How much Torque is needed to do Skid Steering

[vsming]

Hello, I need help in figuring out how much weight can Lynxmotion A4WD1 (lynxmotion.com/p-657-a4wd1-combo-kit-for-autonomous.aspx) carry
if it is equipped with four GHM-04 motors (lynxmotion.com/p-96-gear-head-motor-72vdc-501-175rpm-6mm-shaft.aspx)
but still be able to “smoothly” perform Skid Steering (a.k.a. Tank Steering)
such as Spin (In Place Pivot) and Hard Turn (Circular Pivot) according to beam-wiki.org/wiki/Steering_Techniques

The Lynxmotion A4WD1 is equipped with four 4.75" (0.12065 m) RC truck tires
(lynxmotion.com/p-108-off-road-robot-tire-475d-x-2375w-pair.aspx)
and its total mass is around 2.1kg (wheels + chassis + electronics).

Below are some of GHM-04 motor’s specs (worst-case scenario):
Rated Voltage = 7.2 V
Rated Torque/Load = 1.0000 kgf-cm = 0.0981 Nm
Stall Torque = 7.1000 kgf-cm = 0.6963 Nm
Speed at Rated Load = 131.4 RPM = 2.19 RPS
Efficiency at Rated Load = 40% to 45%

I would like the A4WD1 to carry a payload of at least 4.9kg (giving a total mass of 7kg),
and I estimate its expected efficiency to be 30%,because GHM-04’s efficiency is already around 40%, so (40% * 75% = 30%)

Using RobotShop’s calculator in robotshop.ca/dc-motor-selection.html
with the given input (7kg, four 0.0603m radius tire, 30% efficiency)
to produce the desired torque of 0.0981 Nm (GHM-04’s rated torque),
and using the RMF equation in societyofrobots.com/mechanics_dynamics.shtml
I obtain the following performance:

Under an incline of (0 degree ), A4WD1 can accelerate (0.2788 m/s^2) to a velocity of (0.8299 m/s)
Under an incline of (1 degree ), A4WD1 can accelerate (0.1075 m/s^2) to a velocity of (0.8299 m/s)
Under an incline of (1.628 degrees), A4WD1 can accelerate (0 m/s^2)

Which are obtained by rearranging the Torque relation in robotshop.ca/drive-motor-tutorial.html
to solve for acceleration as a function of incline angle (units omitted below):

T = (100/e)(a + gsin@)MR/N
0.0981 = (100/30) * (a + 9.81sin@) * 7 * 0.0603 / 4
0.0981 = (a + 9.81sin@) * 0.35175
0.2788 = a + 9.81sin@
a = 0.2788 - 9.81sin@

And using this acceleration into the RMF equation in societyofrobots.com/mechanics_dynamics.shtml
to solve for velocity (units omitted below):

Torque * RPS >= Mass * Acceleration * Velocity * (100/efficiency%) / (2PI) / #Wheels
Where:
Acceleration = a + 9.81sin@ = 0.2788 - 9.81sin@ + 9.81sin@ = 0.2788
Gives:
Torque * RPS >= Mass * 0.2788 * Velocity * (100/efficiency%) / (2PI) / #Wheels
0.0981 * 2.19 >= 7 * 0.2788 * Velocity * (100/30) / (2PI) / 4
0.2148 >= Velocity * 0.2588
0.8299 >= Velocity

**This seems to suggest that A4WD1 is able to carry a total mass of 7kg,
and still achieve an acceleration of 0.2788 m/s^2 (at best)
without overheating the four GHM-04 motors under its 0.0981 Nm rated torque/load.

However, I believe this calculation is only valid when A4WD1 is “travelling in straight lines”
and I am unsure of how to calculate for the case when A4WD1 needs to perform Skid Steering
such as Spin (In Place Pivot) and Hard Turn (Circular Pivot).

I am sincerely hoping for some advice on how to calculate the amount of Torque needed to do Skid Steering…**because based on my experience, 4 Wheeled Robots are unable to “turn smoothly”
where the main cause seems to be due to friction, according to both websites below:
ikalogic.com/tut_mech_1.php
gizmology.net/tracked.htm
But I am unsure of how to take friction into account, and am sincerely hoping for detailed guidance on this…

[bboyes]

Wow - I like your thorough analysis of the problem! Also a huge factor in my experience is the surface material. Carpet has way more friction than a smooth wood floor; enough to noticeably bog down the motors of small tracked robots if the nap is long enough.

[cbenson]

Hi vsming,

You are entirely correct that in order to turn, there are many factors which need to be considered. Even the weight the robot can theoretically support does not factor in IF the robot can actualy support it. In the case of the Lynxmotion 4WD, the wheels are not “inflated” and would flatten under a 7Kg load. As you have also seen, when travelling on a relatively flat surface there is not much torque required to move in a straight line.

Calculating the torque required to turn using skid steering would take into consideration:

]Static and dynamic friction coefficients/:m]
]Load on each wheel (you would need the pressure and contact area)/:m]
]Which wheels are driven and which are idle (not all robots with four wheels have all four driven)/:m]
]Distance between the wheel centers and wheel radius/:m]We suggest a torque balance looking down at the robot from above. Assuming all four wheels are driven, the robot should pivot about its center. You will need to pay close attention to the direction of the forces, especially friction. Friction will likely act an angle to each wheel.

In order to do a force balance, you might set the robot’s angular velocity and angular aceleration to zero and do a force balance using static friction, solving for the force applied per wheel. Therefore anything above this value will cause the robot to turn. There are of course other ways to do this, such as introducing an angular velocity and/or acceleration, the results of which will still have an error.

There are several papers and articles online going into greater detail which may be of interest as well and we suggest continuing your search. Should there be additional interest in this topic, we may consider looking more thoroughly into the subject.

Hope this helps,