Hello, I need help in figuring out how much weight can Lynxmotion A4WD1 (

http://www.lynxmotion.com/p-657-a4wd1-combo-kit-for-autonomous.aspx) carry

if it is equipped with four GHM-04 motors (

http://www.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

http://www.beam-wiki.org/wiki/Steering_TechniquesThe Lynxmotion A4WD1 is equipped with

four 4.75" (0.12065 m) RC truck tires (

http://www.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

http://www.robotshop.ca/dc-motor-selection.htmlwith 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

http://www.societyofrobots.com/mechanics_dynamics.shtmlI 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

http://www.robotshop.ca/drive-motor-tutorial.htmlto solve for acceleration as a function of incline angle (units omitted below):

T = (100/e)*(a + g*sin@)*M*R/N

0.0981 = (100/30) * (a + 9.81*sin@) * 7 * 0.0603 / 4

0.0981 = (a + 9.81*sin@) * 0.35175

0.2788 = a + 9.81*sin@

a = 0.2788 - 9.81*sin@And using this acceleration into the RMF equation in

http://www.societyofrobots.com/mechanics_dynamics.shtmlto solve for velocity (units omitted below):

Torque * RPS >= Mass * Acceleration * Velocity * (100/efficiency%) / (2*PI) / #Wheels

Where:

Acceleration = a + 9.81*sin@ = 0.2788 - 9.81*sin@ + 9.81*sin@ = 0.2788

Gives:

Torque * RPS >= Mass * 0.2788 * Velocity * (100/efficiency%) / (2*PI) / #Wheels

0.0981 * 2.19 >= 7 * 0.2788 * Velocity * (100/30) / (2*PI) / 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:

http://www.ikalogic.com/tut_mech_1.phphttp://www.gizmology.net/tracked.htmBut I am unsure of how to take friction into account, and am sincerely hoping for detailed guidance on this...