
http://www.societyofrobots.com/calculator.shtml (http://www.societyofrobots.com/calculator.shtml)

Did you try using the RMF calculator?
http://www.societyofrobots.com/calculator.shtml (http://www.societyofrobots.com/calculator.shtml)
Yeah but I was looking directly for a torque calculation. Does the calculation make sense? I don't really see where I messed up, but the torque value seems rather large.

acceleration to max velocity = 1.3m/s^2
[...] but this results in a large torque
Sure, when you wanna reach your top speed in about 1 second, you need torque.

Your acceleration of 1.3m/s^2 is huge for your small robot! Make it 0.13m/s^2.

The torque I have calculated doesn't take into effect the acceleration to the max velocity. The maximum velocity is so small that the acceleration to that point is barely a transient in steady state.
The torque I've calculated is based strictly on the weight of the vehicle. I just wanted someone to verify it for me.

The torque I have calculated doesn't take into effect the acceleration to the max velocity. The maximum velocity is so small that the acceleration to that point is barely a transient in steady state.
The torque I've calculated is based strictly on the weight of the vehicle. I just wanted someone to verify it for me.
Soooo . . . what acceleration did you use?
If you used 1.3, even if its an instantaneous very short lived acceleration, you still need that high torque to reach that acceleration for that short period of time.

Its all stated in the first post how I went about the calculation. The acceleration of the motor from 0 m/s to 0.31 m/s was neglible. The resultant torque was calculated from the result of the acceleration due to gravity.
All I want to to know is how everyone calculates the torque of a load before picking out a motor.

You should follow my equations and calculator.
On a flat plane, the acceleration of gravity has nothing to do with the wheel motor torque of your robot (with the exception of friction, of course). If the robot was going up/down a slope, then there is a constant gravitational acceleration applied to your robot that can be calculated by using trigonometry (you factor in the angle of the slope).
My calculator accounts for slopes, of which has a slope angle you can set.

The resultant torque was calculated from the result of the acceleration due to gravity.
All I want to to know is how everyone calculates the torque of a load before picking out a motor.
The accelerate on a level surface gravity doesn't really come into the calculation. Only the Mass of the Bot.
The forward force required is: F = ma (mass of bot * desired acceleration) = 2kg * 1.3m/s/s = 2.6 Newton
The find the required torque on the wheel's axial use:
Torque = wheel radius (moment arm) * Force = 0.0381m * 2.6N =0.099Nm = 0.010kilogram meter = 14 ounce inch
If the Bot needs to accelerate up a ramp than the required torque increases by mg * sin(ang) so the total F = ma + mg*sin(ang)
where:
g is the acceleration of gravity (9.8m/s/s on Earth's surface)
m is the robot mass (kilogram)
ang is the angle of the incline
So for a 7.65° incline, a 2kg bot, 0.0381 radius wheels, acceleration of 1.3m/s/s the Force is:
F = 2*1.3 + 2*9.8*sin(7.65°) = 2.6+2.61 = 5.21N
Torque = 0.0381*5.21 = 0.20 Nm = 0.020 kilogram meter = 28 ounce inch
Then to account for losses divid by 75% (0.75) so Torque = 37 ounce inch
Note:
I use an online conversion app for the different torque units here:
http://www.onlineconversion.com/torque.htm (http://www.onlineconversion.com/torque.htm)

The resultant torque was calculated from the result of the acceleration due to gravity.
All I want to to know is how everyone calculates the torque of a load before picking out a motor.
The accelerate on a level surface gravity doesn't really come into the calculation. Only the Mass of the Bot.
The forward force required is: F = ma (mass of bot * desired acceleration) = 2kg * 1.3m/s/s = 2.6 Newton
The find the required torque on the wheel's axial use:
Torque = wheel radius (moment arm) * Force = 0.0381m * 2.6N =0.099Nm = 0.010kilogram meter = 14 ounce inch
If the Bot needs to accelerate up a ramp than the required torque increases by mg * sin(ang) so the total F = ma + mg*sin(ang)
where:
g is the acceleration of gravity (9.8m/s/s on Earth's surface)
m is the robot mass (kilogram)
ang is the angle of the incline
So for a 7.65° incline, a 2kg bot, 0.0381 radius wheels, acceleration of 1.3m/s/s the Force is:
F = 2*1.3 + 2*9.8*sin(7.65°) = 2.6+2.61 = 5.21N
Torque = 0.0381*5.21 = 0.20 Nm = 0.020 kilogram meter = 28 ounce inch
Then to account for losses divid by 75% (0.75) so Torque = 37 ounce inch
Note:
I use an online conversion app for the different torque units here:
http://www.onlineconversion.com/torque.htm (http://www.onlineconversion.com/torque.htm)
Thanks for the reply! Ok my only question is this. When the acceleration is zero (aka the velocity is constant), isn't there a torque due to the weight imposed on the wheels? If there isn't, then the force is zero and there is no torque needed to turn the wheels. That is why initially I was using a sum of gravity + acceleration of the motor combined when computing the total force.

When the acceleration is zero (aka the velocity is zero),
Velocity and acceleration are independent terms. It's possible to have zero acceleration but very high velocity.
If there is no acceleration, then there is no torque.
However, your robot will always experience acceleration due to friction from air resistance, the wheels touching the ground, and from the gears in the motors rolling against each other.

I don't understand. It seems that the weight should play into the equation somehow. If when the wheel is moving at a constant rate, and the torque then is zero as you say, then according motor principles the motor should stop spinning since torque is relative to speed.

The mass plays into momentum and friction, but on a level plane, gravity does not.
If when the wheel is moving at a constant rate, and the torque then is zero as you say, then according motor principles the motor should stop spinning since torque is relative to speed.
uhhhh no, the motor shouldn't stop spinning if torque is zero. Torque is relative to acceleration, not velocity :P

If the acceleration is zero and friction is zero the motor/wheels will turn at a constant speed determined by the voltage applied and the motor characteristics. The motor will also draw the minimum current. Any friction (a Force in the direction opposite to the motion) will require an addition force from the motor and the motor will draw more current (and slow down some).
Try running a motor that is not connected to anything and apply a voltage. The motor will not stop spinning eventhough there is not any torque be to apply to accelerate a mass. The motor will just run at its fastest speed and draw the minimum current. This is known as the No Load current. The Torque produced by a motor is proportional to the motor current. Here is a good tutorial on motor calculations and characteristics:
http://www.micromo.com/n390432/n.html (http://www.micromo.com/n390432/n.html)
If there isn't a load on the motor then there can not be any Torque. See the second diagram in the dynamic tutorial and the graph in the MicroMo tutorial. The Force to move the Bot is opposite and equal the Force of Friction without acceleration as per the equations. This Force is provided from the Torque out of the motor. If this Frictional Force goes to zero then the Torque also goes to zero.
Think of what happens if the Bot is going down an incline. Gravity can then be the Force that over comes the Frictional Force and the motor isn't required to produce Torque.
The Acceleration of Gravity only comes into the equations when the Bot is on an incline and to calculate the Friction Force.

If the velocity is zero and friction is zero the motor/wheels will turn at a constant speed determined by the voltage applied and the motor characteristics.
I'm just pointing out a typo, it should be 'If the acceleration is zero and friction...'
:P

The force you apply to keep moving at a steady velocity must be equal to the force of friction.
The weight plays into it because the more weight you have, the more friction you have (generally).
You can get a close approximation of the amount of friction your bot has by taking a hanging type scale (like the ones you weigh fruit with in a grocery store) and attaching it to the front of your bot. Make sure your wheels can move freely. Pull on the scale at the speed you want your robot to move. The scale should read very close to the force of friction.

If the velocity is zero and friction is zero the motor/wheels will turn at a constant speed determined by the voltage applied and the motor characteristics.
I'm just pointing out a typo, it should be 'If the acceleration is zero and friction...'
:P
Yep...thanks for the correction.
I edited my post above to fix the typo so that it reads correctly.